TRANSFORMATION OF SURFACES BY BENDING. 463 



15. To find the values of Si and St . 



In the annexed figure let abed represent the small 

 quadrilateral on the surface of the sphere. The exte- 

 rior angles a,b,c,d are equal to those of the four facets 

 which meet at the point P of the surface, and the 

 sides represent the angles which the planes of those 

 facets make with each other ; so that 



ab = /, be = t, cd <= I + Si, da = t + St, 

 and the problem is to determine Si and St in terms of the sides I and t and the angles abed. 

 On the sides ba, be complete the parallelogram abed. 



Produce ad to p, so that ap = aS. Join Sp. 

 Make Cq = cd and join dq, 

 then Si = cd— ab, 

 - cq - cS, 

 m — (qo + oS). 

 Now qo = qd tan qdo 



= cd sin qcd cot qod, 

 but cd = I nearly, sin qcd = qcd = (c + b + tt) and qod = <j>; 

 .-. qo = I (c + b - tt) cot <p. 



pS 



Also oS = 



sin Sop 

 aS (Sap) 



sin <p 



1 



= Z'(a + b - tt) 



sin (^ 

 Substituting the values of a, b, c, d from Art. (11), 



Sl= - (qo + oS) 



= -I- — COt 03tt - J' - — ! . Su. 



t du ' r du sin <p 



Finally, substituting the values of/, /', and Si from Art. (14), 



_£ / * fA s s ' = cot d«' i ds . . , l ds l ds' 



du \p sin <p du) p sin du 7 r' fa p'siri'cf) farfa" 



which may be put under the more convenient form 



d , d / i ds'\ 1 ds p 1 ds l 



:r; (log p) = — log -— - — + _ —cot + £ - — ^— ; 

 «« ow \sm (p du') r du ^ p r du sm <p 



and from the value of St we may similarly obtain 



d n ,. d I l ds\ 1 ds' j p' l ds l 



57 ^^ - d^ l0g Utf, fa) + ; dl? COt ^ + -p 7 A? shT^ 

 Vol. IX. Part IV. 60 



