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The Scheme being constructed, there is

P formed the Oblique Triangle C ZP, in which there is given ZP = 39° 04', the Co-latitude ; the An. gle P = 30° oo'; H and the Side CP(= AC + AP=20° 34' +90) = 110° 34'; to find the Side CZ, the Co-altitude required ; which is done by Cafe 4, thus. The Segment L P is found as in the foregoing Case of this Problem to be 35° 06'; hence CL will be 75° 28'. Then fay,

As the Co-sine of LP=35° 06' Co. Ar. c.0871672 To the Co-fine of Z P = 399 04' 9.8900929 So is the Co-sine of CL= 75° 28' 9.3995754 To the Co-sine of Z C = 760 14' 9.3768355

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Whose Complement to a Quadrant is BC= 132 46', and is the Altitude of the Sun, for the given Time and Declination.

Case 3. . Suppose the Declination North 200 34' as before, but the Time 5 in the Morning, or z at Night ; I demand the Altitude of the Sun for those Moments in the Latitude of Chichester 50, 56' ?

By

B В

La
By these Data,
construct the Scheme,

P
and you will have
formed the Oblique
Triangle CPZ,
wherein there will be

0 given the Side ZP=

H
39° 04'; the Side
CP= 69° 26; and
the Angle included
ZPC= 1050 oo';
to find the Side ZC;
the 'Complement of

N
the Altitude required.

Let fall the Perpendicular Z L on the Side CP continued out ; and then to find P L, fay by Case 4, and Analogy to Variety 1.

As Radius

10.0000000 Is to the Co-fine of P = 1050 00

9.4129962 So is the Tangent of ZP = 399 04? - 9.9094022 To the Tangent of P L= IIS 52

9.3223984 To which add CP

6926 The Sum is

81 18'

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CL =

Now say again ;
As the Co-sine of PL=11° 52' Co. Ar. 0.0093810
Is to the Co-sine P Z = 399 04' 9.8900929
So is the Co-fine of CL= 810 181

9.1797265 To the Co-fine of CZ = 83° 07' 9.0792004

Or the Sine of BC = 62 53' = Altitude of the Sun required.

PRO

PROBLEM VI.

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A

The Latitude of the Place, Declination and Altitude of the Sun being given, to find the Hour of the Day.

Practice. Suppose I observe the Declination of

P the Sun to be 200 Æ 34' North, and his Altitude to be 51° 35' in the Latitude

B of Chichester 50° 56', I demand what Hour

H of the Day the Obfervation was made ?

Having prepared the Diagram, there

S will be formed the

N
Oblique Triangle
CZP, in which there is given all the three sides ;
for the Side CZ = 38° 25' = the Co-altitude ; the
Side ZP= 399 04' = the Co-latitude ; and the
Side CP = 6926' = the Co-declination ; to find
the Angle at P = the Hour from Noon when the
Observation was made.

This is done by Case 5, or Theorem 39, thus.

Here it is CPZP = AM = 30° 22'.

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2

2

Then (the Sine of CP= 69° 26' Co. Ar. 0.0286016

the Sine of ZP= 39°04' Co. Ar, 0.2005049 {

= 349 24' 9.7520231 the Sine of CZTAM

40 01 8.8453874 The Sum of all these is

18.8265170 The of which is the Sine of P = 159 9.4132585

The Double whereof is 30o = P, the Angle of the Hour fought.

This 30° reduced into Time gives just two Hours, so that the Observation was made either at 10 aClock in the Morning, or at 2 in the Afternoon.

After this Manner the Angle P of the Hour may be found when the Sun hath South Declination, as in the Triangle to Case 2 of the last Problem ; and also if the Time be before fix in the Morning, or after six at Night, as in Cafe 3, of the foregoing Problem ; supposing all the three sides given there as here.

PPOBLEM XVII.
Given the Latitude, Declinacion and Altitude of
the Sun, to find his Azimuth.

Practice.
Suppose these Data
the same as in the last
Problem ;

N

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38° 25', and Z P= 390 04'; Whence we find the Angle at 2 = the Azimuth, as the Angle P was found in the last Problem ; but for Variety's fake, I shall here use the Method by a Perpendicular, according to Prob. 5, of Chap. 18. Making therefore Z P the Base, the Perpendicular let fall thereon is CI; and let LZ = L M; then fay,

As the Tang. of IZP=190 32 Co. Ar. 0.4500489 To the Tang of CZ+CP=53° 55' 10.1374113 So is the Tang. of CZ-CP=15° 31' 9.4434786 To the Tangent of PM 470 oz' 10.0309388

Hence P M=94°04'; wherefore P M-Z P= ZM=550 001; and the half of Z M is ZL=270 30'.

Therefore in the Right-angled Triangle LZG there's given the Side Č Ž = 38025', and L Ż 27° 30', to find the Angle LZC= the Azimuth from the South. Thus

As the Tangent of CZ = 38° 25' 9.8993082 To the Tangent of L2 = 279 302 9.7164767 So is Radius To the Co-line of LZC = 489 59' 9.8171685

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Wherefore the Angle C Z P=131° 01'; and consequently the Azimuth of the Sun, or Point of the Compass he is then upon, is 3 Points nearly from the East towards the South ; that is, on the Point S E and by E in the Morning ; or SW and by W in the Afternoon.

Thus alfo may the Azimuth be found for a South Declination, and for any Time of the Day.

This Problem is very useful to a Diallist, in finding the Plane's Declination; and to the Navigator in

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