**Note:** This translation is a work in progress by Andrew Leahy. The Latin original
may be found here.
A paper explaining some of the modern interpretations of this work can
be found here. Another paper
specifically discussing the Fundamental Theorem of Calculus may be
found in the online journal Convergence. Another paper discussing James Gregory's take on the Pappus-Guldin theorem may be also be found at Convergence.

It has been observed by the geometers of our generation that mathematics was wrongly divided by the ancients into geometry, arithmetic, etc., and that it would be better that mathematics be divided into the universal and the particular. The universal part of mathematics, which is commonly (even if perhaps incorrectly) called geometry and to which the analysis of more recent times is associated, discusses the portion of mathematics apart from any type of quantity. The particular part of mathematics is divided: into geometry (thus appropriately named), which is nothing other than the universal part of mathematics restricted to shape; into arithmetic, which is the same as the universal mathematics restricted to number; and into statics, which is the same as the universal mathematics restricted to motion; etc. Unless I am mistaken, I see the same thing in geometry. Indeed, when I observed that the most general analytical methods served every problem provided that it was possible to apply those methods and that analysis is nothing except the examination of unknown quantities until they are finally reduced to equations with known quantities, I believed that the defect of analysis (which is especially apparent in the measurement of curved quantities) would be able to be made good in some respect if only from a given essential property of a figure a method were found for transforming that figure into another equal figure having known properties, and this into another and so forth until finally a transformation were made into some known quantity. Thus, the sought-after measurement of the proposed quantity would be exhibited exactly as in the resolution of an analytical equation. And I do not believe that my opinion is a disappointment. Indeed, I think that this book contains the universal part of geometry to such an extent that it excludes no particular figure of any sort which has been considered by geometers up until now. But if other figures are proposed for consideration, this knowledge will be moved forward. Indeed, just as there are infinite types of figures, this part of geometry--just as all others--will also be infinite. Nevertheless, it will be more brief and more elegant to apply a universal doctrine according to the properties of a figure to any particular case than to publish an entire volume concerning one figure. One interested in this method ought to be versed before anything in analysis for, without it, examining the proposed properties of any figure will be beyond men of any temperament.

I do not deny that I have read many vestiges of such a method among outstanding geometers, but these are mostly demonstrated either very particularly or not sufficiently geometrically. Let the reader who compares this book with the books of others judge those things which are mine and which belong to others. Indeed, I assert nothing lest I seem to ascribe to myself things which were discovered by others previously (certainly with me not knowing). Unless I am in error, I make use of a method for demonstrating peculiar to me certainly much more brief than the Archimedean method and no less geometrical. I also make use of the Cavallierian method, which is also reduced with no trouble to the Archimedean method or to our method, in more obvious propositions. But if a geometer, after the diligent application of this method according to the properties of the figure, discovers no end to the problems, it should be reduced to a convergent series, the termination of which is itself an unknown figure or to that figure in a given ratio. In fact, because of this ratio, I have endeavored to reduce the proportions of other figures to a proportion between plane figures. Indeed, I believe that the doctrine of convergent series is easier in these. Nevertheless, I do not dare to assert that a converging series is always able to be found. In fact, I suspect that this method is insufficient for finding all non-analytical proportions. However, we would like our book about the true quadrature of the circle and the hyperbola to be the ultimate recourse of our method, for the doctrine of converging series, which exhibit a possible solution from the properties of the invented figure, is general. Accordingly, let us be satisfied here with the objections against our doctrine.

First, there is an objection about the title--namely, that my book is
badly named *The true quadrature of the hyperbola and the
circle*, since it argued more that this is impossible. I respond
that, if it were impossible, no proportion between the circle and the
square of the diameter could be given and thus Definition Five of Book
Five of Euclid would be false. On the other hand, if it is possible,
our error ought to be demonstrated if our claim is not true. Others
object thusly: This is no quadrature, since a proportion between the
circle and the square of the diameter is not designated. To this
objection, I respond that, with the diameter b of the circle having
been assumed, the circle itself will be the termination of the
converging series whose first terms are b^{2}/ 2 and
b^{2} and whose second terms are √(b^{4}/2) and
2 b^{4}/ (b^{2} +√2 b^{4}).
Consequently, the square of the diameter is to the circle as
b^{2} is to the previously-mentioned termination, which is in
no manner more unknown to us than the cube root of the number 40.
Others say that it was not well demonstrated (in the scholium to
Proposition Five) that the sector ABIP is the same as the termination
of the converging series whose first terms are the triangle ABP and
the trapezium ABFP and whose second terms are the rectilinear figures
ABIP and ABDLP. Consequently, I display here a complete
demonstration: If the sector and the previously-mentioned termination
are not equal, let Z be the difference between them and let the
converging series be extended until the difference of the converging
terms (namely, P and Q) is less than Z. Indeed, it is evident (from
Proposition Six) that this is able to be done. With these things
having been assumed, it is evident that the sector and the termination
of the series are between the terms P and Q. Thus, there are four
quantities, the largest and smallest of which are P and Q. Moreover,
the sector and the termination of the series are in between and the
difference of the intermediates will be more than the differences of
the extremes, namely Z, which is absurd. Therefore, there is no
difference between the sector and the termination of the series.
Hence they are equal, which is what ought to have been demonstrated.
Others object about Proposition 11 thusly: if a^{3} is added
to the term a^{3} + a^{2} b and to the term b
a^{2} + b^{2} a, the strength of each demonstration
would be weakened. I respond that a^{3} is an indefinite
quantity and that other indefinite quantities in addition to the
converging terms themselves are not able to enter upon the
composition, because the analysis in not able to be hidden. Others
object: In this same manner it is able to be demonstrated that between
two indefinitely commensurable quantities P and Q a mean proportional
commensurable with these is not able to be found. But this is
nevertheless false if P and Q are planosimilar. I respond that there
is a very great discrepancy between the extraction of the root and
this sixth operation. For in the extraction of the root, when the
divisor measures the dividend--which occurs in the
previously-mentioned case--the extraction of the root never coincides
with the four prior operations. However, the sixth operation, since
it is infinite by its own nature, never coincides with the previous
operations. A certain geometer not familiar with the circle objects
that the quadrature was done by the operations of analysis with the
aid of the quadratrix. I deny this entirely to the one asking by
asserting: As he designates those operations analytic, indeed, I judge
that the base of the quadratrix is no more able to be designated by
analytic operations than the square root of two is able to be
designated by the first four operations of arithmetic.

With these objections against our doctrine which are either offered by others or able to be imagined by me having been addressed, let us also satisfy those who are interested in mechanical operations. If anyone should wish to square the circle or to divide an angle into a given ratio mechanically, I do not believe that any method is more simple than a common quadratrix curve, described by planar and solid material accurately and point-by-point. Because it pertains to the quadrature of the hyperbola, I believe that it is sufficiently easy to exhibit in lines the ratio of an inscribed triangle or trapezoid (indeed, they are equal between themselves) to the circumscribing triangle, to continue the sought-after converging series from these lines, and to hold forth the approximation until there will be a convergence. All things which can be desired about logarithms and the compouding of ratios can be found with no trouble with the help of the following curve.

Let DE be a straight line and let AB, on which there is a moving point C, be another line normal to it. Let the point C move in a ratio so that, while the perpendicular moves toward E or D always normal to the line ED, the segments between the point C and the line DE are to the line CB itself in ratios multiplied between themselves in the ratio of the motion of the perpendicular line AB. For instance, let AB be moved to KM and NP. It is desired that the point C be moved by the law that the ratio of CP to CB is multiplied with the ratio of CM to CB in the ratio of PB to MB, With AB having been supposed to be moved in the same manner toward the point D to IG and LO so that the point C is moved by the law that the ratio of CO to CB is multiplied with the ratio of CG to CB in the ratio of OB to GB. From the line DE and two points of the sought after curve having given as you please by the motion of the described point, it is easy to describe the curve itself pointwise. For instance, having supposed that the curve passes through the lines AB and KM in the point C, it is desired to find the other points of the curve: Let BM be divided in half at H and let the perpendicular HC be the mean proportional between BC and MC. I say that C is one of the sought-after points. Indeed, the ratio of CM to CB is the duplicate ratio of CH to CB, and the line MB is twice the line BH. Therefore, C is on the sought-after curve. Next, let the line SM be equal to the line MB and as CB is to CM thus let CM be to the perpendicular CS. It is demonstrated as before that C is a point on the sought-after curve. And by this method points are able to be found wherever you like and however much of the curve you like is able to be produced.

Certain special properties of this curve which are discerned with no trouble ought to be noted. First, the curve is able to be extended in each direction ad infinitum. Second, in one direction, namely F, although it always becomes closer to the line ED, nevertheless it never intersects the line, yielding a space on the FD side finite in quantity even though it is infinite in length. Third, with one of the perpendiculars--or the ordinates--assumed the location of unity and the location of the remaining numbers having been situated, the line segment on DE--or the asymptote--between unity and the number is always the logarithm of the number. For instance, with CO having been situated in unity and CG double it, likewise CB triple it, and CH quadruple it, etc., OG will be the logarithm of the double, OB of the triple, and OH of the quadruple, etc. If this curve should be defined precisely in a plane solid, not only will however many mean proportionals be found between the two lines by means of a ruler and compass but all problems imaginable concerning the composition of ratios are also able to be done easily. But I omit these things deliberately, since they are easy, reminding the reader for the moment that the space contained by the portion of the previously-mentioned curve, its own asymptote, and the two ordinates, does not have an analytic ratio to the rectangle inscribed in it, as is able to be demonstrated from Proposition 11 of the previously--mentioned tract.

These operations ought not to be thought of as from geometry, since they are not performed by ruler and compass alone, just as the most keen mathematician D. Carolus Renaldinius observes best in his own ``geometry promota'' when he treats these curves, which he calls Medicean. But so that this might be more clear I will endeavor here to show that in fact no cubic equation is solved with the aid of a ruler and compass alone: Each cubic equation has either one alone or three real roots. If these were found only with the help of a ruler and compass--or if they were the intersection of a circle and a straight line--the straight line would cut the circle either in one point or in three, either of which is very absurd. For a similar reason, a cubic equation having three real roots is never able to be reduced to a pure equation which has only one. For in these equations a reduction succeeds in no manner, since it is impossible with straightedge and compass to change an imaginary root into a real root or vice versa.

Let KB be any simple and non-winding curve--that is, from K to B always drawing closer to a certain line MP given in position or always receding from it. Let (as in the first figure) K be the point on the curve KB closest to the line MP. From any two points, say, C and B, on the curve KB let perpendiculars CO and BP hang down to the line MP. From the point C on the curve let a line CA tangent to the curve at C be drawn as close as possible to the point K and intersecting the extended line PB in A. I say that the line CA is greater than the curve CB.

Let a line BRD be drawn tangent to the curve at B and intersecting the line CA in R and the extended line OC in D. Since the curve KB is always moving away from the line PM in the advance from K to B, therefore the line BD, tangent to the curve in the point B and inclining back toward K, extends to an intersection with the line PM. Therefore, the angle ABR is obtuse, being certainly greater than the right angle MPB, into which it runs when PM is extended. Therefore, the angle ABR is greater than the angle RAB and the side AR is greater than the side BR. By adding the common part, the line AC is more than the lines BR and RC. But BR and RC, touching the curve in the points B and C, respectively, are more than the arc BC. Therefore the line AC is much more than that same arc CB, which it was desired to demonstrate.

Second, I say that the line BD is less than the curve BC.

By what has been demonstrated up until now the angle ABD is obtuse and likewise, because AP and DO are parallel, angle CDB equals angle ABD. Consequently with the subtending line CB having been drawn, the angle BDC is more than the angle BCD and the line DB less than the line BC. But the line BC is less than the curve BC and therefore the line BD is much less than the curve BC.

From the extremal points K and B let perpendiculars KM and BP be sent down to the line MP. Next, let the line MP be divided into how many so ever equal segments MN, NO, and OP. From the points P, O, N, and M erect perpendiculars PA, OE, NH, and ML, respectively, cutting the curve in the points B, C, F, and K, from which, respectively, are drawn toward B lines KH, FE, and CA tangent to the curve and terminating at the nearest perpendicular. Let the tangent BD at the point B terminate at the closest perpendicular toward K in the point D, and let KI be parallel to the line BD: It is manifest that KH is more than the line KI because the obtuse angle HIK is equal to the angle ABD.

I say that all of the lines KH, FE, and CA together are more than the curve KB and that the excess of the line HK over the line KI is more than the excess of all of the lines KH, FE, and CA together over the curve KB.

Let the tangents AC and EF be extended to L and G, respectively, on the nearest perpendiculars toward K. From the first result, AC is greater than the curve BC, EF is greater than the curve FC, and HK is greater than the curve KF. Consequently, together all of KH, FE, and CA are more than the unbroken curve KB. From the second result, BD (that is, KI) is less than the curve BC, CG (that is, CA) is less than the arc FC, and LF (that is, FE) is less than the arc KF. Consequently all of KI, FE, and CA together are less than the entire curve KB and hence the excess of the lines KH, FE, and CA over the lines KI, FE, and CA is more than their excess over the curve KB. But the lines FE and CA are common to each sum of lines and consequently the excess of the line KH over the line KI is equal to the excess of the sums of the lines KH, FE, and CA over the sum of lines KL, FE, and CA, which demonstrates that this excess is greater than the excess of the lines KH, FE, and CA above the curve KB, which is what ought to have been demonstrated.

If the actual curve slopes down toward the line MP, as in the second figure, K ought to be the point on the curve most remote from the line MP and some words in the demonstration ought to be changed, as the industrious reader will understand for himself.

Let 79CD be any simple and non-winding curve (if indeed it were winding, it is reasonable to divide it into many simple curves) above which is imagined the surface of a right cylinder whose altitude is the line X. From some point of the curve, say, 9, let a line 93 be dropped down perpendicular to a line, say, 26. Also, let a line 96 cutting the curve normally in the point 9 be drawn to the line 26. Let the line 39 be extended to a point S such that 3S is equal to the line 96. It should likewise be supposed that this is done in all points of the curve 79CD in such a way that from the lines perpendicular to the curve extended normally to the line 26 through their own points a region RV δ2 composed of the curve RSTV and the lines R2, Vδ, and δ2 is produced. Next, let the altitude X of the cylinder be to 3N as 93 is to S3. Likewise, we suppose that this is done in all the other lines of the region RVδ2 perpendicular to the line 2 δ, so that a region PHδ2 comprised of the curve PH and the lines Hδ, P2, and 2δ is filled out.

That is, I understand that two regions have been described, the first of which has the property that the perpendicular S3 sent down to the line 26 out of whatever point S of its curve is equal to the normal to the curve 79CD at the point 9 of the intersection, specifically, the line 96. However, the second has the property that the perpendicular 3N dropped down to the line 26 out of whatever point N of its curve is to the altitude X of the cylinder as the part intersecting with the first area S3 is to the part of N3 intersecting the given curve 93.

I say that the second curved area PHδ2 is equal to the surface of the right cylinder whose base is the given curve 79CD and which has line X as the altitude.

If the area of PHδ2 is not equal to the surface of the cylinder mentioned before, there will be a difference between them which will be a planar figure α. Let the planar figure α be applied to the line X and let the side β be the breadth. Let a line 7A be drawn from the point 7 tangent to the curve D7 at the point 7. From the point D let a line Dγ be drawn touching the curve in the point D and let line 7ν be parallel to the line Dγ. From the points 7 and D let perpendiculars 72 and δD be dropped to the line 26 and let the line 2δ be divided into so many equal parts that from N3, the last perpendicular erected, the excess of the tangent 7A beyond the parallel line 7ν is less than the line β. Indeed, it is manifest that this is always able to be done if the tangent 7A is not perpendicular to the line 2δ and the given curve is non-winding. From the points 2, 3, 4, and δ of the partition of 2δ let perpendiculars 2P, 3N, 4L, and δH be erected, cutting the given curve in the points 7, 9, C, and D, the first invented curve in the points R, S, T, and V, and the second invented curve in the points P, N, L, and H. Then, let lines 7A, 9B, CF and Dγ be drawn tangent to the given curve at the points 7, 9, C, and D and terminated at the closest perpendiculars. Let 78, 90, CE, D5, YP, NK, LG, HI, LM, and NQ be lines parallel to and equal to the corresponding parts of 2δ and terminated in the nearest perpendicular.

The angle B96 is right, because 9B is a tangent and 96 is a perpendicular to it. The angle O93 is also right. Consequently, with the common angle O96 being taken away, an angle O9B equal to the angle 693 remains. Therefore, the right triangles O9B and 693 are similar and hence as 93 is to 96, so is 9O to 9B. But as 93 is to 96 so is X to 3N. Consequently, as 9O is to 9B thus is X to 3 N. Therefore, the rectangle formed by 9O and 3N--namely, N4--is equal to the rectangle formed by 9B and X. By this same method it is demonstrated that the rectangle P3 is equal to the rectangle formed by 7A and X and that the rectangle Lδ is equal to the rectangle formed by CF and X. Consequently, the rectangle formed by the line X and the lines 7A, 9B, and CF together is equal to the entire rectilinear figure 2PYNKLGδ. Therefore, the rectangle formed by X and the lines 7A, 9B, and CF together is more the area of the curved figure PNLHδ2.

Let DΠ be the normal to the curve at the point D. Therefore, the angle ΠDγ is right. But the angle δD 5 is also right. Consequently, by taking away the common angle, namely δD γ, it follows the angles δDΠ and γD 5 are equal. Therefore, the right triangles δD Π and γD 5 are similar. Consequently, as δD is to DΠ--that is, as D5 is to Dγ--so is X to δH. Therefore, the rectangle formed by δH and D5--that is, the rectangle H4--is equal to the rectangle formed by X and Dγ--that is, by X and 7ν.

But the rectangle M4 is equal to the rectangle Lδ--that is, to the rectangle formed by X and CF--and the rectangle Q3 is equal to the rectangle N4--that is, to the rectangle formed by X and 9B. Therefore, the rectilinear figure QNMLIHδ2 is equal to the rectangle formed by X and the lines FC, B9, and ν7 together. Therefore, the rectangle formed by X and the lines FC, B9, and ν7 together is less than the curve PNLHδ2. On the other hand, the lines FC, B9, and A7 together are more than the curve 79CD and the lines FC, B9, ν7 together are less than the same curve, each of which is clear from the preceding result. Therefore, the rectangle formed by X and the lines FC, B9, and A7 is more than the surface of the cylinder above the curve 79CD and the rectangle formed by X and the lines FC, B9, and ν7 is less than that same curve.

But it was demonstrated that the rectangle formed by X and the lines FC, B9, and A7 is more than the area of PNLHδ2 and that the rectangle formed by X and the lines FC, B9, and ν7 is less than that same area. Consequently, the difference between these rectangles is greater than the difference between the surface of the cylinder and the area. But the difference of these rectangles is the rectangle formed by X and the difference of the lines 7A and 7 ν. Moreover, the difference of the lines 7A and 7ν is less than the line β by supposition. Consequently, the difference between the rectangle formed by X and FC, B9, and A7 together and the rectangle formed by X and FC, B9, and ν7 together is less than the quantity α--the rectangle formed by X and β, which is, by supposition, the difference between the surface of the cylinder and the curved area. This is absurd. Therefore the surface of the cylinder and the area are equal, which is what ought to have been demonstrated.

If the curve 7D were not simple but winding, it should be divided into many simple pieces and the demonstration carried out individually on each.

To be sure, if the tangent 7A were perpendicular to the line 78, the curved region PNLHδ2 would be extended infinitely to the point P. Since this does not stand in the way, I say to this that the area of PNLHδ2 is equal to the surface of the right cylinder about the curve 79 CD, whose altitude is X. If they are not equal, let (if possible) the curved region be more than the surface and let the area NLHδ3, equal to the surface above the curve 79CD, be cut off by the line N3 parallel to Hδ. Indeed, without a doubt this is able to be done. By the same method as before, it is demonstrated that the area NLHδ3 is equal to the surface above 9CD. Therefore, the surface of the right cylinder whose altitude is X over the curve 9CD is equal to the surface of the right cylinder whose altitude is X over the curve 79CD, which is absurd. Therefore, the area of PNLHδ2 is not more than the said surface of the cylinder. Suppose (if it is possible) that it is less, and let the curve 9D be cut away so that the surface of the right cylinder existing above 9D is equal to the area of PNLHδ2. Let a line 39N be drawn parallel to the line Hδ. It is demonstrated as before that the surface of the right cylinder existing over 9D is equal to the area of NLHδ3. But by supposition the same surface of the right cylinder is equal to the area of PNLHδ2. Therefore the areas PNLHδ2 and NLHδ3 are equal among themselves, which is absurd. Consequently, the curved region is not less than the surface, but in fact it has been demonstrated that neither is is more. Therefore the surface of the right cylinder above the curve 7D whose altitude is X is equal to the area PNLHδ2, even when the tangent 7A is perpendicular to the line 78, which is what ought to have been demonstrated.

From this demonstration, it is manifest that the area of PNLHδ2 and the surface of the cylinder about the curve 79CD are proportional quantities in magnitude and weight, since the same equality which is demonstrated about their entireties is demonstrated in the same manner about their proportional parts. Consequently, by this same method their centers of equilibrium divide the line 2 δ. But the curve 7D itself is proportional in magnitude and weight with the surface of the cylinder. Consequently, the curve is in fact proportionate in magnitude and gravity with the curved region as well as with the remaining ones having the center of equilibria on the line 2 δ. It is also observed that the area of PNLHδ2 is to the rectangle formed from X and 2δ as the curve 79CD is to the line 2 δ.

With the same things being assumed as in the previous result, let it be supposed that the surface of the right cylinder above the curve 79CD is cut by a plane passing through the line 2δ and inclining toward the plane Dδ2 in a 45 degree angle. We call the lower part of the surface of the cylinder cut by the plane the surface of the trunk. I say that the surface of the trunk is equal to the curved region RSTVδ2.

Let the curve 79CD be supposed to be extended into a line ΓΛΣΔ equal to itself and let a line Δp equal to the line X be joined so that a rectangle hpΔΓ which is necessarily equal to the surface of the cylinder and to the curved region PNLHδ2 is filled out. Let Γε be equal to the line 27 and from the point ε let a curve ερψΩ be drawn of such a nature that, when any line ΓΛ equal to some small part of the curve, say, 79, has been assumed, the perpendicular to the line ΓΛ at the point Λ on the curve εΩ, namely Λρ, is equal to the perpendicular from the point 9 to the line 2δ--namely, 93. It is manifest that the curved region ερψΩΔΓ is equal to the surface of the trunk, since the inclination of the cutting plane is assumed to be a 45 degree angle.

Therefore, our proof is to demonstrate the equality of the curved regions RSTVδ2 and ερψΩΔΓ. First, we will demonstrate this equality in the latter figures, where we suppose that the curve RV always draws closer to the line 2δ in the progression from R to V and likewise that the curve 7D draws closer to that same line in the progression from 7 to D. Therefore, the curve εΩ draws closer to the line ΓΛ in the progression from ε to Ω , since when 7D draws closer to the line 2δ the curve εΩ draws to the line ΓΛ in the same manner. If the curved surfaces mentioned above are not equal, let α be the difference between them and let the curved region RVδ2 be divided into so many lines parallel to the line R2--namely, S3, T4, Vδ, and R2--that when the perpendiculars Rξ, Sγ, Sλ, Tθ, and Tμ, and Vη have been drawn from the intersections R, S, T, and V to the nearest parallels on either side there are two rectilinear regions, namely, RξSλTμδ2 above the curved region and γSθTηVδ2 within the curved region, the difference of which is less than α. It is evident that this is able to be done. Let the lines S3 and T4 be extended so that they intersect both curves PH and 7D in the points N, L, 9, and C, and let the lines PY, NQ, NK, LM, LG, and HI, parallel to 2δ and terminated by the dividing lines of the curved region, be joined.

Next, let the rectangle hΔ be divided by the lines hΓ, kΛ, mΣ, and pΔ, parallel to the line pΔ into the rectangles hΛ equal to the curved region PN32, kΣ equal to the curved region NL43, and mΔ equal to the curved region LHδ4. From the intersections of the lines dividing the rectangle hΔ with the curve εΩ, namely, ε, ρ, ψ, and Ω, let perpendiculars be drawn in both directions to the nearest dividing lines, namely, εζ, ρπ, ρσ, ψτ, ψω, Ωφ, so that πρτφΩΔΓ is a rectilinear figure within the curved region and εζρσψω ΔΓ is a rectilinear figure circumscribing the curved region. It is apparent from the previous proposition that P2 is to X--or hΓ--as R2 is to 72--or εΣ. By permuting, as P2 is to R2, thus is hΓ to εΓ. Therefore, as P3 is to R3, thus is hΛ to εΛ. But the rectangle hΛ is equal to the curved region PN32. Therefore, by permuting, as P3 is to the region PN32, thus is R3 to εΛ. Since P3 is greater than the curve, R3 will be greater than εΛ. In this same manner it is demonstrated that the rectangle S4 is more than ρΣ and that Tδ is more than ψΔ. Therefore, the rectilinear figure RξS λTμδ2 is more than the rectilinear figure εζρσψωΔΓ. In the same manner, as N3 is to X--or kΛ--thus is S3 to 93--or ρΛ. By permuting, as N3 is to S3 thus is kΛ to ρΛ. That is, as Q3 is to γ3, thus is hΛ--or the region PN32--to πΛ. By permuting, as Q3 is to the region PN32, thus is γ3 to πΛ. But Q3 is less than the curve. Therefore, γ3 is less than πΛ. In the same manner, it is demonstrated the θ4 is less than τΣ and ηδ is less than φΔ. Therefore, the rectilinear figure γS θTηVδ2 is less than the rectilinear figure πρτψφΩΔΓ.

