Archdeacon Pratt on the Stability of Arches. 265 



In the first case the arch will be stable, because the slight dis- 

 placement raises the centre of gravity ; and as this will fall, the 

 openings will close up again when the arch is left to itself. In 

 the latter case the centre of gravity descends, owing to the dis- 

 placement of the parts of the arch, and will go on descending 

 when the arch is left to itself, and therefore the arch will fall. 



When any design of an arch is determined upon, it is an easy 

 matter to take different portions of the load L and find all the 

 quantities involved in the above formula by construction and 

 measurement by a scale ; and the character of the proposed arch 

 for stability can be determined in this way. 



Example. — Suppose W 1 ,W 2 ,W 3 ,W 4 areas 10, 6, 5, 9 indepen- 

 dently of L, and sin B = sin 55° = 0'819, sin D = sin 45°=0-707, 

 sin F= sin 80° = 0*985 : also let 



h 1 _l 

 a-, ~ 6 



H, 



a 



9^ 



Putting these in the above formula of comparison, the two sides 

 are 7*65 and 7*89. The first is less than the second, and there- 

 fore the arch would be in itself stable. Now introduce L, and 

 suppose the centre of gravity to be over B, and half added to Wj 

 and half to W 2 . Then the two sides of the comparison are 



7-65+0-75 Land 7-89, 



and therefore the arch will be at its limit between stability and 

 instability when L = 032, or about g ¥ th part of the weight of 

 the whole bridge. Practical men must decide whether this would 

 be safe. 



II. Suppose the load L is near the crown of the arch, and 

 that the arch is forcibly held in the position marked in fig. 3. 







Fig. 3. 









J> 









^^B 





~--\^ 



.i/x 









JB%^\ 



E 



By reasoning precisely as before, and observing that in this 

 case #2 :=a i an d «3=«^ we obtain 



