Then, the Triangles K GE, KHL, are fimilar, becaufe LH is parallel to EG; the Triangles KLH FLE,are fimilar, because the Angles K LH, FLE,are equal, and the Angles KHL, FEL, are both Right Angles; and the Triangles FLE, FEI, are fimilar, because the Angles FEL, FIE, are both Right Angles, and the Angle F is common: Therefore the Triangles FIE and K GE are fimilar, whence, KG (or/rss): KE (or r) :: FI (or s): FE (or a) i

Now rss r 2r 873 16r5 12877 &c. by extracting the Square Root :

And if rs be divided by that Series, the Quotient,

ss t

5253545

2rr

+

874 +

+
16r6 128,8'

c. will be the

Fluxion of the Arc; therefore the Fluent thereof, viz.

c. will be equal to the Arc of a Circle whofe Radius is r, and Sine's. But if r be put equal to Unity, then

Let it be required to find the Length of the Arc of 30 Degrees to 6 Places of Decimals, the Radius being Unity.

Here's, and ss; whence the Operation may be as follows:

Hence the Length of the Arc of 30 Degrees is 2523598+. Now if this Arc be multiplied by 6, we fhall have the Length of the Arc of the Semicircle in fuch Parts as the Radius is 1, or of the whole Circumference in fuch Parts as the Diameter is 1, viz. 3,14159+.

But there is no Series so easy to be retained in the Memory, and fo readily put in Practice, for obtaining the Ratio of the Diameter of the Circle to its Circumference, as that which is derived from the Tangent. For if t be put equal to the Tangent of any Arc, then a = t = 1/2 1 3 + 1 ts3 — — 12 + 1 13, &c.

Now

4

Now the Radius being Unity the Sine of 30 Degrees, and confequently the Cofine; and because the Cofine is to the Right Sine, as the Radius to the Tangent; it will be✔::::/, the Tangent of 30° oo't, whence tt. Wherefore, if the Root of be divided continually by 3 feveral Quotients by the odd Numbers fucceffively, viz. the first by 3, the fecond by 5, &c. the Sum of the affirmative Quotients made lefs by all the negative ones, will be the Arc of 30 Degrees.

And because the Arc of 30 Degrees is Part of the Semicircumference, if inftead, of ✓ be taken 6√3

12, we fhall have the Semicircumference in fuch Parts as the Radius is Unity; or the whole Circumference, the Diameter being Unity.

The OPERATION ftands thus:

Whence 3,546231,4046403,141591 the fame as before. The Impoffibility of expreffing the exact Proportion of the Diameter of a Circle to its Circumerence, by any received Way of Notation, has put the moft celebrated Men, in all Ages, upon approximating the Truth as near as poffible; there being a Neceffity of a near Quadrature, inasmuch as it is the Bafis upon which the most useful Branches of the Mathematics are built. And after the famous Van Ceulen, who carried it to 36 Places of Decimals (which he order'd to be engraven on his Tomb ftone, thinking he had fet Bounds to farther Improvements), the firft that attempted it with Success was the most indefatigable Mr. Abraham Sharp, who, by a double Computation, viz. from the Sine of 6 Degrees one Way, and from the Sine and Cofine of

12 Degrees another Way, carried it to twice the Number of Places that Van Ceulen had done, viz. 72.

And fince that time Mr. Machin, late Profeffor of Aftronomy in Gresham College, and Secretary to the Royal Society, by a different Method of Computation, has carried it to 100 Places, almoft triple the Number that Van Ceulen had done, which not only confirms Mr. Sharp's Quadrature, but fhews us, that, if the Diameter be 1,00000, &c. the Circumference will be 3,14159.26535.89793.23846.26433.83279. 50288. 41971. 69399. 37510. 58209. 74944. 59230. 78164. 06286. 20899. 86280. 34825. 34211. 70680 + of the fame Parts.

Which is a Degree of Exactnefs far furpaffing all Imagination; being, by Eftimation, more than fufficient to calculate the Number of Grains of Sand that may be comprehended within the Sphere of the Fixed Stars.

The late Mr. Cunn's Series for determining the Periphery of an Ellipfis (who was my Predeceffor in the Mathematical School, erected by Frederic Slare, M. D. and established by a Decree of the High Court of Chancery for qualifying Boys for the Sea Service) being new and curious, this Opportunity is taken of making it public.

Let A be equal to a Quadrant of the Circle circumfcribingtheEllipfis,whofe Periphery is required. Then Ax 1.3-5-7-5-8

I

L

1.3.1

2.2 2.4.8

1.3.5.1

2.4.6.16 2.4.6.8.128

1.3.5.7.9.7 10 1.3.5.7.9.11.21

2.4.6.8.10.256

2.4.6.8.10.12.1024

ez, &c. is

the Periphery of a Quadrant of the Ellipfis, where

=

11-, t being the Semi-tranfverfe Diameter,

tt

and c the Semi-conjugate.

When this Series came to hand, it was imperfect, inafmuch as there were only the first five Terms without the Law of Continuation: But, being defirous of rendering it compleat, after fome Confideration, I found the Law to be as follows: It is plain by Inspection, that the Numerators and Denominators of each Term are compos'd of Numbers that run in arithmetical Progreffion, except the laft in each Term, viz. 1, 4, 6, 128, c. and those being found by the continual Multiplica

I

tion

1

tion of thefe Fractions, ×××ק×7×1, &c. the Law of continuing the whole Series as above, is evident. Whence, by a well known Method of subftituting Capital Letters for each Term refpectively, the 1.32 following Series is deduced, viz. AXI - ÷ee - 1·3,2 B

4.4

10.10

12.12

6.6 8.8 where the Law of Continuation is evident alfo, fince each Capital Letter is equal to its precedent Term, viz. R=4ee, C=132 B, &c. and without doubt

4.4

in Practice is preferable to the former Series: But the Investigation of that, on which this laft depends, is omitted; purely on account of its being foreign to the present Subject.

But to return; if the Series expreffing the Length of the Arc, viz. 5+ 53 + 355, &c. be reverfed, we fhall have the Value of s in the Terms of a, and confequently a direct Method for finding the Sine of any Arc from its Length given. Thus,

If a=s+53 + 2055 + 73257, &c.

40

120
as

2.3 2.3.4.5 2.3.4.5.6.7

3

For puts Aa+ Ba3 + Cas, &c.)

Then s

And 3s=

A3 a3+A Ba,&c. =a.

+ Asas &c.)

3 40

And confequently A a=a; whence A=1:

Alfo, B+ A3=0; or B=(-A3) :
Again C+ A B+A=0; o C=—A1B—A3:
That is C (XIX -

Wherefore A 1, B = —, C, &c, and =

confequently, sa

120 120

which three Terms the Law of Continuation is

eafily discovered: Therefore