Therefore, both rectilinear figures εζρσψωΔΓ and πρτψφΩΔΓ are between the two rectilinear figures RξSλTμδ2 and γS θ TηVδ2. That is, the greater of the first rectilinear curves is less than the greater of the second curves and the lesser of the first curves is more than the lesser of the second. The curved region ερψΩΔΓ is between the two first rectilinear figures and therefore is likewise between the two latter rectilinear figures, namely RξSλTμδ2 and γS θTηVδ2. But the curved region RSTVδ2 is likewise between these two rectilinear curves. Therefore, the difference between the rectilinear curves RξSλTμδ2 and γSθTηVδ2 is more than the difference of the curved regions RSTVδ2 and ερ ψΩΔΓ. But by supposition, the difference of these rectilinear figures is less than α. Therefore, the difference of the curved regions is much less than α, which is absurd. In fact, equality is shown. Therefore, the curved regions RSTVδ2 and ερψΩΔΓ do not differ. In fact, they are equal, which it was desired to demonstrate.

Second, we will demonstrate the same equality in the first figures, where we suppose the curve V always draws closer to the line 2δ in the progression from R to V and, in contrast, the curve 7D in the progression from 7 toΔ is protracted more from the same line. Therefore, on that account, the curve εΩ is more extended from the line ΓΔ in the progression from ε to Ω, since when 7D is extended from the line ΓΔ in the same manner the curve εΩ is extended from the line ΓΛ. If the curved regions are not equal, let their difference be α. Next, let the curved regions RSTVδ2 and let ερψΩΔΓ be divided by lines perpendicular to their bases 2δ and ΓΔ. It is done entirely as in the preceding demonstration, however, with the stipulation that the difference of the rectilinear figures RξS λTμδ2 and γSθTηVδ2 and also the difference of the rectilinear figures πρτψφ ΩΔΓ and εζρσψωΔ Γ together are less than the quantity α. It is manifest that this is possible, since this division is able to be done infinitely. With the same method which we used in the previous demonstration, it is demonstrated that the rectilinear figure RξS λTμδ2 is more than the rectilinear figure πρτψφΩΔΓ and the rectilinear figure γ SθT ηVδ2 is less than the rectilinear figure εζρσψωΔΓ.

With these things having been understood, if the given curved regions are not equal, let RSTVδ2 be more than the other. It is manifest that the excess of the rectilinear figure RξSλTμδ2 above the rectilinear figure εζρσψω ΔΓ is equal to all the rectangles γξ, θλ, ημ, πζ, τσ, and φω together with the excess of the rectilinear figure πρτψφ ΩΔΓ above the rectilinear figure γSθTηVδ2 having been removed. But the excess of the greater curved region above the lesser curved region is less than the previous-mentioned excess of the rectilinear regions, since the greater curved region is less than the greater rectilinear figure and the lesser curved region is more than the lesser rectilinear figure. Therefore, the excess of the greater curved region above the lesser curved region is less than the previously-mentioned rectangles together with the excess of the rectilinear figure εζρσψωΔΓ above the rectilinear figure γSθTηVδ2. But, by hypothesis, the rectangles together are less than the quantity α. Therefore, the rectangles together with the excess having been removed are much less than the quantity α. Hence, the greater curved region exceeds the lesser by much less excess than α, which is absurd. In fact, it is asserted that there is an excess of the greater curved region about the lesser. Therefore, the curved region RSTVδ2 is not more than the curved region εζρσψωΔΓ.

Let (if it is able to be done) it be less. It is manifest that the excess of the rectilinear region πρτψφΩΔ Γ above the rectilinear curve γSθTηVδ2 is equal to all of the rectangles πζ, πσ, φω, γξ, θλ, and νμ together with the excess of the rectilinear curve RξSλTμδ2 above the rectilinear curve εζρσψωΔΓ having been removed. But the excess of the greater curved region ερψΩΔΓ above the lesser curved region RSTVδ2 is less than the previously-mentioned excess of the rectilinear curves, since the greater curved region is less than the greater rectilinear figure and the lesser curved region is more than the lesser rectilinear figure. Therefore, the excess of the greater curved region above the lesser is less than the previously-mentioned rectangles together with the excess of the rectilinear figure RξSλT μδ2 above the rectilinear curve ερψΩΔΓ having been removed. But together the rectangles are, by hypothesis, less than α. Therefore, the rectangles together with the excess having been removed are much less than α. Hence, the greater curved region exceeds the lesser curved region by much less than α, which is absurd. In fact, it is asserted indeed that α is the excess of the greater curved region above the lesser curved region. Therefore, the curved region ερψΩΔΓ is not more than the curved region RξSλTμδ2. But it was also demonstrated that it is not less. Therefore, the curved regions ερψΩΔΓ and RξSλTμδ2 are equal among themselves, which is what ought to have been demonstrated.

There are other cases of this theorem, to all of which this second demonstration is able to be applied. However, I have wished to employ the preceding proof, since it appears more simple to me even if it is not general. Nevertheless, I remind the reader that the preceding proof is able to handle the most general case, namely in the first figures, with the remaining things staying the same while the curve RV is extended more from the line 2δ in the progression from R to V.

From this demonstration, it is manifest that the curved region RSTVδ2 and the surface of the trunk are proportional quantities in magnitude and in weight, since each equality which is demonstrated in the entirety likewise is demonstrated by the same method in their proportional parts. Therefore, their centers of equilibrium in the same manner divide the line δ2.

I do not reckon that there is a need to remind the reader that, with this one given surface of a trunk having been presented, all others whose cutting plane cuts the base in that same line (extended if needed) are given. In fact, such surfaces of trunks among themselves are as their altitudes or as the tangents of the inclining cutting planes, as is commonly and easily demonstrated.

Let OIGE be any simple and non-winding curve. Let two lines, AD and Lξ, be drawn in any way whatever parallel between themselves and let the line RV be normal to both of these. Let the curved region RVYBD be of such a nature that, with a line GSB drawn wherever you please parallel to the lines AD and Lξ, the line SB extended between the line RV and the curve VYBD is always equal to the line GM tangent to the curve at the point G and extended to the line Lξ. Above the curve OIGE (whose curvature is such that the farther it is extended from the point O the less it is separated from the line RV) let the surface of a right cylinder be imagined cut by a plane passing through the line Lξ and cutting the base of the cylinder in a 45 degree angle. I say that the area of RVYBD is equal to the surface of the lower trunk of the cylinder sectioned by the plane.

If they are not equal, let the difference between them be λ and let the line RV be divided into so many equal parts at the points R, S, T, and V, that, when the lines SC, Tδ, and VZ have been drawn parallel to the line RD and the rectangles RC, RB, Sδ, SY, and TZ have been filled out, the difference of the rectilinear figure Rα BXYT inscribed in the curved region with the rectilinear figure RDCBδYZV circumscribing the curved region is less than the quantity λ. Indeed, it is manifest that this is able to be done, since the line RV is able to be divided into more and more parts ad infinitum. Let the parallel lines DR, CS, δT, and ZV be extended to the points E, G, I, and O on the constructed curve. Let the lines EL, GM, IN, and 04, tangent to the curve at the points E, G, I, and O and which intersect the closest parallels in either direction at the points F, H, N, ω, θ, and μ, be extended to the points L, M, N, and O on the line Lξ. Let EQ, GK, and IP be perpendiculars to the nearest parallel lines.

It is manifest (because the lines GK and SV are parallel) that GK is to GH as SV is to GM--or SB. Consequently, the rectangle Sδ--that is, the rectangle formed by GK and SB--is equal to the rectangle formed by GH and SV. But the rectangle formed by GH and SV is more than the portion of the trunk of the surface above the curve GI, since the line GH is more than the curve GI and--because the plane cuts the base at a 45 degree angle through the line Lξ--the line SV is equal to the highest altitude of the portion of the surface of the cylinder above the curve GI. Therefore, the rectangle Sδ is more than the portion of the trunk of the surface above the curve GI. In the same manner, the rectangle RC is shown to be more than the portion of the trunk of the surface above the curve GE and the rectangle TZ is more than the surface of the trunk above the curve OI. Therefore, the rectilinear figure RDCBδYZV circumscribing the curved region is more than the entire surface of the trunk.

Next, because the lines IP and SV are parallel, as IP is to IN--or GK or ST is to θI--thus is TV to IN--or to TY. Consequently, the rectangle SY--that is, the rectangle formed by ST and TY--is equal to the rectangle formed by θI and TV. But the rectangle formed by θI and TV is smaller than the portion of the surface of the trunk above the curve IG, since the line θI is less than the curve GI and--because the plane cuts the base in a 45 degree angle through the line Lξ--the line TV is equal to the smallest altitude of that same portion trunk of the surface above the curve GI. Therefore the rectangle SY is less than the portion of the trunk of the surface above the curve GI. In the same manner, the rectangle RB is shown to be less than the portion of the trunk of the surface above the curve GE. Consequently, the rectilinear figure TYXBαR inscribed in the curved region is less than the entire surface of the trunk.

But, by supposition, the difference between the inscribed rectilinear figure and the circumscribing figure is less than the quantity λ. Therefore, the difference between the truncated surface and the curved region is much less than λ, since either of the two is demonstrated to be more than the inscribed rectilinear area and less than the circumscribed rectilinear area, which is not able to be done. It is supposed that λ is the difference between the surface of the trunk and the curved region. Therefore, there is no difference between the curved region and the surface of the trunk. Consequently, they are equal, which is what ought to have been demonstrated.

With these same things being assumed, let the curved region VYBD23ξ be of such a nature that from any point you please, say, G, on the curve EO, when the line, say, G2, has been drawn parallel to the line Lξ, the segment B2 between the two curves VD and Dξ is equal to the tangent to the given curve at the point G extended to the line AD, namely, the line Gμ. I say that the curved region VYBD23ξ is equal to the upper surface of the trunk of the cylinder given above whose altitude is the line RV.

Let G7, cutting the line RV at the point 7, be the line normal to MGμ at the point G. Because of the similarity of the triangles SG7 and GKH, as GS is to G7 thus is GK to GH. And as GK is to GH, thus is SV to GM--or SB--and also SR to Gμ--or B2. Therefore, GS is to G7 as RV is to S2. In this same manner it is able to be demonstrated that IT is to I8 as RV is to T3. And since this holds at all points of the curve EO, it is manifest from Proposition Two that the curved region RD2 3ξV is equal to the surface of the right cylinder above the curve EO and whose altitude is RV. But the surface of the lower trunk is equal to the curved region RVYBD and consequently the surface of the upper trunk is equal to the curved region DBYVξ3 2D, which it was desired to demonstrate.

Likewise, from this it is manifest that the surface of the lower trunk and the region VYBD are proportional quantities in magnitude and in weight, since this same equality which was demonstrated for the entirety is demonstrated in the same manner for their proportional parts. It is also manifest that the surface of the upper trunk and the region VYBD 2eξ are proportional quantities in magnitude and in weight. In fact, the curved region RVξ3 2 D is proportional in magnitude and weight to the entire surface of the right cylinder and the curved region RVYBD is proportional in magnitude and weight to the surface of the lower trunk. Therefore, (because it is left over) the curved region DBYVξ3 2 D is proportional to the remaining surface of the upper trunk.

The cases of this proposition are diverse, but in all of them the preceding conclusion is able to be verified in the same manner.

Let two curves AE and AD be drawn to the line AF and let the line AF be perpendicular to the line FD cutting the curves in the points E and D. Let the lines GE and CD be drawn tangent to the curves. I say that the lines EG and DC are not parallel.

Let them (if possible) be parallel and let the line AB be drawn parallel and equal to the line ED. Then through the points B and D let a curve agreeing in all parts with the curve AE be drawn, with the provision that the point A is placed above the point B and the point E is placed above the point D. It is manifest that the curve BD cuts the curve AD and that moreover the line CD parallel to the line GE is tangent to the curve AD. But by supposition CD is also tangent to the curve BD, which is absurd, since the curves AD and BD cut each other. Therefore, the lines CD and GE are not parallel, which is what ought to have been demonstrated.

It ought to be noticed that we show this important result for those simple curves which, the longer the distance from A, the greater the line ED cut off. For on this supposition depends the strength of the demonstration.

To find a curve which has the same ratio to its axis as any exhibited figure has to a rectangle inscribed in it and applied to a given line or axis of the sought-after curve.

Let the exhibited figure be ABSO and let the inscribed rectangle be ABRO. Let the curve BS be simple or non-winding, but if it is not, the curve ought to be divided into many simple parts and the demonstration carried out separately. Next, let the curve AFLP be of such a nature that when any line IN is drawn normal to the line AO and cutting the curve AFLP in L, the square on IN will equal a sum of the squares of both IL and IM. Next, let the curve AEKQ be drawn of such a nature that when any line IM is drawn perpendicular to the line AO and cutting the curve AEKQ in K and AFLP in L, the rectangle MIK is equal to the curved region IAFL. I say that the figure ABSO is to the rectangle ABRO as the curve AEKQ is to the line AO.

For let K be a point on the curve AEKQ, through which is drawn the
line IN perpendicular to the line AO and cutting the curves AFLP, BR,
and BHNS in the points L, M, and N. Let IK be to IC as IL is to IM
and let KC be drawn. The line KC either cuts or is tangent to the
curve AQ in the point K. If possible, let it cut the curve at K and
let it will fall within the curve at a point E toward the vertex A.
Through the point E let a line DH be drawn parallel to IN and cutting
AQ, AP, BR, and BS in the points E, F, G, and H, and the line KC in
α. Also, complete the rectangle ILZC, whose side LZ cuts the
line DH at X. Since IL is to IM as IK is to IC, the rectangle MIK--or
the curved region IAFL--will be equal to the rectangle IZ. Since the
rectangle GDE is equal to the curved region DAF, as IK is to DE thus
will the curved region IAFL be to the curved region DAF. But IK has a
greater ratio to DE than to Dα. Thus, the curved region IAFL has
a greater ratio to the curved region DAF than IK has to Dα--or IC
has to DC. Therefore, the curved region IAFL has a greater ratio to
the curved region DAF than the rectangle IZ has to the rectangle DZ,
and, *per conversionem rationis*, the curved region IAFL has a
smaller ratio to the curved region IDFL than the rectangle IZ has to
the rectangle IX. By permuting, the curved region IAFL has a smaller
ratio to the rectangle IZ than the curved region IDFL has to the
rectangle IX. Since the rectangle IZ is equal to the curved region
IAFL, the rectangle IX will be less than the curved region IDFL. But
it is also more than the curved region IDFL, which is absurd.
Consequently, the line KC does not fall within the curve AQ toward the
vertex.

If possible, let the line CK fall within the curve toward the base,
with the remaining things holding as in the previous situation. As IK
is to DE, thus will the curved region IAFL be to the curved region
DALF. But IK has a greater ratio to DE than to Dα. Therefore,
the curved region IAFL has a greater ratio to the curved region DALF
than IK has to Dα--or IC has to DC. Therefore, the curved region
IAFL has a greater ratio to the curved region DALF than the rectangle
IZ has to the rectangle DZ. Through inverting, *conversionem
rationis*, and inverting in turn, the curved region IAFL has a
greater ratio to the curved region IDFL than the rectangle IZ has to
the rectangle IX. By permuting, the curved region IAFL has a greater
ratio the rectangle IZ than the curved region IDFL has to the
rectangle IX. Since the curved region IAFL is equal to the rectangle
IZ, the rectangle IX will be more than the curved region IDFL. But it
is also less, which is absurd. Therefore, the line CK does not fall
within the curve AQ toward the base. Thus, the line KC is tangent to
the curve AQ in the point K.

Let the line KT, intersecting the line AO in T, be perpendicular to the line CK. It is manifest that CI is to CK as IK is to KT. But CI is to CK as MI is to NI, since the lines IN, IM, and IL are a right triangle similar to the triangle CIK, with sides IM and IN corresponding to the sides CI and CK. Consequently, as IK is to KT, thus is IM to IN. Since it is done in the same manner for all points of the curve AQ, it is manifest from Proposition Two that the line AO is to the curve AQ as the rectangle OB is to the figure ABSO, which it was desired to demonstrate.

The inverse of this proposition is also easily demonstrated. To be sure, if the line AO is to the curve AQ as the rectangle OB is to the figure ABSO--likewise, if the curve AP is of such a nature that when IN is drawn perpendicular to some line AO, its square will be a sum of the squares of IL and IM--the rectangle MIK will be equal to the curved region IAFL. If it is not thus, let a curve AVY be drawn of such a nature that the rectangle MIV is equal to the curved region IAFL. It will be demonstrated that the lines (which touch the curves AY and AQ in the points V and K) are parallel between themselves, which is a contradiction of the preceding proposition.

In fact, there are various cases of this proposition. But when this case is understood, no difficulty remains in the remaining cases.

To draw a line tangent to a given curve at a given point of the curve if the curve is from the category which Descartes calls Geometrical.

Let the curve BHC be a hyperbola whose diameter is the line AK and
whose ordinates EH and KC are of such a nature that the solid formed
by the square on BE together with AE is to the solid formed by the
square on BK together with AK as the cube on EH is to the cube on KC.
Let the given AB be a, let BE be b, and let the ratio of the solid
from the square on BE together with AE be to the cube on EH as
a^{3} is to c^{3}. It is desired to find a point F so
that the line FH touches the hyperbola in the point H.

From the given lines AB and BE, it is given that AE is a + b and EH is
^{3}√(a b^{2} c^{3} + b^{3}
c^{3})/a. Let EF be z and DE be nothing or o. Therefore, BD
will be b - o, AD will be a + b - o, and FD will be z - o. Therefore,
DG will be

^{3}√(c^{3} a b^{2} - 2 c^{3} a
b o + c^{3} a o^{2} + c^{3} b^{3} - 3
c^{3} b^{2} o + 3 c^{3} b o^{2} -
c^{3} o^{3})/ a.

Since we suppose that the ordinate DG falls upon the curve in the same point G where the line FH runs into the same curve (if it is able to be done), as EH is to EF thus will DG be to DF. Therefore, the rectangle formed by DF and EH, namely,

^{3}√(b^{2} c^{3} a z^{3} - 3
b^{2} c^{3} a z^{2} o + 3 b^{2}
c^{3} a zo^{2} - b^{2} c^{3} a
o^{3} + b^{3} c^{3} z^{3} - 3
b^{3} c^{3} z^{2} o + 3 b^{3}
c^{3} z o^{2} - b^{3} c^{3}
o^{3})/ a

will be equal to the rectangle formed by EF and DG, namely,

^{3}√(c^{3} a b^{2} z^{3} - 2
c^{3} a b z^{3} o + c^{3} a z^{3}
o^{2} + c^{3} b^{3} z^{3} - 3
c^{3} b^{2} z^{3} o + 3 c^{3} b
z^{3} o^{2} - c^{3} z^{3}
o^{3})/ a.

By taking away the denominators because they are equal and also by cubing each end of the equation and taking away the equal things, the equation

3 b^{2} c^{3} a z o^{2} - 3 b^{2}
c^{3} a z^{2} o - b^{2} c^{3} a
o^{3} - 3 b^{3} c^{3} z^{2} o + 3
b^{3} c^{3} z o^{2} - b^{3}
c^{3} o^{3}

= c^{3} a z^{3}
o^{2} - 2 c^{3} a b z^{3} o - 3 c^{3}
b^{2} z^{3} o + 3 c^{3} b z^{3}
o^{2} - c^{3} z^{3} o^{3}

results. By dividing everything through by o, this is

3 b^{2} c^{3} a z o - 3 b^{2}
c^{3} a z^{2} - b^{2} c^{3} a
o^{2} - 3 b^{3} c^{3} z^{2} + 3
b^{3} c^{3} z o - b^{3} c^{3}
o^{2}

= c^{3} a
z^{3} o - 2 c^{3} a b z^{3} - 3 c^{3}
b^{2} z^{3} + 3 c^{3} b z^{3} o -
c^{3} z^{3} o^{2}.

And by rejecting the quantities in which o or a power of o is found,

-3 b^{2} c^{3} a z^{2} - 3 b^{3}
c^{3} z^{2} = -2 c^{3} a b z^{3} - 3
c^{3} b^{2} z^{3}

is left over. By adding the defects and by dividing everything
through by c^{3} b z^{2}, the equation is 3 b a + 3
b^{2} = 2 a z + 3 b z. Consequently, the line EF, namely, z =
(3 b a + 3 b^{2})/(2 a + 3 b), which is what ought to
have been found.

Let ADIM be a curve whose axis is AL and let AFKO be another curve of such a nature that when any line HIK is drawn perpendicular to the line AL the curve AI is to the line IK as P is to Q. It is desired to draw a line tangent to the curve AFKO at the point K.

Let a line BI be drawn tangent to the curve ADIM at the point I (we suppose in fact that this is able to be done) and let the line IB be equal to the curve AI. Let the line BK be drawn. I say this line is tangent to the curve AFKO in the point K.

If it does not touch the curve, let it fall within the curve and let G
be a point on the line within the curve in the direction of the
vertex. Let a line GFEDC be drawn parallel to KH. It is manifest
that IK is to EG as IB is to EB and, *per conversionem
rationis*, IK is to the difference between IK and EG as IB is to
IE. By permuting, as IK is to IB (or the curve IA)--that is, as Q is
to P--thus the difference between IK and EG is to EI. But as Q is to
P, thus DF is to the curve DA. Therefore, as IK is to the curve IA
thus DF is to the curve DA. By permuting, as IK is to DF, thus the
curve IA is to the curve DA. *Per conversionem rationis*, as
IK is to the difference between IK and DF, thus the curve IA--or the
line IB--is to the curve ID. But the difference between IK and EG is
more than the difference between IK and DF, since the point G is
supposed to lie within the curve. Hence, IK has a smaller ratio to
the excess beyond EG than to the excess beyond DF. Therefore, IB is
in a smaller ratio to IE than to the curve DI. Therefore, IE is more
than the curve DI, which is absurd. (See Proposition One.) Therefore,
the line BK does not fall within the curve AFKO toward the vertex.

Let the line fall inside the curve in the direction of the base at the point R (if it is able to be done). IK is to NR as IB is to NB, and IK is to the difference between IK and NR as IB is to IN. By permuting, as IK is to IB (or to the curve IA)--that is, as Q is to P, thus the difference between IK and NR is to IN. But as Q is to P, thus MO is to the curve MA. Therefore, as IK is to the curve IA thus MO is to the curve MA. By permuting, as IK is to MO thus the curve IA is to the curve MA. As IK is to the difference between IK and MO thus the curve IA--or IB--is to the curve IM. But the difference between IK and NR is less than the difference between IK and MO since we suppose that R falls within the curve. Hence, IK has a greater ratio to the difference between IK and NR than to the difference between IK and MO. Therefore, IB is in a greater proportion to IN than to IM. Consequently, IN is less than IM, which is absurd. Hence, the line BK does not fall within the curve toward the base and hence touches the curve in the point K, which it was desired to demonstrate.

Through this proposition curves of all cycloids are able to be compared with their own axes or bases following the method of Proposition 2.

Let AEIO be a curve whose axis is AR and let AGMT be another curve of such a nature that when any line NIM is drawn perpendicular to the line AR, the curve AI is to the line NM as P is to Q. It is desired to draw a line tangent to the curve AGMT at the point M.

Let a line IC be drawn tangent to the curve AEIO at I (we suppose in fact that this is able to be done) and intersecting the line AD parallel to NM at C. Let MZ be to the line IC as Q is to P and let ZD be parallel to the line AR. Let the line DM be drawn. I say this line is tangent to the curve AGMT at the point M.

If it were not tangent, let it fall within the curve and let H be a point within the curve toward the vertex. Let a line be drawn parallel to NM and cutting the remaining lines as in the figure. AE is to BG as P is to Q and AI is to NM as P is to Q. Consequently, as AE is to BG, thus AI is to NM. By permuting, as AE is to AI, thus BG is to NM, and as AE is to EI thus BG is to KM. By permuting, as AE is to BG--that is, as P is to Q--thus EI is to KM. Therefore, as CI is to ZM--that is, as P is to Q--thus FI is to LM, which I demonstrate as follows: The ratio of CI to ZM is compounded of the ratio of CI to DZ and the ratio of DZ to ZM, and the ratio of FI to LM is compounded of the ratio of FI to HL--or CI to DZ--and the ratio of HL to LM--or DZ to ZM. Consequently, as EI is to KM, thus FI is to LM. By permuting, as EI is to FI thus is KM to LM. But since we suppose that H falls within within the curve, KM is less than LM and consequently EI would be less than FI, which is absurd. (See Proposition One.) Therefore, the line DM does not drop within the curve toward the vertex.

Let it fall within the curve in the direction of the base at the point V (if it is able to be done). AI is to NM as P is to Q, and AO is to RT as P is to Q. Therefore, AI is to NM as AO is to RT. By permuting, AI is to AO as NM is to RT, and as AI is to IO thus NM is to XT. By permuting, as AI is to NM--or as P is to Q--thus IO is to XT. Consequently, as CI is to ZM--or P is to Q--thus IS is to XV (which is proved as in the previous case). Therefore, as IO is to XT thus IS is to XV. By permuting, as IO is to IS, thus XT is to XV. But (since we suppose that V falls inside the curve) XT will be more than XV. Therefore IO will be more than IS, which is absurd. (See Proposition One.) Therefore, the line DM does not fall inside the curve in the direction of the base. Consequently, DM is tangent to the curve at the point M, which it was desired to demonstrate.

Through this proposition, the curve of any surface of a truncated right cylinder expanded into a plane is able to be compared with its axis or the base with the aid of Proposition Two, if the tangent line to the base of the cylinder is able to be drawn at a given point.

