For example, in _Case 1_, _Example 3_ there are given _n_ = .020, _Q_, and _d_. Find the intersection of the vertical line for _n_ = .020 with the sloping diameter line for _d_ = 6 inches. Project the intersection horizontally to the right to the vertical “diameter” line. Place a straight-edge at this point and at _Q_ = 1.0 on the quantity scale. The required value of _S_ is read at the intersection of the straight-edge and the slope scale and is equal to 0.13. The intersection of the straight-edge in this position with the velocity scale is not the required value of the velocity since the velocity scale is made out for _n_ = .015 and not .020. It is necessary to change the position of the straight-edge so that it may lie on _Q_ equal 1.0 and on _d_ equal 6 inches for _n_ equal .015. The value of _V_ is shown in this position as 5 feet per second. The reverse process for Fig. 15 and 16 is illustrated _by Case 4_, _Example 2_ in which _n_ = .011 and _Q_ and _V_ are also given. When _Q_ and _V_ are given the value of _d_ is fixed independent of all other factors. Therefore the value of _d_ can be read from the scale with _n_ = .015 and is found to be 12 inches. Now find the value of _d_ = 12 inches on the scale for _n_ = .011 and project on to the “diameter” line. Place the straight-edge at this point and at _Q_ = 2. The required slope is read as .0022. Fig. 17 is prepared for the solution of problems in which _n_ = .015 only. For problems in which _n_ has some other value it is necessary to transform the data to equivalent conditions in which _n_ = .015. This is done by means of the conversion factors shown in Fig. 18. The given slope or velocity is multiplied by the proper factor to convert from or to the value of _n_ = .015. For example in _Case 1_, _Example 4_ there are given _n_ = .020, _Q_, and _d_. With _Q_ and _d_ given the value of _V_ can be read from Fig. 17 without conversion. The corresponding value of _S_ for _n_ = .015 is .0065. It is now necessary to use the transformation diagram Fig. 18. The hydraulic radius of the given pipe is one foot. On Fig. 18 at the intersection of the slope line for _R_ = 1.0 foot and _n_ = .020 the value of the factor is read as 1.92. Since the given _n_ is for rougher material than that represented by _n_ = .015 the required slope must be greater than for _n_ = .015 to give the same velocity. It is therefore necessary to multiply .0065 × 1.92 and the required slope is .0125.