Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.

CHAP. VI. Of the Proportions of Magnitudes of divers sorts com∣pared together.

Proposition XXXVII.

THE Parallelogram ABCD (Fig. 77. N. 1.) is to the Triangle BCD upon the same base DC, and of the same heighth as 2 〈◊〉〈◊〉 1. This has been already Demonstrated in Consect. 3. De••••∣nit. 12. Here we shall give you another

Demonstration.

Suppose, 1. the whole Base CD divided into four equ•••• parts by the transverse Parallel Lines EG, HK, LN, then w•••• (by reason of the similitude of the ▵▵ DGF, DKI, DN•• DCB) GF be 1, KI 2, NM 3, CB 4; and having further∣more continually Bisected the Parts of the Base, the Indivisible or the Portions of the Lines drawn transversly thro' the Trian∣gle will be 1, 2, 3, 4, 5, 6, 7, 8, &c. ad infinitum, all a∣long in an Arithmetical Progression, beginning from the Poi•••• D, as 0; to which the like number of Indivisibles always an∣swer in the Parallelogram equal to the greatest, viz. the Li•••• BC. Wherefore by the 4th Consect. of Prop. 16. all the In••••¦visibles of the Triangle, to all those of the Parallelogram take•••• together, i. e. the Triangle it self to the Parallelogram, is as 〈◊〉〈◊〉 to 2. Q. E. D.

SCHOLIUM.

NOW if any one should doubt whether the Triangle 〈◊〉〈◊〉 Parallelogram may be rightly said to consist of an in••••¦nite number of Indivisible Lines, he may, with Dr. Wa••••••

[illustration]

Pag. 115.

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instead of Lines, conceive infinitely little Parallelograms of the same infinitely little Heighth, and it will do as well. For ha∣ving cut the Base (N^o 2) into 4 equal parts by transverse Pa∣rallels, there will be circumscribed about the Triangle so many Parallelograms of equal heighth, being in the same Proportion as their Bases, by Prop. 28. i. e. increasing in Arithmetical Progression. In the following Bisection, there will arise 8 such Parallelograms approaching nearer to the Triangle, in the next 16, &c. so that at length infinite such Parallelograms of in∣finitely less heighth, and ending in the Triangle itself, will constitute or make an infinite Series of Arithmetical Propor∣tionals, beginning not from 0 but 1; to which there will an∣swer in the Parallelogram infinite little Parallelograms of the same heighth, equal to the greatest. Whence it again follows, by Consect. 9. Prop. 21. that the one Series is to the other, i. e. the Triangle to the Parallelogram as 1 to 2; which being here thus once explained, may be the more easily applied to Cases of the like nature hereafter.

CONSECTARYS.

I. SInce in like manner in the Circle (Fig. 79.) the Periphe∣rys at equal intervals from one another, as so many Ele∣ments of the Circle, increase in Arithmetical Progression; the Sum of these Elements, i. e. the Circle it self will be to the Sum of as many Terms equal to the greatest Periphery, i. e. to a Cylindrical Surface, whose Base is the greatest Periphery, and its Altitude the Semidiameter, as 1 to 2.

II. Hence the Curve Surface of a Cylinder circumscribed a∣bout a Sphere, i. e. whose Altitude is equal to the Diameter, is Quadruple to its Base.

III. Also the Sector of the Circle bac, to a Cylindrical Sur∣face, whose Base is the Arch bc, but its Altitude the Semidia∣meter ac, is as 1 to 2.

IV. And because the Surface of the Cone BCD is to its cir∣cular Base, as BC to CA, i. e. as the √2 to 1. by Schol. Prop. 17. the Cylindrical Surface, the Conical Surface and the Cir∣cular

one we have hitherto made use of will be as 2, √2 and 1, and consequently continually Proportional.