Let AD be any curve and let BN be any line. From any two points A and D of the curve let two parallel lines be drawn to the line BN, namely, AB and DE, and let the line AD be brought together and extended to H. Let the lines DG and AN be drawn tangent to the curve in the points A and D. Next, let the parallelograms ABGO and DENS be completed and let the lines AO and NS be extended so that they intersect at Q. I say that the trapezoid ADEB is more than the curved region ADLO.

Let HK be drawn parallel and equal to the line AB. It is manifest that the trapezoid ADEB is equal to the trapezoid ADMK. Also, the trapezoid ADMK is more than the trapezoid ADLO and therefore is much more than the curved region ADLO. Therefore, the proposition is clear, namely, that the trapezoid ADEB is more than the curved region ADLO.

Let the line DG be extended to C. I say that the rectilinear region ABEDC is less than the curved region ADSQ.

Let the line CF be drawn parallel to the lines AB and DE and let CR be drawn parallel to the lines AQ and DS. It is clear that the trapezoid ABFC is equal to the trapezoid ACRQ and that the trapezoid CFED is equal to the trapezoid CDLI. Consequently, the rectilinear region ABEDC is equal to the rectilinear region ACDLIRQ, which is less than the rectilinear region ACDSQ. Therefore, the rectilinear region ABEDC is much less than the curved region ADSQ, which is what ought to have been demonstrated.

Let ABKI be any region contained by the curve BK, the line AI, and two parallel lines BA and KI. Let the curve MY be of such nature that (with some point C on the curve BK having been assumed and out of that point a line CE having been drawn parallel to the line AB and a line CZ having been drawn tangent to the curve BK and terminating at the line AI, extended to Z if necessary) the line EZ is always equal to the line CS parallel to the line AZ and terminated at the curve YM. I say that the curved region BKMY, contained by the curves BK and MY and the lines BY and KM parallel to the line AZ, is equal to the curved region BAIK.

If they are not equal, let X be their difference and let the curvilinear figure BKMY be divided by so many lines CS, GP, and KM, parallel to the line BY, that (with the lines OM, QN, TR, and YV drawn parallel to the line AB) all the parallelograms ON, QR, and TV together are less than X. This is able to be done by the indefinite number of parallels. Let the subtending lines BC, CG, and GK and the tangents, BD, DF, FL, and KL at the points B, C, G, and K be drawn. It is manifest from the preceding proposition that the trapezoid ABCE is more than the curved region BCST, the trapezoid CEHG is more than the curved region CGPQ, and the trapezoid GHIK is more than the curved region GKMO. Therefore, the rectilinear figure ABCGKI is more than the curved region BKMOPQST. It is also clear from the preceding proposition that the rectilinear figure ABDCE is less than the curved region BCVY, the rectilinear figure ECFGH is less than the curved region CGRS, and the rectilinear figure HGKI is less than the curved region GKNP. Consequently, the rectilinear figure ABDFLKI is less than the curved region BKNPRSVY. Therefore, since the curved region BAIK is between the rectilinear figures ABCGKI and ABDFLKI and the curved region BKMY is between the curved regions KMOPQSTB and KNPRSVYB, and likewise the rectilinear figures ABCGKI and ABDFLKI are between the curved regions BKMOPQST and BKNPRSVY, it is manifest that the curved regions ABKI and KMYB differ by a smaller quantity than the curved regions BKMOPQST and KNPRSVYB. But, by supposition, the difference of these is less than X. Therefore, the difference of the areas ABKI and BKMY is much less than X, which is absurd. It is supposed that their difference is more than X. Therefore, there is no difference between the curved regions ABKI and BKMY. Hence they are equal, which is what ought to have been demonstrated.

The demonstration of the two preceding results would be nearly the same even if the convexity of the curve BK were in the direction of the line AI.

Let ABFE be a figure comprised of the parallel lines AB and EF, the line AE normal to the parallels, and any curve BF you wish. Likewise, let the figure GHK by comprised of the lines GH and GK (in such a way that GH is equal to the line AB and GK is equal to the line EF) and the curve HK which is equal to the curve BF by the rule that, with any curves BD and HI having been assumed, the joined line GI is equal to the line DC perpendicular to the line AE. I call the figure GHK the involute of figure ABFE and the figure ABFE the evolute of the figure GHK.

I also call the points B and H, or D and I, or F and K mutually relative to themselves.

I call the point G the center of involution.

I call the angle HGK the angle of involution.

I call the line AE the axis of the evolute figure; also, a CD perpendicular to that line I call an ordinate to the axis.

Let AFGB be the rectangle which, when involuted, makes the sector of the circle BEG. Likewise, let ADHB be the rectangle which, when involuted, makes the sector BCH. I say that the angle of involution CBH is more than the angle of involution EBG.

In fact, as BG is to BH thus the arc EG--or the arc CH--is to the arc OH. But BG is more than BH and therefore CH is more than OH. Thus, the angle CBH is more than the angle EBG, which ought to have been demonstrated.

Let ABMI be a figure which, when involuted, makes the figure NPX. Let PT be the arc of a circle from the center of involution N. I say that the arc PT is less than the line AI, the axis of the evoluted figure.

Let BK be drawn parallel and equal to the line AI, and let EGLI be a rectangle such that the rectilinear figure ABFGLI is inscribed in the figure ABMI. If ABMI is involuted, the angle of involution will be less than the angle of involution of the rectangle ABKI. Now this is noted thusly: The rectilinear figure ABFGLI involuted is the same as the rectangle ABFE involuted together with the rectangle EGLI involuted. The rectangle ABKI involuted is the same as the previous rectangle ABFE involuted together with the rectangle EFKI involuted. But from the preceding result the angle of involution of the rectangle EGLI is less than the angle of involution of the rectangle EFKI. Therefore, the angle of involution of the involuted rectilinear figure ABFGLI is less than the angle of involution of the involuted rectangle ABKI. In exactly the same manner (if the lines QR and SV are drawn parallel to the lines AB and IM and the lines Rθ and VY are drawn parallel to the axis AI so that they might fill up the rectilinear figure ABQRθGSVYI) it will be demonstrated that the angle of involution of this figure is less than the angle of involution of the rectilinear figure ABFGLI. Consequently, it will be much less than the angle of involution of the rectangle ABKI--namely, γNT--for we assumed that the sector γNT is the rectangle ABKI involuted.

Finally, it is always demonstrated in the same manner that the less the inscribed rectilinear figure differs from the figure ABMI, the more will the excess of the angle γNT always be above the angle of involution of the rectilinear figure. Therefore, the angle γNT of the figure ABMI itself exceeds by much more the angle of involution PNX. Consequently, the axis AI of the evolute figure--that is, the arc γT--exceeds the arc PT, which ought to have been demonstrated.

Second, let OX be the arc of the circle from the center of involution N. I say that the arc OX is more than the line AI--that is, the axis of the evoluted figure PNX.

Let the line MD be drawn parallel and equal to the line AI and let AB be extended to D. Let ACGE be a rectangle such that the rectilinear figure ABMI is circumscribed in the figure ACGHMI. If ACGHMI is involuted, the angle of involution will be more than the angle of involution of the rectangle ADMI. Now this is noted thusly: The rectilinear figure ACGHMI involuted is the same as the rectangle EHMI involuted together with the rectangle ACGE involuted, and the rectangle ADMI involuted is the same as the previous rectangle EHMI involuted together with the rectangle ADHE involuted. But from the preceding proposition, the angle of involution of the rectangle ACGE is more than the angle of involution of the rectangle ADHE. Consequently, the angle of involution of the rectilinear figure ACGHMI involuted is more than the angle of involution of the rectangle ADMI. In exactly the same manner (if the lines Rβ and VZ are drawn parallel to the lines AB and IM and the lines Rα and VX are drawn parallel to the axis AI so that they fill up the rectilinear figure AαRβGXVZMI) it will be demonstrated that the angle of involution of this figure is more than the angle of involution of the rectilinear figure ACGHMI. Consequently it will be much more than the angle of involution of the rectangle ADMI--that is, δNX--for we suppose that the sector δNX is the rectangle ADMI involuted. Finally, it is always demonstrated in the same manner that the less the inscribed rectilinear figure differs from the figure ABMI, the more will the excess of the angle of involution of the rectilinear figure always be above the angle δNX. Therefore, the angle of involution of the figure PNX itself exceeds by much more the angle δNX of the figure ABMI itself. Consequently, the axis AI of the evolute figure--that is, the arc δX--is much less than the arc OX, which it was desired to demonstrate.

From a given involuted figure, to find its axis of evolution.

Let LBK be the involuted figure, the axis of evolution of which it is desired to find. Let MK be the arc of a circle with center L, and let the figure OP78 be contained by the parallel lines OP and 87, by the line 08 cutting those lines normally, and by the curve P7 of such a nature that (with any line LC in the involuted figure extended to N, and likewise the line SV in the figure OP78 drawn perpendicular to the line O8 and cutting it in the ratio of MN to NK) 87 is to SV as LK is the LC. I say that the circumscribed rectangular figure OR78 is to the figure OP78 as the arc MK is to the axis of the figure LBK evoluted.

If it were not thus, let OR78 be to OP78 thus as MK is to α, which differs from the axis of the figure LBK evoluted by the quantity δ. Next, from the center L let similar circular arcs AC, EG, and HK be circumscribed around the involute figure LBK and let just as many similar circular arcs BD, CF, and GI be inscribed within it so that the difference between the inscribed arcs and the circumscribed arcs is less than δ.

Next, let the line 08 be divided into as many equal parts OS, S4, and 48 as the arc MK by the extended lines LC and LG. Let lines SY and 4Z be drawn perpendicular to O8 itself and cutting the curve P7 in the points V and 2. And let PT, QV3, and X26 be joined parallel to the line O8. It is manifest from the description of the figure OP78 that SY is to SV as LN is the LC--that is, as the arc MN is to the arc AC. Therefore, as the rectangle OY is to the rectangle OV thus the arc MN is to the arc AC. In the same manner it is proved that as the rectangle SZ is the rectangle S2 thus the arc NH--that is, the arc MN--is to the arc EG, and as the rectangle 47 is to the rectangle 47 thus the arc HK is to the arc HK.

Since all the firsts are equal among themselves and all the thirds are equal among themselves, as all firsts--that is, the rectangle O7--are to all seconds--that is, the rectilinear figure OQVX2Z78--thus all thirds--that is, the arc MK--will be to all fourths--that is, the arcs AC, EG, and HK. Moreover, as 07 is to the figure OP78 thus MK is to α. But the rectilinear figure OQVX2Z78 is more than the figure OP78. Therefore the arcs AC, EG, and HK together are more than α. But the arcs AC, EG, and HK are also more than the axis of the evolute figure LKB, which is proved thusly: The axis of the entire figure LBK evoluted is equal to the axes of the figures BLC, CLG, and GLK evoluted. But from the preceding proposition the axis of the figure LBC evoluted is less than the arc AC, and the axis of the figure CLG evoluted is less than the arc EG, and likewise the axis of the figure GLK evoluted is less than the arc HK. Therefore, the axes of all the partial figures taken together--that is, the axis of the figure LBK evoluted--is less than all the arcs AC, EG, and HK taken together.

Next, from the description of the figure OP78, as OR is to OP--or OY is to OT--thus LM is to LB--that is, MN to BD. In the same manner it is demonstrated that as SZ is to S3 thus NH is to CF, and as 47 is to 46 thus HK is to GI. Since the firsts are equal among themselves and the thirds are equal among themselves, as all the firsts--namely, O7--are to all the seconds--namely, the rectilinear figure OPTV3268--thus all the thirds--namely, MK--will be to all the fourths--namely, the arcs BD, CF, and GI. Since O7 is to the figure OP78 as MK is to α and the rectilinear figure OPTV3268 is less than the figure OP78, all the arcs BD, CF, and GI will together be less than α. But the arcs BD, CF, and GI are also less than the axis of the evolute figure LBK, which is proved thusly: the axis of the entire figure LBK evoluted is equal to the axes of the figures BLC, CLG, and GLK evoluted. But from the preceding proposition, the axis of the figure LBC evoluted is more than the arc BD, the axis of the figure CLG evoluted is more than more than the arc CF, and the axis of the figure GLK is more than the arc GI. Therefore, the axis of all the partial figures taken together--or the axis of the evolute figure LBK--is more the all the arcs BD, CF, and GI taken together.

Therefore, it is evident that there are four magnitudes--namely, the first all the arcs BD, CF, and GI together, the second the axis of the figure LBK evoluted, the third α, and the fourth all the arcs AC, EG, and HK together--the maximum and minimum of which are the arcs AC, EG, and HK together and all the arcs BD, CF, and GI together. Therefore the difference of these is more than the difference of the two remaining quantities--namely, the axis of the figure LBK and the quantity α, which is absurd, for it is supposed even less. Therefore, there is no difference between α and the axis of the evolute figure LBK. Therefore, they are equal, which it was desired to demonstrate.

Hence it is manifest that it is possible from the given involuted figure to find the evolute of the same figure, for from this proposition the axis of the evolute figure LBK or some part BLC of this (when evoluted) is given, and the ordinates are given since they are the same as the lines between the center of involution L and their respective points in the evolute figure LBK.

In the preceding figure it is desired to find the ratio between the sector MLK and the figure BLK.

Let the figure OP78 be contained by the parallel lines OP and 87, with the line O8 cutting them normally, and let the curve P7 be of such a nature that (with any line LC extended to N in the assumed involute and likewise with the line SV drawn in the figure OP78 perpendicular to the line O8 and cutting it in the ratio of MN to NK) 87 is to SV in the duplicate ratio of the line LK to LC. I say that the circumscribing rectangle OR78 is to the figure OP78 as the sector LMK is to the involute LBK.

If it is not thus, as OR78 is to OP78 thus let MLK be to α, which differs from the involute BLK by the quantity δ. Next, let similar circular sectors LAC, LEG, and LHK be circumscribed around the involuted figure BLK and let as many similar circular sectors LBD, LCF, and LGI be inscribed in the same figure so that the differences between the inscribed figure LBDCFGI and the circumscribed figure LACEGHK is less than δ. Next, let O8 be divided into as many equal parts OS, S4, and 48 as the arc MK by the extended lines LC and LG, and let lines SY and 4Z be drawn perpendicular to O8 itself and cutting the curve P7 in the points V and 2. Let PT, QV3, and X26 be joined parallel to the line O8. It is manifest from the description of the figure OP78 that SY is to SV--that is, OY is to OV--in the duplicate ratio of LN to LC--that is, as the sector LMN is to the sector LAC. It is proved in the same manner that SZ is to S2 as LNH is to LEG, and 47 is to 47 as LHK is to LHK. Since all the firsts are equal among themselves and all the thirds are equal among themselves, as all the firsts--namely, the rectangle O7--are to all the seconds--namely, the rectilinear figure OQVX2Z78--thus all the thirds--namely, the sector MLK--will be to all the fourths--namely, the figure LACEGHK. Moreover, as O7 is to the figure OP78 thus the sector MLK is to α. But the rectilinear figure OQVX2Z78 is more than the figure OP78. Therefore, the figure LACEGHK is more than α.

Next, from the description of the figure OP78, OR is to OP--that is, OY is to OT--in the duplicate ratio of LM to LB--that is, as MLN is to BLD. It is proved in the same manner that SZ is to S3 as NLH is to CLF and 47 is to 46 as HLK is to GLI. Since the firsts and thirds are equal among themselves, all the firsts--namely, 07--will be to all the seconds--namely, the rectilinear figure OPTV3268--thus as all the thirds--namely, MLK--will be to all the fourths--namely, the figure LBDCFGI. But the rectilinear figure OPTV3268 is less than the figure OP78. Therefore, the figure LBDCFGIL is less than α. Therefore, it is evident that there are four magnitudes--namely, the first the figure LBDCFGI, the second the involute LBK, the third α, and the fourth the figure LACEGHK--the maximum and minimum of which are the figures LACEGHK and LBDCFHI. Therefore, the difference of these is more than the difference of the two remaining--namely, α and the involute LBK--which is absurd, for it is supposed less. Therefore, there is no difference between α and the involute BLK. Therefore, they are equal, which it was desired to demonstrate.

Each evolute figure is twice its involute.

Let LBK be the involuted figure, which, when evoluted, makes the figure OP78. I say that the figure OP78 is double the figure LBK.

If it is not thus, let OP78 be the double of the quantity α, which differs from the involute LBK by the quantity δ. Let the figure LBDCFGI be inscribed in involute LBK and let the figure LACEGHK be circumscribed around the same so the difference of those is less than δ. Let B, C, G, and K be points on the evolute relative to P, V, 2, and 7, and let the rectangles OV, OT, S2, S3, 47, and 46 be filled out. The line LC is equal to the line SV and the curve AC is more than the line QV. Therefore, the sector of the circle LAC is more than half the rectangle OV. In the same manner, it is proved that the circular sector LGE is more than half the rectangle S2 and the circular sector LHK is more than half the rectangle 47. Therefore, the figure LACGHK is more than half the figure OQVX2Z78. Consequently, it is much more than half of the evolute OP78--namely, α.

Furthermore, the line OP is equal to the line LB and the curve BD is less than the line PT. Therefore, the sector of the circle LBD is less than half the rectangle OT. In the same manner it is demonstrated that the sector LCF is less than half the rectangle S3 and the sector LGI is less than half the rectangle 46. Therefore, the figure LBDCFGI is less than half the rectilinear figure OPTV3268. Consequently, it is much less than half the evolute--namely α. Therefore, it is evident that there are four quantities--namely, the first the figure LBDCFGI, the second the involute LBK, the third α, and the fourth the figure LACFGHK--the maxima and minima of which are the figures LACEGHK and LBDCFGI. Therefore ,the difference of these is more than the difference of the two remaining--namely, α and the involute LBK--which is absurd, for it is assumed less. Therefore, there is no difference between α and the involute LBK. Therefore, they are equal, which it was desired to demonstrate.

Let ABG be an involuted figure. Let the line AG be extended and let the line BK, which falls completely outside the curve BG, be perpendicular to it. I say that the line BK is not less than the axis of the evolute of the figure ABG.

If it is able to be done, let it be less than the axis and let the axis of the evolute of the figure ABC be less than the excess of the axis of the evolute of the figure ABG above the line BK. From the center A let the circular arcs CI and HO be drawn. It is manifest that the arc CI is more than the axis of the evolute of the figure CAH and that HO is more than the axis of the evolute of the figure HAG. But the line DI is more than the arc CI and the line IK is more than the arc HO. Therefore, the line DK is much more than the axis of the evolute of the figure CAH together with the axis of the evolute of the figure HAG.

That is, the line DK is more than the axis of the evolute of the figure CAG. But the axis of the evolute of the figure BAG is above the line BK by a greater excess than the axis of the evolute of the figure BAC. Consequently, the axis of the evolute of the figure BAG minus the axis of the evolute of the figure BAC--that is, the axis of the evolute of the figure CAG--is more than the line BK. But it is also less than the line DK, which is absurd. Therefore the line BK is not less than the axis of the evolute of the figure BAG, which ought to have been demonstrated.

From the point G let the line GF be perpendicular to the line AB. I say that AB is less than the axis of the evolute of the figure ABG.

Let GE be a circular arc from the center A. GE is more than the line GF and less than the axis of the evolute of the figure ABC. Therefore, the line GF is much less than the axis of the evolute of the figure ABG, which it was desired to demonstrate.

Let ACRQ be an evoluted figure, in which OI is any given ordinate. Next, with the line OI remaining, let the figure ACRQ be involuted into the involuted figure IBOP. I say that the curve BOP falls between the curve AOQ and its axis CR. Consequently, the curves AOQ and BOP are mutually tangent at the point O.

If it is able to be done, let the curve OP fall outside the curve OQ at the point N: I suppose that the curve OQ (when closer to Q) has a greater distance from the axis IR. Therefore, the line NM perpendicular to the extended line IO falls outside the curve ON. Consequently (from the preceding result) the line NM is not less than the axis of the evolute of the figure ION. In the evolute, let the ordinate VS be equal to the line IN. It is manifest that the figure OVSI is the evolute of ION.

But IN--that is, SV--is more than LT and therefore SI is more than IL. That is, the axis of the evolute figure ION is more than MN, which is absurd. Consequently, OP does not fall outside of OQ. In the same manner it is proved that OP does not coincide with OQ. Therefore, it falls within it, which it was desired to demonstrate.

Second, if it is able to be done, let the curve OB fall outside the curve OA at the point D. I suppose that the curve OA (when closer to A) has a lesser distance from the axis CI. Let the perpendicular DF fall on the line OI within the curve DO. Therefore, DF is less than the axis of the evolute of the figure IDO. In the evolute, let the ordinate EH equal the line DI. It is manifest that the figure OEHI is the evolute of the figure IOD, but ID--that is, HE--is more than the line KG and therefore GI is more than HI. In particular, DF is more than the axis of the evolute figure IDO, which is absurd by the preceding proposition. Therefore, BO does not fall outside of AO. In the same manner it is proved that OB does not coincide with OA. Therefore, it falls within it, which it was desired to demonstrate.

If the perpendicular DF falls on the line IO extended, it is not possible that IOB is the involute of the figure IOAC, since ID will be more than any ordinate of the figure IOAC.

Since the figures CAOQR and IBOP are mutually tangent in the point O, it is manifest that the line tangent to one of these figures at the point O is also tangent to the other at the same point. And hence this is evidently a method for drawing a line which is tangent to the involute at a given point, if only a method for drawing the line which is tangent to the evolute in a point is given, and conversely. Indeed, the line touching the evolute inclines at the same angle toward the ordinate as the line touching the involute inclines to the same ordinate in the center of involution coinciding with the remaining things.

All things mentioned previously about the involute of the figure are demonstrated in the same manner when the evolute curve is turned toward the axis, although in our figures the curve of the evolute is concave toward the axis.

Let AB be a figure above which is imagined a right cylindrical figure cut by a plane passing through the line FG and cutting the plane of the base AB seminormally. Let ML be a line parallel to the line FG and above the line ML let MOLN be a region of such a nature that (with any line EDCNO having been drawn normal to the line FG and cutting the figures AB and LOMN in the points D, C, N, and O) a designated line P is to the arithmetic mean between DE and CE as DC is to NO. I say that the right cylinder whose base is MOLN and whose altitude is the line P is equal to the lower trunk of the cylinder above AB cut as stated above.

Indeed, since P is to the arithmetic mean between DE and CE, as DC is to NO, therefore the rectangle formed by P and NO is equal to the trapezoid formed by the lines DC, CE, and DE, with right angles at D and C. But such a trapezoid is the common section of the plane above the line EO perpendicular to the base AB with the lower trunk of the cylindrical figure. Since it will always be this way wherever the line EDCNO is drawn, it is manifest from the doctrine of Gregory of Saint Vincent that the right cylindrical figure whose base is MOLN and altitude is P is equal to the lower trunk of the right cylindrical figure of the line above AB cut as stated above, which it was desired to demonstrate.

Hence it is also apparent that the cylinder whose base is MOLN and altitude is P is proportional in magnitude and weight to the previously-mentioned trunk. Since the cylinder is proportional to its own base, it is also apparent that the trunk is proportional in magnitude and weight to the same base.

With the same things having been assumed as in the preceding result, let FZ be a line normal to FG, above which is a region QTVR of such a nature that (with any line RTDS drawn normal to the line FR and intersecting the figures QTVR and AB in the points T, D, and S) a designated line P is to DS as the line RF is to the line RT. I say that the right cylinder whose base is QTVR and altitude is P is equal to the lower trunk of the cylinder above AB cut as stated above, namely, passing through the plane in the line FG and cutting the base AB seminormally.

Since P is to DS as RF is to RT the rectangle from P and RT will be equal to the rectangle from RF and DS. But the rectangle from RF and DS is the common intersection of the lower trunk of the cylinder with the plane above the line RS normal to the base AB.

Therefore, the rectangle formed by P and RT is equal to that same common section. Since it will always be this way wherever the line RS is drawn, it is manifest from the doctrine of Gregory of Saint Vincent that the right cylinder whose base is QTVR and altitude P is equal to the lower trunk of the right cylinder above AB cut as stated above, which it was desired to demonstrate.

Hence, it is also evident that the cylinder whose base is QTVR and altitude is P is proportional in magnitude and weight to the previously-mentioned trunk of the right cylinder. Consequently, the base of the right cylinder QTVR is also proportional to the same trunk in magnitude and weight.

If it is assumed that the squares and centers of gravity of all the figures have been given, it will be easy to find the cubature and center of gravity of any truncated cylinder you please, or conversely, from this and the preceding proposition. In the same manner from the second, third, and fourth propositions, with the squares and centers of gravity of all figures having been given, it is not difficult to find the quadrature and center of gravity of the surface of any trunk of a right cylinder or conversely, which it is sufficient to have recalled here.

Let any solid of revolution be cut through the diameter AD by a plane normal to the circular base BOCDN, which has diameter BC and intersection with the plane making the figure ABC. Let the solid of revolution ABC be cut by another plane FNGO normal to the plane AB in such a way that the common intersection of the solid and the plane makes the figure FNGO, whose intersection with the plane ABC is the line FG. Let FK be drawn parallel to the line BC and intersecting the line AD in I. Let a solid of revolution be conceived having the diameter FG and consisting of circles whose radii are all the perpendiculars from the line FG to the curve FO in such a way that the diameter FG of the figure passing through the centers of those infinitely many circles inclines toward all of them in an angle equal to FGB itself. From these givens, the solid ABC and the points F and G, it is desired also to display the measure of the other solid of revolution on FOGN.

