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

Proposition XVIII.

THE whole spiral(α) 1.1 space comprehended under the second right line EA and the second spiral EGIA (see Fig. 141. n. 2.) is to the second circle as 7. to 12.

Demonstration.

For having divided the circumference of the circle first into three equal parts, there will be drawn to the second spiral four right lines BE, BG, BI and BA being as 3, 4, 5, 6, and but only three sectors circumscrib'd, viz GBg, IBi and ABa, which proceed according to the squares of the three latter lines, viz. 16, 25, 36, so that the sum is 77, while the sum of three equal to the greatest is 108, and so the one to the other (dividing both sides by 9) as 12 to 8 〈 math 〉〈 math 〉▪ Having moreover bisected the arches and parts of the line BE, so that that shall be 6, the second BF will be 7, and so the other five 8, 9, 10, 11, 12; and the sectors answering to them (excepting the first) 49, 64, 81, 100, 121, 144, so that their sum shall be 559, while the sum of six equal to the greatest, i. e. the whole circle is 864, and so one to the other (dividing both by 72) as 12 to 7 〈 math 〉〈 math 〉. In the other bisection of the arches and the parts of the line BE, so that the one shall be 12, the second 13, &c. to the thirteenth BA which will be 24, the sum of twelve sectors will be found to be 4250; and the sum of

as many equal to the greatest 6912, and so the one to the other (dividing both sides by 576) as 12 to 7 〈 math 〉〈 math 〉 s. 〈 math 〉〈 math 〉. Therefore the proportion will be

• I. In the first case 12 to 7+1+½+〈 math 〉〈 math 〉 viz. 〈 math 〉〈 math 〉.
• II. In the second case 12 to 7+½+¼+〈 math 〉〈 math 〉 viz. 〈 math 〉〈 math 〉.
• III. In the third case 12 to 7+¼+⅛+〈 math 〉〈 math 〉, &c.

The first and second fractions thus decreasing by ½ the latter by ¼. Wherefore the proportion of the second circle to the second spiral space will be as

12 to 7+1+½+〈 math 〉〈 math 〉

−½−¼−〈 math 〉〈 math 〉

−¼ &c.−⅛ &c.−〈 math 〉〈 math 〉 &c.

By vertue of Consect. 3. and 8.=0=0 Prop. 21. lib. 1. i. e. as 12 to 7. Q. E. D.

CONSECTARY I.

BEcause the second circle is to the first spiral space as 12 to 1, by Consect. 1. of the preceding Prop. and to the se∣cond spiral space as 12 to 7, by the present. it will be to the second space without the first (viz. BCDEAIGE) as 12 to 6 i. e. as 2 to 1.

CONSECTARY II.

THerefore the second space separately to the first is as 6 to 1.

CONSECTARY III.

SInce in the trisection of both these circles, first and second, there arise six lines, and as many sectors, viz. three lines BC, BD, BE, i. e. 1, 2, 3, to which there answer three ar∣ches in the same progression within the second circle, and also as many equal to its greatest; therefore the sum of all the un∣equal arches will be 21, but the sum of the equal ones of both circles (each of which in the first are equivalent to 3, in the second to 6) will be 27. Wherefore the sum of both the Pe∣ripheries to the sum of all the circumscrib'd arches will be as

27 to 21, i. e. (dividing both sides by 9) as 3 to 2 ⅓. More∣over bisecting the arches of the circles and the parts of the line BA, there will arise six circumscribed unequal arches within the first circle, which are as 1, 2, 3, 4, 5, 6, and as many within the second 7, 8, 9, 10, 11, 12; the sum of all which is 78, while the sum of as many equal ones on both sides is 108. Wherefore the one will be to the other, i. e. the sum of both the peripheries to twelve circumscribed arches ta∣ken together, is now as 108 to 78, i. e. (dividing both sides by 36) as 3 to 2 ⅙. And making yet another bisection, the proportion will be found to be as 3 to 2 〈 math 〉〈 math 〉, &c. and hence at length may be evidently inferr'd; that the sum of both the peripheries will be to the sum of all the arches circumscribible ad infinitum, i. e. to the whole helix as

3 to 2+⅓

−⅙

−〈 math 〉〈 math 〉 &c.=0. that is, as 3 to 2. Q. E. D.

