A 7. I 



tJie eaft and the foiitli, then tlic ftar'» azimnth is faid to be 

 45'^' caiUvard of the fouth. It is the complement of the 

 callcni orweftern amphtudc to a quadrant. 



The azimuth is found tvigonomctrically, by this propor- 

 tion ; as radius is to the tangent of the latitude, fo is the 

 tangent of the fun's altitude to the cofine of the a/.iniuth 

 from the fouth at the time of the equinox. Otherwife, — 

 fuppofe the latitude of the place, and the fun's declination 

 to be given, and let it be required to hnd the fun's alti- 

 tude and azirauth at 6 o'clock. E. G. Let London be the 

 place in N. lat. 51° 32', and let his dechnation be 23° 28', 

 as it is on the longeft day ; then to find his altitude and 

 •azimuth at 6 o'clock in the morning and evening, conllruft 

 a figure in the following manner. Defcribe the meridian 

 {P/utell. j-IJlroiwmy, fig. 20.), draw the horizon HR, and 

 prime verticalZN; make RP = latitude 51° 32' N. ;'draw 

 the 6 o'clock femicircle PS, the equator EQj the 23^ 28' 

 N. parallel of declination /;«;, interlcAing the 6 o'clock fe- 

 micircle PS in ; and through Z, ©, N, defcribe the 

 azimuth circle Z0N, interleiting the horizon in A ; tlien 

 the triangles Z0P and T^OA are fupplemental triangles 

 to one another. Li the fphci'ical triangle Z0F, right- 

 angled at P, we have 



Given the co-lat. ZP = 38^ 28' 

 theco-dcclin. ©P = 66° 7,2' 

 Required the co-altitude ZO 



the azimuth ^0ZP. 

 'Or, in the fpherical triangle ryA©, right-angled at A, 

 Given the lat. Aty© = 51" 32^ 

 the declin. ty © = 23° 28' 

 Required the altitude A© 

 the co-azimuth tyA 

 To find the altitude A©. 

 As Radius - - . . 10.00000 



To fin. declin. ;= 23" 28' - - 9.60012 



So fin. lat. = 51° 32' - - 9-89'575 



To fin. alt. =; 18° 10' - . 9-49387 



A Z I 



tion and altitude ; required the hour from noon, and the 

 fun's azunuth. E.G. In the latitude of 51^ 32' N. the fun's 

 altitude was obferved to be 46° 20', when his declination 

 was 23° 28' N. ; what was the fun's azimuth, and the 

 hour when the obfervation was made? 



Let the primitive circle ZRNH (fg. 21.) reprefcnt the 

 meridian of London, HR the horizon, and ZN the prime 

 vertical; make IIP =^51° 32' the height of the pole at 

 IvOndon ; draw the axis PS, and the equator EQ^; lay off 

 the declination En, Qm, 23° 28' N. the altitude Hr, Rj, 

 46° 20' ; and defcribe the parallel of declination «m, and 

 the parallel of altitude rs, interfedling one another in ©, 

 the place of the fun at that time ; through Z, ©, N, defcribe 

 an azimuth circle ZQN, and through P, ©, S, defcribe an 

 hour circle POS ; then the angles ©ZP, ©PZ, being mea- 

 fured, will give the azimuth and hour from noon required ; 

 or, they may be computed in the following manner. 



In the oblique-angled fpherical triangle POZ, 

 Given the co-lat. ZP = 38° 28' 



the co-alt. or zenith diftance Z© ;= 43'' 40' 

 the co-declin. or polar diftance ©P =; 66° 32' 

 Required the azimuth, Z.0ZP, 



and the hour from noon Z. Q PZ. 



To find the azimuth, or angle ©ZP. 

 Here Z© =43'' 40' 

 ZP = 38^ 28' 



To find the azimuth AR. 

 As Radius . . - 



To Cof. lat. = 51° 32' 

 So tang, declin. =: 23° 28' 

 To CO. tang, azimuth ^ 74° 53' 



10.00000 



9-79383 

 9-63761 

 9.43144 



For the arc AR meafures the .^RZA, the azimuth. 



On the ftiortfd day at London the parallel of fouth decli- 

 nation cuts the 6 o'clock hour-circle below the horizon; 

 and as the triangles rA© and Va© are congruous, the 

 depreflion below the horizon, on the fiiorteft day at 6 o'clock, 

 will be equal to the altitude at the fame hour on the longeft 

 day; and the azimuth will alfo be equal, if eftimated 

 from the fouth. Thus, on the 2 ill of June, the fun will 

 bear N. 74° 53' E. at 6 o'clock in the morning, and N. 

 74° 53' W. at 6 m the evening ; but on the 21ft of De- 

 cember, at the fame hours, it will bear S. 74° 53' E., and 

 S. 74° 53' W. From this problem, it appears, that as the 

 dechnation mcreafes, the altitude incrcafes and the azimuth 

 lefllns, and the contrary happens while the declination is 

 decreafing ; fo that on the days of the equinoxes, when the 

 fun has no dechnation, the altitude at 6 o'clock will be 

 nothmg, or the fun will be in the horizon ; and the azimuth 

 being then 90°, the fun will be due eaft in the mornino-, 

 and weft in the evening ; that is, on the days of the equl- 

 moxes, the fun nfes and fcts at fix, in the call and weft points 

 or the horizon. 



Again. Given the latitude of a place, the fun's declina- 



©Z— ZP = 5° 12' = D. 

 P© =66= 32' 



2) 



71° 44' 



6j° 20' 



35° 5^' 

 30° 40' 



Then co-arith. fin. co-lat. =: 38° 28' 

 co-arith. fin. co-alt. = 43° 40' 

 fin. I fum. co-declin. and D = 35° 52' 

 fin. i diff. co-dechn. and D = ?o° 4.0' 



The fum of the four logs. 

 The \ fum gives 56° ^i\' 



0.20617 

 0.16086 

 9.7678J 

 9.70761 



19.84246 



9.92123 



And 56° 3 li' doubled gives 1 13° 3' for the azimuth fought", 

 reckoning from the north. 



To find the hour from noon, or Z. ©PZ. 

 Here P© = 65° 32' 

 PZ = 38° 28' 



P© — PZ = 28° 4' = D 

 OZ = 43° 40' 



^)T 



7'° 44' 



5° 36' 



Then co-arith. fin. co-declin. = 66° 

 co-arith. fin. co-lat. = 38° 



fin. i fum. co-alt. and D = 35° 

 fin. I diff. co-alt. andD = 7° 



The fum of the four logs. 



The I fum gives 21° 55' 



This 21° 55' doubled gives 43° 50' for the mcafure of 

 the hour from noon, which is 2'' 55" 20'. 



Hence it appears that the obfervation was made either at 



9" 4" 