Let AD cut the line FG in E and let the line XEZ parallel to the line BC be drawn through E. From the given points F and G, the point E is given, since the figure ABC is assumed to be given. Below XZ let a plane be drawn parallel to the base BOCN and making the lines PQ and RS the common intersection with the planes ABC and FOGN. It is manifest that RS is the radius of the circle in the solid of FOGN passing through R. And on account of the revolution of the solid ABC and the normal section of the plane ABC to the base, the point S is on the semicircle whose diameter is PQ. Therefore, the square on the line RS is equal to the rectangle PRQ, and the circle with radius RS is equal to the circular ring on PRQ. [EDITOR'S NOTE: Use Euclid II.5.]

Since it will always be this way between E and G, the portion of the solid of revolution on FOGN beneath E is equal to all the circular rings between the circle XEZ and the ring on BGC--that is, to the hollowed-out solid equal to the portion of the solid of revolution on XZCB above the removed cone whose vertex is E and whose base is the radius DG.

It is proved in the same manner that the portion of the solid of revolution FOGN between E and F is equal to the hollowed-out solid on FKZX above the removed cone whose vertex is E and whose base radius is IF. Therefore, if from the given portion FKCB of the solid of revolution four times as much toward the vertex of the given cones FEI and DEG were removed (NOTE: his translation.), the sought-after solid of revolution FOGN will be left behind.

Let the center of gravity of the solid of revolution ABC and the center of the removed figure AFK be given. Also, let the center of gravity of the portion on FKCB be given. With these having been given together with the center of gravity of the removed cone, the center of gravity of the hollowed-out portion, which is proportionally analogous with the solid of revolution FOGN, is also given, as is evident from the demonstration. Therefore, the center of gravity of the solid of revolution on FOG is also given.

There are even more cases of this proposition, but with this having been understood, no difficulty remains in the other cases.

Let a solid of revolution be cut through the diameter AE by a plane normal to the circular base in the diameter BC and making the figure BAC in the intersection with the plane. Let FAEG be a parallelogram and let a curve ALD be described whose nature is such that with any line HL parallel to the base BC having been drawn, the lines HK, IK, and KL are in a continuing proportion. From the given ratio of the solid of revolution on ABC to the given cylinder, it is desired to find the quadrature of the figure ALDE.

Let any line HIKLN be drawn. The square on the side IK--that is, the rectangle HKL--is a quarter part of the square on IN. Therefore, the rectangle HKL is to the circle on the diameter IN in a ratio compounded from the ratio of 1 to 4 and the square of the diameter of a circle to the circle. But the point K was assumed arbitrarily. Consequently, the right cylindrical figure with base ALDEK and altitude HK is to the solid of revolution ABC in the ratio compounded from the ratio of 1 to 4 and the square on the diameter of a circle to the circle. Therefore, four times the previously-mentioned cylindrical figure is to the solid of revolution ABC as the square on the diameter of a circle is to the circle--that is, as a rectangular parallelpiped is to the cylinder inscribed in the same altitude. By permutando, four times the original cylindrical figure is to the parallelpiped as the solid of revolution is to the circular cylinder. Consequently, the ratio of four times the cylindrical figure to the parallelpiped is given. Therefore, the cubature of the cylindrical figure and the quadrature of its base ALDEK are given.

From this demonstration, it is also manifest that the solid of revolution ABC is proportional to the figure ALDEK both in magnitude and in weight, since those same things which are demonstrated concerning the entirety are able to be demonstrated in the same manner concerning their proportional parts. Therefore, the center of gravity of the solid of revolution is the center of equilibrium of the figure.

Hence, it is also manifest that the cylindrical figure whose base is AKED and whose altitude HK is double the trunk of the right cylindrical figure whose base AIBE is cut by the plane inclining in a semi-right angle and cutting the base plane through the line AE. Moreover, given this truncated figure, any other trunk cut off through that same line AE is given, since such truncated figures are to themselves as their altitudes, or as the inclination of their tangents, which is demonstrated easily.

The converse of this problem--namely, from a quadrature and center of gravity of a given figure to find a solid of revolution with a given proportion to a cylinder and its center of gravity--is also proved with no difficulty.

If a right cylindrical figure above a given figure is cut by a plane, the trunk of this cylindrical figure will be to the solid of revolution arising from the rotation of its base around the common intersection of the extended (if necessary) base and the cutting plane as the altitude of the cylindrical figure is to the circumference of the circle whose radius is the radius of rotation.

Let ABDC be a right cylindrical figure above a figure DCF which is cut by a plane KINM in such a way that the figure RGHQ is the common intersection of the plane with the cylinder.

Let the cutting plane be extended until it cuts the base plane DCF in the line KI and the plane of AB in the line MN. From any point, say, L, on the line IK let a plane OLP be drawn perpendicular to IK and cutting the planes IKMN and DFC normally in the lines OL and LP. Let the line OP be perpendicular to LP. We assume that KI is the axis of rotation. We call the line PL normal to this the radius of rotation. I say that the trunk RQDC of the cylinder is to the solid of revolution arising from the rotation of the figure DEFC around the axis of rotation IK as the altitude OP of the cylinder is to the circumference of the circle whose radius is the radius of rotation LP.

Let EF be any line terminated on both sides of the circumference of the base at E and F and drawn through the base DEFC, which, when extended, intersects the axis of rotation normally in the point V. From the points E and F let EH and FG be sent out perpendicular to the base plane and terminated in the cutting plane at H and G. Since the cylinder is right, these are necessarily on the surface of the trunk. Let the line VH be drawn, which is necessarily in the cutting plane IKMN. It is manifest that the triangles OLP and HVE, which have right angles at the points P and E, are similar (since the angles OLP and HVE are equal to the inclination of the cutting plane IKMN). With the line GV having been drawn, the triangles HEV and GFV are similar for the same reason. Since GF is parallel to the line HE and EF intersects EV, the lines GV and HV lie on one line in the plane IKMN. GHEF will be the common intersection of the plane GFV, parallel to the plane OPL, with the truncated cylinder RQDC. Therefore, it is apparent that OP is to PL as HE is to EV. Therefore, as OP is to the circumference of the circle whose radius is PL, thus HE is to the circumference of the circle whose radius is EV. And as OP is to the circumference of the circle whose radius is PL thus the triangle HEV is to the circle whose radius is EV (by drawing the remaining boundaries to the same altitude, namely, half of the line EV). It is demonstrated in the same manner that as OP is to the circumference of the circle whose radius is PL, thus the triangle GFV is to the circle whose radius is FV. Therefore, the entire triangle GFV is to the entire circle with radius FV thus as the cut-off triangle HEV is to the cut-off circle whose radius is EV. Consequently, the remaining trapezoid GFEH will be in the same ratio to the remaining annulus arising from the revolution of the line FE around the axis of rotation IK, namely, in the ratio of OP to the circumference of the circle whose radius is LP. But this proportion is demonstrated in the same manner for all lines drawn in the base DEFC which, when extended (if necessary), meet the axis of rotation IK normally. But the base DC consists of all such lines, the trunk RQDC consists of all of the trapezoids above these lines, and the solid of revolution arising from the revolution of the base around the axis of revolution IK consists of all of the annuli produced by the revolution of these lines. Consequently, as one antecedent is to one consequent--namely, the altitude OP of the cylinder is to the circumference of the circle whose radius PL is the radius of rotation--thus, all antecedents--namely, all trapezoids--that is, the trunk RQDC--are to all consequents--namely, all annuli--that is, the solid of revolution arising from the rotation of the figure DC around the axis IK, which it was desired to demonstrate.

This theorem is demonstrated in the same manner for the upper trunk if the figure AB is conceived to be rotated around the line MN.

It is apparent from the demonstration that the trunk RQDC and the solid of revolution arising from the revolution of the base DC around the axis of rotation IK are proportional quantities in magnitude and weight, since each proportion which is demonstrated among the entireties is demonstrated in the same manner concerning their proportional parts.

In the following results it ought to be noted that (when we speak about the surface of the cylindrical figure or of the trunk) we understand a surface alone without bases. That is, we never consider figures which are the base of the cylinder, nor do we consider the common intersection of the plane cutting the cylinder.

With the same things having been supposed as in the preceding proposition, the surface of the trunk will be to the surface of the solid of revolution arising from the rotation of the base around the common intersection of the extended (if necessary) base and cutting the plane as the altitude of the cylinder is to the circumference of the circle whose radius is the radius of rotation.

Let the figure and preparation be just as in the preceding proposition. I say that the surface of the trunk RQDC is to the surface of the solid of revolution arising from the rotation of the figure DEFC around the axis of rotation IK as the altitude OP of the cylinder is to the circumference of the circle whose radius is the radius of rotation LP.

In the previous proposition it was demonstrated that OP is to the circumference of the circle whose radius is PL as HE is to the circumference of the circle whose radius is EV. But this proportion is demonstrated in the same manner for all lines on the surface of the trunk RQDC perpendicular to the base DEFC to all circumferences of circles described by their lowest points in the rotation. But the surface of the trunk itself consists of all these lines. And the surface of the solid of revolution arising from the rotation of the base around the axis IK consists of all the circumferences of the circles out of the bottom points of the lines--that is, from the revolution of the described base. Therefore, as one of the antecedents is to one of the consequents--namely, the altitude OP of the cylinder to the circumferences of the circle whose radius is the radius of rotation LP--thus all the antecedents--that is, the surface of the trunk RQDC--are to all the consequents--that is, the surface of the solid arising from the rotation of the base DEFC around the axis of rotation IK--which ought to have been demonstrated.

This theorem is also demonstrated in the same manner for the upper trunk if the figure AB is conceived to be rotated around the line MN.

It is clear from the demonstration that the surface of the trunk RQDC and the surface of the solid of revolution arising from the rotation of the base DEFC around the axis of rotation IK are proportional quantities in magnitude and weight, since the same proportion which is demonstrated between the entireties is demonstrated in the same manner to be between their proportional parts.

With the same things having been assumed, I say that by supposing that the angle of inclination of the cutting plane with the extended (if necessary) base of the cylinder is semi-right the square on the radius of the circle equal to the surface of the solid of revolution is double the surface of the trunk.

Indeed, by the preceding proposition, as the altitude of the cylinder is to the circumference of the circle whose radius is the radius of rotation, thus the surface of the trunk is to the surface of the solid of revolution. Moreover, in this case, because the angle of inclination is semi-right, the altitude of the cylinder is equal to the radius of rotation. Therefore, in our case, as the radius is to its own circumference, thus the surface of the trunk is to the surface of the solid of revolution. Moreover, as the radius is to the circumference, thus half of the square of the radius is to the circle. Therefore, as half of the square of the radius of a circle is to the circle, thus the surface of the trunk is to the surface of the solid of revolution. By convertendo and permutando, a circle is to the surface of the solid of revolution as half of the square on the radius of the circle is to the surface of the trunk. But the circle is assumed equal to the surface of the solid of revolution. Therefore, half of the square of the radius on that circle is equal to the surface of the trunk. Therefore, the square on the radius is double the surface of the trunk, which ought to have been demonstrated.

These two propositions are demonstrated in nearly the same manner for surfaces of revolution arising from the rotation of one or more curves or lines, or of curves or lines not enclosing a figure. Indeed, the surface of revolution arising from the curve or curves always have the previously-mentioned ratios to the surface of the right cylinder above the curve or curves.

With the same things being assumed as in the preceding proposition, I say that the cube on the radius of the sphere equal to the solid of revolution is to the trunk as three is to two.

In fact, the trunk (in this case) is to the solid of revolution as a radius is to its own circumference--that is, as the double of the square on a radius is to the quadruple of the circle--that is, as 2/3 of the cube on a radius is to the sphere. Therefore, the trunk is to the solid of revolution as 2/3 of the cube of the radius is to the sphere. By convertendo and permutando, the solid of revolution is to a sphere as the trunk is to 2/3 of the cube on the radius of the sphere. But the sphere is assumed equal to the solid of revolution. Therefore, 2/3 of the cube on the radius of the sphere is equal to the trunk. But the cube on the radius has the same ratio to 2/3 of itself as 3 has to 2. Therefore, the cube on the radius of the sphere has the same ratio to the trunk, which ought to have been demonstrated.

If two given right cylindrical figures of equal height are both cut by a plane into two trunks, the proportion of the solid of revolution arising from the rotation of the base of the cylinder around the common intersection of the extended (if necessary) base with the cutting plane to the solid of revolution arising from the same rotation of the base of the other cylinder is compounded directly from the proportion of the radii of rotation and the proportion of the lower trunks of the cylinder.

Let ABCD and NMOYXZ be two right cylindrical figures of equal height with bases DC and XYZ which are cut by planes together into two trunks. Specifically, let ABCD be cut by the plane KHFG into trunks AB43 and 43DC and let NMOYXZ be cut by a plane PSVR into trunks NMO765 and 765ZXY. Let the lines KH and GF be the intersections of the plane HFGK with the planes of the extended (if necessary) parallel bases DC and AB of the cylinder, and let the lines RP and VS be the intersections of the plane RVSR with the planes of the extended (if necessary) parallel bases NMO and ZXY of the cylinders. Let there be planes normal to the lines FG and SV which intersect the cutting planes in the lines IE and QT and the planes of the extended (if necessary) DC and ZXY bases in the lines LE and WT, and let the angles ILE and QWT be right.

It is manifest from Proposition 23, with HFGK and PSVR the assumed cutting planes and GF and VS the axes of rotation, that LE and WT are the radii of rotation. Therefore, I say that the solid of revolution arising from the rotation of the figure DC around GF is to the solid of revolution arising from the rotation of the figure ZXY around VS in the ratio compounded from the proportion of the trunk 34CD to the trunk 567YXZ and from the proportion of the radius of rotation LE to the radius of rotation WT.

The ratio of the solid of revolution arising from the figure DC to the solid of revolution arising from the figure ZXY is compounded from the ratio of the solid of revolution arising from DC to the trunk 34CD, from the ratio of the trunk 34CD to the trunk 567YXZ, and from the ratio of the trunk 567YXZ to the solid of revolution arising from YXZ. But the ratio of the solid of revolution arising from DC to the trunk 34CD is equal to the ratio of the circumference of the circle described by the radius LE to the altitude IL of the cylinder. (See Proposition 23.) The ratio of the trunk 567YXZ to the solid of revolution arising from YXZ is equal to the ratio of the altitude QW--that is, IL--of the cylinder to the circumference of the circle described by the radius TW. Consequently, the ratio of the solid of revolution arising from DC to the solid of revolution arising from ZXY is compounded of the ratio of the trunk 34CD to the trunk 567YXZ, from the ratio of the circumference of the circle described by the radius LE to the line IL, and from the ratio of the line IL to the circumference of the circle described by the radius TW. But these last two ratios compound to the ratio of the circumference described by the radius LE to the circumference of the circle described by the radius TW, which is the same as the ratio of the radius LE to the radius TW. Consequently, the solid of revolution arising from the rotation of the figure DC around the axes FG is to the solid of revolution arising from the rotation of the figure XYZ around the axis VS in a ratio compounded from the proportion of the lower trunk 34CD to the lower trunk 567YXZ and from the proportion of the radius of rotation EL to the radius of rotation TW, which ought to have been demonstrated.

With the same things having been assumed as in the preceding proposition, the proportion of the surface of the solid of revolution arising from the rotation of the base of the cylinder around the common intersection of the extended (if necessary) base with the cutting plane to the surface of the solid of revolution arising from the similar rotation of the other cylinder is compounded from the direct proportion of the radii of rotation and the direct proportion of the surfaces of the lower trunks.

Let the figures and the preparation be the same as in the preceding proposition. I say that the surface of the solid of revolution arising from the rotation of the figure DC around GF is to the surface of the solid of revolution arising from the rotation of the figure ZXY around VS in a ratio compounded from the proportion of the surface of the trunk 34CD to the surface of the trunk 567YXZ and the proportion of the radius of rotation LE to the radius of rotation WT.

The ratio of the surface of the solid of revolution arising from the figure DC to the surface of the solid of revolution arising from the figure ZXY is compounded from the ratio of the surface of the solid of revolution arising from DC to the surface of the trunk 34CD, the ratio of the surface of the trunk 34CD to the surface of the trunk 567YXZ, and the ratio of the surface of the trunk 567YXZ to the surface of the solid of revolution arising from ZXY. But the ratio of the surface of the solid of revolution arising from DC to the surface of the trunk 34CD is equal to the ratio of the circumference of the circle described by the radius LE to the altitude IL of the cylinder. (See Proposition 24.) The ratio of the surface of the trunk 567YXZ to the surface of the solid of revolution arising from YXZ is equal to the ratio of QW--that is, of the altitude IL of the cylinder--to the circumference of the circle described by the radius TW. Consequently, the ratio of the surface of the solid of revolution arising from DC to the surface of the solid of revolution arising from ZXY is compounded from the ratio of the surface of the trunk 34CD to the surface of the trunk 567YXZ, from the ratio of the circumference of the circle described by the radius LE to the line IL, and from the ratio of the line IL to the circumference of the circle on the radius TW. But these last two ratios compound the ratio of the circumference of the circle described by the radius LE to the circumference of the circle described by the radius TW, which is the same as the ratio of the radius LE to the radius TW. Consequently, the surface of the solid of revolution arising from the rotation of the figure DC around the axes FG is to the surface of the solid of revolution arising from the rotation of the figure XYZ around the axes VS in a ratio compounded from the proportion of the surface of the lower trunk 34CD to the surface of the lower trunk 567YXZ and from the proportion of the radius of rotation EL to the radius of rotation TW, which ought to have been demonstrated.

This theorem also has the same place and is demonstrated in the same manner in surfaces of revolution arising from the rotation of lines or curves not enclosing a figure.

If a right cylindrical figure above a figure symmetric around an axis is envisioned which is cut by a plane into two trunks such that a plane drawn through the axes on opposite bases of the cylinder is normal to that cutting plane, then one trunk will be to the other trunk as the reciprocal of the parts of the radius of rotation having been cut by the center of gravity of the figure.

Let the right cylinder ABDMLK above a figure KLM symmetric around the axis KN be cut into trunks ABDZY2 and ZY2KLM by a plane ETVG normal to the plane ACNK drawn through the axes AC and KN of the opposite bases of the cylinder. Let P and O, which are joined by a line PO, be the centers of gravity of the opposite bases. Let the cutting plane be extended until it intersects the extended (if necessary) axes AC and KN in the points F and S. Let FI and SQ be perpendiculars to those same extended (if necessary) axes. It is manifest that FISQ is a rectangular parallelogram, also that FI is an altitude of the cylinder, and that IS is a radius of rotation which is certainly perpendicular to the intersection of the cutting plane and the base LKM--that is, to the line TV--since it is drawn in the plane FQSI, which is normal to both the cutting plane and the base LKM. I say that the trunk ABDZY2 is to the trunk 2YZMLK as the reciprocal of the ratio of IO to OS.

From the midpoints of the lines FI and QS--namely, H and R--let the lines HS, RF, and HR be drawn, with HR cutting OP in X. Thus FI, PO, and QS; and likewise FQ, HR, and IS; and likewise FR and HS will be parallel and equal among themselves. Since FR bisects QS, it will also bisect all the lines in the triangle FQS equidistant to QS itself. Consequently, it will bisect all the diameters of the rectangles in the trunk ABDZY2 which are cut normally by the plane FISQ. Therefore, it will pass through the centers of gravity of all of those rectangles.

Since the trunk consists of all of those rectangles, the line FR will pass through the center of gravity of the trunk. Suppose this point is 3. It is demonstrated in the same manner that the center of gravity of the trunk YZMLK2 is on the line HS. Therefore, since the midpoint X of the line OP is the center of gravity of the entire cylinder, if the line 3X4 is drawn through X from 3 until it intersects the line HS in 4, 4 will be the center of gravity of the trunk YZMLK2. Because the triangles XR3 and XH4 are similar on account of RF and HS being parallel, as X4 is to X3 thus XH will be to XR--that is, IO will be to OS. But as X4 is to X3 thus the trunk ABDZY2 will be to the trunk ZYMKL2. Consequently, as the trunk ABDZY2 is to the trunk YZKMK2 thus IO will be reciprocally to OS, which ought to have been demonstrated.

Consequently, by componendo the entire cylinder ABDMLK is to the lower trunk YZMLK2 as the radius of rotation IS is to the distance between the center of gravity of the original figure and axis of rotation of the original figure--namely, OS.

With the same things having been supposed as in the preceding proposition, the surface of one trunk is to the surface of the other reciprically as the parts of the radius of rotation cut by the center of gravity of the perimeter of the figure.

Let the figure and the preparation be the same as in the preceding proposition, with this one exception: O and P and are now assumed to be centers of gravity of the perimeters of the opposite bases. I say that the surface of the trunk ABDZY2 is to the surface of the trunk 2YZMKL reciprocally as IO is to OS.

From the midpoints of the lines FI and QS--namely, H and R--let lines HS, RF, and HR be drawn, with HR cutting OP in X. Thus, the lines FI, PO, and QS; and likewise FQ, HR, and IS; and likewise FR and HS will be parallel and equal between themselves. Since FR bisects QS, it will also bisect all the lines in the triangle FQS equidistant to QS itself. Consequently, it will bisect all the diameters of the rectangles in the trunk ABDZY2 which are cut normally by the plane FISQ. Therefore, it will pass through all the centers of gravity of the opposite sides perpendicular from those rectangles to the cylindrical base, since the center of gravity of the opposite sides is in the middle of the diameter. Since the surface of the trunk itself is made up of all of those opposite sides perpendicular to the base of the cylinder, therefore the line FR will also pass through the center of gravity of the surface of the trunk itself. Suppose this is 3. It is demonstrated in the same manner that the center of gravity of the surface of the trunk YZMLK2 is on HS. Therefore, since the midpoint X of the line OP is the center of gravity of the entire surface of the cylinder, if the line 3X4 is drawn through X from 3 until it intersects the line HS in 4, 4 will be the center of gravity of the surface of the trunk YZMLK2. Because the triangles XR3 and XH4 are similar on account of RF and HS being parallel, as X4 is to X3 thus XH will be to XR--that is, IO to OS. But as X4 is to X3 thus the surface of the trunk ABDZY2 will be to the surface of the trunk YZMLK2. Therefore, as the surface of the trunk ABDZY2 is to the surface of the YZMLK2, thus reciprocally IO will be to OS, which ought to have been demonstrated.

By componendo the entire surface of the cylinder ABDMLK is to the surface of the lower trunk YZMLK2 as the radius of rotation IS is to the intersection of the center of gravity of the perimeter of the original figure and its axis of rotation--namely, OS.

These results are also demonstrated in the same manner for trunks consisting of a curve or curves not encompassing a figure, if only they are symmetric around an axis.

The two preceding propositions are also true in each cylindrical figure, but the construction serving for the general demonstration is intricate. Therefore, we will demonstrate our intentions in another manner.

If there are two figures symmetric around an axis which are rotated in such a way that the axes of rotation of each of the figures are normal to the axis of each figure, then the ratio of one solid arising from such a rotation to the other solid arising from the same rotation is compounded directly from the ratio of the first figure to the other figure and from the ratio of the segment between the center of gravity and the axis of rotation of the first figure to the similar segment of the other figure.

Let ABC and HIL be any two figures symmetric around the axes BF and IN, which are rotated around the lines EG and MO cutting the extended (if necessary) axes BF and IN normally at the points F and N. Let D and K be the centers of gravity of the figures ABC and HIL. I say that the ratio of the solid arising from the figure ABC rotated around the line EG to the solid arising from the figure HIL rotating around the line MO, is compounded from the ratio of the figure ABC to the figure HIL and from the ratio of DF to KN.

Above the figures ABC and HIL let right cylindrical figures of equal height be cut by planes passing through the lines EG and MO, each one into two trunks, namely, an upper and a lower trunk. The ratio of the solid of revolution arising from ABC to the solid of revolution arising from HIL is compounded from the ratio of the lower trunk of the cylinder above ABC to the lower trunk of the cylinder above HIL and from the ratio of the radius of rotation of the figure ABC to the radius of rotation of the figure HIL.

But the lower trunk of the cylinder above ABC is to the lower trunk of the cylinder above HIL is in a ratio compounded from the ratio of the lower trunk of the cylinder above ABC to the entire cylinder above ABC, from the ratio of the entire cylinder above ABC to the entire cylinder above HIL, and from the ratio of the entire cylinder above HIL to its lower trunk. But from the convertendo of the Consequence to Proposition 29 the lower trunk of the cylinder above ABC is to the entire cylinder as FD is to the radius of rotation of the figure AB. Also, the cylinder above ABC is to the cylinder above HIL as the figure ABC is to the figure HIL. Similarly, by the Consequence to Proposition 29, the cylinder above HIL is to its own lower trunk as the radius of rotation of the figure HIL is to KN. Consequently, the ratio of the lower trunk of the cylinder above ABC to the lower trunk of the cylinder above HIL is compounded from the ratio of the line DF to the radius of rotation of the figure ABC, from the ratio of the figure ABC to the figure HIL, and from the ratio of the radius of rotation of the figure HIL to the line KN. Therefore, the ratio of the solid arising from the rotation of the figure ABC to the solid arising from the rotation of the figure HIL is compounded from the ratio of the figure ABC to the figure HIL, from the ratio of the line DF to the radius of rotation of the figure ABC, from the ratio of the radius of rotation of the figure ABC to the radius of rotation of the figure HIL, and from the ratio of the radius of rotation of the figure HIL to the line KN. But the last three ratios compound to the ratio of DF to KN. Therefore, the ratio of the solid arising from the rotation of the figure ABC around EG to the solid arising from the rotation of the figure HIL around MO is compounded from the ratio of the figure ABC to the figure HIL and from the ratio of the segment between the center of gravity of the figure ABC and its axis of rotation--namely, DF--to the segment between the center of gravity of the figure HIL and that same axis of rotation--namely, KN--which ought to have been demonstrated.

