a Mechanical and Physiological point of view. 



27 



Tjsina^ 



T 2 sin a 2 



-fW, . 



. . (7) 



T 2 sina 2 = 



T 3 sin a a 



+ w, . 



. (8) 



Tgsin a 3 = 



T 4 sin a /M 



+w, . 



. (9) 



T 4 sin a 4 = 



T 5 sin * 5 



+w, . 



■ • (10) 



T 5 sin« 5 = 



T 6 cin« 6 



+w, . 



■ ". (11) 



T 6 sin« 6 = 



fT 7 sin a7 l 



i =0 r 



■•+'W. . . 



. (12) 



The twelve unknown quantities in these equations, which 

 denote the inclinations and tensions of the first six portions of 

 the rope, may be all expressed as follows, after a few reductions, 

 in terms of X, the horizontal strain on the funicular polygon : — 



T^v^W^X 2 , tana, 



T 3 =v/l6W 2 + X 2 , 



T 4 = V / 9W 2 +X 2 , 

 T 5 =</4W 2 + X 2 , 

 T 6 = v / W 2 + X 2 , 



6W 



tan a 2 = 



5W 



tan a 3 = 



4W 



tana 4 = 



3W 



IT' 



tan a 6 = 



2W 



T 3 



tana 6 = 



w 



X* 



(13) 



"'} 



(14) 



If a v ff 2 , a 3 , a 4 , « 5 , # 6 denote the lengths of the six portions of 

 the rope, we obtain, from geometrical considerations, the follow- 

 ing equation : — 



«! cos a x -f a 2 cos « 2 + a 3 cos a 3 -f a 4 cos « 4 + a b cos 



■f a 6 cos a 6 = const., 



in which equation the constant depends on the span of the poly- 

 gon from the pillar of the hall to the top of the vaulted kitchen. 

 If we were to substitute in equation (14) the values of the 

 cosines of the angles found from (13), it would become an equa- 

 tion ultimately of a high degree, the real root of which would 

 give the solution of the problem sought ; but it is not necessary 

 to trouble ourselves with this equation, as we know that the value 

 of X must be real and positive, and by (13) we have 



T 1= \/(6W) 2 -t-X 2 . 

 In this equation T x denotes the pull on the rope that Telema- 



