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831. Let L be the given co-latitude, D the sun's co-declination, and x his co-altitude at the time required. Also let the hour angle from noon be measured by the < 0; then L, x and D make a spherical A, which gives

cos. L. cos. D cos. 8 =

1
sin. L. sin. D.

sin. 6
.. dx = d. sin. L. sin. D... (2)

sin.
But from the same À we have

sin. 0 : sin. (azimuth) :: sin. r. sin. D .. dx = di x siu, L. sin. azimuth. Hence the variation of the sun's altitude is greatest when his azimuth is 90°, or when he is crossing the prime vertical. Again, in this case

cos. D - cos. X. cos. L cos. azimuth =

sin. x. sin. L

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832. Let the moon's horizontal parallax be P; her periodic time t; the apparent and real semidiameters of the sun a and r', the radius of the earth r; the length of the sidereal year t', and the distances of the sun and moon from the earth p and s'; then we have (see 453)

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g's

:

ť sin.'Pt2 sin.3a
But if d, d' be the densities of the earth and sun,

Hiu :: pd : g'id'
.. d:ď :: t sin. 3a : t sin. "P.

833. In consequence of the daily revolution of the earth about its axis, the sun appears to describe, with an uniform motion, a circle whose plane is parallel to that of the equator. Let S, S', S", S (Fig. 107,) be the intersection of the sun's path with the surface of the earth, P P being the earth's axis, Q 12Q the equator, and PSP the meridian of the place. Then, at the hours of 12, 1, 2, if the sun be supposed to be in positions corresponding to S, S, S", and the earth to be hollow, the shadow of the axis PC will be thrown upon the equator in the directions CQ, Ci, C2, which are the intersections of the circles of latitude passing through S, S, S" with the equator; and since the motion of the sun is uniform, those circles will be at equal intervals, and will consequently give

Q1 - 12 = &c. Hence, if at Z, the place of the spectator, there be placed a circular plate qa' with its plane parallel to the equator, and with a gnomon cp I to it; the shadow of this gnomon will uniformly describe the circumference of the plate, and will therefore indicate the time of the day according as the graduation may be effected. If hours merely are to be noted, at q write 12, take ql = 15o = 12=23, &c. Then when the shadow is at q, 1, 2, &c., the hour will be 12, 1, 2, &c., respectively.

This is called the Equatorial Dial.

Now, instead of the shadows being cast upon the plane parallel to the equator, which is the simplest case, let them fall upon the horizon, as in Fig. 107. In this case Hi, 12, 23, &c., are not equal. Their magnitudes must be calculated by the resolution of the right-angled A PIH, P2H, &c. Let L be the known latitude of the place; then

PH = 1
And < IPH = 15°
Hence, by the rules of spherical trigonometry

tan. Hi = sin. L. tan. 15°
tan. H2 = sin. L. tan. 2 X 150
tan, H3 = sin. L. tan. 3 x 15°

&c. = &c. which give, by the tables,

H1, H2, H3, &c. and therefore the hour HC1, 1C2, 2C3, &c.

Upon the horizontal plate hh let hh' be the meridian line, cp parallel to the earth's axis, and from the centre c draw c1, c2, c3, &c., making the hcl, 1c2, &c. = HCi, iC2, &c. The shadows of the goomon cp will indicate the hours of the day.

To construct a Vertical North or South Dial. The hour circles P1, P2, &c., (Fig. 108,) will intersect the prime vertical ZN, making the several right-angled A P'Ni, P'N2, &c., and, as before, we have

tan. Ni = sin. P'N. tan. 15o = cos. L. tan. 15°
tan. N2 = cos. L. tan. 2 x 15°.

&c. = &c. whence the 2 NC1, 1C2, &c., or the 2 nci, 1c2, &c., may be found, and the dial constructed.

If P be the north pole, the figure represents the position of the gnomon for a dial facing the south. The construction for the Vertical North Dial is nearly the same.

To construct a l'ertical East or West Dial. Let cp (Fig. 109,) parallel to the earth’s axis be the gnomon, and 6, 11 the dial plate parallel to the meridian of the place, also

let cp be on the eastern side of the plate; then the shadows upoo the plate will evidently be all parallel to the gnonom and to one another. Moreover, at six o'clock the sun, being due east, will cast the shadow perpendicularly upon the dial, and the Lp611 therefore is a right 2. Hence we bave

67 = p6 x tan. 15 = a x tan. 15°
68 = a x tan. 2 x 15°

&c. = &c.
which give the positions of the hour lines.

Similarly may be constructed a Vertical West Dial.

In the problem we are required to find the 2 between the hour lines of 12 and 3 in the horizontal dial. We have already found that

tan. H3 = sin. L x tan. 45°.

= sin. L

which gives the required by the tables.

834. By the preceding problem we have

6 11 = a tan. 5 X 15° and

10 = a tan. 4 x 15°
:: 3 = 11 10 = a x (tan. 5 x 15° tan. 4 X 150
which gives

3
2N 3

1
2 3 2

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835. Let C (11) (Fig 108) be the position of the substyle ; then by the question in the right-angled A NP(11), NP the co-latitude and Z NP' (11) = 15° are known, and we may therefore find the ZPN (11), which is the inclination of the dial to the meridian. By trigonometry we have

cos. P'N = cot. 15° x cot. P'N(11)
or sin. L = cot. 15° X cot. I.

cot. 15°
.. tan. I =

sin. L which gives the required positi.

836. The sun being due east and due west at 6 o'clock A.M. and P.M., will not shine upon a Vertical South Dial before the one time, nor after the other, in any latitude, or on any day. Consequently the greatest number of bours that can be indicated by such a dial is twelve ; and this evidently happens at both equinoxes, or when the sun rises and sets at 6 o'clock ; and from this time during the whole summer, since the sun never reaches the east and west points, except when under the horizon, the number of hours indicated by the Vertical South Dial will be the same as that of daylight.

1)

837. By 833, we have

tan. H4 = sin. L x tan. 4 X 150 and tan. H3 = sin. L. tan. 3 x 15° tan. H2 = sin. L. tan. 2 x 15°

tan. (H4) tan. (H3)
.. tan. (3, 4) =

1+ tan. H4 X tan. H3
sin. L x (tan. 60°-tan. 459)
1 + sin. ’L tan. 60°. tan. 450
sin. L X (2x 3
2 + sin. 'L X N 3

sin. L (tan. 45
Also tan.(2, 3) =

tan. 30)
i+sin. 'L, tan. 45. tan. 30
sin. L (V3 - 2)

20W 3,+ sin. 'L)
:. by the question
2/3 - 1

3-2
m :n ::

:
2 + 3 sin. ?L 2( 3 + sin. 'L)
which gives, when reduced,
sin. L =

m 3
In 3 - m.

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838. Let the < 2PH = 0 (Fig. 10) and generally suppose the difference between this and the < 10 PH = a a given quantity. Then by 833' we get

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