CONSECTARY IV.

THerefore, since the periphery of the second circle is dou∣ble of the first, that alone will be equal to the whole spiral.

CONSECTARY V.

THerefore, if the periphery of the second circle be 2, the periphery of the first will be 1, and the first spiral line ½ by Consect. 2. of the anteced. Prop. wherefore the second spiral alone will be 1 ½, and so the periphery of the second circle alone will be to the second spiral alone as 2 to 1 ½ i. e. as 4 to 3; and to the first alone as 4 to 1.

SCHOLIƲM I.

BUT as Consect. 4. may be also deduced after another way, viz. by comparing only the arches of the second circle with the correspondent circumscripts, but considering them as taken twice (because that circle is twice turned round while the whole helix or spiral is described) and finding in the first

trisection the proportion of double the second periphery to all the circumscripts as 12 to 7; and in the succeeding bisection as 12 to 6 ½; in the second bisection as 12 to 6 ¼, &c. and at length by inferring, that the second periphery is double of all the arches circumscribible about the whole helix ad infinitum, that is to the helix it self.

as 12 to 6+1

−½

−¼ &c.=0. i. e. as 12 to 6;

and consequently the simple second periphery will be to the whole helix as 6 to 6: Thus the 5. Consect. may be separate∣ly had after the same manner, if instead of the first trisection, you only bisect; (vid. Fig. 141. n. 3.) for so in the first bi∣section the arches circumscribed about the second spiral line would be separately two semi-circles Dd, 3 and Aa, 4, (for as the line BC is one, BE, 2, BD, 3, BA, 4; so the arch described by the radius BD is 3 and described by the radius BA=4,) and their sum 7; while the sum of two equal to the greatest is 8. In the second bisection (when BE is 4) BF and its arch is made 5, the arch BD 6, the arch BG 7, the arch BA 8, the sum 26; while the sum of so many quadrants equal to the greatest is 32. Thus in the third bisection the sum of eight Octants circumscrib'd about the second helix will be found to be 100, the sum of so many = to the greatest 128, &c. Wherefore the periphery of the second circle in the first case will be to the arches circumscrib'd about the se∣cond helix as 4 to 3+½; in the second as 4 to 3+¼; in the third as 4 to 3+⅛, &c. and so to all the arches circumscri∣bible in infinitum, i. e. to the second helix it self as

4 to 3+½

−¼

−⅛ &c. = 0. i. e. as 4 to 3. Q. E. D.

By the same method you may easily find the proportion of the third circle to the third spiral space, and of that periphery either to the whole spiral, or separately to the third, as will be evident to any one who trys.

I. For the third spiral space.

(Fig. 142)

BC 1	BF 4	BI 7	49
BD 2	BG 5	BK 8	64
BE 3	BH 6	BA 9	81

are the three first sectors circumscrib'd about the parts of the third helix. The sum of these three sectors is 194; and the sum of so many equal to the greatest 243. Therefore the first proportion of the one sum to the other will be as 243 to 194, i. e. (dividing both sides by 9) as 27 to 21 〈 math 〉〈 math 〉.

In the first bisection there will be seven lines:

BH 12,	BL 13	169	Sectors circumscribed about the parts of the third helix.
	BI 14	196
	BM 15	225
	BK 16	256
	BN 17	289
	BA 18	324
	Sum	1459

; while in the mean time the sum of as many equal to the greatest is 944, and so the second pro∣portion as 1944 to 1459 i. e. (dividing both sides by 72) as 27 to 20 〈 math 〉〈 math 〉.

