100 



TITK ARCir. 



at each anglo being exactly equalized ; that is, tliat tlie rod 

 AB leans or thrusts against BC with a force which BC 

 resists with perfect accuracy, because tliey are balanced ; for 

 if the thrust of either were in the slightest degree altered, 

 the equililirium would bo destroyed and tlio arcdi would 

 tumble down at once. 



In this diagram B e and 0/ are supposed to be vortical 

 lines, and the other lines Ce parallel to A B, eg to B C, 

 B/to CD, and//i to B C, 



Now, in the parallelogram gJiGe wo have a parallelo- 

 gram of forces. If the lengths of the linos ;/ B and BC 

 represent the opposing thrusts against the jjoiiit 1? in those 

 directions, then B e must represent the vertical load on B 

 (or suspended from B, which is^ the same thing). Also at 

 the point C, BC and Gh represent the opposing thrusts in 

 those directions, and C/ the vertical load on C ; and it is 

 evident to the eye, from the comparative acuteness of the 

 angle C, that there is a greater vertical load on C tlian 

 onB. 



T;ot US draw the horizontal lines B k and h I perpendicu- 

 lar to C/, then it is evident, with regard to the thrust B C 

 against the angle B, it is made tip of the horizontal thrust 

 B k combined with the downward vortical thrust C k. 

 Again, the thrust of C 7i against h (or its continuation to D) 

 is composed of the horizontal thrust I h and the vortical 

 C I ; but B A; is equal to I h, because the two triangles are 

 equal ; therefore B k and C I, which represent the hori- 

 zontal thrust, are the same in both cases. In like manner, 

 if from the points C and g, perpendiculars were drawn to 

 the line B e produced, the horizontal thrust there would be 

 also equal to B k. We see, therefore, that the horizontal 

 thrust is the same in all three sections of the arch, notwith-- 

 standing the over-varying vertical loads; this principle iS 



