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42. tang (B — A) — cot ✯ C
46. cos c
47. sin a
45. cos ccos a cos b
cot A cot B
sin c sin A
sin 1⁄2 (b — a)
length of a =
sin § (b − a)
cos 1⁄2 (b
For a right angled spherical triangle in which C is the right angle, and the opposite side c, the hypotenuse, as in Fig. 59.
cos (B – - A)
51. If any small arc or angle a, not exceeding, or not much exceeding a degree, be expressed in seconds, and if
206264".8 we have,
sin a= very nearly.
48. tang a= sin b tang A
For the sine of a small arc is very nearly equal to the length of the arc itself; and to obtain the length of an arc, expressed in seconds, we have this proportion. As the number of seconds in the whole circumference is to the seconds in the arc, so is the length of the circumference to the length of the arc. Hence,
sin a = length of a =
1296000": a :: 6.2831853 : length of a,
As the circumference of a circle, divided by 6.283, &c., gives the radius, it is evident that 206264′′.8 are the seconds in the radius.
Cor. The number of seconds in an arc is equal to the product of a by the length of the arc; the radius being unity.
INVESTIGATIONS OF VARIOUS FORMULE.
52. To find formulæ for the values of the rectangular co-ordinates CD and AD, of a place A, on the earth's surface, Fig. 4, and its distance AC from the centre, in terms of the latitude and eccentricity.
Then, we have, CD =p cos p', and AD
Let the semicircle eBq be described on eq, and let DA be produced to meet it in B. Then, by Conic Sections, DA: DB,
AC, the equatorial radius Cq being assumed 1.
zCq, the geocentric latitude, earth's eccentricity.
√ (1 − e2): 1 :: p sin p': DB = p sin q'.
DB2 + DC2 =
Cq = 1.
Hence, p2 sin '.
+ p2 cos2 p' = 1,
or, (1 — e3)
But (76 A), tan3 p′
Again (App. 2), p sin p'
tan3 q′ + 1
1 — e2 sino
e3 sin3 + cos2 &
Whence, p cos p'
p2 cos2 p′
e2) sin &
p2 cos2 o'
√ (1 — e2 sin2 † )
p cos p'tan p' p cos p tan ¢ (1
e2 sin2 )
p2 cos2 p'
e2 sin2 )
or, p sin p'
Adding together the squares of the equations (A) and (B) we have,
e2)2 sin2 + cos3 &
2 e2 sin2 & + ea sin2 & + cos2 ∞
1 e2 sin2 p
1 — 2 e2 sin3 † + ea sin3 p
or, p = √
+ sin L sin D cos L cos D
The value of p, multiplied by the number of miles in the equatorial radius, gives the value of AC in miles.
€3) e2 sin3 o
53. To find the times of longest and shortest twilight at a given place. Let HZG, Fig. 32, be the meridian of the place, Z its zenith, HR its horizon, FG, parallel to HR, 18° below, P the elevated pole, AB the part of sun's diurnal path included between HR and FG, PA and PB arcs of declination circles, and ZA and ZB arcs of vertical circles. Put L PH 18°. Then,
sun's declination, and 2a
latitude of the place, D (App. 34), we have, COS AZ
e2 sin2 p
cos PZ cos AP sin PZ sin AP
=±tang L tang D
PZ cos BP
cos (ZPA + APB) = cos ZPB
cos (90° + 2a) — sin L cos (90° ± D) cos L sin (90° ± D)
or, (App. 23),
2 sin APB sin (ZPA † † APB)
Let now, H ZPA, and x APB, when AP is greater than 90°, that is, when the declination is of a different name from the latitude, and H' ZPA, and x' APB, for an equal declination, when of the same name with the latitude. Then, it is evident from the expression for cos ZPA, that H will be less than 90°, and that H' will be the supplement of H. Hence, we have,
2 sin x sin (H + x)
2 sina sin (180° - H+x')
2 sin x′sin (H. — § x'′) cos L cos D Consequently, sin' sin (H-x)=sinx sin (H + or (App. 18), 1⁄2 cos (H cos H or, cos (H — x') + cos (H + x)
sin 2a cos L cos D
tang L tang D.