SCHOLIUM

ALL which may also abundantly appear this way, viz. by putting for the Diameter of the Circle a, for the Semidiameter ½ a and for the Circumference ea, you'l have the Area of the Circle ¼ eaa by Consect. 1 Definit. 31. and Multi¦plying ••he heighth of the Cylinder AB i. e. ½ a by the Periph•• ea yo••'l have the Cylindrical Surface ½ eaa by Consect. 6. Definit 1•• as now is evident also by Consect. 1, 2 and 3. Now if y•••• would also have the Surface of the Cone, since it's side by the Pythagorick Theorem is 〈 math 〉〈 math 〉 and the half of that 〈 math 〉〈 math 〉 i. e. (b•• N^o 2 of Schol. Prop. 22.) 〈 math 〉〈 math 〉 and this half being multi¦plied by the Periphery of the Base ea, you'l have (by virtue o•• Consect. 4. Definit. 1.8.) the Su••f••ce of the Cone 〈 math 〉〈 math 〉 i. e. (b•• the Schol. just now cited) 〈 math 〉〈 math 〉: So that now appears also th•• 4th Consect. of this; because the Rectangle of those Extreme•••• eaa and ¼ eaa is ⅛ eaa{powerof4} as well as the square of the mean.

Proposition XXXVIII.

A Parallelepiped(α) BF (Fig. 77. N^o 3.) is to a Pyra•••••• ABCDE upon the same Base BD and of the same heighth, 〈◊〉〈◊〉 3 to 1. This was D••monstrated in Consect. 3. Definit. 17. bu•• here we shall give you another.

Demonstration.

Suppose 1 the whole Al••itude BE divided into 3 equal Pa•••• by transverse P••ains Parallel to the Base, then will (by reason 〈◊〉〈◊〉 the Similitude of the Pyramids abcd ACBD•• and AECDE) the Bases abcd ABCD and ABC•• be by Consect. 2. Prop. 34 and Consect. 3. Defi•••••• 17 in duplicate Proportion of the Altitudes i. e. in duplicate Arithmetical Progression 1, •• 9, mo••eover 2, bisecting the parts of the Altitude, the qu••¦d••angular Sections now double in Number (as the Indivisibles o•• Elements of the proposed Pyramid) will be as 1, 4, 9, 16, 2••

36, &c. ad Infinitum, all along in a duplicate Arithmetical Pro∣portion; while in the mean time there answer to them as many Elements in the Parallelepiped equal to the greatest ABCD, wherefore by Consect. 10. Prop. 21. all the Indivisibles of the Pyramid taken together will be to all the Indivisibles of the Parallelepiped also taken together, i. e. the Pyramid it self to the Parallelepiped, as 1 to 3. Q. E. D.

CONSECTARY.

THis Demonstration may be easily accommodated to all other Pyramids and Prisms, and also Cones and Cylinders,(α) since here also (Fig. 78.) the circular Planes ba, BA, and BA are as the squares of the Diameters, and so as 1, 4, 9. and so likewise all the other Elements of the Cone by continual bisection are in duplicate Arithmetical Progression; when in the mean time there answer to them in the Cylinder as many Ele∣ments equal to the greatest BA &c.

Proposition XXXIX.

A Cylinder is to a Sphere inscribed in it i. e. of the same Base and Altitude as 3 to 2.

Demonstration.

Suppose 1 (Fig. 80.) the half Altitude GH (for the same pro∣portion which will hold when demonstrated of the half Cylinder AK and Hemisphere AGB, will also hold the same of the whole Cylinder to the whole Sphere) to be divided into 3 equal parts, then will AH, C1, E2, be mean proportionals between the Segments of the Diameter by Prop. 34 Schol. 2 N^o 3, and so by Prop. 17. the Rectangles LHG, L1 G, L2 G equal to the Squares AH, C1, E2, being in order as 9, 8 and 5. and also 2dly, having bisected the former parts of the heighth, the six Squares cutting the Sphere Cross ways will be found to be as 36, 35, 32, 27, 20, 11. &c. in the progression we have shewn at large in Consect. 12. Prop 21. Wherefore since all the Indivisibles of the Hemisphere, viz. the circular Planes answering to the Squares of the said ••aniverse

Diameters have the same proportion of Progression, by Pr•••• 32. and there answer to them the like number of Elements i•• the Cylinder equal to the greatest AH: All thes•• these taken together will be to all the other take•• together i. e, the whole Cylinder AK to the whol•• Hemisphere AGB by vertue of the aforesaid Co••¦sect. 12. as 3 to 2(α) Q. E. D.