In the second Bisection there will be thirteen lines, viz. BH 24, the rest 25, 26, &c. but the sum of the sectors, i. e. of the square numbers answering to the twelve latter will be found to be 11306; while in the mean time the sum of as ma∣ny equal to the greatest will be 15552, so that you will have the third proportion of this sum to the other, viz. as 15552 to 11306, i. e. (dividing both sides by 576) as 27 to 19 〈 math 〉〈 math 〉.

Therefore the I. proportion will be as 27 to 19+2+〈 math 〉〈 math 〉 i. e.

to 19+2+½+〈 math 〉〈 math 〉

II. — as 27 to 19+1+〈 math 〉〈 math 〉 i. e.

to 19+1+¼+〈 math 〉〈 math 〉

III. — as 27 to 19+〈 math 〉〈 math 〉—i. e.

to 19+½+⅛+〈 math 〉〈 math 〉.

Therefore the proportion of the third circle to the third spi∣ral space will be

as 27 to 19+2+½+〈 math 〉〈 math 〉

−1−¼−〈 math 〉〈 math 〉 i. e. as 27 to 19.

−½ &c. −⅛ &c. −〈 math 〉〈 math 〉 &c.

= 0. = 0. = 0. Q. E. D.

II. For the third spiral line.

If instead of the first trisection (as less commodious for the end proposed) you make use here also, as before, of bisecti∣on in the same figure, there will come out six lines from the point B to the helix, viz. Bm, 1, BE, 2, Bn, 3, BH, 4, Bo, 5, BA, 6; to which there answer as many semicircular arches in the same progression, and to the greatest of the two as many equal to 2, 4, 6; so that the sum of the unequal ones is 21, and of the equal ones 24, and so the proportion of three peripheries together to all the circumscripts together will be as 24 to 21 (and dividing both by 6) as 4 to 3 ½. In the se∣cond bisection the twelve unequal lines and arches make the sum 78, and as many equal to the greatest of the four will give the sum 96; so that the second proportion will be 96 to 78, i. e. (dividing both sides by 24) 4 to 3 ½. In the third bisection the proportion will come out as 384 to 300, i. e. (dividing both sides by 96) as 4 to 3 ½, &c. Therefore the proportion of the three circles together to the whole Helix will be as 4 to

3+½

−¼

−⅛ &c.=0. i. e. as 4 to 3 or 12 to 9.

Q. E. D.

CONSECTARY VI.

NOW, if the periphery of the first circle be made 2, the second will bee 4, the th••rd 6, and consequently the sum 12; it will be manifest that the third periphery separately will be to the whole helix as 6 to 9, i. e. as 2 to 3.

CONSECTARY VII.

AND because the second periphery (which is 4) is equal to the first and second helix together, by the above Con∣sect. 4. the remaining third spiral will be 5, and so the pro∣portion of the third periphery to it as 6 to 5.

CONSECTARY VIII.

WHerefore the proportions of each of the peripheries to their correspondent spirals will be in a progression of ordinal numbers, viz. so that the latter of every two will denote the periphery of a circle, and the former an inscribed spiral; and consequently the spiral lines will be in an arith∣metical progression of odd numbers, and the peripheries of the circles in a progression of even ones.

• 1—The first Spiral,
• 2—The first Periphery,
• 3—The second Spiral,
• 4—The second Periphery,
• 5—The third Spiral,
• 6 &c.—The third Periphery, &c.

SHCOLIƲM II.

THE seventh Consectary may also be easily deduced sepa∣rately this way: In the first bisection the line BA and its periphery is 6, the line Bo and its periphery 5, the sum of the circumscribed Peripheries 11; the sum of as many equal to the greatest 12. Therefore the periphery of the third circle will be to the two circumscripts as 12 to 11, i. e. as 6 to 5 ½. In the second bisection the four circumscribed quadrants will be 12, 11, 10, 9, their sum 42; and the sum of four equal to the greatest, i. e. the periphery of the third circle 48. Therefore the proportion is now as 48 to 42, i. e. (dividing both sides by 8) as 6 to 5 ¼. Thus you will have the third proportion as 192 to 164, i. e. (dividing both sides by 32) as 6 to 5 ⅛. Wherefore the proportion of the third periphery to the third helix or spiral is

as 6 to 5+½

−¼

−⅛ &c.

