Question Here is a question I recently sent you followed by an answer from another source; Do you agree with the conclusion?
QUESTION: Thank you for assisting a non engineer with a problem.
10,000 pounds of thrust is moving in a vertical path at the rate of one foot per
second.
This is attached to a horizontal shaft (size not determined) by what ever means
necessary to transfer the maximum energy.
What is the maximum foot pounds the shaft can deliver (by whatever combination
of gears or other devices) if the desired RPM is between 50 and 100 RPM.
What combination of gears or other devices would be needed
Is there any way to increase the torque by gearing?
Given,
Thrust = 10000 lb
Speed of thrust = 1 foot/sec
Desired RPM = 50 – 100
Target: To maximize torque.
The thrust is able to deliver a power, which is constant. The power delivered by the thrust is calculated as follows:
Power from thrust = Thrust x Speed of thrust = 10000 lb x 1 foot/sec = 10000 lb-foot/sec
The following arrangement of shaft can transfer the power to rotational energy
The thrust transfers the power to the shaft, which in turn gets transferred into rotational energy by the bearing. The shaft rotates about the bearing at a speed calculated below
Assuming the energy is transferred to the shaft without any loss.
Power received by shaft = power transmitted by thrust
i.e. Power received by shaft = 10000 lb-foot/sec
The power of the shaft is expressed as,
Shaft power = Shaft speed x Torque
The shaft power is constant, as calculated above. To get maximum torque, the speed should be minimum. This is visible from the above expression.
Hence taking the minimum speed, i.e. 50 rpm
Shaft speed = 50 RPM.
From the above expression,
Shaft power = Shaft speed x Torque
i.e. 10000 lb-foot/sec= 50 RPM x Torque
i.e. 10000 lb-foot/sec = (50x2x3.14/60) rad/sec x Torque
i.e. Torque = (10000 x 60) / (50 x 2 x 3.14)
i.e. Torque = 1910.83 lb-foot
This is the maximum amount of torque generated using this system.
DO YOU AGREE WITH THIS CONCLUSION?
Answer Your solution is correct. You also ask what gearing should be used. This depends on the rpm of the driving motor. If this motor runs at 1720 rpm, the gear ratio needed is 34.40:1. This is an awkward number. For worm gears, the ratio must be a whole number. So you must adjust the torque to make this possible.
I looked a bit further into this problem. At 50 rpm, the radius from the center of the shaft to the center of the cable holding the load is 1.592 inches. This is a reasonable size to hold a 10,000 pound load and 1.909.85 ft lb torque. The cable must hold 10,000 pounds, so multiple cables must be used. Multistrand steell cables come to mind.
