Simple gear pair: efficiency and output torque

[liao]

Hi. I have a question about output torque and efficiency of a simple gear pair as shown on the picture. So, I have a pinion and a gear. I give an input torque Tp in the clockwise direction. Therefore, the pinion will rotate with ωp angular velocity in clockwise and the gear ωg in counter-clockwise. Then, I apply a load Tl in the opposite direction of the gear motion (clockwise direction). The gear is still rotating in the counter-clockwise. I neglect the friction between the gear mesh, but I consider the bearing friction both in pinion cp*ωp and gear cg*ωg. The moments of inertia of the pinion and the gear are Ip and Ig, respectively. My question is what the output torque To is because I want to find the efficiency of this gear pair. Thank you very much.

 

[liao]

View attachment 207233
Hi. I have a question about output torque and efficiency of a simple gear pair as shown on the picture. So, I have a pinion and a gear. I give an input torque Tp in the clockwise direction. Therefore, the pinion will rotate with ωp angular velocity in clockwise and the gear ωg in counter-clockwise. Then, I apply a load Tl in the opposite direction of the gear motion (clockwise direction). The gear is still rotating in the counter-clockwise. I neglect the friction between the gear mesh, but I consider the bearing friction both in pinion cp*ωp and gear cg*ωg. The moments of inertia of the pinion and the gear are Ip and Ig, respectively. My question is what the output torque To is because I want to find the efficiency of this gear pair. Thank you very much.

@CWatters if To = Tl, I would not get a correct efficiency η.
η = Po/Pi = To ωg/(Tp ωp) = Tl ωg/(Tp ωp)

we know that rg ωg = rp ωp and ωg/ωp = rp/rg, so

η = Tl rp/(Tp rg)

In this equation, we can see that Tl/Tp has to be rg/rp in order to have η = 1. If I don't have Tl/Tp close to rg/rp, the efficiency could be much less than 1, for example 0.2, which doesn't make sense.

 

[CWatters]

In this equation, we can see that Tl/Tp has to be rg/rp in order to have η = 1. If I don't have Tl/Tp close to rg/rp, the efficiency could be much less than 1, for example 0.2, which doesn't make sense.

If Tl <> TO the system accelerates (decelerates) putting energy into (removing energy from) the angular momentum of the gears. Load also takes more/less power. That's where the missing energy goes (comes from) if n <> 1.

What happens if Tl >> TO ? Which end is driving which?

 

[liao]

If Tl <> TO the system accelerates (decelerates) putting energy into (removing energy from) the angular momentum of the gears. Load also takes more/less power. That's where the missing energy goes (comes from) if n <> 1.

What happens if Tl >> TO ? Which end is driving which?

Thanks again for your reply. If Tl >> To, I believe the gear will drive the pinion in the load direction. However, I'm only interested in the case where the load is small such that the pinion still drives the gear. Just to confirm, did you think that my analysis on the efficiency when To = Tl correct?

 

[CWatters]

did you think that my analysis on the efficiency when To = Tl correct?

Yes.
.

 

[CWatters]

Regarding friction..

Without friction
To = Tp * rg/rp

With friction in the bearings..
To = ((Tp - cp*ωp) * rg/rp) - cg*ωg

 

[liao]

Regarding friction..

Without friction
To = Tp * rg/rp

With friction in the bearings..
To = ((Tp - cp*ωp) * rg/rp) - cg*ωg

Thanks for your reply. I simulated your To with friction in the bearings in Simulink, but To is incorrect. Here are the results. So, I drive the pinion for a SFUDS (simplified federal urban driving schedule). rg/rp =3. The first figure is the angular velocities of both pinion and gear. We can see that the pinion turns 3 times faster than the gear. The second figure is the torque. Let me know your thoughts. Thanks again. I really appreciate your help! :)

 

[CWatters]

Sorry I've never used Simulink. Clearly something is wrong as you are applying an input torque and getting no (or very little) output torque. Is there a load?

 

[Nidum]

The problem is over defined for the steady state case and under defined for the transient case .

Just for starters which case is it that are you looking at ?

In any case I strongly suggest that you use simple hand calculations for this problem rather than Simulink .

 

[liao]

Sorry I've never used Simulink. Clearly something is wrong as you are applying an input torque and getting no (or very little) output torque. Is there a load?

