cos. (m - D) – cos. D. cos. L

sin. D. sin. L which, by the arithmetic of sines will give D, and therefore

z (=m - D.)

860. If the moon's velocity were augmented, her periodic time would be diminished, and the monthly motion of the node of her orbit, which is due to the disturbing force of the sun, would also be diminished. But the difference between the sidereal and synodical periods, is owing to the motion of the nodes. Therefore, &c. &c.

861. Let z, z' be the two given zenith distances of the sun, and a, a' the given azimuths; also let L denote the co-latitude, and D the co-declination, which is supposed to undergo no variation between the times of observation. Then from the A whose sides are

L, 2, D; Lz'D; we have

cos. D=cos, z cos. L + sin. z. sin. L. COS. a

cos. D = cos.ž. cos. L + sin. z'. sin. L. cos. . cos. z + sin. z.cos. a. tan. L = cos. z' + sin. z' .cog. a' x tan. L

cos. z'
.. tan. L =
sin. z' , cos. á

sin. z which gives L, and therefore 90° – L.



COS. a

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862. Let g, g' be the distances of the sun from the earth and Jupiter; then the apparent magnitudes of the sun's diameter are as

1 1

or as é, po

863. Let a denote the earth's mean distance, a – de his least distance; then since the apparent diameter o inversely as the distance, by the question, we have

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864. Let L denote the co-latitude of the place; D, D the co-declinations of the stars subtending their common azimuth of 90°; h the given difference of their right ascensions; and H, H + h the hour angles from noon. Then, by Napier's Rules for the resolution of right angled spherical A, we get

tan. L = cos. H . tan. D.

tan. L = cos. (H + 1) tan. D' Hence, by the arithmetic of sines, H being eliminated, we get tan. D

tan. L=cos.h tan. L+sin.h. (tan.”D-tan.:D) tan. D Moreover, if a be the given distance of the stars when on the prime vertical from the A, whose sides are D, D'and that distance, we get

cos. D cos. D cos. h

sin. D.sin. D' and this, together with the above equation involving D, D', will give, when reduced, the declinations required.

COS. a

865. Let the given sum of the azimuth () and hour 2 from noon H, be

m =a + H. Also let D, and z, be the giyen co-declination and zenith distance

of the sun, and L the co-latitude of the place. Then from the A, whose sides are D, z, L, we have cos. H =

cos. z — cos. D. cos. L

sin, D. sin. L

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But cos. H = cos. (m – a) = cos. m . cos. a + sin. m . sin. a. - cos. D.cos. L

COS. m X sin. D. sin. L

sin. L

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cos. D


cos. L

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sin. z

sin. m X

cos. z . cos. Ly}}


{ sin. z . sin.' L + cos. D sin. z sin. L which, being reduced, will give L.

866. Let h be the difference of the times of setting, H, H +h the hour; 4 from noon, D, D the co-declinations of the two stars, and L the co-latitude of the plane; then from the A whose sides are

90°, D, L; 90°, D', L

we have

cos. 90°

cos. D. cos. L cos. H =

cot. D. cot. L sin. D. sin. L and cos. (H + h) = cot. D. cot. L

:: cot. D' cot. L = cos. h . cot. D . cot. L + sin. h . (1 cot.' D. cot. L); which being reduced and adapted to logarithmic computation will give L, and therefore the required latitude.

867. If I denote the longitude of the sun, l that of the star, and a its latitude; then (Woodhouse, p. 299) the aberration in latitude is

20'. 25 x sin. (l ?) sin. a which is therefore 0 when

1=l', which makes known the longitude of the sun, and therefore the day of the year.

But since the longitudes of the sun and star are equal on this day, their right ascensions are also equal, and they consequently cross the meridian at the same moment, or at twelve o'clock.

The problem may be easily generalized; by stating it, Required at what time of the day a star, whose longitude and latitude are given, crosses the meridian, on that day of the year when its aberration is latitude is a given quantity a.

In this case the longitude of the sun may be found from the formula

a = 20'. 25. sin. (l' – 1) sin. a and thence the right ascension of the sun (A) from

tan. A = cos. O x tan. I, O being the obliquity of the ecliptic.

Moreover, the right ascension (A) of the star is obtained from

tan. A' = cos. O x tan. ľ and A- A' the time required measured from noon is therefore known.

868. This problem relates to navigation—to oblique sailing.

Since the vessel is steered constantly to the same point of the compass, her path along the surface of the sea cuts the meridians at the same L, viz., 45°. Hence, if ds denote the constant increment of her route, the corresponding increment of latitude is

ds ds x cos. 45o =

which is also constant.

Hence the whole path described from the equator to the pole is

90° X 7 2.

Again, to find the difference of the longitude, suppose the degrees of longitude to be of the same length for every parallel of latitude, (as in Mercator's Projection, then the degrees of latitude must increase as the secant of the latitudes,

Hence, if a denote the latitude, its increment, in order to make that of the longitude the same for all parallels, must be

dx d. a =

cos. a

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Hence, if a' be calculated for any two latitudes of the vessel, viz., a, , the difference of the longitudes will evidently be D= (a - 6') tan. 8o

(2) O measuring the distance of the point of the compass from the north or south, or the constant inclination of the route to the meridians.

In short, all problems of this kind are resolved by the two formulæ

R= a sec. 0


tan. (450 + and D=1.

tan. (459 +

x tan .

R being the length of the route.

Ex. 1. In the problem, a = 90°, je = 0, and 0 = 45.

:: R = 90° X N 2, as before. and D=1 x tan. 90o

So that the ship would never actually reach the pole, although she would go on for ever approaching more nearly to it.


Ex. 2. Generally, let a = 90°, M = 0. Then

R = 90°. sec. 0, and Dl.

tan. 90°

x tan. }

tan. 45 which shews that in whatever direction, other than direct north or south, a ship might sail, she would never by possibility reach the pole. If she steer to the north or south, then we have 0 = 0 and ..

R = 90°

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