2 cos H.
sin 2a cos L cos D
cos L cos D
cos (H+ x),
x) + cos (H + x) = 2 cos H cos x. cos (H
2 cos H (1
Now, since H is less than 90°, the second member is affirmative. Consequently, cos (H x') is greater than cos (H — x), and therefore, x' is greater than x; that is, the twilight is longer when the latitude and sun's declination are of the same name, than when they are respectively of the same values, but of different names. And it is easy to perceive from equations (A) and (B), that the longest twilight at a place, occurs when the declination is greatest and of the same name with the latitude.
For the shortest twilight. Let the triangle BPC, having the side BP AP, have also, PC PZ, and BC = AZ 90°; then its three sides being respectively equal to those of the triangle APZ, we have the angle CPB ZPA. Taking CPA from each, we have APBZPC. Hence, the twilight will be shortest when the angle ZPC is least.
Let ZD be a vertical circle through C, and PE an arc of a great circle bisecting the angle ZPC. Since PZ and PC are constant, the angle ZPC will be least when ZC is least; and since BZ and BC are constant, ZC will be least when the angle ZBC is least, that is, when it becomes Zero, or when BZ and DZ coincide. But, when BZ and DZ coincide, we must have DC BC 90°. Hence, as DZ 90° +2a, it follows, that, when the twilight is shortest, ZC 2a.
Now, as the triangle ZPC is isosceles, and PE bisects the vertical angle, it must also bisect the base ZC and be perpendicular to it. Hence, in the right-angled triangle ZEP, we have (App. 47),
Twice the angle ZPE converted into time, gives the duration of shortest twilight. From the right-angled triangles ZEP and DEP, we have (App. 45),
As cos PD is negative, PD, the sun's distance from the elevated pole, must be more than 90°, and, consequently, the declination must be of a name contrary to that of the latitude.
When PD has been found from the above expression, the sun's declination is known; and by a Nautical Almanac, the times in the year when the sun has this declination are easily found.
54. To find the annual variations of a star in right ascension and declination.
Referring to Fig. 20, described in previous Articles (126 and 128), as it is evident that mg does not sensibly differ from EG, and as the difference between the complements of two arcs is the same as the difference between the arcs themselves, we have,
E'G' mg = E'm + gG',
E'G' — EG
or, ann. var. in R. Ascen.
ann. var. in Declin.
Draw P'r perpendicular to the declination circle PsG. Then, as PP' and P'r are very small, we may, without material error, regard PpP' as a spherical triangle, right angled at P, and P', PrP' and EmE as right angled plane triangles, and sP' sr. Put
right ascen. of the star
QEC = obliq. of the ecliptic.
Then, taking small arcs or angles instead of their sines, and observing that PpP' CC′ = EE′ = 50′′.2 (126, Cor.), and P'Ps 90° GPQ EQ — GQ EG A, we have, from the triangles PpP' and PrP', PP' PpP' sin pP 50".2 sin ε, PP' sin P'Ps = 50′′.2 sin & sin A,
PP' sin A
PP' cos P'Ps
· PP' cos A
50".2 sin ε cos A.
50′′.2 sin ɛ cos A. (C)
and, Consequently, Ps P's PsWe have, also, from the right angled spherical triangles P'rs and gG's, the latter of which may be regarded as right angled at g as well as at G', we have,
gsG' cos G's, and gG′
P'r tang G's 50′′.2 sin & sin A tang G's.
But, since the quantity which is multipled by tang G's is small, we may, without sensible error, put tang Gs or tang D, instead of tang G's. We then have,
Substituting, in formula (A and B), the values of E'm, gG' anl
Ps - P's (E, D, and C), we have,
Ann. var. in R. Ascen.
Ann. var. in Declin.
gG' 50′′.2 sin & sin A tang D From the triangle E'mE, we have,
E'm E'E cos mE'E — 50′′.2 cos ɛ
50′′.2 cos ε + 50′′.2 sin & sin A tang D
50".2 sin cos A.