38

Feet.

Let the radius of curvature at the vertex - - = c

the height of the key stone - - - = n

the thickness at any other voussoir - - = l

the breadth of the arch from face to face - - = b

the semi-span of an arch - - - - = y

the versed sine or height of an arch - - = x

lbs. avoird.

the weight of one cubic foot of the materials / _composing the arch - ~ ~ S ‘

the weight of aprismatick column one foot squaredof the given materials standing vertically, >necessary to cause fracture at its base Jthe weight of a prismatick column of the "same'material, having its section equal to onesquare foot, and equal in weight to the ^weight to be placed upon or to be borne byeach superficial foot at the vertexthe angle which any joint makes with a horizontalline = <p

f

w

Since it is shown in Tract 3. page 85, that ( ng + tv) t= (ng-\- tv) c sec. = the tension at the point of'suspen-sion when the breadth is considered as I; by multiply-ing by b, we have the like tension when the breadth is b— (ng -j- w) be sec. <p, hence changing tension into com-pression, and referring this expression to the vertexwhere, p = 0, or sec. <p = 1, we have the absoluteforce of compression at the vertex of such an arch,taking into consideration the effect of incumbentweight, = (ng + w) be.

Also because the compression at the vertex isalways equal to the force of horizontal thrust, wehave the force of the horizontal thrust at any pointin such an arch, taking into consideration the effectof incumbent weight = (ng + w) be.