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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.”

Paoblem 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 cor‘rectly measured; according to the method proposed by Professor or (Edinburgh Transactions, vol. v.), which is as below. Let the ellipse Pepa represent a terrestrial meridian passing through the poles P, p, and cutting the equator in E, Q. Let c be the centre of the earth, cq the radius of the equator = a, and Pc, half the polar axis = b. Let AB.be any small arc of - * * the meridian, having its centre of curvature in H ; join HA, HB, intersecting cq in K and L. Let Q be the measure of the latitude of A, or the measure of the angle- QKA, expressed in decimals of the radius 1; not in degrees and minutes. Then, the excess of the angle QLB above QKA, that is, the angle LHK or BHA will be = 32. Also, if the elliptic are QA = 2, then will AB = 2x = 22 x AH. which latter, because sin” p = 4 (1 — cos 22), becomes 32 = (a - 2c) 90 + 3 cop — ; c cos 20%. Consequently, by integration, z = (a oc) Q #c sin 29 = ap c (p + # sin 29); an expression which needs no correction. Let MN be any arc of the elliptic meridian, © the latitude of M, one of its extremities, and p" that of N, its other extremity; we have QM = ago oc (@' + 4 sin 20") QN = ap" – #c (?" + 3 sin 20") and their difference QN – QM, that is, MN = a (p" — «») – oc [(p” – p") + 3 (sin2.0" — sinop')]. 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 l be the length of an arc measured, m the coefficient of a, and n of c, computed by the last formula; also, if l 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 ma no = l, and m'a n'c. = l'.

* 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.”

Now it is shown by the writers on conic sections, that the radius of curvature at, A, that is, -

2

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

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arc of 8°20'24" measured from Dunkirk to Perpignan,

l found; F 500 nearly. And from the same theorems,

I, by comparing the are in Peru with the arc from Dunnose to Clifton, in Yorkshire, amounting to 2°50'

- I - - 234", obtained -oo: 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 v 3 or v 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,

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11. Demonstrate that the tangent of the sum of any number of arcs will be represented by - A – c + E – 6 &c.

* . . . .

TTDT. S.C. the sum of all the tangents of the separate arcs being denoted 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 v 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°.

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