38 BELL SYSTEM TECHNICAL JOURNAL 



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 (G2 — Gi)V4(Gi2 + G'cd), 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 

 equivalent 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 = con- 

 stant, 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 p are given ; the total length of wire L and the portion traversed 

 by the slider S are then calculated. The arc employed, for each 

 half of the slider of Fig. 9, extends equally both ways from the vertex 

 to the points where the values of x and y are (L ± S)/2, on the hyper- 

 bola xy = {U — 5^)/4, Substituting for L and 5 the values given 

 in the paper, it will be found that this range of x actually gives the 

 range of conductance {G ± g)/2, as required. 



(9) The exact defect in conductance is 



11 r r ( ^ r 



R R-hr RiR-i-r) R^ 



(i-i + 



(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 — nY R/10 at the point 

 connected to C? 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 - ny/250R between =0^, ^(?, d'C re- 

 spectively. The sum of the first two is constant and the first de- 

 creases by equal steps, of l/25i? 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 lOi?, in the 

 fringe Hi?, and in the largest single fringe 2.hR. With the complete 

 fringe the total required is (10 + 11) i? = 2\R\ 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 con- 

 ductance steps, these resistances are 21/25 and 1/2. 