SCHOLIUM. I.

HOnora••us Fabri elegantly deduces this Prop. a priori, in a g••¦netick Method in his Synopsis Geom. p. 318. (which a•• Carotus Renaldinus performs from the same common Foundationi•• lib. 1. de Compos. and Resol. p. 301, and the following, but aft•••• a more obscure way and from a demonstration further fetch'd Fabri's is after this Method: The whole Figure (81) AL being turned round about BZ, the Quadrant ADLBA will describe a Hemisphere, the Square AZ a Cylinder and the triangle BM•• a Cone all of the same Base and Altitude. Since therefore Circl•• are as the squares of their Diameters by Prop. 32. and the Squa•• of GE=to the Squares of GD and GF taken together (f•• the Square of GF i. e. GB×□ GD is = □ BD or BA or G•• by the Pythag. Theor.) and so the Circle described by GE •••• be = to 2 Circles described by GD and GF taken together then taking away the Common Circle described by GF there w•••• remain the circle described by GF within the Cone equal t•• the Annulus or Ring described by DE about the Sphere. A•• since this may be demonstrated after the same way in any othe•• case, viz. that a circle described by g f, will be equal to an A••¦nulus described by de; it will follow, that all Rings or An•••• described by the Lines DE or de (i. e. all that Solid that conceived to be described by the trilinear Figure ADLM turne round) will be equal to all the Circles described by GF or gf (i•• to the Cone generated by the Triangle BLM;) and so as th•• Cone i•• ⅓ part of the Cylinder generated by AL, by the C••••sect. of Prop. 38. so also the Solid made by the Triline•• ADLM (viz. the Excess of the Cylinder above the Sphere will be ⅓ of the Cylinder, and consequently the Hemisphere Q. E. D.

CONSECTAYS.

HEnce you have a further Confirmation of Consect. 2. Prop. 32. and Prop. 36. N. 3.

II. Hence also naturally flows a Confirmation of Consect. 2. Definit. 20. and consequently the Dimension of the Sphere both as to its solidity and Surface. For putting a for the Dia∣meter of the Sphere and circumscribed Cylinder, and Ea for the Circumference, the Basis of the greatest Circle will be ¼ eaa, and ••hat multiplied by the Altitude, gives ¼ ea{powerof3} for the Cylinder. Therefore by the present Proposition, ⅙ ea{powerof3} gives the Solidity of the Sphere (by making as 3 to 2 so ¼ to ⅙) This divided by ⅙ a, will give, by vertue of Consect. 1. of the aforesaid Def. 20. and Consect. 3 Definit. 17. the Surface of the Sphere eaa.

III. Therefore the(α) Surface of the Sphere eaa, is manifest∣ly Quadruple of the greatest Circle ¼ eaa.

IV. The Surface of the Cylinder, without the Bases, made by multiplying the Altitude a by the Circular Periphery of the Base ea, will be eea, equal to the Surface of the Sphere.

V. Adding therefore the 2 Bases, each whereof is ¼ eaa, the whole Surface of the Cylinder 1 ½ eaa, will be to the Sur∣face of the Sphere eaa as 3 to 2.

VI. The Square of the Diameter aa to the Area of the Cir∣cle ¼ eaa, is as a to ¼ ea, i. e. as the Diameter to the 4th part of the circumference.

