Commencing at the point 0, we draw the segments of the trial equilibrium polygon parallel with the rays in the force diagram which run from the point O1 ,to the load line, and obtain the trial equilibrium polygon OB1 By drawing from O1 the line o1n parallel to the line OB1 ,we obtain the point n on the load line, from which we draw an indefinite horizontal line which will be the locus of the pole of the true equilibrium polygon. A vertical from the point o1 intersects the horizontal from n in the point o2, and this would be the pole of a trial equilibrium polygon whose closing line is horizontal, and whose vertical ordinates are equal to those of the corresponding vertical ordinates of the trial equilibrium polygon 0B1 It is only necessary to find the proper ratio by which these several ordinates should be multiplied, in order to find the corresponding ordinates of the special equilibrium polygon. It is also necessary to shift the entire trial equilibrium polygon up or down, so that the line v'"m"' which corresponds to it shall coincide with the line vm which has already been drawn. The special line v'"m'" corresponding to this trial equilibrium polygon, is found by satisfying Equation 56; but since dx is in this case a constant, it is found by determining the average value of z' ", which is the distance from any point in the trial equilibrium polygon to the proper position of the line v'"m"r corresponding to this equilibrium polygon. If the trial equilibrium polygon had been again redrawn with O2 as a pole, it should terminate in the point B; but it is practically unnecessary to do this, since we may draw the line vm' parallel to 0B1 , and measure the distance from vm' down to the various points of the trial equilibrium polygon 0B1 These various distances in the column headed z'" are as given in the accompanying tabular form (page 414).

We have here the rather unusual case that the trial equilibrium polygon is entirely below the line vm'; and the ordinates are all negative, instead of being partially positive and partially negative. In any case, the algebraic sum should be taken, which should be divided by the number of ordinates. In this case we find that the mean value of z'" is - 3.44. Drawing the line v'"m'" parallel to vm', and at a vertical distance of 3.44 feet below it, we find the position of the vm line for the trial equilibrium polygon 0B1 This line intersects the trial equilibrium polygon at the points h" and k". The student should note that, it is a mere accidental coincidence that the point k" comes almost exactly on the intrados. By drawing verticals from h" and k" to the line vm, we obtain the points h and k, which will be two points in the true equilibrium polygon.