I am trying to figure out the torque needed for an auger, but I seem to run into a wall. The auger is 6.00 inches in diameter and 44 feet long and on a 17% incline. The weight of the product going through the auger is 29lbs per cubic foot. So I calculated the cubic feet of the auger, 3.14 * r^2*H = 3.14*0.5^2*44=34.57 cubic feet. Then 34.57ft^3*29lbs*ft^3 = 1002lbs. I am having a difficult time figuring out how to calculate the torque needed, even if it is a conservative value to know how large of a gearbox we would need with a 40:1 gear ratio and if we can use a 1HP motor on it. The only other information that I have is that the flow rate is 1200 bushels per hour. Any help is greatly appreciated.