VII. A Cone of the same Base and Altitude with the Sphere and Cylinder, will be by Consect. 2. of this, Prop. and the Con∣sect. of Prop. 38. 〈 math 〉〈 math 〉 ea{powerof3}, and of the Cylinders ¼ or 〈 math 〉〈 math 〉 ea{powerof3}. There∣fore a Cone, Sphere, and Cylinder, of the same heighth and dia∣meter, are as 1, 2, 3. The Cone therefore is equal to the Excess of the Cylinder above the Sphere; as is otherwise evident in Scholium 1. of this.

SCHOLIUM II.

AND thus we have briefly and directly demonstrated the chief Propositions of Archimedes, in his 1. Book de Sphaer•• & Cylind. which he has deduced by a tedious Apparatus, and 〈◊〉〈◊〉 ••••directly. And now if you have a mind to Survey the 〈◊〉〈◊〉 and more perplext way of Archimedes, and compare it with this shorter cut we have given you; take it thus: Archimedes thought it necessary first of all to premise this Lemma; Th•••• all the Conical Surfaces of the Conical Body made by Circum∣volution of the Polygon, or many angled Figure A, B, C, D, E, &c. (Fig. 82.) inscribed in a Circle, according to Definit. 19. I say, those Conical Surfaces taken all together, will be equal to a Circle, whose Radius is a mean Proportional between the Diameter AE and a transverse Line BE, drawn from one ex∣tremity of the Diameter E to the end of the side AB next to the other extremity. This we will thus demonstrate by the help of specious Arithmetick: Since BN, HN are the Right Sines of equal Arches, CK and CK whole Sines, &c. and the Lines BH, GC, &c. parallel; having drawn obliquely the transverse Lines HC, GD, all the angles at H, C, G, D, &c. will be equal by Consect. 1. Definit. 11. and consequently all the Tri∣angles BNA, HNI, ICK, &c. equiangular, both among them∣selves, and to the ▵ ABE; since the angle at B is a Right one, by Consect. 1 Prop. 33. and the angle at A common with the ▵ BNA. Wherefore as

• BN to NA
• or HN to NI

so

• CK to CI.
• & GK to KL.

and so

• DM to ML
• FM to ME

so EB to BA; and so by making BN, HN, DM, FM=a CK and GK=b, EB=c, for NA, NI, ML and ME, you may rightly put ea for KI and KL eb for AB, ec. Which being done you may easily obtain the Conical Surfaces of the inscrib'd Solid, and the Area of a Circle whose Radius shall be a mean Proportional between AE and FB, and it will be evidently manifest, that these two are equal. For, 1. (for Conical Surfaces) the Diameter of the Base BH=2a, and the side of the Cone AB=ec: Therefore (making here o the name of the Reason between the Diameter and Circumference) the circumference will be 2 oa

which multiplied by half the side ½ ec, gives the Conical Surface oaec, by Consect. 4. Definit. 18. And since the Circumference BH is as before 2 oa, and the circumference CG = 2ob, half of the sum 2oa+2ob, viz. oa+ob is the equated Circum∣ference: which multiplied by the Side BC=ec gives the Sur∣face of the truncated Cone BHGC=oaec−obec, by Con∣sect 5. of the aforesaid Def. since, lastly, the Surface of the truncated Cone DFGC is equal to one, and likewise the conical Surface EDF to the other, by adding you'l have the Sum of all 40oaec+2obec.

2. (for the Area of the Circle) the Diameter AE is =4ea+2eb, and BE=c: the Rectangle of these is =4eac+2 ebc = (which also is evidently equal to the Rectangle of all the transverse Lines BH, CG, DF into the side AB, as Ar∣chimedes proposes in the matter) = to the Square of the Radius in the Circle sought, because the Radius is a mean Proportio∣nal between AE and BE, and so equal to 〈 math 〉〈 math 〉, so that the whole Diameter is 〈 math 〉〈 math 〉. There∣fore, 2. the circumference of this Circle will be 〈 math 〉〈 math 〉, i. e. 〈 math 〉〈 math 〉: which multiplyed by half the Semidiameter, i. e. by 〈 math 〉〈 math 〉 gives the Area of the Circle sought 〈 math 〉〈 math 〉. But this Root extracted is 〈 math 〉〈 math 〉, equal to the superiour Sum of the conical Surfaces. Q. E. D.

