upon the Antique Dials. 83 



Then, cos n . AW = cos (90 n . WD) = sin n . WD 



cos n . AZ = cos (90 + n AZ) = sin n.AZ = sin n. WD. 



Hence the formula (A D L ) becomes 



tan D = tan I sin n . WD for point W, and 

 tan D = tan I sin n . WD for point Z : 



and the branches ID, DS at equal distances from the common extremity, 

 are hence equal in all respects, but reversed in position. 



IV. 



The same great circle will be a tangent to both branches of the curve at 

 D, but will be on different sides of it in the two hemispheres. For we 

 have just seen, that the curve is composed of equal and similar branches, but 

 reversed in position, and therefore we may at once infer, that any great 

 circle through D will cut off equal portions from each side of it. Hence, 

 when the circle YDY' ceases to cut one branch, it ceases to cut the other, 

 and it becomes simultaneously a tangent to each ; and the curve has its 

 point of contrary flexure at its intersection with the equator. 



The same conclusion might, however, be readily obtained from the for- 

 mula. For, consider Y a point in the hectemorion, and join DY by a great 

 circle. Then, to find the spherical angle YDW, we have the sides DW, 

 W Y of the right-angled triangle WDY. Hence, 



sin WD = cot WDY tan WY, or 

 in /9 L \ _ cot WDY tan D 



= cot WDY tan I cos ni,, or 



cot WDY = sin ~ raL Vot I sec L. 



/ 



g n 



n 



If in this general expression of the value of WDY we make wL = 90 ; 

 then 



sin cot I 



cot DWY = : 



cos 90 - 



