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.
In _Case 6_, _Example 1_ there are given _n_ = .018, _d_, and _S_. The remaining factors are to be solved by Fig. 17. Solve first as though _n_ = .015 in order to find an approximate value of _d_ or _R_. In this case it is evident that _d_ is greater than 57 inches. The value of _R_ is therefore about 1.25. Referring to Fig. 18 the conversion factor for the slope for _n_ = .018 is about 1.52. Since the given slope for _n_ = .018 is .001, for an equal velocity and for _n_ = .015 the slope should be less. Therefore in reading Fig. 17 it is necessary to use a slope of .001⁄1.52 = .00066. The diameter is found to be about 80 inches. Since this is nearer to the correct diameter the value of the conversion factor must be corrected for this approximation. The hydraulic radius for an 80 inch pipe is 1.67 feet, and the conversion factor from Fig. 18 is about 1.48. The slope for _n_ = .015 should be therefore .001⁄1.48 = .000675 and from Fig. 17 the required diameter and quantity are read as 80 inches and 185 second-feet, respectively.
_If n is not given_ but must be solved for, the solution on Fig. 15 and 16 is relatively simple. The desired value of _n_ is read at the intersection of the sloping diameter line representing the known diameter and the horizontal projection of the intersection of the straight-edge with the vertical “diameter” line.
For example in _Case 7_, _Example 1_ there are given _Q_, _d_, and _S_. Lay the straight-edge on the given values of _Q_ = 3 and _S_ = .002. At the point where the straight-edge crosses the vertical “diameter” line project a horizontal line to the sloping diameter line for _d_ = 18 inches. The vertical line passing through this point represents a value of _n_ = .019. In order to find the value of _V_ lay the straight-edge on _Q_ = 3 and _d_ = 18 inches for _n_ = .015. The value of _V_ is read as 1.7.