Having thus demonstrated the Lemma, we will easily demon∣strate with Archimedes (tho not after his way) That the Sur∣face of any Sphere, is Quadruple of the greatest Circle in it, which is already evident from the 3d Consect. For since all the conical Surfaces of the inscribed Solid taken together, by the preceding Lemma, are equal to the Area of a Circle whose Ra∣dius is a mean Proportional between the Diameter AE and the Transverse EB; and this mean Proportional approaches always so much nearer to the Diameter AE, and those Surfaces so much nearer to the Surface of the Sphere, by how many the more sides the inscribed Figure is conceived to have, by Con∣sect. 1 and 2. Def. 18. if you conceive in your mind the Bi∣section of the Arches AB, BC, &c. to be continued in Infinitum, it will necessarily follow, that all those conical Surfaces will at length end in the Surface of the Sphere it self, and that mean

Pro∣portional in the diameter AE, and so the Surface of the Sphere will be equal to a Circle, whose Radius is the Diameter AE, But that Circle would be Quadruple of the greatest Circle in the present Sphere, by Prop. 35. Therefore the Surface of the Sphere is Quadruple of that Circle also. Q. E. D.

Hence also it would be very easie to deduce with Archime∣des (tho again after another way) that celebrated Proposition, which we have already demonstrated from another Principle in the Prop. of this Schol. viz. That a Cylinder is to a Sphere of the same Diameter and Altitude, as 3 to 2. For by putting a for the Diameter and Altitude, and ea for the Circumference, the Area of the Circle, will be ¼ eaa; and this Area being multiplied by the Altitude a, gives ¼ ea{powerof3} for the Cylinder, by Consect. 5. Definit. 16. and the same Quadruple, i. e. eaa mul∣tiplied by ⅙ a gives ⅙ ea{powerof3} for the Sphere, by Consect. 1. Definit. 20. and Consect. 3. Definit. 17. Wherefore the Cylinder will be to the Sphere as ¼ to ⅙, i. e. in the same Denominator as 〈 math 〉〈 math 〉 to 〈 math 〉〈 math 〉, i. e. as 6 to 4, or 3 to 2. Q.E.D.

Whence it is evident, that the Dimension of the Sphere would be every ways absolute if the Proportion of the Diame∣ter to the Circumference were known; which now with Ar∣chimedes, we will endeavour to Investigate.

Proposition. XL.

THE Proportion of the Periphery of a Circle(α) to the Diame∣ter, is less than 3 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 to 1. and greater than 3 〈 math 〉〈 math 〉 to 1.

Demonstration.

The whole force of this Proposition consists in these, that, 1. Any Figure circumscribed about a Circle, has a greater Peri∣phery than the Circle, but any inscrib'd one a less. 2. The Peri∣phery of a circumscribed Figure of 96 sides, has a less Pro∣portion to the Diameter, than 3 〈 math 〉〈 math 〉 to 1.

To demonstrate this second, we will enquire 3. the Proportion of one side of such a Figure whether circumscribed or inscribed after the fol∣lowing way.

For the first part of the Proposition.

Suppose the Arch BC (Fig. 84. N. 1.) of 30 degrees, and its Tangent BC making with the Radius AC a Right Angle, to make the Triangle ABC half of an Equilateral one, so that AB shall be to BC in double Reason, viz. as 1000 to 500; which being supposed, AC will be the Root of the Difference of the Squares BC and AB, i. e. a little greater than 866, but not quite 〈 math 〉〈 math 〉.