=0. i. e. as 6 to 5. Q. E. D.

CONSECTARY IX.

AS Consect. 8. supplies us with a rule to determine the pro∣portion of every spiral of every order to the periphery of the correspondent circle, viz. if the number of the order be doubled for the periphery of the circle, and the next antece∣dent odd number be taken for the spiral line; so what we have hitherto demonstrated supplys also another rule, to define the proportion of the spiral space in any order to its circle. For since the circles are in a progression of Squares 1, 4, 9, 16, &c. but the first circle is to the first space as 3 to 1 (i. e. a, 1 to ⅓) by Prop. 17. and the second to the second as 12 to 7 (i. e. as 4 to 2 ⅓) by Prop. 18. the third to the third as 27 to 19 (i. e. as 9 to 6 ⅓) by Schol. 1. of this. And contem∣plating both these series one by another,

• Of the circles, 1, 4, 9.
• Of the spaces, ⅓, 2 ⅓, 6 ⅓.

We see the numbers of the spaces are produced, if from the spuare numbers of the circles you substract their roots, and add to the remainder ⅓. Wherefore, if, e. g. we were to de∣termine the proportion of the fourth circle to the fourth spiral space; the square of 4 viz. 16 would give the circle; hence substracting the root 4, there wiill remain 12, and adding ⅓ you would have the fourth spiral space 12 ⅓; and in like man∣ner the spiral space 20 ⅓ would answer to the circle 25, &c. And that this is certain is hence evident, that if we mul∣tiply these numbers 16 and 12 ⅓, also 25 and 20 ⅓ by 3, that we may have those proportions in whole numbers, 48 and 37, 75 and 61, these are those very numbers Archimedes had hinted at in the Coroll. of Prop. 25.

[illustration]

Pag. 217.

142

143

144

145

146

CONSECTARY X.

NAY what we have now said, is that very Coroll. com∣prehending also that 25th. Proposition, viz. that a spi∣ral space of any order is to its correspondent circle, as the re∣ctangle of the semidiameters of this and the preceding circle together with a third part of the square of the difference be∣tween both semi-diameters to the square of the greatest semi∣diameter. For, if e. g. the proportion of the third spiral space to the third circle be required, since the semidiameter of this third circle is as 3, and the semidiameter of the second precedent one is 2, and so the difference 1; the rectangle of 2 into 3 i. e. 6, together with ⅓ of the square of the diffe∣rence will define the third spiral space 6 ⅓; since the third cir∣cle may be defined by the square of the semidiameter of the greater, viz. by 9, and so in the rest; as the numbers we have found shew, or further that may be found according to given Rules which may be here seen in the following Table.

Orders.	I	II	III	IV	V	VI	VII	VIII	XI	X
Circles.	1	4	9	16	25	36	49	64	81	100
The whole spaces, the preced. ones being included.	⅓	2 ⅓	6 ⅓	12 ⅓	20 ⅓	30 ⅓	42 ⅓	56 ⅓	72 ⅓	90 ⅓
Separate spaces the preced. ones be∣ing excluded.	⅓	2	4	6	8	10	12	14	16	18

CONSECTARY XI.

OUT of which table it is obvious to sight, that the se∣cond space excluding the first is sextuple of the first, as we have already deduced in Consect. 2. Prop. 18. and the third separate space double of the second, and the fourth triple of the same second, and the fifth quadruple, and so onwards.

SCHOLIƲM III.

AND this shall suffice for spirals, which comprehends not only the chief Theorems of Archimedes of spiral spaces, but also the chief of spiral lines (whereof Archimedes has left nothing.) If any should have a mind to carry on our me∣thod further, he may easily demonstrate after the same way what remains in Archimedes, and what Dr. Wallis in his A∣rithmetick of Infinites from Prop. 5. to the 38, and what o∣thers have done on this Argument.

Notes

• (α) 1.1

  Archim. Prop. 25.