With the same things being assumed as in the preceding proposition, the ratio of the surface of one solid arising from such a rotation to the surface of the other solid arising from the same is compounded directly from the ratio of the perimeters of the figures and from the ratio of the segments between the centers of gravity of the perimeters and the axes of rotation.

Let the figure and the preparation be the same as in the preceding proposition, with this one exception: that the points D and K are now assumed to be the centers of gravity of the perimeters of the figures. I say that the ratio of the surface of the solid arising from the figure ABC rotated around the line EG to the surface of the solid arising from the figure HIL rotating around the line MO is compounded from the ratio of the perimeter of ABC to the perimeter of HIL and from the ratio of DF to KN.

Above the figures ABC and HIL are understood right cylinders of equal height cut by planes passing through the lines EG and MO, each one into two trunks, namely, an upper and a lower trunk. The ratio of the surface of the solid arising from ABC to the surface of the solid arising from HIL is compounded from the ratio of the surface of the lower trunk of the cylinder above ABC to the surface of the lower trunk of the cylinder above HIL and from the ratio of the radius of rotation of the figure ABC to the radius of rotation of the figure HIL. But the surface of the lower trunk of the cylinder above ABC is to the surface of the lower trunk of the cylinder about HIL in the ratio compounded from the ratio of the surface of the lower trunk of the cylinder above ABC to the entire surface of the cylinder above ABC, from the ratio of the entire surface of the cylinder above ABC to the entire surface of the cylinder above HIL, and from the ratio of the entire surface of the cylinder above HIL to the surface of its own lower trunk. But by the convertendo of the consequence of Proposition 30 the surface of the lower trunk of the cylinder above ABC is to the entire surface of the cylinder as FD is to the radius of rotation of the figure ABC. The surface of the cylinder above ABC is to the surface of the cylinder above HIL as the perimeter of ABC is to the perimeter of HIL. Likewise, the surface of the cylinder above HIL is to the surface of its own lower trunk as the radius of rotation of the figure HIL is to KN. Consequently, the ratio of the surface of the lower trunk of the cylinder above ABC to the surface of the lower trunk of the cylinder above HIL is compounded from the ratio of the line DF to the radius of the rotation of the figure ABC, from the ratio of the perimeter of ABC to the perimeter of HIL, and from the ratio of the radius of rotation of the figure HIL to the line KN. Therefore, the ratio of the surface of the solid arising from the rotation of the figure ABC to the surface of the solid arising from the rotation of the figure HIL is compounded from the ratio of the perimeter of ABC to the perimeter of HIL, from the ratio of the line DF to the radius of rotation of the figure ABC, from the ratio of radius of rotation of the figure ABC to the radius of rotation of the figure HIL, and from the ratio of the radius of rotation of the figure HIL to the line KN. But the last three ratios compound to the ratio of DF to KN. Therefore, the ratio of the surface of the solid arising from the rotation of the figure ABC around EG to the surface of the solid arising from the rotation of the figure HIL around MO is compounded from the ratio of the perimeter of ABC to the perimeter of HIL and from the ratio of the segment between the center of gravity of the perimeter of ABC and its axis of rotation--namely, DF--to the segment between the center of gravity of the perimeter of HIL and the axis of rotation of the same--namely, KN--which ought to have been demonstrated.

These theorems also have the same place and are demonstrated in the same manner for surfaces of rotation arising from the rotation of a curve or curves not enclosing a figure and symmetric around an axis.

If there are any two figures which are rotated around a given axis, the ratio of the one solid arising from such a rotation to the other solid arising from the same rotation will be compounded from the direct ratio of one figure to the other figure and from the direct ratio of the segment between the center of gravity and the axis of rotation of the one figure to the similar segment of the other figure.

Let ABC and NQP be any two figures which are rotated around the lines EF and Y4, and let their centers of gravity be D and R, which are sent down lines DG and RZ perpendicular to the axes of rotation EF and Y4. I say that the ratio of the solid arising from the figure ABC rotated around the line EF to the solid arising from the figure NQP rotated around the line Y4 is compounded from the ratio of the figure ABC to the figure NQP and from the ratio of DG to RZ.

Let the lines BH and Q2 touching the figures ABC and NQP at B and Q be drawn parallel to the lines DG and RZ. Let the figures ABC and QNP be conceived to be revolved around the lines BH and Q2, like axes, until, attaining the plane on the other part of the axes, they make the figures BLM and QTX, equal and similar to themselves and having exactly the same position toward the lines BH and EF, and Q2 and Y4. Let O and V be the centers of gravity of the figures BLM and QTX. Let the lines OI and V3 be drawn perpendicular to the lines EF and Y4. Also, let the lines DO and RV, intersecting the lines BH and Q2 in the points K and S, be joined.

It is manifest that the point K is the center of gravity of the figure BACBLM symmetric around the axis BH and likewise that the point S is the center of gravity of the entire figure QNPQTX around the axis Q2. It is also apparent that the line DG, KH, OI and also RZ, S2, V3, are equal among themselves. Since the figures BACBLM and QNPQTX are symmetric around the axes BH and Q2, which are normal to the axis of rotation, therefore the solid of revolution arising from the rotation of the figure BACBLM around the line EF is to the solid of revolution arising from the rotation of the figure QNPQTX in the ratio compounded from the ratio of the figure BACBLM to the figure QNPQTX and from the ratio of KH to S2 by Proposition 31. But the solid arising from the figure BACBLM rotated around EF is twice the solid arising from the figure BAC rotated around the same EF. Likewise, the solid arising from the figure QNPQTX rotated around the line Y4 is twice the solid arising from the figure QNP rotated around the same Y4. Also, the figure BACBLM is twice the figure BAC and the figure QNPQTX is twice the figure QNP. Since halves are in the same ratio with their own doubles, the solid of revolution arising from the figure ABC rotated around the line EF will be to the solid of revolution arising from the figure NQP rotated around the line Y4 in the ratio compounded from the ratio of the figure ABC to the figure NQP and from the ratio of KH to S2--or DG to RZ--which it was desired to demonstrate.

It follows that if the centers of gravity of the figures are equally distant from the axes of rotation, the solids of revolution arising from the rotation of figures are in a direct ratio to the figures themselves. If the figures themselves are equal, it follows that the solids of revolution arising from their rotation are in a direct ratio to the segments between the centers of gravity and the axes of rotation. If the segments and figures are equal, the solids of revolution arising from them will be equal even if the figures are dissimilar between themselves.

From these results, it is manifest that between any two figures there are three ratios--namely, of the one figure to the other figure, of the solid of revolution arising from the rotation of one figure to the solid of revolution arising from the rotation of the other figure, and of the segment between the center of gravity and the axis of rotation of the one figure to the similar segment of the second figure--giving two of which always discloses the unknown third.

If there are any two figures which are rotated around an axis, the ratio of the surface of one solid of revolution arising from such a rotation to the surface of the other solid arising from the same rotation is compounded from the direct ratio of one perimeter to the other perimeter and from the direct ratio of the segment between the center of gravity of the perimeter and the axis of rotation of the one figure to the similar segment of the other figure.

Let ABC and NQP be any two figures which are rotated around the lines EF and Y4, and let the centers of gravity of the perimeters be D and R, from which perpendiculars DG and RZ to the axis of rotation EF and Y4 are sent down. I say that the ratio of the surface of the solid of revolution arising from the figure ABC rotated around the line EF to the surface of the solid of revolution arising from the figure NQP rotated around the line Y4 is compounded from the ratio of the perimeter ABC to the perimeter NQP and from the ratio DG to RZ.

Let the lines BH and Q2 be drawn parallel to the lines DG and RZ, touching the figures ABC and NQP in B and Q, Let the figures ABC and QNP be conceived to be revolved around BH and Q2, like axes, until in the other part of the axes on the same plane they make the figures of revolution BLM and QTX , equal to themselves, similar, and having completely the same position to the lines BH and EF, and Q2 and Y4. Let O and V be the centers of gravity for the perimeters BLM and QTX. Let the lines OI and V3 be drawn perpendicular to the lines EF and Y4, and let the lines DO and RV be joined, intersecting the lines BH and Q2 in the points K and S. It is manifest that the point K is the center of gravity of the entire perimeter BACBLM. Also, the figure BACBLM is symmetric around the axes BH. In the same manner, the point S is the center of gravity of the entire perimeter of the figure QNPQTX symmetric around the axis Q2. It is also apparent that the lines DG, KH, OI, and likewise RZ, S2, V3 are equal between themselves. Since the figures BACBLM and QNPQTX, are symmetric around the axes BH and Q2 normal to the axes of rotation EF and Y4, therefore the surface of the solid of revolution arising from the rotation of the figure BACBLM around the line EF is to the surface of the solid of revolution arising from the rotation of the figure QNPQTX in the ratio compounded from the ratio of the perimeter of BACBLM to the perimeter of QNPQTX and from the ratio of KH to S2. But the surface of the solid arising from the figure BACBLM rotated around EF is double the surface of the solid arising from the figure BAC rotated around the same EF. Likewise, the surface of the solid of revolution arising from the figure QNPQTX rotated around the line Y4 is double the surface of the solid arising from the figure QNP rotated around the same line Y4. Also the perimeter of BACBLM is double the perimeter of BAC and the perimeter of QNPQTX is double the perimeter of QNP. (That is, if the figures themselves in turn touch in one point. But if they touch in curves, it desirable that there be imagined a little distance between the unions of the figures so that the demonstrated is general.) Since the halves are in the same ratio with their doubles, the surface of the solid of revolution arising from the figure ABC rotated around the line EF will be to the surface of the solid of revolution arising from the figure NQP rotated around the line Y4 in the ratio compounded from the ratio of the perimeter of ABC to the perimeter of NQP and from the ratio of KQ to S2--or DG to RZ--which it was desired to demonstrate.

Hence it follows that if the centers of gravity of the perimeters are equally distant from the axes of rotation, the surfaces of the solids of revolution arising from the rotation are in the direct ratio of the perimeters themselves. If the perimeters themselves are equal, the surfaces of the solids of revolution arising from the rotation are in the direct ratio of the segments between the centers of gravity of the perimeters and the axes of rotation. If the segments and the perimeters are equal, the surfaces of the solids of rotations are always equal.

Therefore, from these results, it is manifest that between any two figures there are three other ratios different from the preceding three--namely, of the perimeter to the perimeter, of the surface of the solid of revolution arising from the rotation of the one figure to the surface of the solid of revolution arising from the rotation of the other figure, and of the segment between the center of gravity of the perimeter and the axis of rotation of one figure to the similar segment of the other figure--giving two of which always discloses the unknown third.

All these things are demonstrated universally In the same manner for every curve or curves not enclosing a figure thus so of all geometrical demonstrations these are maximally universal.

Each solid of revolution is equal to a right cylindrical figure whose base is the figure out of the rotation of which the solid is produced and whose altitude is the circumference of a circle in which the center of gravity of the figure is revolved.

Let AB be a figure whose center of gravity is C. Let a solid of revolution be made from the rotation of the figure AB around the line DF. I say this solid of revolution is equal to the cylinder whose base is the figure AB and whose altitude is the circumference of the circle in which the center of gravity C is revolved.

Let HGKI be a rectangle whose center of gravity is L.

Let a cylinder be conceived from the rotation of the rectangle HGKI around the side HI. From the centers of gravity C and L let perpendicular lines CE and LO, which are the radii of rotation, be sent down to the axes of rotation DF and HI. It is manifest that the cylinder arising from the rotation of the rectangle HK around HI is equal to the solid whose base is a circle with radius IK and whose altitude is HI--that is, to the solid whose base is the rectangle on IK and the semicircumference of the circle with radius IK, or the entire circumference of the circle with radius OL, and the altitude HI. This is the same as the solid whose base is the rectangle HK and whose altitude is the circumference of the circle on the radius OL. But the cylinder from the rotation of HK around HI is to the solid of revolution from the rotation of AB around DF in a ratio compounded from the proportion of the figure HK to the figure AB and from the line OL to the line EC--or the circumference on the radius OL to the circumference on the radius EC. (See Proposition 33.) But the solid whose base is HK and whose altitude is the circumference on the radius OL--or the cylinder generated by the rotation of the figure HK around HI--is in the same ratio to the solid whose base is AB and whose altitude is the circumference of the radius EC. Therefore, the cylinder from the rotation of the figure HK around HI is to the solid of revolution from the rotation of the figure AB around DF as the same cylinder is to the solid whose base is AB and whose altitude is the circumference of the circle on the radius EC. Consequently, the solid of revolution from the rotation of the figure AB around DF is equal to the cylinder whose base is AB and whose altitude is the circumference of the radius EC, which it was desired to demonstrate.

Every surface of a solid of revolution is equal to a rectangle whose base is the perimeter of the figure out of the rotation of which the solid is generated and whose altitude is the circumference in which the center of gravity of the perimeter of the figure is revolved.

With the same things having been assumed as in the preceding proposition, let the centers of gravity of the perimeters of the figures HK and AB be the points L and C. I say that the surface of the solid of revolution arising from the rotation of the figure AB and DF is equal to the rectangle whose base is the perimeter of the figure AB and whose altitude is the circumference of the radius EC.

It is manifest that the surface of the cylinder arising from the rotation of the rectangle HK around HI is equal to the rectangle whose base is the circumference on the radius IK and whose altitude is GK together with IK--that is, to the rectangle whose base is the circumference on the radius OL and whose entire altitude is the boundary HG+ GK+ KI + IH of the figure. This is the same as rectangle whose base is the border of the rectangle HK and whose altitude is the circumference on the radius OL. But the surface of the cylinder arising from the rotation of the rectangle HK around HI is to the surface of the solid of revolution generated by the rotation of the figure AB around DF in the ratio compounded from the proportion of the boundary of the figure HK to the boundary of the figure AB and from the proportion of the line OL to the line EC--that is, the circumference on the radius OL to the circumference on the radius EC. (See Proposition 34.) But the rectangle whose base is the boundary of the figure HK and whose altitude is the circumference on the radius OL--that is, the surface of the cylinder arising from the rotation of the rectangle HK around HI--is in the same ratio to the rectangle whose base is the boundary of the figure AB and whose altitude is the circumference of the radius EC. Therefore, the surface of the cylinder arising from the rotation of the rectangle HK around HI is to the rectangle whose base is the boundary of the figure AB and whose altitude is the circumference of the radius EC, as the same surface is to the surface of the solid of revolution arising from the rotation of the figure AB around the axis of rotation DF. Consequently, the surface of the solid of revolution arising from the rotation of the figure AB around DF is equal to the rectangle whose base is the boundary of the figure AB and whose altitude is the circumference on the radius EC, which it was desired to demonstrate.

This proposition is demonstrated in nearly the same manner even if the perimeter of the figure is not enclosed.

If a right cylindrical figure is cut by a plane seminormal to the base, the lower trunk will be equal to a cylindrical figure whose base is the same as the base of the original cylinder and whose altitude is the line segment between the center of gravity of the base and the common intersection of the base and the cutting plane.

Let the figure be the same as in the preceding proposition. Above the figure AB (whose center of gravity is C) let a right cylinder be conceived cut by a plane cutting the base AB seminormally in the line DF. I say that its lower trunk is equal to the cylinder above the base AB having altitude EC.

The solid of revolution arising from the rotation of the figure AB around the line DF is equal to the cylinder whose base is AB and whose altitude is the circumference on the radius EC. (See Proposition 35.) But that same solid of revolution is to the lower trunk of the right cylinder above AB which has been sectioned by a plane intersecting the base plane AB seminormally in the line DF as the circumference on the radius on CE is to the radius CE. (See Proposition 23.) But the cylinder whose base is AB and whose altitude is the circumference on the radius CE--or the previously-mentioned solid of revolution--is to the cylinder whose base is AB and whose altitude is EC as the circumference on the radius EC is to EC. Consequently, the lower section of the trunk is equal to the cylinder whose base is AB and whose altitude is EC, which it was desired to demonstrate.

If a right cylindrical figure is cut by a plane seminormal to the base, the surface of the lower trunk will be equal to the rectangle whose base is the perimeter of the base of the cylinder and whose altitude is the line segment between the center of gravity of the perimeter of the base and the common intersection of the base plane and the cutting plane.

With the same things being assumed as in the preceding proposition, let C be the center of gravity of the perimeter of the base AB. I say that the surface of the lower trunk is equal to the rectangle whose base is the boundary of the figure AB and whose altitude is the line EC.

The surface of the solid of revolution arising from the rotation of the figure AB around DF is equal to the rectangle whose base is the boundary of the figure AB and whose altitude is the circumference of the radius EC. (See Proposition 36.) But that same surface of the solid of revolution is to the surface of the lower trunk of the right cylinder above AB which has been cut by the plane intersecting the base AB seminormally in the line DF as the circumference on the radius CE is to CE. (See Proposition 24.) But the rectangle whose base is the boundary of the figure AB and whose altitude is the circumference on the radius EC--or the surface of the solid of revolution arising from the rotation of the figure AB around DF--is to the rectangle whose base is the boundary of the figure AB and whose altitude is EC as the circumference on the radius EC is to EC. Consequently, the surface of the lower section of the trunk is equal to the rectangle whose base is the boundary of the figure AB and whose altitude is EC, which it was desired to demonstrate.

This proposition is demonstrated in the same manner if the surface of the right cylinder consists of a curve or curves not enclosing a perimeter.

In the same manner as the two preceding results, it is demonstrated, however the cylinder is cut, first that the lower trunk is equal to the cylinder having the same base as the proposed cylinder and altitude equal to the line perpendicular to the base which is constructed from the center of gravity all the way to the cutting plane; second that the surface of the lower trunk is equal to the rectangle having base equal to the perimeter of the base of the proposed cylinder and altitude equal to the line perpendicular to the base of the proposed cylinder constructed from the center of gravity of its perimeter all the way to the cutting plane.

If a right cylindrical figure which has a base terminated in one part by a line is cut by a plane through that line and if the base of the cylinder is rotated around that same line so that some portion of a solid of revolution results, the center of gravity of that portion will be the same as the center of gravity of the arc of the circle in which the center of equilibrium of the lower trunk of the cylinder is revolved in the base of the cylinder.

Above the figure ABC, terminated in one part by the line BC, let there be a right cylinder which, cut by a plane passing through the line BC, results in a lower trunk BCAG whose center of equilibrium on the base is E. From the rotation of the figure ABC around the line BC let there be a portion of a solid of revolution. I say that the center of gravity of that portion is the same as the center of gravity of the arc of the circle in which the point E is revolved.

If they are not the same, let the line β be the difference between them. Let a solid polyhedron be imagined circumscribing that portion of the solid of revolution and consisting of 2, or 4, or 16, or 32, or however many (as long as it is a number in the progression of doubles from two) equal lower trunks cut by a plane having the line BC as the common intersection with the base. In that same portion of the solid of revolution let there be inscribed another polyhedron similar to the first in such a way that their centers of gravity differ by an interval less than the line β. Indeed, this is able to be done since the more sides the polyhedra have the closer they are to the portion of the solid of revolution ad infinitum. Consequently, through the center of equilibrium E let a plane be drawn normal to the line BC and: making a common intersection with the trunk the triangle HFD, which is right at F; likewise making a common intersection with the portion of the solid of revolution the sector of the circle IVLO, with the circumscribed polyhedron a regular polygon circumscribed around the sector; likewise with the inscribed polyhedron a similar portion of a regular polygon inscribed in the same sector.

It is manifest that the radius IV is equal to the line DF. Let IY9 be the sector of the circle with radius IY equal to the line DE, and let the polygon TS8 similar to the polygon KMN circumscribe the sector IY9. Likewise let the inscribed polygon Y79 be similar to the circumscribed polygon. Let the line IL be drawn cutting the four sides of the similar polygons 7Y, ST, PV and MK in half in the points γ, R, Q, and L. It is manifest that the triangle ILK, right at L, is the common section of the cutting plane with one of the equal trunks out of which the polyhedron circumscribing the portion of the solid of revolution consists.

Let us suppose that this trunk is the same as the trunk ABCG (which we are able to do without absurdity, since all of the lower trunks whose cutting plane cuts the base in BC, have the same center of equilibrium in the base). Consequently, it is apparent that the center of equilibrium of the trunk having common intersection LIK with the cutting plane is the point R. In the same manner it is proved that the point R is the center of equilibrium in the base for the trunk whose intersection with the cutting plane is MIL.

Therefore, the center of gravity of these trunks joined together is the point R. In the same manner it is proved that, in any two trunks on the entire circumscribed polyhedron joined at one surface, the center of gravity of both together is in the point where the line from the vertex I cutting the base of the triangle (which is the common section of cutting plane with the two joined trunks) intersects the arc of the circle 9Y, cutting it in half. It is also proved easily that the point γ is the center of gravity of the two joined trunks whose common intersection with the cutting plane is the triangle PIV. Likewise, the point X is the center of gravity of the two joined trunks whose common intersection with the cutting plane is the triangle φIP. Let the line Xγ, which is cut in half in the point 5, be joined. It is manifest that the point 5 is the center of gravity of the four joined together trunks whose common intersection with the cutting plane is the trapezoid IVPφ. But the point 5 is also the center of gravity of the lines Z7 and 7Y; together. In the same manner it is proved that α is the center of gravity of the four joined together trunks whose common intersection with the cutting plane is the trapezoid IφδO and also of the lines 9ν and νZ together. Let the centers of gravity be joined in the line α5 which is cut in half by the line IZ in the point 3. It is manifest that the center of gravity 3 of the inscribed polyhedron is also the center of gravity of all the inscribed lines of the arc 9Y. In the same manner it is proved that the center of gravity 1 of the circumscribed polyhedron is the same as the center of gravity of all of the circumscribed lines together on the arc 9Y. Let the center of gravity of the portion of the solid of revolution be 4 and the center of gravity of the arc 9Y be 2, which is necessarily on the line IZ between the points 1 and 3. Let the lines 14, 24, and 34 be joined. By construction, 13 is less than 24. Of the angles 124 and 423, let 124 be the greater or certainly not less. Therefore, the side 14 is more than the side 24. But 24 is more than 13. Consequently, 14 is more than 13. That is, the center of gravity of the portion of the solid of revolution is at a distance from the center of gravity of the polyhedron circumscribed in it in an interval more than the center of gravity of of its circumscribing polyhedron is distant from the center of gravity of its inscribing polyhedron, which is absurd. Indeed, the centers of gravity of the inscribed and circumscribed polyhedra are at a greater distance than the centers of gravity of the portion of the solid of revolution and the polyhedron of the circumscribed. Consequently, the centers of gravity of the portion of the solid of revolution and the arc of the circle 9Y are distant by no interval. Therefore, their center of gravity is the same, which it was desired to demonstrate.

It ought to be noted in the following that we always understand the surface of the portion of the solid of revolution apart from the the planes in which it is terminated.

With the same things having been assumed as in the preceding, the center of gravity of the surface of the portion of the solid of revolution will be the same as the center of gravity of the circular arc in which the center of equilibrium of the surface of the lower trunk of the cylinder in the distinguished base of the cylinder is revolved.

With the same things be supposed as in the preceding, let the center of equilibrium of the surface of the trunk BCAG in the base be E. I say that the center of gravity of the surface of portion of the solid of revolution is the same as the center of gravity of the circular arc in which the point E is revolved.

If they were not the same, there would be a line β which is the difference between them. Let it be conceived that a solid polyhedron, consisting of 2, or 4, or 8, or 16, or however many(as long as the number is in the progression of doubles from two) lower trunks equal to the proposed right cylinder cut by a plane having common intersection with the base line BC. And in this same portion of the solid of revolution let be inscribed another polyhedron similar to the previous one in such a way that the centers of gravity of those surfaces are at a distance of an interval less than β. Indeed, this is possible since the more sides such polyhedra have the less is the discrepancy between them ad infinitum, thus so that their distance is able to be less than any proposed quantity. Consequently, through the center of equilibrium E let a plane normal to the line BC be drawn making with the trunk ABCG the common intersection the triangle HFD, right at F, likewise with the portion of the solid of revolution a common intersection the sector IVLO of the circle, with the polyhedron circumscribed to the portion a regular polygon circumscribing the sector, likewise with the inscribed polyhedron a similar portion of a regular polygon inscribed in the same sector. It is manifest that the radius IV is equal to the line DF. Let the sector of the circle Iγ9 with radius Iγ be equal to the line DE. Let the polygon TS8I similar to the polygon IKMN be circumscribed around the sector Iγ9, likewise let the polygon Iγ79 be inscribed similar to the circumscribed. Let the line IL be drawn dividing the four parallel sides of the similar polygons 7γ, ST, PV, and MK in half in the points Y, R, Q, and L. It is manifest that the triangle ILK, right at L, is the common section of the cutting plane with one of the equal trunks from which the polyhedron circumscribing the portion of the solid of revolution consists. Let us suppose that this trunk is the same as the trunk ABCG (which we are able to do without absurdity since all of the surfaces of the lower trunks whose cutting planes cut the base in the line BC have the same center of equilibrium E in the base). Consequently, it is apparent that the center of equilibrium of the base of the trunk, having common intersection LIK with the cutting plane, is the point R. In the same manner it is proved that the point R is the center of equilibrium in the base of the surface of the trunk whose intersection with the cutting plane is MIL. Therefore, it is proved in the same manner that the center of gravity of the surface of those trunks joined together in any two trunks of the entire polyhedron circumscribed to one surface is at the point where the line from the vertex I cutting the base of the triangle (which is the common section of the cutting plane with the two joined trunks) in half intersects the circular arc 9γ. It is also prove easily without a problem that the point Y is the center of gravity of the surface of two joined trunks whose common intersection with the cutting plane is IPV. Likewise, the point is the center of gravity of the surface of the two joined trunks whose common intersection with the cutting plane is the triangle φIP. Let the line XY, which is cut in half by the point 5, be joined. It is manifest that the point 5 is the center of gravity of the surface of the four trunks joined together, whose common intersection with the cutting plane is the trapezium IVPφ. The point 5 is also the center of gravity of the lines Z7 and 7φ together. In the same manner it is proved that the α center of gravity of the lines 9ν and νZ is the center of gravity of the four surfaces of the trunk joined together, whose common intersection is the trapezium IφδA. Let the centers of gravity α and 5 be joined by the line α5, which is cut in half at 3. It is manifest that the center of gravity 3 of the inscribed polyhedron of the surface is also the center of gravity of all the lines inscribed in the arc 9γ. In the same manner it is proved that the center of gravity 1 of the surface of the circumscribed polyhedron is the same as the center of gravity of all of the lines circumscribed around the arc 9γ. Let the center of gravity of the surface of the portion of the solid of revolution be 4 and the center of gravity of the arc 9γ be 2, which is necessarily in the line IZ between the point 1 and 3. Let the lines 14, 24, and 34 be joined. By construction 13 is less than 24. Of the angles 124 and 423, let 124 be more or not even less. Therefore, the side 14 is more than the side 24, but 24 is more than 13. Consequently, 14 is more than 13. That is, the center of gravity of the surface of the portion of the solid of revolution is at a distance from the center of gravity of the surface of the polyhedron circumscribing it in a greater interval than the center of gravity of the surface of its circumscribing polyhedron is from the center of gravity of the surface of the similar inscribed polyhedron, which is absurd. Indeed, the centers of gravity of the inscribing and circumscribing polyhedrons are at a greater distance than the center of gravity of the surface of the portion of the solid of revolution is from the center of gravity of the surface of the polyhedron either inscribed or circumscribed in it. Consequently, the centers of gravity of the surface of the portion of the solid of revolution and of the arc 9δ are distant by no interval. Therefore, their center of gravity is the same, which ought to have been demonstrated.