Then continually Bisecting the Angles BAC by AG, GAC by AH, HAC by AK, KAC by AL, BC is half the side of a circumscribed Hexagon, GC the half side of a Dodecagon (or 12 sided Figure) HC of a Polygon of 24 sides, KC of one of 48; lastly, LC of one of 96 sides; and by N. 3. Schol. 3. Prop. 34. GC will be to AC as BC to BA+AC, and also HC to AC as GC to GA+AC, &c. Wherefore

In the first Bisection, of what parts GC is 500, of the same will AC be 1866 and a little more, and AG (which is the Root of the Sum of the □ □ GC and AC) 1931 〈 math 〉〈 math 〉+.

In the second Bisection, of what parts HC is 500, of the same will AC be found to be 3797 〈 math 〉〈 math 〉+and AH 3830 〈 math 〉〈 math 〉+.

In the third Bisection, of what parts KL is 500 of the same will AC be 7628 〈 math 〉〈 math 〉+ and AK 7644 〈 math 〉〈 math 〉+.

In the 4th Bisection, of what parts LC is 500 of the same will AC be 15272 〈 math 〉〈 math 〉+.

Now therefore LC taken 96 times, will give 48000 the Semi-periphery of the Polygon, which has the same Proportion to the Semi-diameter AC 15272 〈 math 〉〈 math 〉 as the whole Periphery to the whole Diameter. But 48000 contains 15272 〈 math 〉〈 math 〉, 3 times, and moreover 2181 〈 math 〉〈 math 〉 remaining parts, which are less than 〈 math 〉〈 math 〉 part of the division, for multiplied by 7 they give only 1526 〈 math 〉〈 math 〉+.

Therefore it is evident, that the Periphery of this Polygon (and much more the Periphery of a less Circle than that) will have a less Proportion to the Diameter, than 3 〈 math 〉〈 math 〉 to 1. Which is one thing we were to demonstrate.

For the 2d Part of the Prop.

SUppose the Arch BC to be (N^o 2.) of 60 Degr, that is the Angle BAC at the Periphery of 30 Since the Angle at B is a right one by Consect. 1. Prop. 33. the Triangle ABC will be again half an Equilateral one, and BC the whole side of an Hexagon, and GC of a Dodecagon, &c. So that putting for BC 1000 (as before we put 500 for the side of the Hexagon) let AC be 2000 and AB the Root of the difference of the Squares BC and AC i. e. less than 1732 〈 math 〉〈 math 〉 viz. 1732 and not quite 〈 math 〉〈 math 〉.

Then bisecting continually the Angles BAC, GAC, &c. since the Angles at the Periphery BAG, GAC, GCB, standing on equal Arches BG and GC, are equal by Prop. 33. and the Angle at C, (common to the Triangles GCF and GCA) and the o∣thers at H, K, L are all right ones by Consect. 1. of the afore∣said Prop. these 2 Triangles CGF and CGA are equiangular and consequently by Prop. 34. the Perpendicular GC in the one will be to the Perpendicular GA in the other as the Hypothenuse CF in the one to the Hypoth. AC in the other i. e. (by the foun∣dation we have laid in the former part of the Demonstration of N^o 3, Schol. 3. Prop. 34.) as BC to AB+AC; and in like man∣ner in the following HC will be to HA as GC to AG+AC, &c. Wherefore.

In the first Bisection, of what parts GC is 1000 of the same AG will be a little less then 3732 〈 math 〉〈 math 〉; and AC (which is the Root of the Sum of the □ □ AG and GC) will be a little less then 3863 〈 math 〉〈 math 〉.

In the second Bifection, of what parts GC is 1000 of the same will AH be a little less then 7595 〈 math 〉〈 math 〉 and AC a little less then 766 〈 math 〉〈 math 〉.

In the third Bisection, of what parts KC is 1000 of the same will AK be a little less then 15257 〈 math 〉〈 math 〉 and AC a little less 15290 〈 math 〉〈 math 〉.

In the fourth Bisection, of what parts LC is 1000 of the same will AL be a little less then 30547 〈 math 〉〈 math 〉; and AC a little less then 30564, and consequently if LC be put 500, AC will be less then 15282.

