THE AUCH. 



101 



tnio of all arches wliatovor tlioir span— tlio liorlzoiital tkrust 

 is always tho same in all parts of tho arch. 



Again, if tlie diagram is txirnod upside down, it will then 

 represent an inverted or suspended arch, and it will he 

 found that the same law of tho parallelogram of forces will 

 equally, and in the same manner, prove all the problems 

 above mentioned, the only dilFeronce being that the thrnsts 

 become tensions and the tensions thrusts; but the propor- 

 tional amounts of these forces, represented by the various 

 linos, remain exactly the same as before, in the direction of 

 those lines : tho horizontal tension liero also remaining con- 

 stant throughout the length of the arch, while tlie vortical 

 load continually varies. 



This proves tho truth of what has been before stated, 

 that if an arch be truly eipiilibrated in construction, it will 

 be equally true in suspension, and that every constructed 

 arch may be considered to liavo its suspended counterpart, 

 in which the forces in every part are precisely the same 

 both in energy and in direction, and that reasonings based 

 upon tho one will always and equally apply to tho other. 



Having now learnt what we havo from Diagram 1, which 

 represents the simplest possible form of arch, lot us proceed 

 to learn what we can from Diagram 2. 



We may hero suppose that we have an equilibrated arch 

 of six rods or stones, A B, BO, CD, D E, EP, and F Ct, 

 hetwoen tlio abutments A and G, supporting a horizontal 



roadway TR; that those rods meet at the angles B, C, D, 

 E, and F, of whicli D is tho central, and is also the middle 

 point in the arch ; and that AT, B 8, etc., are vertical lines 

 from tho respective angles up to that road line, D being 

 the central of these lines, and its length, termed tho " crown 

 thickness " of the bridge. These lines must of course repre- 

 sent tho vortical loads on each angle, if all the material 



