we knew the value of the belt angle, A, but this being an unknown quantity we will have to find some equivalent of it in known terms. A =? From the diagram we have 15. = A B°E [equation 5 (a)]. But B and Z° = ? From the triangle KBG we have From the triangle KGb (Gf being drawn at right angles to To find the radius, R, of the directing circle we have As R is supposed to be a constant we may calculate its value by using the ratio and belt angle of the first pair of pulleys; hence we may write Before R can be determined we must know the value of A. From the triangle KIE we have Or since this equation is true for any pair of pulleys, Now, when the two pulleys are equal, then D1 = d1, r1 = 1, and A, 0, in which case the above equation reduces to 22. . R1 R1 = C(a=de) a- =aC - D1 [equation 2 (b)]. 2 R may be either negative or positive, depending upon the relative numerical values of the positive and negative terms in equations 19 and 22. When it is negative then E° will be nega tive, equation 17, and the sign in equation 15 will be positive, as indicated in equation 5. The constant a was determined empirically, and was found to give the best results by giving it two values: the first 0.314, to be used in the equations as long as the belt angle, A, is below 18°; and the second, 0.298, for angles between 18° and 30°. For greater angles than 30° the method was not tested, as it is supposed that a greater angle will very seldom, if ever, be used in practice. Since equation 19 contains the quantity a, it will be necessary to calculate two values for R in any case where the belt angles lie on both sides of 18°. The easiest way to calculate the second value of R (R) is indicated in Fig. 51. When A, is less than 18°, then R1 is represented by the line GS, its center being at G, from which the arc WS is described tangent to HI, the belt line of D and d. R is represented by the line mU. The arc UX, described with this radius from the center m, is tangent to the line US, which is tangent to the first arc, WS, and is drawn at an angle of 18° with the center line EF. All the belt lines whose angles are less than 18° are tangent to the arc WS, and all those greater than 18° are tangent to UX. Now, from the diagram we have 23... R, R1.0152C [equation 2 (d)]. = When A, is greater than 18° then R1 = m U, and R2 = GS, whence 24. . . R1 = R1 + .0152C [equation 2 (c)]. Equation 23 should be used when the belt angle, A, passes the 18° point from a smaller to a greater angle, and equation 24 when it passes 18° from a greater to a smaller angle. Care should be taken, however, to assign to R1 its proper sign, as it may be either positive or negative, as previously explained. When R, and R2 are both positive, the centers G and m will both be outside of the belt line, and the diagram will take the form of Figs. 36, 41, 49, 50. When R and R2 are both negative, the centers will both be inside the belt line, as in Figs. 37, 42 and 44 (see also example 1). When one is positive and the other is negative, then they will lie on opposite sides of the belt line, as shown in Fig. 43. Example 4 illustrates this case. 2 All the formulæ are now complete with the exception of equation 14. This reduces to an indeterminate form when the two pulleys become equal (A = 0 and r = 1), and will give us no result. To provide for this case we can find the proper value of d from the formula for the length of belt. Referring to Fig. 52, the length of an open belt is Substituting these values in the first equation, we have 25. L=2 C′ cos_A ̧+ .01745 d (180 + (r− 1) (90 + 4)) (Equation 8), This equation, 25, will give the length of the belt mathematically correct, the value of A being obtained from equation 20. Solving equation 25 for d we have = L-2 C cos A (equation 9). Making A 0 and r = 1, this equation reduces to |