In the same manner, this propostion is demonstrated concerning any surface of revolution made by rotation of one line or many lines around an axis as long as the radius of rotation does not cut them in many points.

If a right cylindrical figure is cut by any plane and the base of the cylinder is rotated around the common intersection of the base and the cutting plane so that it makes a portion of a solid of revolution, the center of gravity of the portion will be the same as the center of gravity of the circular arc in which the center of equilibrium in the base of the lower trunk of the surface of the cylinder is revolved.

Let a right cylindrical figure conceived above any base KECD be cut by a plane having the line AB as the common intersection with the base. Let the center of equilibrium in the base of the lower trunk be denoted as the point F. From the rotation of the figure KECD around the line AB let there be a portion of a solid of revolution. I say the center of gravity of this is the same as the center of gravity of the circular arc in which the point F is revolved.

Let the lines CA and KB be joined as you please. Suppose that the right cylinder above the base ACDKB is cut by the above-mentioned plane through the line AB. Suppose also that the center of equilibrium of the lower trunk in the base is the point H, and, likewise, that the center of equilibrium in the base of the trunk above the figure ACEKB is the point G. From the rotation of the figure ACDKB around the line AB (in such a way that its extreme part--namely, the figure CDKE--describes the portion of the previously-mentioned solid of revolution) let there be a portion of the solid of revolution whose center of gravity is M. It is manifest that the portion of the solid of revolution arising from the rotation of the figure ACDKB is equal to the portion of the solid of revolution arising from the rotation of the figure CDKE (whose center of gravity is L) and the solid of revolution arising from the rotation of the figure ACEKB (whose center of gravity is N)). Likewise, the trunk above the figure ACDKB is equal to the trunk above the figures CDKE and ACEKB. Consequently, it is clear that the points F, H, and G are in one line and, likewise, FH is to HG as the trunk above ACEKB is to the trunk above CDKE. Likewise, the points L, M, and N are in one line, and LM is to MN as the portion of the solid of revolution arising from the figure ACEKB is to the portion of the solid of revolution arising from the figure CDKE. But the portions of the solids of revolution between themselves are in a direct ratio to the trunks, as is easily gathered from Proposition 23. Therefore, as FH is to GH, thus LM is to MN. Through the points F and L let the line FLO be drawn, and through the points H and M let the line HMP be drawn. Likewise, through the points G and N let the line GNQ be drawn. It is sufficiently clear that lines HMP and GNQ (since they are radii of circles in whose circumferences the points H and G are revolved) and, therefore, also the line FLO are parallel among themselves and normal to the line AB. Therefore, as PM is to PH--or QN is to QG--thus OL is to OF. But M and N are centers of gravity of similar circular arcs in which the points H and G are revolved around the centers P and Q. Therefore, L is also the center of gravity of the similar circular arc in which the point F is rotated around the center O. But L is also the center of gravity of the portion of the solid of revolution arising from the rotation of the figure CDKE around the line AB, which it was desired to demonstrate.

If a curve or curves are rotated around a line so that from these a portion of a surface of revolution is generated, even if the radius of rotation cuts them in a number of points, the center of gravity of the portion of the surface of revolution will be the same as the center of gravity of the circular arc in which is revolved the center of equilibrium in the base of the surface of the trunk of the right cylinder cut by a plane having common section with the base the axis of rotation.

Let ABO be a curve or curves which are assumed to be revolved around the axis IK and let these curves be cut in two points by the radius of rotation. Above the curves of ABO imagine the surface of a cylinder perpendicular to the base, which is cut by a plane also cutting the base in the line IK. Let the center of equilibrium in the base of the surface of the lower trunk be E. From the rotation of the curves ABO, let there be a portion of a surface of revolution whose center of gravity is F. I say that F is the same as the center of gravity of the circular arc in which the point E is revolved.

Let B be the point in which the radius of rotation touches the curves ABO. Let the center of equilibrium in the base of the portion of the surface of the trunk above the curve or curves AB be C and the center of equilibrium in the base of the portion of the surface of revolution arising from AB be D. Let the center of equilibrium in the base of the surface of the trunk above OB be G and the center of equilibrium of the portion of the surface of revolution arising from the same line be H. It is manifest that the points C, E, and G are in the same line. Likewise, CE is to EG as the surface of the trunk above OB is to the surface of the trunk above AB. In the same manner it is clear that the points D, F, and H are in the same line. Likewise, DF is to FH as the portion of the surface of revolution arising from OB is to the portion of the surface of revolution arising from AB. But the portions of the surfaces of revolution are between themselves in the direct proportion of the surfaces of the trunks, as is gathered from Proposition 24. Therefore, as CE is to EG, thus DF is to FH. Let the lines CD, EF, and GH be extended until they cut the line IK in the points L, M, and N. Since the points D and H are centers of gravity of the circular arcs in which the points C and G are revolved, it is manifest that the lines CDL and GHN are normal to the axis of rotation IK and parallel among themselves. Therefore, the line EFM is normal to the same IK, since as CE is to EG thus DF is to FH. (See Proposition 40.) Consequently, as LD is to LC--or NH is to NG--thus MF is to ME. But D and H are centers of gravity of the similar circular arcs in which the points C and G are rotated. Therefore, the point F is also the center of gravity of the similar circular arc in which the point E is rotated, which ought to have been demonstrated.

In the same manner, this theorem is demonstrated when the radius of rotation cuts the lines in three points by supposing this proposition just as this proposition supposed Propostion 40, and thus one after another (when the line is cut in many points) by always supposing the previous demonstrations.

With the same things having been assumed as in Proposition 6, I say that the solid of revolution arising from the rotation of the figure ABSO around the line AB is to the cylinder arising from the rotation of the rectangle ABRO around the same AB as the surface arising from the rotation of the curve AQ around that same AB is to the circle from the radius AO.

The figure ABSO is analogous proportionally to the curve AQ. (See Proposition 2.) Therefore, the centers of gravity of the figure ABSO and the curve AQ are distant by the same interval from the line AB. Let this interval be K. The centers of gravity of the line AO and of the rectangle ABRO are also apart by the same interval from the line AB, namely BR/2. But the figure ABSO is to the rectangle ABRO as the curve AQ is to the line AO. (See Proposition 2.) Therefore, by drawing the first and third in the circumference whose radius is K and the second and forth in the circumference whose radius is BR/2, the cylinder whose base is ABSO and whose altitude is the circumference on the radius K--namely, the solid of rotation arising from the rotation of the figure ABSO around AB--is to the cylinder whose base is ABRO and whose altitude is the circumference on the radius BR/2--namely, the cylinder arising form the rotation of ABRO around AB--just as the rectangle whose base is AQ and whose altitude is the circumference of the radius K--namely, the surface generated by the rotation of AQ around AB--is to the rectangle whose base is AO and whose altitude is the circumference of the radius BR/2--namely, the circumference from the radius AO--which ought to have been demonstrated. (See Propositions 35 and 36.)

Let BCE be any figure balanced from the line BE with the circumscribed rectangle ABED. I say that the moment of the rectangle ABED is to the moment of the figure BCE as the cylinder made from the revolution of the rectangle ABED around BE is to the solid of revolution made from the revolution of BCE around the same line BE.

Let FH be any line parallel and equal to AB and cutting the figure BCE in G. Let the lines FH and GH be bisected at the points O and P. It is manifest that the moment of the line FH is to the moment of the line GH in a ratio compounded from the ratio of FH to GH and from the ratio of OH to PH. That is, in the duplicate ratio of FH to GH--or as the circle on the radius FH is to the circle on the radius GH. Since this always happens thus, both the first and third are always the same.

As the moments of all FH's--namely, the moments of the entire rectangle ABED--are to the moments of all GH's--namely, the moment of the figure BCE--thus all of the circles on the radius FH--namely, the cylinder from the rotation of the rectangle ABED--are BE are to all the circles on the radius GH--namely, the solid of revolution arising from the rotation of the figure BCE around the same BE--which it was desired to demonstrate.

With the same things being assumed as in the preceding proposition, if the solid of revolution arising from the rotation of the rectangle ABED and of the figure BCE around BE is cut by a plane through the line BE which is perpendicular to the horizontal and if the semisolids are balanced from the line BE, I say that the moment of the semisolid ADEB is to the moment of the semisolid BCE as all the cubes from FH are to all the cubes from GH.

Let the centers of gravity of the semicircles on FH and GH be O and P. It is manifest that the moment of the semicircle FH is to the moment of the semicircle GH in the ratio compounded from the ratio of the semicircle FH to the semicircle GH--namely, from the duplicated ratio of FH to GH--and from the ratio of OH to PH--or FH to GH. Therefore, the moment of the semicircle FH is to the moment of the semicircle GH in the triple ratio of FH to GH--or as the cube from FH is to the cube on GH. Since it always turns out thus, both the first and the third are always the same. As the moments of all of the semicircles FH--namely, the moment of the semisolid BADE itself--are to the moments of all of the semicircles GH--namely, the moment of the semisolid BCE itself--thus are all of the cubes from FH to all of the cubes from GH, which it was desired to demonstrate.

This theorem recently found by the most intelligent geometer R. P. Stephanus de Angelis will be able to be demonstrated in the same manner concerning any part of a solid of revolution you please. cut by two planes in the axis of the solid of revolution by cutting them in turns.

With these things having been said concerning the transmutation and measurement of curved quantities in general, now let us undertake the application in particular cases. This is able indeed to be done with no difficulty through analysis in certain easy cases. But we, being eager for variety, will demonstrate the proposition sometimes purely with geometry, sometimes purely with analysis, and sometimes partly with geometry and partly with analysis. Hence indeed the keen reader will perceive easily what ought to be done for the demonstration in each case.

To find a circle equal to the surface of a parabolic conoid.

Let a parabolic conoid be generated by the revolution of a semiparabola ABC around the axis AB. It is desired to find a circle equal to this surface. Let AB be extended to D so that AD is a forth part of the latus rectum, and let the line AE equal to half of the latus rectum be tangent to the parabola at the vertex A. From the axis DB let a parabola DEK be drawn with the ordinate BC extended to K and let the line X be the diagonal of a square equal to the segment AEKB of the parabola. I say that the line X is the radius of a circle equal to the surface of the parabolic conoid generated from the rotation of the semiparabola ACB around the axis AB.

From any point on the parabolic curve AC--namely, M--let the ordinate MG, which is extended until it intersects the exterior parabolic curve DEK in N, be dropped down to the axis. Let GH be a line equal to half of the latus rectum. It is apparent from the demonstrations of many that the line MH touching the curve in the point M cuts it normally. Consequently, from the doctrine of asymptotes of parabolas passed down by Gregory of St. Vincent and others, it is manifest that the parabolas ABC and DBK are asymptotic.

Therefore, (from this same doctrine) the square of the line GN is equal to the squares of the lines GM and AE--that is, to the squares of the lines GM and GH. Moreover, the square of the line MH is equal to these squares. Therefore, the lines HM and GN are equals. Since this will always be the case in all points of the parabolic curve AMC, it is manifest from Proposition 3 that the parabolic segment AEKB is equal to the surface of the trunk above the curve AMC cut by a plane passing through the line AB and inclined in a semi-right angle with the parabola--or the base of the trunk. Therefore, since the square with sides the line X is the double of the surface of the trunk, X will be the radius of a circle equal to the surface of the conoid generated from the rotation of the parabolic curve AMC around the axis AB, which ought to have been demonstrated. (See Proposition 25.)

Hence it is also manifest that the center of gravity of the surface of a parabolic conoid is the same as the center of equilibrium of the parabolic segment AENKB on the axis AB. Indeed, by Proposition 3 the segment AEKB is proportional in magnitude and in weight with the surface of the trunk, and by Proposition 24 the surface of the trunk is proportional in magnitude and in weight with the surface of the conoid. But this proportion in magnitude and in weight follows from this proposition itself, since the same things which are demonstrated concerning the entirety are demonstrated in the same manner concerning their proportional parts--v.g., in the same manner it is demonstrated that the section of the surface of the parabolic conoid arising from the revolution of the curve MC is equal to the circle, the square of whose radius is double of the segment GNKB, which the preceding proposition demonstrated.

To find a circle equal to the surface of an oblong spheroid.

Let there be an oblong spheroid arising from the revolution of a semiellipse EFT around the longer axis ET. It is desired to find a circle equal to the surface of this spheroid. Let the lines TR and ED equal to half of the latus rectum touch the ellipse in the vertices T and E. Next, let an ellipse DFR with center G and vertex F be drawn. Let the square whose diagonal is X be equal to the elliptical segment DFRTE. I say that X is the radius of a circle equal to the surface of the proposed oblong spheroid.

Let the ellipse DFR be extended until it intersects the extended axis ET in the points Z and B. It is easily apparent that the lines GZ and FG are the conjugate semiaxes of the semiellipse BFZ. Let the lines BA, EC, TQ, and ZY, all equal to the line FG, be drawn touching the semiellipses in the vertices B, E, T, and Z. Let the line ACFQY be joined. Also, let GQ, GR, and GY, which cuts QT in S, be drawn. From any point N of the given semiellipse let NP be perpendicular to the axis ET. This should be extended so that it cuts the line AY in I and the lines GY, GR, and GQ in the points O, M, and L, and the ellipse BFZ in K. Let the line PH be equal to the line PM and let HN be joined. It is manifest (from the van Schooten commentaries on Descartes, page 214) that HN cuts the tangent to the ellipse normally in the point N.

Because FRZ is an ellipse and FGZY is a rectangle on the semiaxis whose diameter is GY, the square of the line QT is equal to the squares of the lines RT and TS. Since the lines LP, MP, and OP are in the same ratio as the lines QT, RT, and ST, the square of the line LP will be equal to the squares of the lines MP and OP. But the square of the line IP is equal to the squares of the lines LP and NP on account of the above-mentioned property of ellipses. Therefore, the square of the line IP is equal to the squares of the lines MP, NP, and OP. But the square of the line IP is also equal to the squares of the lines OP and KP. Consequently, the squares of the lines MP, NP, and OP are equal to the squares of the lines KP and OP and, by taking away equals, the square of the line KP is equal to the squares of the lines MP and NP--or HP and NP . But the square of the line HN is equal to those same squares. Therefore, the lines HN and KP are equals. Since this will always be the case in all points of the elliptical curve EFT, it is manifest from Proposition 3 that the elliptical segment EDFRT is equal to the surface of the trunk above the curve EFT cut by a plane passing through the line ET and inclined in a semi-right angle toward the base. Therefore, since the square with sides equal to the line X is double the surface of the trunk, X will be the radius of a circle equal to the surface of a spheroid generated by the rotation of the curve EFT around the axis ET, which it was desired to demonstrate.

Hence it is also apparent that because the surface of the spheroid is in magnitude and in gravity proportional with the segment EDRT of the ellipse, indeed the same thing which is demonstrated concerning the entirety is demonstrated by a not dissimilar method concerning its proportional parts--v.g., in the same manner it is demonstrated that the surface arising from the rotation of the curve FN is equal to the circle, the square of whose radius is the double of the elliptical segment, which the previous proposition demonstrated. But it also ought to be understood in these three following results, since the ratio for this and for those is the same, that I always repeat the same thing.

To find a circle equal to the surface of an oblate spheroid.

Let there be an oblate spheroid generated from the rotation of the semi-ellipse MCR around the smaller axis MR. It is desired to find a circle equal to the surface of this. Let the lines MA and RD, equal to half of the latus rectum, touch the ellipse in the vertices M and R. Next, let a hyperbola ACD with vertex C and center P be drawn through the points A and D. Let the square whose diagonal is the line X be equal to the hyperbolic segment ACDRM. I say that X is the radius of a circle equal to the surface of the proposed oblate spheroid.

Let PF be the hyperbolic asymptote cutting the line DR in F, let BCG be a line touching the hyperbola at the vertex, and let the lines PD and PG be joined. From any point I on the ellipse let IQ, cutting the lines CG, PD, PF and PG in the points H, K, O, and L and likewise intersecting the hyperbola in E, be perpendicular to the axis MR. Let the line QN be equal to the line QK. It is manifest (from the commentary of van Schooten on Descartes, page 214) that the line NI, touching the ellipse in the point I, cuts it normally. Since PF is a hyperbolic asymptote and CG is a line touching the hyperbola in the vertex and since GR is equal to the half-axis CP, the square on the line DR will be equal to the squares on the lines RF and RG. And since the lines KQ, OQ, and LQ are proportional to the lines DR, FR, and GR, the square of the line KQ will be equal to the squares on the lines OQ and LQ.

But the square on the line EQ is equal to the squares on the lines HQ and OQ, and the square on the line HQ is equal to the squares on the lines IQ and LQ. Therefore, the square of the line EQ is equal to the squares on the lines IQ, LQ and OQ--that is, to the squares on the lines IQ and KQ--or IQ and QN. But the square on the line IN is equal to those same squares. Therefore, the lines EQ and IN are equal. And since this will be the case at all the points of the elliptical curve MCR, it is manifest from Proposition 3 that the segment of the hyperbola ACDRM is equal to the surface of the trunk above the elliptical curve MCR cut by a plane passing through the line MR and inclined in a semi-right angle with the base of the trunk. Therefore, since the square with sides equal the line X is the double of the surface of the trunk, X will be the radius of the circle equal to the surface of the oblate spheroid generated by the rotation of the elliptical curve MCR around the axis MR, which ought to have been demonstrated.

To find a circle equal to the surface of a hyperbolic conoid.

Let there be a hyperbolic conoid generated by the revolution of the semi-hyperbola FST around the axis FS. It is desired to find a circle equal to this surface. Let the line FH equal to half of the latus rectum touch the hyperbola in the vertex F. Next let the hyperbola, passing through the point H and having the same center--namely D--and the same conjugate semiaxis--namely DE--as the proposed hyperbola FST, be described. Let the described hyperbola be BHR, cutting the extended ordinate ST in R. Let the square whose diagonal is X be equal to the hyperbolic segment FHRS. I say that X is the radius of a circle equal to the surface of the proposed hyperbolic conoid.

Let the asymptote of the proposed hyperbola be DV, cutting the extended line FH in K, and let DY be the asymptote of the created hyperbola, cutting the extended FH in C. Let the line DH be extended. From any point Q of the proposed hyperbola let the line QG, which intersects the extended lines DH, DV, and DY in the points Z, O, and L, and the created hyperbola in the point P, be an ordinate from the axis. Let GA be equal to the line GZ. It is manifest (from the van Schooten commentaries on Descartes, page 216) that the line QA touching the hyperbola in the point Q cuts it normally. Since the line FK, touching the hyperbola in the vertex, is extended to its asymptote in the point K, FK will be equal to the conjugate semiaxis DE. Because BHR is a hyperbola and DY its asymptote, the square on the line FC is equal to the squares on the lines FH and DE--that is, to FH and FK.

And since the lines GZ, GO, and GL are in the same ratio with the lines FH, FK, and FC, the square on the line GL will be equal to the squares on the lines GO and GZ. But because FQT is a hyperbola and DV its asymptote, the square on the line GO is equal to the squares on the lines GQ and DE. Therefore, the square on the line GL will be equal to the squares on the lines GZ, GQ, and DE. But the square on the line GL is also equal to the squares on the lines GP and DE. Therefore, by taking away equals, the square on the line GP is equal to the squares on the lines GZ and GQ--that is, to the squares on the lines GA and GQ. Moreover, the square on the line AQ is equal to these squares. Therefore, the lines GP and AQ are equal. And since this will always come about at all the points in the hyperbolic curve FQT, it is manifest from Proposition 3 that the hyperbolic segment FHRS is equal to the surface of the trunk above the hyperbolic curve FQT cut by a plane passing through the line FS and inclined in a semi-right angle with the base of the trunk. Therefore, since the square with side the line X is the double of the surface of the trunk, X will be the radius of the circle equal to the surface of the hyperbolic conoid generated by the rotation of the hyperbolic curve FQT around the axis FS, which ought to have been demonstrated.

If a hyperbolic curve is rotated around its conjugate axis, to find a circle equal to the surface arising from the rotation of that hyperbolic curve.

Let αC be a hyperbolic curve which generates a surface when rotated around its own conjugate axis FZ. It is desired to find a circle equal to this surface. Let Z be the center, α the vertex, and ZD an asymptote of the hyperbolic curve. Let the line αK, equal to half of the latus rectum of the same hyperbola, touch the hyperbola in the vertex α. Let HK be a line parallel and equal to the line Zα which, extended indefinitely, cuts the hyperbola in R and its asymptote ZD in M. Suppose HV is a a segment of this line whose square is equal to the squares on the lines HR and HK. Next, let αVA be a hyperbola with center Z and vertex α which is extended until it intersects the line FCA parallel to the line Zα at the point A. Let the square whose diagonal is X be equal to the hyperbolic segment FAVαZ. I say that X is the radius of the sought after circle.

Let ZB, cutting the line HV in L, be the asymptote of the created hyperbola. Let TI and Tβ, perpendicular to each axis (extended if necessary), be dropped down out of a given point T in the original hyperbola. Let Gγ be a line cutting the hyperbolic curve normally in the point T and cutting each axis (extended if necessary) in the points G and γ. Let the lines Tβ and TI be extended (if necessary) and intersect the line ZKY in the points Y and O. It is manifest from the place cited in van Schooten that the lines βY and βγ are equal.

Moreover, as βγ is to βT--or EO--thus IT--or βZ--is to IG. Therefore, as Yβ is to OE, thus Zβ is to IG. But as Yβ is to OE, thus Zβ is to ZE--or IO. Therefore, the lines GI and IO are equal.

Let the line IT be extended so that it intersects the created hyperbola in Q and its asymptote in P, and let its intersection with the asymptote of the original hyperbola be the point N. By construction, the square on the line HV (that is, the square on the lines HL and Zα) is equal to the squares on the lines HR and HK--or the squares on the lines HM, HK, and Zα. Therefore, by taking away the same on each side, the square on the line HL equal to the squares on the lines HK and HM remains. And since the lines IP, IN, and IO are in the same ratio with the lines HL, HM, and HK, the square on the line IP will also be equal to the squares on the lines IN and IO. Therefore, by adding the square on Zα to each, the squares on the lines IP and Zα (that is, the square on the line IQ) is equal to the squares on the lines IO, IN and Zα--that is, to the squares on the lines IO and IT (or IG and IT)--that is, to the square on the line TG. Therefore, the lines QI and TG are equal. And since this always occurs in all the points of the hyperbola αTC, it is manifest from Proposition 3 that the hyperbolic segment FAαZ is equal to the surface of the trunk above the hyperbolic curve αTC cut by a plane passing through the line FZ and inclined in a semi-right angle with the base of the trunk. Therefore, since the square on the line X is double the surface of the trunk, X will be the radius of a circle equal to the surface arising from the rotation of the hyperbolic curve αTC around the conjugate axis ZF, which ought to have been demonstrated. (See Proposition 25.)

To find a line equal to a parabolic curve.

Let GN be a parabolic curve having the vertex G and the axis GO. It is desired to find a line equal to this. Let the axis GO be extended to A so that GA is equal to half of the latus rectum, and let GK be a line touching the parabola in the vertex and intersecting the line NK parallel to the axis in K. Next, let GD be a line dividing the right angle AGK in half and let AC be a hyperbola having vertex A, center G, and asymptote GD. Let the line NK be extended until it cuts the hyperbola in the point C, and let AE be parallel to the line GK. Finally, as the rectangle AEKG is to the segment AGKC of the hyperbola, thus let the line GK be to the line X. I say that the line X is equal to the parabolic curve GN.

