Reaction Torque from an Ice Auger

[alkaspeltzar]

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

 

[russ_watters]

Assuming steady state, auger is already drilling and up to speed, no acceleration, is it equal and opposite torque of the auger bit?

Yes, but only on average. From my vast experience of watching one get used once, 30 years ago, they tend to "catch" on the ice and judder like a hammer drill, so the torque is not constant.

 

[alkaspeltzar]

Yes, but only on average. From my vast experience of watching one get used once, 30 years ago, they tend to "catch" on the ice and judder like a hammer drill, so the torque is not constant.

Only thing I don't understand, is what force and lever arm creates the torque that keeps the motor from spinning?

So as the motor applies torque to the drive gear, that would make the motor revolve around A, but that doesn't happen. What torque am I not seeing? See picture below. You can assume the motor and gear share same footprint/diameter and that their are directly connected via shaft. Motor ultimately is connected to the auger frame.

 

[Baluncore]

What torque am I not seeing?

The operator applies a torque to the auger frame that is equal and opposite to the torque of the auger in the ice.

 

[alkaspeltzar]

The operator applies a torque to the auger frame that is equal and opposite to the torque of the auger in the ice.

I understand that, but I am curious what torque overcomes internally the motor reaction torque so it doesn't want to spin around itslef. See above

 

[Twigg]

alkaspeltzar, the motor's stator is rigidly connected to the frame that the user is torquing. The rotor is on bearings and connected to the drive gear.

Edit: oops, I should've paid more attention to the pictures and noticed it's not battery powered but gas powered. Baluncore covered it correctly below.

 

[Baluncore]

The auger shaft is connected to the output shaft of the gearbox.
The gears rotate in a gearbox that is bolted to the motor, which is held by the operator.
The side forces on the shaft bearings (crank and gears) provide the couple that becomes the torque.
The gas pressure generates a force between the piston and cylinder head that is converted to a torque by the crank and gearbox.

 

[russ_watters]

Only thing I don't understand, is what force and lever arm creates the torque that keeps the motor from spinning?

Torque is torque. It's a conserved quantity. The lever arm is irrelevant if you already know the torque.

I understand that, but I am curious what torque overcomes internally the motor reaction torque so it doesn't want to spin around itslef.
[separate]
So as the motor applies torque to the drive gear, that would make the motor revolve around A, but that doesn't happen. What torque am I not seeing?

I guess I'm not sure what you are saying here. The motor (stator) doesn't spin around the rotor because you are holding onto it, applying an anti-torque to it, to hold it in place.

 

[alkaspeltzar]

Torque is torque. It's a conserved quantity. The lever arm is irrelevant if you already know the torque.

I guess I'm not sure what you are saying here. The motor (stator) doesn't spin around the rotor because you are holding onto it, applying an anti-torque to it, to hold it in place.

I was trying to see how the torque balances all the way thru, such that the motor doesn’t move, but the whole engine/gearbox/frame relative to the auger.I think i figured it out, as i was counting the motor torque twice and confusing myself.

During steady state, all the torque provided by the motor to the input gear is canceled by the matting gear. That force causes the whole head to have a torque of 60 in-lbs in my problem. But the auger motor also received a reaction torque of 10inch pound, making the overall remain torque 50 in lbs.This proves that the torque felt by the operator is equal to the auger drill torque. And that all internal forces/torque balance out.

 

[Dr.D]

In order to understand the way the torque is transferred through the machine, draw a complete FBD for each element in the power transmission path. This will show you exactly what balances what.

 

[alkaspeltzar]

In order to understand the way the torque is transferred through the machine, draw a complete FBD for each element in the power transmission path. This will show you exactly what balances what.

essentially i did that and put it together. but it was a lot more math and makes asking thequestion confusion LOL

 

[Dr.D]

Would it be better to just attribute the total reaction to the fairies? That seems like the best alternative if the FBD approach is unsatifactory.

 

[alkaspeltzar]

Would it be better to just attribute the total reaction to the fairies? That seems like the best alternative if the FBD approach is unsatifactory.

probably, but my brain likes to make things harder. THat is why i come here, you guys are a good double check

 

[sophiecentaur]

Would it be better to just attribute the total reaction to the fairies? That seems like the best alternative if the FBD approach is unsatifactory.

Not all of us are familiar with using a FBD and it's very easy to leave out the relevant forces.

What torque am I not seeing?

The only torque that you need to worry about is the torque that could cause the mount to spin round on the ice, with you hanging onto it. Your feet are the anchor. The force your feet need to apply will be reduced by the ratio of the auger radius to (the radius of those handles plus the distance from the handles to your feet).
I would imagine that the radius of the auger and the handle distance will have been selected (designed) to get it right. If the auger suddenly sticks then you would either stall / slip a clutch on the drive or be spun round. How slippery is the ice when you start the operation?

 

[alkaspeltzar]

All, I did draw a picture, FBD, with all the relevant forces and torques.

If you look above you'll see originally I showed how the motor creates a torque on drive gear A, which then thru the contact forces causes a torque on auger gear B.

MY questions was how does the overall operator feel only 50lb-inches of torque to keep the entire assembly from spinning(based on my easy numbers) and trying to understand all the internal forces/torques thru the gear.

Correct me if i am wrong, but as drive gear A produces 10 lbs of force at 5 inches on drive gear B, that creates 50lb-inches of torque to the auger. This causes a reaction force of 10 lbs back on drive gear A, which cancels the input torque on A, plus causes the force couple to be unbalanced and if you sum the forces about B, creates a torque of 60lb inch. HOWEVER, this is all connected to the chasis at the pivot points and the motor applies an opposite reaction torque of 10lb-inches, so overall, after everything cancels out, there is 50 lbs-inches of torque that the operator has to provide to make the drilling effective.

Essentially the engine torque cancels out, leave the output to the auger of 50lb-inches. And with gear trains, if there is 50lb-inches going out, the operator HAS to provide the equal torque.

 

[berkeman]

Thread closed temporarily for Moderation...

 

[berkeman]

An off-topic conversation has been deleted. Thread is re-opened.

 

[alkaspeltzar]

An off-topic conversation has been deleted. Thread is re-opened.

Berkeman, can you review post #19 and tell me if you agree? I see you liked my previous comments, i tried to be more clear and am just looking for verification at this point. Thank you

 

[berkeman]

Berkeman, can you review post #19 and tell me if you agree? I see you liked my previous comments, i tried to be more clear and am just looking for verification at this point. Thank you

I think you mean post #15 now that the off-topic posts have been deleted. I'll let others who are more familiar with your thread comment for now.