The relative velocity of the periphery of the wheel to the velocity of the current should be 50 per cent. with curved blades for best effect. The most useful and convenient sizes for power purposes are from 10 to 20 feet, and from 2 to 20 feet wide, although, as before stated, there is scarcely a limit under 100 feet diameter for special purposes. In designing this class of wheels special attention should be given to the concentration and increase of the velocity of the current by wing dams or by the narrowing of shallow streams; always bearing in mind that any increase in the velocity of the current is economy in increased power, as well as in the size and cost of a wheel for a given power. The blades in the smaller size wheels should be 1/4 of the radius in width, and for the larger sizes up to 20 feet, 1/5 to 1/6 of the radius in width and spaced equal to from 1/4 to 1/3 of the radius. They should be completely submerged at the lowest point. For obtaining the horse power of a current wheel, the formula is Area of 1 blade × velocity of the current in ft. per sec. ---------------------------------------------------------- 400 × by the square of difference of velocities of current and wheel periphery = the horse power; or A × V 2 ------ × (V - v) = h. p. 400 [TEX: \frac{A \times V}{400} \times (V - v)^2 = h. p.] in which A equals the area of blade in square feet, V and v velocities of current and wheel periphery respectively, in feet per second. Thus, for example, a wheel 10 feet in diameter with blades 6 feet long and 1 foot in width, running in a stream of 5 feet per second--assuming the wheel to be giving as much power as will reduce its velocity to one half that of the stream--the figures will be 6' × 5' 2 ------- × 2.5 = 0.468 400 [TEX: \frac{6' \times 5'}{400} \times 2.5^2 = 0.468] horse power of the wheel. The total power of the stream due to the area of the blade equals the Square of the velocity of the stream ------------------------------------ × Twice gravity (64.33) volume of water in cubic feet per second × 62.5 (weight of 1 C') = the value or gross effect in pounds falling 1 foot per second. This sum divided by 550 = horse power. Thus, as per last example, 2 5 ------ × 30 × 62.5 64.33 ---------------------- = 1.32 the horse power of the current 550 [TEX: \frac{\frac{5^2}{64.33} \times 30 \times 62.5}{550} = 1.32 \text{ the horse power of the current}] due to the area of the blades of the water wheel. For the efficiency of this class of wheel, with slightly curved and thin blades, divide the horse power of the wheel by the horse power of the current area, equals the percentage of efficiency. As in the last case, 0.468 / 1.32 = 0.35½ per cent. efficiency of the water wheel.