....

cos. x + cos. y = 2 cos. m. cos. L . . . . (4) Whence by the arithmetic of sines

x, y and

may be found; and these being known, it is easy to find

90° Z PH2

840.

Let a be the angular distance of the shadow of the gnomon from the meridian at a second past 4 o'clock, and ẞ that at 4 o'clock; then, by 833, we have

But by the question the angle due to a second at noon is

841.

Let ZHZ (Fig. 110,) be the meridian of the place, PP' the earth's axis, QQ' the equator, SS' the path of the sun, and HH' the horizon, whose intersection with SS' is s. Then sPH measures the time of sunrise, from midnight. Let PH the latitude be denoted by L, Ps the co-declination of the sun by D, and by the rules of trigonometry, we have

which gives the co-declination of the star, and therefore the declination.

843. We will premise the solution of this problem, with a general investigation of the nature of the curve traced by the shadow of a point, elevated above the horizon, upon the horizontal plane.

Let A (Fig. 111, a, b) be the given point, AC CB the shadow at noon, and CP for any azimuth 4.

horizon, From P

let fall PM 1 CB, and make CMx, PM =

=y, CAh, ZP co-lat. of the place = L, PS co-declination of the sun = D, and

ZS co-altitude of the sun = a; then by trig.

cos. Dcos. a x cos. L

sin. a x sin. L

√(x2 + y2)

cos. @=

√(h2 + x2 + y2)

cos. D (h+x2+y)= h cos. L+x sin. L....(a) which gives

y3 cos. D + x3 (cos. D — sin.3 L) - xxh sin. 2L + h2 (cos.3 Dcos.' L) = 0...... (1) the equation to a conic section, which is a parabola, an ellipse, or hyperbola, according as cos2 D-cos.' L is zero positive or negative.

The transverse axis 2a is found by making y = 0; for then

Also the other axis is twice the maximum value of y, which being found by putting the differential of the value of y2 cos. D equal to O, thence getting x and substituting in equat. (1), gives

The curve traced on the horizon being thus determined, let us proceed to deduce from it the curve traced on any plane bpC, which

is to the meridian ACB, and inclined to the horizon at the

These values of x and y being substituted in equation (1) give

y' cos. D + x2 {cos. L

sin. L (1 + sin. ß sin. L) } – hx' {cos. B sin. 2L +

(cos.2 D

-

cos.' L)} + h2 (cos.2

D cos.' L) = 0,

2 sin. 8 ×

.... (6) the equation to a conic section, which is a parabola, an ellipse, or an hyperbola, according as the co-efficient of x" is zero, positive or negative.

The semiaxes of this curve may be found in the same way as those of the horizontal trace were determined.