As even a very small percentage of fine material is capable of getting in the narrow places and reducing the carrying power of the gravel, it is important that all such matters should be carefully removed by washing before putting the gravel in place. In England and Germany gravel is commonly screened for use in revolving cylinders of wire-cloth of the desired sizes, on which water is freely played from numerous jets, thus securing perfectly clean gravel. In getting gravel for the Lawrence filter, an apparatus was used, in which advantage was taken of the natural slope of the gravel bank to do the work, and the use of power was avoided. The respective grades of gravel obtained were even in size, and reasonably free from fine material, but it was deemed best to wash them with a hose before putting them in the filter. To calculate the frictional resistance of water in passing gravel, we may assume that for the very low velocities which are actually found in filters the quantity of water passing varies directly with the head, which for these velocities is substantially correct, although it would not be true for higher rates, especially with the coarser gravels.[6] In the case of parallel underdrains the friction from the middle point between drains to the drains may be calculated by the formula: Total head = (1/2)[(Rate of filtration × (1/2 distance between drains)^2)/(Average depth of gravel × discharge coefficient)]. The discharge coefficient for any gravel is 1000 times the quantity of water which will pass when _h_/_l_ is 1/1000 expressed in million gallons per acre daily. The approximate values of this coefficient for different-sized gravels are as follows: VALUES OF DISCHARGE COEFFICIENT. For gravel with effective size 5 mm _c_ = 23,000 For gravel with effective size10 mm _c_ = 65,000 For gravel with effective size15 mm _c_ = 110,000 For gravel with effective size20 mm _c_ = 160,000 For gravel with effective size25 mm _c_ = 230,000 For gravel with effective size30 mm _c_ = 300,000 For gravel with effective size35 mm _c_ = 390,000 For gravel with effective size40 mm _c_ = 480,000 Example: What is the loss of head in the gravel at a rate of filtration of 2 million gallons per acre daily, with underdrains 20 feet apart, where the supporting gravel has an effective size of 35 millimeters, and is uniformly 1 ft. deep? Total head = (1/2)[(2 × 10^2)/(1 × 390,000)] = .000256 ft. The total friction would be the same with the same average depth of gravel whether it was uniformly 1 foot deep, or decreasing from 1.5 at the drains to 0.5 in the middle, or from 2.0 to 0. The reverse case with the gravel layer thicker in the middle than at the drains does not occur and need not be discussed.