mainder will be the common logarithm of the spherical excess in seconds and decimals.

5. Lastly, when the three sides of the triangle are given in feet; add to the logarithm of half their sum, the logs. of the three differences of those sides and that half sum, divide the total of these 4 logs. by 2, and from the quotient subtract the log 9.3267737; the remainder will be the logarithm of the spherical excess in seconds, &c. as before.

One or other of these rules will apply to all cases in which the spherical excess will be required.*

PROBLEM XII.

To determine the ratio of the earth's axes from the measures of convenient portions of a meridian in any two given latitudes; the earth being supposed a spheroid generated by the rotation of an ellipse upon its

minor axis.

The most accurate way of solving this problem, will be to compare, not merely single degrees measured on different parts of the meridian, but large portions of 5, 6, or 7 degrees, the most extensive that have been correctly measured; according to the method proposed by Professor Play fair (Edinburgh Transactions, vol. v.), which is as below.

Let the ellipse PEPQ represent a terrestrial meridian passing through the poles r, p, and cutting the equator in E, Q. Let c be the centre of the earth, co the radius of the equator = a, and rc, half the polar axis = b.

The intelligent student who wishes to go more minutely into the subject of geodesic operations, especially in reference to the determination of the figure and magnitude of the earth, may consult the chapter on that subject in the 3d vol. of Dr. Hutton's Course, Colonel Mudge's "Account of the Trigonometrical Survey of England and Wales," M. Puissant's works, entitled "Geodésie" and "Traité de Topographie, d'Arpentage, &c." Mechain and Delambre, "Base du Système Métrique Décimal," and chap. xxxv. of Delambre's quarto" Astronomie."

Then, the excess of the angle QLB above QKA, that is, the angle LHK or BHA will be.

Also, if the elliptic arc QAz, then will AB = d2 = de × AH. Now it is shown by the writers on conic sections, that the radius of curvature at A, that is,

b2 sin2 9). - *.

If, therefore, c be the compression at the poles, or = a b, we shall have b2 = a2

2ac + c2; or, rejecting the powers of c higher than the first, because c is very small in comparison to a (that is, about the 300th part), it will be 6 a2 2ac, whence, by substitution, AH=a3 (a 2c) (a2 — a2 sin2 + a2 sin2 4 — 2ac

sin2 ) =a3 (α-2c) (a2 - 2ac sin2)

2c

=&

sin? 4)

3c

sin2) nearly;

a

rejecting, as before, the terms that involve c2, c3, &c.

3c

Hence AH (a− 2c) (1 + sin?)

бъ = (a 2c) do + 3 cdp Consequently, by integration,

z = (a — (c) Q - c sin 29

ap− c ( + sin 20);

an expression which needs no correction.

Let MN be any arc of the elliptic meridian, the latitude of м, one of its extremities, and " that of N, its other extremity; we have

QM = α&

sin 20′)

sin 20′′)

QM, that

is,

and their difference QN MN=a (9′′ — Ø′) - §c [ (9′′ — Ø′) + (sin 20′′ — sin 29′)]. If, therefore, MN be an arc of several degrees of the meridian, the length of which is known by actual measurement, and the latitude of its two extremities M and N also known, the above formula for MN becomes an equation in which a and c are the only unknown quantities. In like manner by the measurement of another arc of the meridian, another equation will be obtained, in which a and c are the only unknown quantities. Therefore, by comparing these two equations, the values of a and c, that is, of the radius of the equator, and its excess above half the polar axis, may be determined.

Thus, if be the length of an arc measured, m the coefficient of a, and n of c, computed by the last formula; also, if ľ be the length of any other arc, m' the coefficient of a, and n' of c, computed in the same manner; then we shall have

Note. From these theorems, Professor Playfair, by comparing an arch of 3° 7′ 1′′ measured in Peru, with an

arc of 8° 20′ 21′′ measured from Dunkirk to Perpignan,

found = nearly. And from the same theorems,

a 300

I, by comparing the arc in Peru with the arc from Dunnose to Clifton, in Yorkshire, amounting to 2° 50′ 23", obtained for the resulting compression.

332.68

(See my Collection of Dissertations and Letters relating to the Trigonometrical Survey of England and Wales, pa. 47.) There is great reason to conclude that the true compression lies between these limits.

SECTION II.

Problems without Solutions.

1. Demonstrate the truth of the following analogy, viz. As the sine of half the difference of two arcs which together make 60° or 90° respectively, is to the difference of their sines; so is 1 to 3 or 2, respectively.

2. Demonstrate that 4 times the rectangle of the sines of two arcs, is equal to the difference of the squares of the chords of the sum and difference of those arcs. 3. Demonstrate that of any arc A,

log sin (A+B) = log sin A + log sin B

+log (cot A+ cot B) - 20.

9. Demonstrate that

11. Demonstrate that the tangent of the sum of any number of arcs will be represented by

ACE- G &c.

1 -B+DF &c.

the sum of all the tangents of the separate arcs being de noted by A, the sum of all their rectangles by B, the sum of all their solids by c, &c.

12. Find the arc whose secant and co-tangent shall be equal.

Ans. Arc whose sine is 5 , or 38° 10′. 13. Find the arc whose sine added to its cosine shall be equal to a; and show the limits of possibility.

Ans. a' must never exceed 2.

14. What arc is that whose tangent and cotangent shall together be equal to four times radius? Ans. Arc of 75° or of 15°.