- #1
alkaspeltzar
- 354
- 37
- TL;DR Summary
- Trying to determine what torque a user has to provide to keep the head of an auger from spinning.
Trying to determine the reaction torque a person would have to supply to keep the auger head from spinning when the ice auger is drilling a hole
So I am an ice fisherman and I got thinking, when you drill a hole with an ice auger, the head always wants to turn opposite of the auger bit. You have to apply a torque to overcome the reaction torque. What is that reaction torque? Assuming steady state, auger is already drilling and up to speed, no acceleration, is it equal and opposite torque of the auger bit?
To figure this out, I drew myself a simplified problem with motor and input gear direct drive, say 1inch radius. Then the driven gear is 5 inch radius. The torque applied from the motor is 10lb-inches, created by a 5 lb couple. Auger bit is firmly on ice cutting.
If I sum the torques about B I get:
Tmotor=10 in-lbs, Treaction on driven gear=50in-lbs, Tmotorreaction= -10 in lbs, Tdrivengear=-50 in lbs. The motor torques cancel out. Leaving there the 50 in lbs on the auger bit, and a 50 in lbs torque back on the housing/user. Is this correct?
Also, what force/torque keeps the motor from spinning CCW about A? I assume the 10lbs force at the driven gear at the distance of 1 inch balance out the reaction torque there?
Thank you for the help
So I am an ice fisherman and I got thinking, when you drill a hole with an ice auger, the head always wants to turn opposite of the auger bit. You have to apply a torque to overcome the reaction torque. What is that reaction torque? Assuming steady state, auger is already drilling and up to speed, no acceleration, is it equal and opposite torque of the auger bit?
To figure this out, I drew myself a simplified problem with motor and input gear direct drive, say 1inch radius. Then the driven gear is 5 inch radius. The torque applied from the motor is 10lb-inches, created by a 5 lb couple. Auger bit is firmly on ice cutting.
If I sum the torques about B I get:
Tmotor=10 in-lbs, Treaction on driven gear=50in-lbs, Tmotorreaction= -10 in lbs, Tdrivengear=-50 in lbs. The motor torques cancel out. Leaving there the 50 in lbs on the auger bit, and a 50 in lbs torque back on the housing/user. Is this correct?
Also, what force/torque keeps the motor from spinning CCW about A? I assume the 10lbs force at the driven gear at the distance of 1 inch balance out the reaction torque there?
Thank you for the help