From any point M on the parabolic curve let ML and MH be dropped down perpendicular to the lines GO and GK, and let OI be a line cutting the parabolic curve normally in the point M and intersecting the lines GO and HK in the points O and I. It is manifest that the line LO is equal to half of the latus rectum, or GA. Let the line MH be extended so that it cuts the hyperbola in the point B and the asymptote GD in F.

It is also manifest, because GD is an asymptote, that the square on the line HB is equal to the squares on the lines AG and HF or, because the angles HGF and HFG are equal, to the squares on the lines LO and GH--or LO and LM. But the square ob the line OM is equal to the squares ob the lines OL and LM. Therefore, the lines OM and HB are equal. Next, because of the similarity of the triangles HMI and LOM, as HM is to MI, thus OL is to OM--or GA to HB. And since this will always be the case in all the points of the parabolic curve GN, by Proposition 2 the rectangle AEKG will be to the segment of the hyperbola ACKG as the line GK is to the parabolic curve GN. Moreover, as the rectangle AEKG was to the hyperbolic segment ACKG, thus the line GK was to the line X. Consequently, the line X is equal to the curve GN, which it was desired to demonstrate.

It is demonstrated in the same manner that the rectangle PEKH is to the hyperbolic segment BHKC as the line HK is to the curve MN. Consequently, the hyperbolic segment ACKG is proportional in magnitude and weight with the curve GN, as is seen from Proposition 2. And since the lower trunk of the right cylinder above ACKG is given easily by Propositions 21, 35, and 37, the center of gravity of the hyperbolic segment ACKG will be given. Therefore, the center of equilibrium of the curve GN on the line GK becomes known. Hence also, the proportions in magnitude and in weight, which lie between the parabolic curve, the surface of the oblate sphere, and the surface arising from the rotation of the hyperbolic curve around its own conjugate axis, as is evident from inspection alone of the figures of Propositions 48, 50, and 51, are worth noting.

If a parabolic curve is revolved around a line touching it in its vertex, to find a circle equal to the surface arising from such a rotation.

Let the parabolic curve IED, whose vertex is D, be rotated around the line BD touching it in its vertex so that a surface of revolution is generated. It is desired to find a circle equal to this surface. Let the latus rectum of the parabolic curve be quadruple the line DR, and let its axis be DK. From the point I let IK be a line perpendicular to the axis DK and let DGL, cutting the extended line IK in L, be a hyperbola with transverse axis RD (to which the conjugate axis is also equal). Next, let a square whose diagonal is X be equal to the semi-hyperbola DKL. I say that X is the radius of the sought after circle.

From any point E on the parabolic curve let a line EF, which intersects the extended hyperbola in the point G, be dropped down perpendicular to the axis. Let EA be a line tangent to the parabola at the point E and intersecting the tangent DB in C and the extended axis at A.

Let CH be equal and parallel to the line DF. It is manifest that FA is double the line DA. Therefore, EF is double the line CD. Therefore, EH is equal to the line HF. The rectangle formed by RF and FD is equal to the square on the line DF together with the rectangle RDF. But the rectangle RDF is equal to the square on EH, since RD is a quarter part of the latus rectum. Therefore, the square on the line CE (which is equal to the squares on the lines EH and HC = DF) is equal to the rectangle formed by RF and FD. But the square on the line FG is equal to the same rectangle. Consequently, the lines EC and FG are equals between themselves. And since it is always this way, it is apparent from Proposition 4 that the semi-hyperbola DKL is equal to the surface of the trunk above the parabolic curve DEI cut by a plane passing through the line DB and inclined at a semi-right angle with the base of the trunk. Therefore, since the square on the line X is double the surface of the trunk, X will be the radius of a circle equal to the surface generated by the rotation of the parabolic curve around the line DB, which ought to have been demonstrated.

It is apparent from this proposition that the semi-hyperbola DKL is proportional in magnitude and weight with the surface. From this proposition, the preceding, and Proposition 38 it is not difficult to find the center of gravity of the parabolic curve or some given portion, which it suffices to have suggested.

To find the center of gravity of a right semi-cylinder cut obliquely.

Let the base of the cylinder be a circle BGQG whose diameter is BQ, and let the line BA touching the circle at A be an altitude of the semi-cylinder equal to the line BQ. In the same way indeed it is as all of the semicylinders above the same base are analogous, which are extended indefinitely to a part B. Let the line PQ be indefinitely long and parallel to it. Also, let AP, CR, and DT be parallel to the diameter BQ and let the line AQ be drawn with the remaining things having themselves as in the figure. Let the figure CRM be of such a nature that (with any line EYKM drawn parallel to the line PQ) EY is to FY as GY is to KM. The figure CRM is proportional in magnitude and gravity to the semi-cylinder. But the center of equilibrium of this figure--or of the semi-cylinder--is found thus: Let CR and CS be equal with the remaining things having themselves as in the figure, and let the figure DOT be of such a nature that (with any line EYKMNO having been drawn) EY is to KL as KM is to NO. And let the rectangle DV circumscribe the figure DOT. Next, as the figure CRM is to the figure DOT thus EY--or RC--is to CZ. It is manifest from Proposition 37 that Z is the center of equilibrium of the figure CMR. Moreover, the figure DOT becomes known in the same manner. The rectangle DV is given, since DT is equal to BQ and DX--or TV--is the fourth part in a continuing proportion whose first is EY and whose second is half the line BQ.

Consequently, NO is always the fourth in a continuing proportion whose first is EY and second is GY, which I prove thus: From the preceding things the following proportions are manifest:

EY: FY = QY :: GY : KM EY: KL = BY :: KM . NO EY^{2}: BY x YQ = GY^{2}:: KM x GY : KM x NO

and therefore,

EY^{2} : GY^{2} :: GY : NO.

and therefore,

GY^{3} = EY^{2} x NO. Therefore, with EY supposed
first and GY supposed second, NO will be the fourth in the continuing
proportion. Therefore, the figure DOT is to the rectangle DV as the
parabolic conoid (whose base is the circle BGQG) is to the parabolic
cylinder circumscribing the conoid. And since the proportion of the
parabolic conoid to the cylinder circumscribing it and the rectangle
DV are given, the figure DOT will also be given. Therefore the point
Z will also be given. Therefore, the center of equilibrium in the
base of the semi-cylinder is given. Consequently, the center of
gravity of the semi-cylinder will be on the given perpendicular to the
base erected from the center of equilibrium. But the same center of
gravity is on the line Qα, drawn between the point Q and the
midpoint α of the line AB from the point perpendicular to the
base BGQ, with it having been supposed of course that the cylinder is
cut by a plane cutting the base plane in the line PG. Therefore, if
as RC is to CZ, thus Qα is to αδ, δ will be the
center of gravity of the semi-cylinder, which ought to have been
found.

In Proposition 62 we will teach another method for finding the proportion between DV and DOT.

Let ABIK be a parallelogram and let the curve ACI be of such a nature that (with any line DE having been drawn parallel and equal to the line AB and cutting the curve in C) the ratio of AB to EC is multiplied with the ratio of AK to AE in the ratio of P to Q. I say the the parallelogram ABIK is to the figure ACIK as P + Q is to Q.

Let the line IB be extended to G and (with a line CF drawn tangent to the curve ACI at a given point C and with the parallels EC and FH having been drawn) let the curve AHB be drawn of such a nature that the line CH is equal to and parallel to the line EF. It is manifest from Proposition 7 that the line AE is to FE--or LC is to HC--as P is to Q. And since it will always be this way, it is evident that the figure ACIB is to the figure ACIGH as P is to Q. But from Proposition 11 the figure ACIGH is equal to the figure ACIK. Therefore, the figure ACIB is to the figure ACIK as P is to Q. By componendo, the parallelogram ABIK is to the figure ACIk as P + Q is to Q, which ought to have been demonstrated.

If P is less than Q, ACIK will be one of the infinite parabolas. If indeed P is more than Q, ACIK will be one of the infinite curves.

From this proposition it is also manifest that the parallelogram ABIK is to the figure ABIC as P + Q is to P, which is also understood thusly: since the ratio of AB to EC is multiplied with the ratio of AK to AE in the ratio of P to Q--that is, the ratio AB to AL is multiplied with the product of the ratio BI to LC in the same ratio P to Q--by convertendo, BI to LC will be multiplied with the ratio of AB to AL in the ratio of Q to P. Therefore, from the proposition itself, the parallelogram ABIK is to the figure ABIC as P + Q is to P, which ought to have been demonstrated.

With the same things being assumed as in the preceding proposition, I say that the ratio of the figure ACIK to the figure ACE is multiplied with the ratio of IK to CE in the ratio P + Q to P.

From the preceding proposition, ABIK is to ACIK as P + Q is to Q. In the same manner, ALCE is to ACE as P + Q is to Q. Therefore, ABIK is to ACIK as ALCE is to ACE. By permutando, ABIK is to ALCE as ACIK is to ACE. But the ratio of ABIK to ALCE is compounded from the ratio of AK to AE and the ratio of IK to CE. Consequently, the ratio of ACIK to ACE is compounded from the same ratios. But the ratio of IK to CE is multiplied with the ratio of AK to AE in the ratio of P to Q. By convertendo, the ratio AK to AE is multiplied with the ratio IK to CE in the ratio Q to P. By componendo, the ratio compounded from the ratios of AK to AE and IK to CE (namely, the ratio of the figure ACIK to the figure ACE) is multiplied with the ratio IK to CE in the ratio P + Q to P, which it was desired to demonstrate.

With the rectangle ABIK having been assumed and with the remaining things above the figures ABIK and ACIK as in the preceding proposition, if right cylinders, whose altitude is AB are supposed and likewise it is supposed that the cylinder above ACIK is cut by a plane semi-right to the bse ACIK and cutting it in the line AK, then I say that the parallelpiped above ABIK is to the lower trunk of the cylinder above ACIK as 4 P + 2 Q is to Q.

Suppose that the plane cutting the cylinder above ACIK is extended so that it also cuts the parallelpiped. It is manifest that its lower trunk is the triangular prism half of the parallelpiped. Next, let AMTK be a rectangle with an inscribed curve AT with the property that (with any point E having been assumed and the line DEO drawn perpendicular to AK and cutting the curves ART and ACI in the points R and C and likewise with a plane normal to AK having been drawn through the point E and cutting the lower cylinder of the trunk in the isosceles right triangles whose bases are DE and CE) OE is to RE just as the triangle above DE is to the triangle above CE. It is apparent that the triangle above DE is to the triangle above CE--or OE is to RE--in the duplicate ratio of DE to CE. But the ratio of DE to CE is multiplied with the ratio AK to AE in the ratio of P to Q. Therefore, the ratio of OE to RE is multiplied with the ratio of AK to AE in the ratio of 2 P to Q. And since this will always be the case wherever the point E is assumed, it is clear from Proposition 54 that the rectangle AMTK is to the figure ARTK as 2 P + Q is to Q. But OE is always to RE as the triangle above DE--namely, the common intersection of the plane normal to AK with the prismatic triangle--is to the triangle above CE--namely, the common intersection of the preceding plane with the lower trunk of the cylinder above ACIK. Also, the preceding quantities--namely, all the lines OE among themselves and all the triangles above DE among themselves--are equal among themselves. Therefore, as all of the lines OE--namely, the rectangle AMTK--are to all the lines RE--namely, the figure ARTK--thus all of the triangles above DE--namely, the prismatic triangle--are to all the triangles above CE--namely, the lower trunk of the cylinder above ACIK. Consequently, the prism is to the trunk as 2 P + Q is to Q. Therefore, the double of the prism--namely, the parallelpiped above ABIK--is to the lower trunk of the right cylinder ACIK as 4 P + 2 Q is to Q, which ought to have been demonstrated.

Also, from Proposition 54, the parallelpiped above ABIK is to the cylinder on the same altitude above ACIK as P + Q is to Q--that is, as 4 P + 2 Q is to (4 PQ + 2 Q Q)/ (P + Q). Therefore, the parallelpiped above ABIK is to the upper trunk of the same cylinder as 4 PP + 6 P Q + 2 QQ is to 3 PQ + QQ. But such an upper trunk is the same as the trunk of the lower cylinder reflected in the plane cutting the base seminormally in the line BI. Therefore, it is also clear that the ratio of this trunk to the parallelpiped is as 3 PQ + QQ is to 4 PP + 6 P Q + 2 QQ.

With the parallelpiped and the cylinder assumed to have altitude AK and the remaining things as in the preceding result, if the right cylinder above ACIK is cut by a plane seminormal to the base and cutting it in the line AB, then I say that the parallelpiped above ABIK is to the lower trunk of the cylinder above ACIK in the ratio of P + 2Q to Q.

Let AMTK be a square with an inscribed curve ART of such a nature that (with any point E having been assumed and the line DEO drawn perpendicular to AK and cutting the line ART in the point R and likewise with a plane drawn through E normal to AK whose intersection with the lower trunk is the rectangle LAEC) AE is to RE just as BA is to CE. It is clear from the construction that the ratio of AE to RE is multiplied with the ratio of AK to AE in the ratio of P to Q. By componendo, the ratio of AK to RE--that is, the ratio of KT to RE--is the product of the ratio of AK to AE with the ratio of P + Q to Q. Therefore, the square AMTK is to the figure ARTK as P + 2Q is to Q. (See Proposition 25.) But as ABIK is to AMTK, thus IK--or AB--is to TK--or AK. Therefore, the rectangle ABIK is to the figure ARTK in a ratio compounded from the ratio of P + 2Q to Q and from the ratio of AB to AK. Since AE is to RE as BA is to CE, the rectangle from AE and CE--namely, AECL--will be equal to the rectangle from AB and RE. And since this will always be the case, the cylinder above ARTK having altitude AB will be equal to the lower trunk of the cylinder above ACIK. Therefore, the parallelpiped above ABIK with altitude AB is to the cylinder above ARTK--or the lower trunk of the cylinder above ACIK--as the base ABIK is to the base ARTK--that is, in the ratio compounded from the ratio P + 2 Q to Q and the ratio of AB to AK. But the parallelpiped above ABIK with altitude AB is to the parallelpiped above ABIK with altitude AK as AB is to AK. Therefore, the parallelpiped above ABIK with altitude AK is to the lower trunk of the cylinder above ACIK as P + 2 Q is to Q, which ought to have been demonstrated.

Also, from Proposition 54, the parallelpiped above ABIK is to the cylinder of the same altitude above ACIK as P + Q is to Q--that is, as P + 2 Q is to (Q P + 2 QQ)/(P + Q). Therefore, the parallelpiped above ABIK of altitude AK is to the upper trunk of the cylinder above ACIK of the same altitude AK as PP + 3 PQ + 2 QQ is to QQ. But such an upper trunk is the same as the lower trunk of its cylinder reflected by a plane cutting the base seminormally in the line IK. Therefore, the ratio of the same parallelpiped to this lower trunk is the same.

From Propositions 29, 56, and 57, it is manifest that the center of equilibrium of the figure ACIK thus divides the line IK so that the part toward I is to the part toward K in the ratio 3 P + Q to P + Q. Likewise, the center of equilibrium of the same ACIK thus divides AK so that the part toward A is to the part toward K in the ratio P + Q to Q.

In the two preceding propositions, even though ABIK was assumed to be a rectangle because it was easy, the conclusions nevertheless are not less true for any parallelogram. Indeed, they are demonstrated with no difficulty from the equality and proportionality in magnitude and weight of the trunks above the rectangles and their figures with the trunks above whatever parallelograms and their figures. I will also wish this to be understood in the following results for figures infinite in length.

Let AEF be a right angle and let the curve ADF be of such a nature that (with any line DC having been drawn perpendicular to AE) the ratio of EF to CD is multiplied with the ratio of AE to AC in the ratio of any odd number to the next least number. It is desired to find a line equal to the curve ADF.

For instance, let the ratio of EF to CD be multiplied in the ratio of AE to AG with the ratio of 5 to 4. Using Proposition 2, as the line AE is to the curve AF thus let any rectangle AGNE be to the region AGLE. Let the latus rectum (namely, the line drawn 4 in the fourth power of the line CD always making the fifth root of the line AC) of the curve AF be l. Also, let AG be b, GI be c, and IL be d, and any increasing length GH be a. Let HO--or CA--be z.

It is manifest from Proposition 7 that BC is 4 z/5 and CD is the
fourth root of z^{5}/l, and BD is z5 /l + 16
z^2/25). Therefore, BC will be to BD as CK is to CO.
That is
. . . . . therefore, it is manifest that the trilineum GIL is
compounded from the addition of the triline, squared, cubed, and
raised to the power 4, of who all of the axes are GI and the bases the
trilines squared . . . cubed and fourth-powered . Thereupon, the
triline GOLI is . ... Therefore the line AE is to the curve AF as bd
or the rectangle AGNE is to bd . . . or the curved region AGLe.
Therefore, the line is given equal to the curve AF, which ought to
have been found. How is the given surface found from the rotation of
the curve AF around the line AG. It is evident from Proposition ?3
since from Propostion 57 it is easy to find the center of gravity of
the curved region AGLE. But if . . . .

Let ABFE be a parallelogram and let FGH be a curve of such a nature that (with any line CG having been drawn parallel to the line AB and cutting the curve FH in G and the line BF in D) the ratio of AB to CG is multiplied with the ratio of BD to BF in the ratio of P to Q. I say that the parallelogram ABFE is to the figure HAEFGH infinite in length as Q - P is to Q.

Let the line BF be extended to M and let the curve MKH be drawn of such a nature that (with the line GI having been drawn touching the curve FGH in any point G and with the parallels GC and KI having been drawn) the line GK is equal and parallel to the line CI. It is manifest from Proposition 7 that the line AC is to CI--or LG is to GK--as P is to Q. And since this will always be the case, it is evident that the figure HFBH is to the figure HFMH as P is to Q. But from Proposition 11 the figure HFMH is equal to the figure HAEFH. Therefore, the figure HAEFH is to the figure HFBH as Q is to P. By conversion of the ratios, HAEFH is to the parallelogram ABFE as Q is to Q - P. By convertendo, as the parallelogram ABFE is to the figure HAEFH thus Q - P is to Q, which ought to have been demonstrated.

With the rectangle ABFE having been assumed and the remaining things having been supposed as in the preceding proposition, if above the figures ABFE and HAEFH right cylinders are supposed, with the altitude of the first AB and, moreover, the altitude of the second infinite, and likewise by supposing that the cylinder above HAEFH is cut by a plane seminormal to the base itself and cutting it in the line AE, I say that the parallelpiped above ABFE is to the lower trunk of the cylinder above HAEFH as 2 Q - 4 P is to Q.

By considering what ought to be considered, the demonstration does not differ from Proposition 56.

With the same cylinders having been assumed as in the previous proposition, but with the altitude BF and with it having been supposed that the cylinder above HAEFH is cut by a plane seminormal to the base and cutting it in the line AH, I say that the parallelpiped above ABFE is to the lower trunk of the cylinder above HAEFH in the ratio 2 Q - P to Q.

First, let P be less than Q. As AB is to CG thus let BD be to DR, extended toward AE. Let the square NBFO be filled out from the portion AE. Likewise from all the points R, let a curve BRO be imagined . It is apparent from the construction that the ratio of BD to DR is multiplied with the ratio of BD to BF in the ratio of P to Q. By dividendo, the ratio of DR to BF--namely the ratio of DR to FO--is multiplied with the ratio of BD to BF in the ratio of Q - P to Q. Therefore, the square NBFO is to the figure BROF in the ratio of 2 Q - P to Q. (See Proposition 54.) But as ABFE is to NBFO thus EF--or AB--is to OF--or BF. Therefore, the rectangle ABFE is to the figure BROF in the ratio compounded from the ratio of 2 Q - P to Q and the ratio of AB to BF. Since as AB is to CG thus BD is to DR, the rectangle from AB and DR will be equal to the rectangle from BD and CG. But the rectangle from AB and DR is the common intersection of the right cylinder above BROF having altitude AB with the plane above RD cutting the base normally. The rectangle from BD and CG is the intersection of the same extended plane as before with the lower trunk of the cylinder above HAEFH. And since these intersections always happen to be equal, it is manifest that the lower trunk of the cylinder above HAEFH is equal to the cylinder above BROF having altitude AB. Therefore, the parallelpiped above ABFE with altitude AB is to the cylinder above BROF--or the lower trunk of the cylinder above HAEFH--as the base ABFE is to the base BROF--that is, in the ratio compounded from the ratio of 2 Q - P to Q and from the ratio of AB to BF. But the parallelpiped above ABFE with the altitude AB is to the parallelpiped above the same ABFE with altitude BF as AB is to BF. Consequently, the parallelpiped above ABFE with altitude BF is to the lower trunk of the cylinder above HAEFH as 2 Q - P is to Q.

But if P is more than Q, as AB is to CG thus let BD be to CR, extended toward LK. Let the square NAEO be filled out from the portion LK. Thus let a curve be imagined from all the points R. It is apparent from the construction that the ratio of BD to CR is multiplied with the ratio of BD to BF in the ratio of P to Q. By assuming the excess of the antecedent above the consequent to the consequent, the ratio of BF to CR--namely, the ratio of EO to CR--is multiplied with the ratio of BD to BF--or AC to AE--in the ratio of P - Q to Q. Therefore, the square NAEO is to the figure HAEOH infinite in length as 2 Q - P is to Q. But as ABFE is to NAEO thus EF--or AB--is to EO--or BF. Therefore, the rectangle ABE is to the figure HAEOH in a ratio compounded from the ratio of 2 Q - P to Q and the ratio of AB to BF. Since as AB is to CG thus BD is to CR, the rectangle from AB and CR will be equal to the rectangle from BD and CG. But the rectangle from AB and CR is the common intersection of the right cylinder above HAEOH having altitude AB with the cutting plane normal to the base above CR. The rectangle from BD and CG is the intersection of the same plane as before with the lower trunk above HAEFH. Since these intersections always happen to be equal, it is manifest that the lower trunk of the cylinder above HAEFH is equal to the cylinder above HAEOH having altitude AB. Therefore, the parallelpiped above ABFE with altitude AB is to the cylinder above HAEOH--or the lower trunk of the cylinder above HAEFH--as the base ABRE is the base HAEOH--that is, in the ratio compounded from the ratio of 2 Q - P to Q and the ratio of AB to BF. But the parallelpiped above ABFE with altitude AB is to the parallelpiped above the same ABFE with altitude BF as AB is to BF. Consequently, the parallelpiped above ABFE with altitude BF is to the lower trunk of the cylinder above HAEFH as 2 Q - P is to Q, which it was desired to demonstrate.

From this and Propositions 29 and 37, it is not difficult to infer that the center of equilibrium of the figure HAEFH on the line AH is thus determined--namely, that AB is to the distance of the center of equilibrium from the point A as 2 Q - 4P is to Q - P. Likewise, the center of equilibrium of the same figure thus divides the line AE so that the part toward A is to the part toward E in the ratio of Q - P to Q.

Let KDOH be a quadrant of a circle and let the line DA be parallel to the radius HK. Let the curve DEH be of such a nature that (with any line GI having been drawn parallel and equal to the line DF and cutting the circular curve in O and the curve DEH in E) the ratio of GI to EI is the triplicate ratio of GI to OI. I say that the quadrant of the circle KDOH is to the figure KDEH as 4 is to 3--that is, that the curve DEH divides the quadrant of the circle in the ratio of 3 to 1.

Let the curve HBD be drawn of such a nature that (with the line EC having been drawn tangent to the curve HED in some point E and with the parallels EF and BC having been drawn) the line EB is equal to and parallel to the line CF. It is manifest from Proposition 7 that the line CF--or BE--is three times the line EO. Since this will always be the case, it is evident that the figure HBDE is three times the figure HODE. But from Proposition 11 the figure HBDE is equal to the figure HEDK. Consequently, the figure HEDK is to the figure HODE in the ratio of 3 to 1. Therefore, the curve HED divides the quadrant of the circle in the ratio of 3 to 1, which it was desired to demonstrate.

If the ratio of GI to EI were the quintuplicate of the ratio of GI to OI, the curve HED would divide the quadrant of the circle in the ratio of 5 to 3. If, moreover, it were the septuplicate, it would divide it in the ratio of 7 to 5. Thus, in general, if the ratio of GI to EI is multiplied with the ratio of GI to OI in the ratio of some odd number to unity, the curve HED will divide the quadrant of the circle in the ratio of the same odd number to the next smallest odd number. All of these are demonstrated just as in this theorem. In nearly the same manner not only is the interpoluation by Wallis with which he measures the cycloidal space in the book on the cycloid demonstrated, but perhaps also all those which are able to be imagined, if the measurement of the first interpolation is supposed. But if the ratio of GI to EI is multiplied with the ratio of GI to OI in the ratio of some even number to unity, the quadrature of the figure KHED is given easily with the help of Proposition 54, as Wallis teaches in his Arithmetic of the Infinite. Therefore, the center of equilibrium of the figure DEHK in the line DK becomes known with the help of Proposition 37. Hence, it is also evident that DEF always has the same ratio to DEO as DEHK has to DEHO, which ought to be noted very easily. For (aside from many very beautiful problems) if the parabolic conoid is cut by a plane parallel to the axis, out of the given distance from the same axis whichever segment of the conoid you please will be given easily.

Let CBA be half of a primary cycloid curve, whose base is CG, and let AEG be the generating circle touching the base in the point G on the diameter GA. I say that the curve CBA is twice the line AG.