Now therefore LC taken 96 times will give 48000 for the Periphery of the inscrib'd Polygon, and 15282 and a little less for the Diameter AC. But 48000 contains 15282 thrice, and moreover a remainder of 2154 parts, which are more then 〈 math 〉〈 math 〉 of the Divisor; for 〈 math 〉〈 math 〉 of this number makes 215 〈 math 〉〈 math 〉 and so 〈 math 〉〈 math 〉 makes 2150 〈 math 〉〈 math 〉 i. e. 2152 〈 math 〉〈 math 〉. Therefore it is evident that the Peri∣phery of this Inscribed Polygon (and much more the Periphery of a Circle greater then that) will be to it's Diameter in a greater proportion then 3 〈 math 〉〈 math 〉 to 1, which is the 2d thing.

SCHOLIUM

IF any one had rather make use of the small numbers of Archimedes, which he chose for this purpose, by putting in the first part of the Demonstration, for AB 306 and for BC 153, in the second for AC 1560 and for BC 780, by the like process of Demonstration, he may infer the same with Archimedes. We like our Numbers best, tho' somewhat large, because they may be remember'd, and are more proportionate to things, and make also the latter part of our Demonstration like the former. The Proportion in the mean while of the Dia∣meter to the Periphery of the Circle by the Archimedean way is included within such narrow Limits, that they only differ from one another 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 parts; for 〈 math 〉〈 math 〉 Subtracted from 〈 math 〉〈 math 〉 leave 〈 math 〉〈 math 〉, as 3 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 and 3 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 if they are reduced to the same Deno∣mination make on the one hand 〈 math 〉〈 math 〉 on the other 〈 math 〉〈 math 〉. Hence it would be easy, having divided the difference 〈 math 〉〈 math 〉 into 2 Parts, to express a middle proportion of the Periphery to the Diameter be∣tween the 2 Archimedean and Extreme ones as in these Numbers, 1561 to 4970, (or by dividing both sides by 5) as 3123 to 994, or (dividing both sides by 7) as 446 〈 math 〉〈 math 〉 to 142, or (by dividing again by 2) as 223 〈 math 〉〈 math 〉 to 71, &c.

While these Numbers become as fit for use as those of Archi∣medes, which we therefore use before any other, particularly in Dimensions that dont require an exact Niceness; where they do those may be made use of exhibited by Ptolomey Vieta, Ludolphus a Ceulea, Metius, Snellius, Lansbergius, Hugeus, &c. as if

The Diameter be

The Circumference will be

10,000,000

31, 416, 666. Ptolomey.

10000,000,000

31, 415, 926, 535. Vieta.

100,000,000,000,000,000000

314, 159, 265, 358, 979, 323, 846 ½, &c. Ludolph. a Ceulen.

Proposition XLI.

THe Area of a Circle has the same proportion to the Square of it Diameter, as the 4th part of the Circumference to the Dia∣meter.

Demonstration.

Though we have before Demonstrated this Truth in Consect. 6. Prop. 39. yet here we will give it you again after another way. Since therefore the Circumference is a little less then 3 〈 math 〉〈 math 〉, and •• little more than 3 〈 math 〉〈 math 〉 Diameters, for this excess putting z if the Diameter be 1 we will call the Circumference 3+z therefore the 4th Part of it will be 〈 math 〉〈 math 〉: And the Area of the Circle (by Multiplying the half Semidiameter) i. e. ¼ by the Circumfe∣rence, you'l have both 〈 math 〉〈 math 〉 and the square of the Diameter =•• Q. E. D.

CONSECTARY.

THerefore if the(α) Proportion of Archimedes be near e∣nough truth to be made use of, viz. 22 to 7; the Ar•••• of the Circle will be to the Square of the Diameter as 11 to 14 because the quarter part of 22, i. e. 5 ½ or 〈 math 〉〈 math 〉 to the Diam. 7 〈 math 〉〈 math 〉 is in the same Proportion.

Proposition XLII.