Yes, there is. Basically I derived the equation of motion of the simple gear by Lagrange's equation, then find the time response. I have tried three other options for To. Could you check it on the attached brief write-up? Thanks a lot CWatters!

 

[liao]

The problem is over defined for the steady state case and under defined for the transient case .

Just for starters which case is it that are you looking at ?

In any case I strongly suggest that you use simple hand calculations for this problem rather than Simulink .

Hi Nidum. Thanks for your reply. Sorry, I should've said that I'm looking at the transient case. Why is it under defined?

I've done simple hand calculations. I use Simulink to validate it.

 

[Nidum]

I don't want to back track through what you have done so far . We'll start afresh using proper engineering principles .

So action the first : Show me two free body diagrams - one for each gear . Show all relevant information on each diagram .

 

[Nidum]

-- and use some good sketching software for the diagrams if you can .

 

[liao]

I don't want to back track through what you have done so far . We'll start afresh using proper engineering principles .

So action the first : Show me two free body diagrams - one for each gear . Show all relevant information on each diagram .

 

[liao]

-- and use some good sketching software for the diagrams if you can .

sorry, I don't have experience of sketching software for these diagrams. I will learn it though. Thanks.

 

[Nidum]

The inertial reaction torques are missing - the gears have moments of inertia and they are accelerating ?

Best not to mix tangential forces and torques . Use only torques when you re-draw the diagrams .

 

[liao]

The inertial reaction torques are missing - the gears have moments of inertia and they are accelerating ?

Best not to mix tangential forces and torques . Use only torques when you re-draw the diagrams .

Yes, the gears have moments of inertia and they are accelerating. Please see the revised diagram, I hope this one is clearer. Tr is the reaction torque.

 

[Nidum]

The reaction torques are in general going to be different for the two gears ?

Also we need a way to define the gear ratio

 

[liao]

The reaction torques are in general going to be different for the two gears ?

Also we need a way to define the gear ratio

Question: The reaction torques are the product of the internal constraint force F and the gear/pinion radius, right? if so, yes, they're different due to different radius.

The gear ratio is ωp/ωg = rg/rp

 

[Nidum]

I was referring to the inertial reaction torques as mentioned in my previous post .

Anyway we have what we need now . It's late here so I am signing off for tonight . Talk to you tomorrow .

 

[liao]

I was referring to the inertial reaction torques as mentioned in my previous post .

Anyway we have what we need now . It's late here so I am signing off for tonight . Talk to you tomorrow .

Thanks a lot Nidum. I really appreciate it! Good night! :)

 

[Dr.D]

The inertial reaction torques are missing - the gears have moments of inertia and they are accelerating ?

Best not to mix tangential forces and torques . Use only torques when you re-draw the diagrams .

I'm sure we could argue about the "inertial reaction torques" until the cows come home, but I would never include such in a FBD. They are exactly analogous to m*a terms, and are not real torques (the cross product of a distance with a force) any more than the m*a terms are force. Neither will obey the third law; they have no reaction to be found anywhere in the universe.

On the second point, regarding using torques rather than forces, I disagree here as well. It is force that interact between the two gears, the contact forces between the teeth. A gear does not exert a torque on a second gear, but rather it exerts a force. I would suggest that the FBDs need to show the contact forces and the bearing forces on each gear, as well as any torques applied through connecting shafts.

 

[liao]

I'm sure we could argue about the "inertial reaction torques" until the cows come home, but I would never include such in a FBD. They are exactly analogous to m*a terms, and are not real torques (the cross product of a distance with a force) any more than the m*a terms are force. Neither will obey the third law; they have no reaction to be found anywhere in the universe.

On the second point, regarding using torques rather than forces, I disagree here as well. It is force that interact between the two gears, the contact forces between the teeth. A gear does not exert a torque on a second gear, but rather it exerts a force. I would suggest that the FBDs need to show the contact forces and the bearing forces on each gear, as well as any torques applied through connecting shafts.

Hi Dr.D. Thanks for your reply and suggestions for the FBDs. What do you think the output torque To is?

 

[Nidum]

@liao has now worked out the final solution to this problem .

 

[Dr.D]

Congratulations to Liao. Why does someone else announce this to us all?