Bambaut — Converting Hour-angle and Declination, fyc. 643 

 To prove this, draw GK parallel to FC and GH, and IK at 



Fig. 1. 



right angles to it. Then 



IK HG + EC 



smIGK = 



IG 



AG 



HG FG EC AC AF 

 " FG ' AG + CD'AF'AG 



= sin S sin + cos S cos # cos t. 



Therefore IGK= h. 



Equation (2) may be obtained from (1) by writing A and h for 

 t and S respectively, and vice versa. We thus see that if FAG=h, 

 and IGK= 8, then ACD = A. 



The instrument which I have to describe consists of a circle M, 

 graduated in hours and minutes on the inner side of the circum- 

 ference, and in degrees on the outer side, to facilitate the readings 

 of hour-angles and azimuths respectively. There is a straight 

 edge corresponding to the line FL in figure 1 graduated, so as to 

 read, at any point G (fig. 1), the angle between the lines AF and 

 A G, and along this slides a piece JYO which carries a pointer, and 

 can be set at any required graduation. 