THE Diameter(β) of a Square AC (Fig. 83.) is incom∣mensurable to the side AB (and consequently also to th•• whole Periphery) i. e. it bears a Proportion to 〈◊〉〈◊〉 that cannot be exactly expressed by Numbers.

Demonstration.

For, if for AB you put 1, BC will be als••

1, and by the Pythagorick Theorem AC will be = √2. There∣fore by Consect. 4. Definit. 30. AC is incommensurable to the side AB, &c. Q.E.D.

CONSECTARY I.

IT is notwithstanding Commensurable in Power; for its Square is to the Square of the Side as 2 to 1.

CONSECTARY II.

NOW if the Proportion of the side or whole Periphery ABCDA, to the Diam. AC is to be expressed by Num∣bers somewhat near, as we have done in the Diameter and Cir∣cumference of a Circle; then making the side AB=100, the Diameter is greater than 141 〈 math 〉〈 math 〉, less than 141 〈 math 〉〈 math 〉.

Proposition XLIII.

THE Area of a Circle is Incommensurable to the Square of the Dia∣meter.

Demonstration.

For dividing the Semidiameter CD (Fig. 85.) into two equal Parts (and consequently the Diameter DF into 4) AC will be 2. viz. √4 and AC √3. by Schol. 2. Prop. 34. N. 3. the sum √4+√3, and the sum of as many equal to the greatest AC 4. Having moreover Bisected the Parts of the Semidiameter, AC will =4 or the √16, ac=√15, AC=√12, AC=√7; the sum 〈 math 〉〈 math 〉; and the sum of as many e∣qual to the greatest AC is =16, &c. And thus the last sums will be the Square Numbers increasing in Quadruple Pro∣portion; but the former Sums will be always composed of the Ra∣tional Root of every such Square, and of several other irrational Roots of Numbers unevenly decreasing; so that it will be im∣possible to express those former Sums by any Rational Number, by what we have said in Schol. 2. Definit. 30. Wherefore all the Indivisibles of the Quadrant ADC are to as many of the Square ACDE equal to the greatest, i. e. the Quadrant it self

ADC to the Square ACDE (and consequently the whole Area of the Circle to this circumscrib'd Square) will be as a Surd Quantity to a true and truly Square Number, i. e. the Area of the Circle will be Incommensurable to the Square of the Diameter, by Consect. 4. of the said Definit. Q. E. D.

CONSECTARY.

AND because the fourth part of the Circumference ha•••• the same Proportion to the Diameter, as the Area of the Circle to the Square of the Diameter, by Prop. 41. There∣fore also that will be Incommensurable to this, and consequently the whole Circumference will be so to the Diameter.

SCHOLIUM.

WHerefore it is somewhat Wonderful, which G. G. Leib∣nitius(α) tells us, that the Square of the Diameter be∣ing 1, the Area of the Circle will be 〈 math 〉〈 math 〉, &c. ad Infinitum. i. e. by adding 1 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉, &c.) to 〈 math 〉〈 math 〉, &c. i. e. to the Sum of infinite Fractions whose common Numerator is 2. But their Denominators Squares les∣sen'd by Unity, and taken out of the Series of the Squares of Natural Numbers by every fourth, omitting the Intermediate ones: Which Sum might seem expressible in Numbers, since all its parts are Fractions redu∣cible to a common Denominaton; while notwith∣standing Leibnitius himself confesses, that the Cir∣cle is not Commensurable to the Square, nor ex∣pressible by any Number.

Notes

• (α)

  Euclid. Prop 7 Coroll. lib. 12.
• (α)

  Euclid. Prop. 10. lib. 12.
• (α)

  A••chim. 32 (al. 31.) lib. I de Sphaer.
• (α)

  Archim. lib. I. de Sph. & Cylind. Prop. 31. (al. 30).
• (α)

  Archim. Cyclom. prop. 2
• (α)

  Archim. Prop. 3. Cy∣clom.
• (β)

  Eucl. last Prop. lib. 10.
• (α)

  Acta E∣rudit. Ann. 82. p. 44 & the following.