Page:The Bell System Technical Journal, Volume 1, 1922.pdf/38

38 by their sum, which is equal to one-fourth of the sum minus the square of the difference divided by four times the sum. The correction due to the difference is thus $$\left(G_2 - G_1\right)^2/4\left(G_{12} + G_{CD}\right)$$, as stated.

(7) These determinants are given at the end of the first paragraph of this appendix. These expressions for the direct capacity are of more special interest in the analytical discussion of networks.

(8) Assume that a wire resistance is to be employed and that a sliding contact is to intercept such an amount of resistance that the equivelent conductance will vary directly with the motion of the slider carrying the contact point. Then if the wire is straight and the intercepted portion is of length x and the slider motion is rectilinear and its extent is y the relation which holds between them is xy = constant, the value of the constant depending upon the units employed.

In the paper it is assumed that the total conductance G, the total shifted conductance g, and the resistance of unit length of the slide wire ρ are given; the total lenght of wire L and the portion traversed by the slider S are then calculated. The arc employed, for each half of the slider in Fig. 9, extends equally both ways from the vertex to the points where the values of x and y are $$\left(L \pm S\right)/2$$, on the hyperbola $$xy = \left(L^2 - S^2\right)/4$$. Substituting for L and S the values given in the paper, it will be found that this range of x actually gives the range of conductance $$\left(G \pm g\right)/2$$, as required.

(9) The exact defect in conductance is

(10) At mid-point the total conductance due to the five resistances (R) on each side, taken in parallel, is 2/5R and to give this same conductance an end fringe must have the resistance 2.5R. Assume a parabolic fringe having the resistance (5 - n)$2$ R/10 at the point connected to and  by resistances nR and (10 - n)R. This gives a Y network and by Fig. 1 the equivalent direct conductances are (10 - n)/25R, n/25R, (5 - n)$2$/250R between  respectively. The sum of the first two is constant and the first decreases by equal steps, of 1/25R each, to zero as the second increases. The parabolic fringe, therefore, gives the required conductances.

The total resistance in the chain of ten resistances is 10R, in the fringe 11R, and in the largest single fringe 2.5R. With the complete fringe the total required is (10 + 11) R = 21R; with a single fringe, subdivided as required, only (10 + 2.5) R = 12.5R is required. Compared with 25R, which would be required for one of the conductance steps, these resistances are 21/25 and 1/2.