Let AILG be a square and let the figure NAGLN be of such a nature that (with some line FM having been drawn parallel to the line AI) the ratio of AI--or AG--to FM is half the ratio of AF to AG. It is manifest from Proposition 59 that the infinite figure NAGLN is double the square AILG. Through any point F on the diameter AG let the line BM be drawn parallel to AI, running into the cycloid and the curve LN in the points B and M and cutting the circle and the line IL as in the figure. Let the lines AE and EG be joined and DB and BH be parallel to them. From Proposition 8 it is apparent that BD is tangent to the cycloid at the point B and BH is perpendicular to the tangent. As BF is to BH, thus EF is to EG--or FA to AE. But the ratio of AF to AG is double the ratio of AF to AE. Therefore, the ratio of AF to AG is double the ratio of BF to BH. But the ratio of AF to AG is also double the ratio of AG to FM. Therefore, as BF is to BH, thus AG--or FK--is to FM. Since it is always thus, it is clear from Proposition 2 that the line AG is to the curve ABC in the ratio of 1/2--namely, as the square AILG is to the figure NAGLM--which it was desired to demonstrate.

Let CAB be a sector of a circle including a spiral AEC of such a nature that (with any line AE having been drawn and, extended, cutting the arc in D) the ratio of AC to AE is multiplied with the ratio of CB to DB in the ratio of P to Q. Let FIK be a right angle and let FHK be a curve of such a nature that (with any line GH having been drawn parallel to the line IK) the ratio of IK to GH is multiplied with the ratio of FI to FG in the ratio of P to Q.

It is apparent from Proposition 14 that the rectangle FNKI circumscribing the figure FIK is to the figure FHKI as the arc BC is to the axis of the evoluted figure of the spiral AEC. Consequently, from Proposition 54, the arc BC is to the axis of the evoluted figure of AEC as P + Q is to Q. As P + Q is to Q thus let the arc BC be to the line FI, to which the line IK equal to the line AC is normal. Let the figure FHKI be of such a nature that (with any line GH having been drawn parallel to the line IK) the ratio of IK to GH is multiplied with the ratio of FI to FG in the ratio of P to P + Q. I say that the figure FHKI is the evoluted area of the spiral AEC.

From any point G on the line FI let GH be normal to the line FI and
intersecting the curve FHK in H. Let AE, equal to the line GH, be
extended to D. The arc BC is to the axis of the evolute of
AEC--namely, FI--as P + Q is to Q. In the same manner the arc EL is
to the axis of the evolute of EL as P + Q is to Q. Therefore, as the
arc BC is to the line FI, thus the arc LE is to the axis of the figure
AE evoluted. By permutando, as the arc BC is to the arc LE thus the
line FI is to the axis of the evoluted figure AE. But the ratio of
the arc BE to the arc LE is compounded from the ratio of the arc BC to
the arc BD and the ratio of the arc BD to the arc LE--or of the line
CA to the line EA. Therefore, the line FI is to the axis of the
evoluted figure AE in the ratio compounded from the ratio of the arc
BC to the arc BD and from the ratio of the line CA to the line EA.
But the ratio of CA to EA is multiplied with the ratio of BC to BD in
the ratio of P to Q. By convertendo, componendo, and convertendo
again, the ratio of AC to AE--or IK to GH--is multiplied with those
ratios (which are compounded from the ratio of the arc BC to the arc
BD and the ratio of the line CA to the line EA)--namely, the ratio of
the line FI to the axis of the evoluted figure AE in the ratio of P to
P + Q. But the ratio of IK to GH is multiplied with the ratio of FI
to FH in the ratio of P to P + Q. Consequently, FG is the axis of the
evoluted figure AE and EA and GH are equal. Therefore, (since it is
always thus) it is manifest from Proposition 14 and its Consequence
that the figure FHKI is the evolute of AEC, which ought to have been
demonstrated.

The figure FHKI is the one described in Proposition 54. Since the line tangent to it in a given point is able to be found with the help of Proposition 7, a line is also able to be drawn touching the spiral AEC in a given point (from the Consequence of Proposition 18). Also very often the line equal to the curve FHK, which is also equal to the spiral curve AEC, is found (as is apparent from Proposition 58).

With the help of Proposition 15 or 16 it is proved with no difficulty that the sector BAC is to the area of the spiral AEC as 2 P + Q is to Q.

Let the sector CAB of a circle including the spiral AEC be of such a nature that (with any line AE having been drawn cutting the arc in D when extended) the ratio of DE to BA is multiplied with the ratio of CD to CB in the ratio of P to Q. Let FN be equal to the line BA. Likewise, let the angle FNK be right and let the curve FK be of such a nature that (with any line HM drawn parallel to the line FN) the ratio of HM to FN is multiplied with the ratio of KM to KN in the ratio of P to Q. Let the rectangle FNKI be filled out and let MH be extended to G. Let the line DA be drawn in such a way that DE is equal to the line HM. Therefore, EA will be equal to the line GH. The ratio of DE to BA--or HM to FN--is multiplied with the ratio of CD to CB in the ratio of P to Q. But the ratio of HM to FN is multiplied with the ratio of KM to KN--or IG to IF--in the same ratio of P to Q. Therefore, as CD is to CB thus IG is to IF, and as IK is to GH thus AC is to AE. Therefore (from Proposition 14) as the rectangle FK is to the figure FHKI-- that is, (from Proposition 54) as P + Q is to P--thus the arc BC is to the axis of the evoluted figure AEC, where FI is to the remaining things as before. Let the curve FTK be of such a nature that (with any line STH having been drawn parallel to the line FI) IF is to ST as the figure FHKI is to the figure FHG. I say that the figure FTKI is the space of the evoluted spiral AEC.

It is manifest from the construction that the line FI is the axis of the evoluted figure AEC. In the same manner (as was shown before) it is demonstrated that the rectangle FH is to the inscribed figure FHG as the arc LE is to the axis of the evoluted figure AE. The ratio of FI--or the axis of the evoluted figure AEC--to the axis of the evoluted figure AE is compounded from: the ratio of the line FI to the arc BC--or the figure FHKI to the rectangle FNKI; the ratio of the arc BC to the arc BD--or the line IF to the line FG (which is proved thusly: as CD is to CB, thus IG is to IF and by dividendo and convertendo as CB is to BD thus IF is to FG)--or of the rectangle FNKI to the rectangle FNMG; the ratio of the arc BD to the arc LE--or the line DA to the line EA--namely, the line GM to the line GH--or the rectangle FNMG to the rectangle FSHG; and the ratio of the arc EL to the axis of the evoluted figure AE--or the rectangle FSHG to the figure FGH. But it is clear that the ratio of the figure FHKI to the figure FHC is compounded from the same ratios. Therefore, as FI is to the axis of the evoluted figure AE, thus the figure FHKI is to the figure FHG--that is, as FI is to ST--or FV. Thus, FV is the axis of the evoluted figure AE. And since the ordinate VT is equal to the line AE and it is always this wherever the points E is assumed, it is manifest (from the Consequence to Proposition 14) that the figure FTKI is the area of the evoluted spiral AEC, which it was desired to demonstrate.

With the help of Proposition 7, the line tangent to the curve FTK is able to be drawn, since it is one of those which Descartes calls geometrical. Consequently, through Proposition 2, it is able to be compared with its own axis or base. If a line is found equal to this, the same line will be equal to the curve AEC, so that at any time the line tangent to the curve AEC in a given point is able to be drawn easily using the Consequence of Proposition 18.

From Propositions 15 and 57 it is found with no difficulty that the
sector BAC is to the area of the spiral AEC as q^{2} + 3 p q +
2 p^{2} is to 2 p^{2}.

With the motion of the earth having been assumed, the line described by the descending of a weight toward the center of the earth will be CEA if only the ratio of P to Q is 2 to 1. A great controversy concerning this has arisen between the most celebrated mathematicians RR. PP. Stephanum de Angelis and John Baptist Ricciolius, which is perhaps eliminated if the most learned Ricciolius considers that, by withdrawing every other motion, the strength of the impulse is in direct proportion to the velocities with which the mobile body draws closer to the resting body. Indeed, to me nothing is more apparent. And there is not anyone skilled in military architecture who does not suppose the preceding axiom in hauling around military artillery. With respect to the purely geometrical part of the controversy, I judge that no one now doubts that a weight falling toward the center of the earth in six hours time will be beyond the semicircle. It is evident that the calculation of R.P. Stephanus de Angelis (in Dialogus Primus, page 19) is correct. Moreover, the demonstration of D. Manfredus (which tried to show the contrary on page 17) is erroneous in this because it tacitly supposes (by assuming the semicircle CHA) that a weight descending to the center of the earth lasts six hours. For (by supposing the observations of Ricciolus and that the radius of the earth is 25,870,000 feet) it arrives at the center in a time of 21 minutes and 53 seconds. And in these the moments or motions of anything will not be interior or exterior. Indeed, geometry and statics abstracts from all physical causes, since they are uncertain and not evident.

Let LRM be a sector of a circle and let NT be an arc with the same center similar to the arc LM. Let the spiral NOM be of such a nature that (with any line RSOK having been drawn) the ratio of the line SO to the line TM is multiplied with the ratio of the arc LK to the arc LM in the ratio of P to Q. Let BC be equal to the line MT and let BH be as you please. Likewise, let CBH be a right angle and let the curve BVI be of such a nature that (with any line EV having been drawn parallel to the line BC) the ratio of EV to BC is multiplied with the ratio of BE to BH in the ratio of P to Q. Let the rectangle BCIH be filled out and let ABHG be a rectangle whose side AB is equal to the line TR. From Proposition 54, it is apparent that the rectangle BCIH is to the figure BVIH as P + Q is to Q. Let the line DEVF be drawn in such a way that EV is equal to the line OS. Therefore RO is equal to the line DV. The ratio of EV to HI--or OS to MT--is multiplied with the ratio of BE to BH in the ratio of P to Q. But the ratio of OS to TM is multiplied with the ratio LK to LM in the same ratio of P to Q. Therefore, as BE is the BH thus LK is to LM, and as DV is to GI, thus RO is to RM. Therefore, from Proposition 14, as ACIG is to ABVIG thus the arc LM is to the axis of the evoluted figure MONR. As ACIG is to ABVIG thus let the arc LM be to the line BH, with the remaining things in the figure ACIG having themselves as before. Let the curve BZI be of such a nature that (with any line XZV having been drawn parallel to the line BH) the figure ABVIG is to the figure ABVD as the line BH is to the line XZ--or A8. I say that the figure ABZIG is the area of the evoluted spiral RNOM.

It is manifest from the construction that the line AG is the axis of the evoluted figure RNOM. In the same manner (as shown previously) it is demonstrated that the rectangle AV is to the figure ABVD as the arc 40 is to the axis of the evoluted figure RNO. The ratio of AG--or the axis of the evoluted figure RNOM--to the axis of the evoluted figure RNO is compounded from: the ratio of the axis of the evoluted figure RNOM to the arc LM--or the figure ABVIG to the rectangle ACIG; the ratio of the arc LM to the arc LK--or the line IC to the line CF, or the rectangle ACIG to the rectangle ACFD; the ratio of the arc LK to the arc 40--or the line KR to the line OR, namely DF to DV, or the rectangle ACFD to the rectangle AXVD; and the ratio of the arc 40 to the axis of the evoluted figure RNO--or the rectangle AXVD to the figure ABVD.

But it is clear that the ratio of the figure ABVIG to the figure ABVD is compounded from the same ratios. Therefore, as the figure ABVIG is to the figure ABVD--or as AG is to A8--thus the axis AG of the evoluted figure RNOM is to the axis of the evoluted figure RNO, which therefore is A8. And since the ordinate 8Z is equal to the line OR and it is always this wherever the point O is assumed, it is manifest (from the Consequence of Proposition 14) that the figure ABZIG is the figure RNOM evoluted, which it was desired to demonstrate.

The curve BZI is one of those which Descartes calls geometrical. Therefore, from Proposition 7 the line tangent to it in a given point is able to be drawn. With the help of Proposition 2 it is also able to be compared with a line. Consequently, the curve NOM is also equal to that and the tangent to it in a given point is able to be drawn.

From Proposition 15 and 56 it will not be difficult to demonstrate
(assuming the ratio of HI to HG is as P is to Y) that the sector RLM
is to the figure RNOM as P is to Y^{2}/P + PQ/(2 P + Q) + (2
PQY)/(P^{2} + 2 PQ + Q^{2}).

It ought to be noted that the demonstration of Propositions 65 and 66 is maximally universal, but with the press running I did not remember to insert it into the appropriate place.

Here let us examine three general spirals. The first of these is the same as those two which R.P. Stephanus de Angeli considered in his book concerning spirals and in the end of book 5 concerning parabolas. The second is also the same as those two which were considered in the book on inverted spirals. The same most learned mathematician also figured out the third and recently shared it with me.

It ought to be observed completely that each figure is able to be conceived as an involute and the same involute is found by our method. Hence geometry is indeed augmented not a little.

This method of ours is so fruitful that it is usually difficult to
propose anything completely impervious to it. But so that we may test
it, let us solve those two problems which Renatus Franciscus Slusius
proposed in Proposition 8 concerning the infinite hyperbolas of
R.P. Stephanus de Angelis. Let AB be b and BD be a, and let the
second term of the sought after be: as b^{2} is to ba -
a^{2} thus a^{2} is to (b a^{3} -
a^{4})/b^{2}, the square of the line DC. Therefore,
from Proposition 54 the sum of all the squares from the infinite DC
will be b^{3}/20. But the sum of just so many of the squares
of the lines AB is b^{3}. Thus, since these sums of the
squares are the doubles of the lower trunks, which are cut in the
right cylinder above ACB and AEFB by a plane seminormal to the base,
(from Proposition 23) as b^{3} is to b^{3}/20--namely,
as 20 is to 1--thus the cylinder from the rotation of AEFB around AB
is to the solid of revolution from the rotation of ACB around the same
AB.

Let the second proposition be as in the same book on infinite
hyperbolas, Proposition 10. Let the ratio of AB to BD be multiplied
with the ratio of AD to DC in the ratio of P to Q, and let the ratio
of AB to BD be multiplied with the ratio of EA to DO in the ratio of P
to Q. It is manifest that EOF is a curve described in Proposition 54.
Therefore, the cube on the side EA--or AB--is to the lower trunk of
the right cylinder above EOBA cut by a plane cutting the base
seminormally in the line EA as P^{2} + 3 PQ + 2 Q^{2}
is to Q^{2}. Moreover, as EA is to DO, thus from the
situation AD--or KH--is to DC. Therefore, the rectangle from EA and
DC is equal to the rectangle from DO and KH. Since this will always
be the case, it is manifest that the cylinder above ACB (whose
altitude is EA) is equal to the lower trunk of the right cylinder
above EOBA cut by a plane cutting the base seminormally in the line
EA. Therefore, as P^{2} + 3 PQ + 2 Q^{2} is to
Q^{2}, thus EABF is to ACB.

These particular problems selected by me, besides certain problems now solved first by me, were found to be very difficult and of importance among geometers. But in fact Archimedes' entire work concerning the sphere and the cylinder is easily demonstrated from Proposition 3 in the manner of Proposition 46 and a few of the following propositions. The book concerning the spherical conoids and all of Luca Valerius' teaching is demonstrated from Proposition 21; all of the work of Guldin, Ioannis de laFaille, and Andreas Tacquet is demonstrated from Proposition 35 and a few of the following propositions. Nevertheless, I admit that I have in no manner been able to find the ratio between the elliptical or the hyperbolic curve and the line--even though in this little book there are many different methods for examining them--so that I therefore easily believe that this ratio is not analytic and that it is above the ratio between the circle and the sphere by one level. But this was not so easily demonstrated. I do not pretend that this method suffices for the resolution of all irregular problems--which are infinite in number and in difficulty, as for example there certainly are concerning the irregular section of bodies and surfaces by planes or other surfaces. Nevertheless among these I show the following sufficiently beautiful result. Indeed, it seems proper to add here certain miscellaneous results which are not at all impractical.

Let AFO be a right cone the axis of which is AM and which is cut by two planes AML and AMG, whose intersections with the surface of the cone are the lines AL and AG. Moreover, let the same cone AFO be cut by a plane making the curve GEBIL the intersection with the surface of the cone, in such a way that from these sections some conical pyramid ALMGEBI is carved out whose vertex is A and whose base figure BILMGEB is contained by the three planes ALM, AMG, and MGEBIL, and the surface of the cone AGEBIL. From the point M on the axis and the common intersection of the three planes let the line MF be perpendicular to the side of the cone. I say that the pyramid (whose base is a plane equal to the surface of the cone AGEBIL and whose altitude is MF) is equal to the conical pyramid ALMGEBI.

If the pyramids are not equal, let their difference be α and let the pyramid consisting of the triangular pyramids MGEA, MEBA, MBIA, and MILA be inscribed in the conical pyramid AMGEBIL. Also let another pyramid consisting of the triangular pyramids MHDA, MDCA, and MCKA circumscribe the same conical pyramid thus so the difference between the inscribed and circumscribed pyramids is less than α. It is manifest that MF is the altitude of the pyramid MHDA: Indeed, by supposition the triangle HDA touches the cone. Therefore, the normal from M to HDA falls normally on the contact line--or the side of the cone. In the same manner it is proved that MF is the altitude, from the vertex M, of all of the remaining triangular pyramids from which the circumscribed pyramid consists. Therefore, the pyramid whose base is equal to all of the triangles AHD, ADC, and ACK and whose altitude is MF is equal to the circumscribed pyramid. Consequently, it is more than the conical pyramid. But it is also more than the pyramid whose base is the plane equal to the surface of the cone AGEBIL and whose altitude is MF, since, having the same altitude with it, it has a greater base because it circumscribes it (as long as it is convex). Next, the triangular pyramid MGEA (with M assumed the vertex) has a smaller altitude than MF, since its base falls within the surface of the cone. In the same manner all the triangular pyramids comprising the inscribed pyramid have an altitude less than MF. Therefore, the pyramid (whose altitude is MF and whose base is equal to the bases of all of the triangular pyramids comprising the inscribed pyramid) is more than the inscribed pyramid. But this pyramid is less than the pyramid having base the plane equal to the conical surface AGEBIL and altitude MF since, having the same altitude, it has a smaller base--as long as the surface is convex. Therefore, the inscribed pyramid is much less than the same pyramid having base plane equal to the conic surface AGEBIL and altitude MF. But the inscribed pyramid is also less than the conic pyramid. Therefore, the inscribed pyramid is less than the conic pyramid and also less than the pyramid having base the conic surface. Indeed, the circumscribing pyramid was shown greater than either of these pyramids. Therefore, the difference between the circumscribed pyramid and inscribed pyramid is more than the difference between the conical pyramid and the pyramid which has the surface of the cone as the base. But the difference between the circumscribed pyramid and the inscribed pyramid is less than α. Therefore, the difference between the conical pyramid and the pyramid (which has the conical surface as the base) is much less than α, which is absurd. Indeed, it was supposed that there is a difference in the spoken of pyramids. Therefore, the conical pyramid AMEBIL and the pyramid whose base is the surface plane equal to the conical surface AGEBIL and whose altitude is the line MF are not different. Therefore, they are equal, which ought to have been demonstrated.

From the vertex A of the cone let the perpendicular line AN be sent down into the base (extended if necessary) of the conical pyramid MGEBIL. It is manifest from this theorem that MF is to AN as the base of the conical pyramid MGEBIL is to the portion of the conical surface of AGEBIL.

If a point is designated in an equilateral figure and from that point lines are sent down perpendicular to all of the sides of the rectilinear figure, the rectangle formed from half of the sum of the perpendiculars and the perimeter of the rectilinear figure will be to the rectilinear figure as the number of sides of the rectilinear figure is to unity.

Let ABCDE be a rectilinear figure. From the point O in the rectilinear figure let perpendicular lines OR, OS, OT, OQ, and OP be sent down to all of the sides of the figure. I say that the rectangle from half of the sum of the perpendiculars and the perimeter of the figure is to the figure as the number of sides is to unity.

From the point O let the figure be divided into triangles AOB, BOC, COD, DOE, and EOA. The bases of these triangles are assumed to be equal among themselves. The triangle AOB is equal to the rectangle from half the perpendicular OR and the base AB. Since all the sides of the figure are equal to the line AB, the rectangle from half of the perpendicular OR and the perimeter of the figure will be to the rectangle from half of the perpendicular OR and the line AB--or the triangle AOB--as the number of sides of the rectilinear figure is to unity. In the same manner in any of the remaining triangles, as the rectangle from half of the perpendicular sent down from the vertex O of the triangle to the base and the perimeter of the figure is to the same triangle thus will the number of sides of the figure be to unit.

Since all of the triangles together are equal to the rectilinear figure itself and all the rectangles mentioned are equal to the rectangle from half the sum of the perpendiculars and the perimeter of the figure, as one of the antecedents--namely, the number of sides--is to one of the consequents--namely, unity--thus will all of the antecedents--namely, the rectangle from half of the sum of the perpendiculars and the perimeter of the figure--be to all of the consequences--namely, the figure itself--which it was desired to demonstrate.

Hence it follows (if two points are designated in some equilateral rectilinear figure and from these points perpendiculars are dropped down to all the sides of the figure) that the perpendiculars dropped down from one point are equal to the perpendiculars dropped down from another point.

Suppose the circumference of a circle is divided into an odd number of equal parts and from some point of the periphery lines are drawn to all of those divisions. When the circle is divided into three equal parts, the sum of the firsts will be equal to the last one. If the circle is divided into five equal parts, the sum of the firsts and last will be equal to the sum of the seconds. If into seven, the sum of the firsts and the thirds will be equal to the sum of the second and the last. If into nine, the sum of the firsts and the thirds and the last will be equal to the sum of the seconds and the fourths. And thus in succession to infinity.

We say, moreover, that the first lines are those which are drawn from the designated point to the closest divisions on either side; the seconds, those lines which are drawn succeeding the first to the divisions from each part; the thirds, those which succeed the seconds, etc. Indeed, the ultimate is that which is drawn most remote to the division from the designated point.

Let the circumference of the circle ABCDE be divided into however many equal parts at the points A, B, C, D, and E and let O be some point on the circumference, from which lines OA, OE, OB, OD, and OC are drawn to all the divisions. I say that the sum of the firsts and the last--namely, OA + OE + OC--is equal to the sum of the seconds--namely, OB + OD.

From the point O let the diameter OF of the circle be drawn and through the point F let lines QFP, FTV, RFS, FXY, and MFN be drawn parallel to the lines OA, OB, OC, OD, and OE. Since the lines OA, OB, OC, OD, and OE make equal angles between themselves, therefore the lines QFP, FTV, RFS, FXY, and MFN will also make equal angles between themselves. Therefore, let F be described as the center of an equilateral and equiangular polygon GHIKL, whose sides are bisected and cut at right angles by lines having been drawn through F.

Consequently, those same sides will be cut normally by lines drawn through O, obviously parallel to the previous lines. Therefore, the lines FP, FS, FN, FV, and FY are equal to the lines O2, OZ, O5, O4, and O3, by the preceding proposition. But the line O2 is beyond the line FP by an excess of AO, the line 05 is beyond the line FN by an excess of OE, and the line OZ is beyond the line FS by an excess of CO. Therefore, the lines O2 + O5 + OZ are above the lines FP + FN + FS by an excess of AO + OE + OC. Also, the line FY is above the line O3 by an excess of OD and the line FV is above the line O4 by an excess OB. Therefore, FY + FV is above the line O3 + O4 by an excess of OD + OB. And since in the series of quantities O2 + O5 + OZ, FP + FN + FS, FY + FV, and O3 + O4, the sum of the extremes--O2 + O5 + OZ + O3 +O4--is equal to the sum of the mediums--FP + FN + FS + FY + FV--the difference between the first and the second--namely, OA + OE + OC--will be equal to the difference between the third and the fourth--namely, OB + OD--which ought to have been demonstrated.

If a circle were to cut a parabola in many points, out of which lines perpendicular to the axis are dropped down from either side, those from one side of the axis will be equal to those from the other side. In particular, if the circle cuts the parabola in two points on each side, similarly the two on one side together will be equal to the two on the other side together.

Let PABCG be a parabola which cuts a circle in the points P, A, B, and C, none of which is the vertex of the parabola. From the points P, A, B, and C, let ordinates AH, BX, MC, and PN be dropped down to the axis. I say that the ordinates on one side of the axis--namely, AH and PN--are equal to the ordinates on the other side of the axis--namely, the lines BX and CM.

Let the lines PA and BC be extended until they run together in D. Therefore, the rectangle on BDC will be equal to the rectangle on ADP. Therefore, the lines touching the parabola PABCG parallel to the lines DC and DP are equal. These lines therefore intersect themselves mutually on the axis of the parabola, making with the same ordinates an isosceles triangle. Therefore, DP and DC, parallel to them, make with their ordinates isosceles triangles--namely, DLB and DGP. Therefore, the angles DBL and DPZ are equal. But the angles ABD and DPC are equal. Therefore, the angles ABO and CPZ are equal. Therefore, the right triangles AOB and CZP are similar. Consequently, if R is the latus rectum of the parabola,

OB: OA :: PZ: ZC

OB: OA :: R: SO

PZ: ZC :: R: SO

PZ: ZC :: R: ZG

R: SO :: R: ZG

Therefore, SO and ZG are equals. Consequently, the difference between the first AH and the second BX--or SX is the same as the difference between the third CM and the fourth PN--or GN. Therefore, the sum of the first and fourth, AH + PN, is equal to the sum of the second and the third, BX + MC, which ought to have been demonstrated.

There are other cases of this theorem, but no difficulty remains in the remaining cases provided that this one is understood.

This theorem serves for the understanding of cubic and quartic equations, excelling moreover for ambiguous equations and the composition of curves. He who needs a complete analysis and doctrine of equations should look for the most complete work of D. Caroli Renaldinius concerning the resolution and composition of mathematics, which is now under the press.

To this point we have carried on purely geometrical speculations. Moreover, so that philosophers might see that geometry is not too abstract and useless, as is commonly believed, we will endeavor to explain geometrically some difficult physicomathematics from the principles of optics.