Question Hi, I taught physics in high school this year, though my degree is in biochemistry.
I took the students to the local grist mill to study the mechanics.
The wheel has 60 buckets each which can hold 200 lbs water. The wheel is 18 feet in diameter. To calculate the hp, I assumed that each bucket moved 18 feet, so each did 200x18' or 3600 ft-lbs work. There were 60 buckets, so a total of 216000 ft-lbs of work was done. The wheel turned at 4.5 rpm, so one rotation took 13.333 sec. I divided the work by 13.33 sec. to get the work/sec, then by 550 ft-lbs to convert to hp, and got 29.5 hp. Was the calculation done correctly.
The next thing we did was to find the torque to start the wheel. 6 buckets had to be filled to start the wheel. Each bucket is 6 degrees from the other (360 deg/60 buckets). Starting at the bucket in the 84 deg. position, with a radius of 9 feet, the lever arm is 9cos(84), or 188 ft-lbs. Repeating the calculation for all 6 buckets led to a total torque of 3,809 ft-lbs.
So it required 3,809 ft-lbs of torque to overcome the static friction of all the wheels in the mill. Knowing the coefficient of friction of bearings (0.004, about--all the gears are steel; the mill was built by Henry Ford) we should be able to calculate the total weight of all the wheels, because F(static)=uF(normal) (u=coefficient of friction). One way to calculate the Fnormal (which corresponds to the weight) is to divide the Fstatic by u.
3,809 divided by .004 is 950,000 lbs or 476 tons, which seems way too high.
But torque is in ft-lbs, so this calc. is probably wrong because the answer is actually 476 ft-tons. The millstone is geared up 24 times. If I assume that 24 represents a lever arm, then I can divide 476 ft-tons by 24 ft, and come out with a normal weight of the waterwheel and all the gears of the mill of about 20 tons, which seems right. What do you think?
Answer Hi Tony: Your first calculation is correct,giving 29.5 hp. Your second calculation is also correct, giving 3,809 ft-lb.
The next step did not go so well. You used a coefficient of friction of .004 which is unrealistically low. 0.16 to 0.2 is more likely. Ref: Machinery's Handbook. Then you try to calculate a frictional torque. But this is a function not only of the normal force and the friction coefficient but of the SHAFT DIAMETER as well. And you confused frictional torque with weight.
