To design an unreinforced sewer arch by the vouissoir method, a desired arch is drawn to scale in apparently good proportions for the loadings anticipated. The arch is then divided into any number of sections of equal or approximately equal length called vouissoirs, and the line of action of the resultant load, including the weight of the vouissoir is drawn above each vouissoir as shown in Fig. 82. The forces are assumed to act as shown in the figure. In symmetrically loaded sewer arches there is no vertical reaction at the crown. The resultant _R_ is assumed to act at the lower middle third of the skewback, which is the inclined joint between the arch and the abutment. The upper horizontal force _H_ is assumed to act at the upper middle third of the middle or crown section. The magnitude of _H_ is computed by equating the sum of the moments of all forces about the point of application of _R_ at the skewback to zero, and solving. The force polygon is then drawn as shown in Fig. 83, and the equilibrium polygon is completed in Fig. 82 with its rays parallel to the corresponding strings drawn from the end of _H_ as origin in Fig. 83. If the equilibrium polygon line, called the resistance line, lies wholly within the middle third of each vouissoir, the arch is satisfactory to support the assumed load without reinforcement. If any portion of the resistance line lies outside of the middle third, an attempt should be made to find a resistance line which lies wholly within the middle third. The true resistance line is that which deviates the least from the neutral axis of the arch. To approximate more nearly the true resistance line find two points at which the resistance line already drawn deviates the most from the neutral axis of the arch. Select points _M_ and _N_ on these joints, _M_ being nearer the crown than _N_. Then let _W_{1}_ and _W_{2}_ be the sum of all the loads between the crown and _M_ and _N_ respectively, _y_ represent the vertical distance from the crown to _N_, and _y′_ represent the vertical distance between _M_ and _N_, and _x_{1}_ and _x_{2}_ represent the horizontal distance from _W_{1}_ and _W_{2}_ to _M_ and _N_ respectively. Then the horizontal thrust, _H_, and _a_, the distance from the crown to the point of application of _H_, are, _H_ = ((_W_{2}x_{2}_ − _W_{1}x_{1})_)⁄_y′_, _a_ = _y_ − (_W_{2}x_{2}_)⁄_H_.[72]