Sun's Pofition, or that Angle which the Sun's Azimuth makes with his Circle of Declination.
S, BP: S, D:: S, PD: S, B
9.634848 25°. 33' or (154°. 27. Ambi.
CASE II.
In the fame Triangle, Given PD, and the Angles D and B: Requir'd BP the Sun's Co-Altitude.
S, B: S,DP:: S, D: S, B P
In the Triangle BPD, given PD the
Complament of the Pole's Elevation
= 38°. 30'. BP the Sun's Co-Alti-
tude 46°. 11'. and the Angle B
= 25° 33'. Requir'd BD the
Complement of the Sun's Declinati-
on. (See Fig. p. 170).
By Cafe 4. R:, B:: T, BP': T, BA
By Rule 2. &, BP : &, BA :: E,DP : &,DA
Draw the Perpendicular A P (by Prob.
6.) from P the Extream of the given
Side BP and oppofite to the given An-
gle at B, which will refolve the Oblique
Angled Triangle into Two Right ones,
which are folv'd by the Two foregoing
Proportions.
2,B= 9.955307
T, BP 10. 017944
T, BA 9.973251 =43°. 14'.
DA=9.911851 35°. 17.
Then here because PA falls within the
Triangle, BA+AD, i. e. 43°. 14' +
35°. 17′ = 78°. 31 BD. required.
In the Oblique angled Triangle ZPS;
Given the Right Afcention, Decli-
nation and Altitude of a Star, fup-
pofe at S: Required it's Azimuth,
viz. Let the Star be Aldebaron,
whofe Right Afcenfion CR 64°.
17'. Declination RS 15°. 48'.
Co-Elevation of the Pole=38°.30'.
as alfo his Altitude AS 30°. are
all given: Required the Angle at Z.?
This is folv'd by Cafe II.
S, ZP = 9.794150 = 38°. 30'
S, ZS = 9.937531 = 60°.
SSP+Diff.x SP-Diff.cr.-19.517286
SqLZ=19. 785605
S, Z = 9.89280251°. 23′
Therefore Z 102°. 46' that is his
Azimuth, is nearly E by S.
|