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<br />Channel Bank Stabilit <br />T max = (.76)(62.4)(.63)(.076) _ <br />T max = 2.27~`/ftz <br />n = 21(2.27) _ .231 <br />62.4(1.65)(2) <br />Assume L = 8 = 4.35, 3:1 sideslope a = tan"~ = 18.430 <br />B = tan- cos 4.35 <br />2 sin 18.43 = sin 18.4 3 <br />.231(.9) <br />B = tan"~ r .997 <br />3.04 + .316 <br />B = 16.5 / <br />n~ _ .231 / 1 + sin (4.35+16.5 ) <br />l\ 2 <br />n~ _ .156 <br />SF = (cos 18.43)(tan 42) <br />.156 (tan 42)+(sin 18.43)(cos 16.5) <br />SF = .949 (.9) <br />(.156)(.9) + (.316)(.959) <br />SF = 1.92 OK <br />Construct the Channel 10 feet wide and 2.0 feet deep with 3h:1V <br />sides slopes. <br />