October 13, 1905.] 



SCIENCE. 



471 



Go from this point in a horizontal direction 

 the distance — sin h cot c. The point thus 

 located is x, y. 



When li is greater than 90°, solve the tri- 

 angle whose given parts are A, V = 180° — &, 

 and c' = 180° — c. 



Given the latitude of the place and the 

 declination of the sun, to find the true azimuth 

 of the sun at any given apparent time. 



Let A denote the latitude of the place 

 and 8 the declination of the sun, north 

 declination being regarded as positive. The 

 product cos \ tan 8 is positive for north 

 declination and negative for south. Its nu- 

 merical value is the vertical distance from 

 the horizontal initial line, measured along the 

 vertical line which is distant cos A from the 

 central line, to the radiating line numbered 8. 



Upon the radiating line which makes the 

 (hour) angle A with the initial direction 

 {-{-x), mark two points, one where it crosses 

 the outer circle, the other where it crosses the 

 circle whose radius is sin A; see Fig. 2. Fol- 



azimuth of the sun is then 103° 57' from the 

 north or 76° 3' from the south. 



A rectangular sheet of waste paper facili- 

 tates the determination of the product cos A 

 tan 8 and the application of this quantity to 

 locating the point x, y. 



If while making a survey A and 8 be re- 

 garded as constant, the azimuth of the sun at 

 any given hour and minute can be obtained 

 with great facility. 



The uses of the azimuth diagram in great 

 circle sailing and in cartography are too ob- 

 vious to require comment. 



Experience shows that if the radius of the 

 outer circle of the diagram is 16§ inches and 

 the circles and the radiating- lines go by half 

 degrees, the azimuth under reasonably favor- 

 able conditions can easily be found to within 

 about three minutes of its true value. 



It is obvious that if two of the three given 

 parts of a triangle are opposites, the unknown 

 part opposite the third given part can readily 

 be ascertained by means of the diagram, be- 



low the horizontal and vertical straight lines 

 until a point is found on a level with the first 

 point and on the vertical passing through the 

 second. Go from this point in a horizontal 

 direction the distance -i- cos A tan 8, and so 

 locate a fourth point {x, y). The angle at 

 the center between the direction — x and 

 this point is the sun's true azimuth (C) from 

 the north in the northern hemisphere and 

 from the south in the southern. In Fig. 2 

 as drawn J. =45° = 3 hours; A = 40° K; 

 8 = 20° K; C = 103°57', showing that the 



cause all such products as sin C sin a, or 

 sin A sin c, or sin A sin & are thereon repre- 

 sented. Such solutions of right-angled tri- 

 angles as involve only sine or cosine factors 

 can therefore be obtained. 



If the X and y of (8) and (9) can not be 

 used on account of the term involving a co- 

 tangent, the required angle can still be de- 

 termined by aid of the diagram, although not 

 as easily as before, because equations (5) and 

 (6) are more complicated than are equations 

 (8) and (9). 



