# Total torque of a multi motor system

• Pinon1977

#### Pinon1977

TL;DR Summary
Trying to estimate the total torque of a multimotor system. This is a theoretical system, but one which I would like to make reality.
So I have four motors, each capable of producing 1 newton meters of torque. They are all running into a singular drive axle at 90degree offsets. There's no gearbox exchange or power transfer medium. It's axle to axle engagement with a pinon gear. Disregarding any sort of losses that may be encountered from the gear mesh, how can I estimate the total torque being witnessed at the primary drive axle? My initial guess would be 4 Nm.

My initial guess would be 4 Nm.
That assumes mitre gears with a 1:1 ratio.

TL;DR Summary: Trying to estimate the total torque of a multimotor system. This is a theoretical system, but one which I would like to make reality.

So I have four motors, each capable of producing 1 newton meters of torque. They are all running into a singular drive axle at 90degree offsets. There's no gearbox exchange or power transfer medium. It's axle to axle engagement with a pinon gear. Disregarding any sort of losses that may be encountered from the gear mesh, how can I estimate the total torque being witnessed at the primary drive axle? My initial guess would be 4 Nm.

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So I have four motors, each capable of producing 1 newton meters of torque. Disregarding any sort of losses that may be encountered from the gear mesh, how can I estimate the total torque being witnessed at the primary drive axle? My initial guess would be 4 Nm.
I believe that your ideal guess is correct.

Because of that arrangement, all the motors and pinions and central shaft will be forced to turn at the same rotational speed as of any one of the single motors.

Each motor will be consuming the same amount of electricity than when working alone; therefore, the input work will be four times greater than if using a single motor.

The output work will need to be similar to that of a single motor times four.
The only way to have that for the same rotational speed is by having increased the output torque four times.

Energy = work = force x distance = torque x rotated angle

Power = work/time = force x velocity = torque x rotational velocity

https://en.wikipedia.org/wiki/Work_(physics)#Units

• Pinon1977
This is a theoretical system, but one which I would like to make reality.
Stall torques (or torques for a forced stable speed) might be additive, but in a real system you'll meet with complicated speed-torque-load (-power) dependencies

You would fare lot better without complications like multiple motors.

• Pinon1977
For any arrangement of motors and transmissions you can think of, this will be true about power ##P##:
$$P_{out} = P_{in}$$
Having 4 motors as inputs and a single drive axle, you get:
$$P_{out} = P_{in\ 1} + P_{in\ 2} + P_{in\ 3} + P_{in\ 4}$$
$$T_{out}\omega_{out} = T_{in\ 1}\omega_{in\ 1} + T_{in\ 2}\omega_{in\ 2} + T_{in\ 3}\omega_{in\ 3} + T_{in\ 4}\omega_{in\ 4}$$
$$T_{out} = T_{in\ 1}\frac{\omega_{in\ 1}}{\omega_{out}} + T_{in\ 2}\frac{\omega_{in\ 2}}{\omega_{out}} + T_{in\ 3}\frac{\omega_{in\ 3}}{\omega_{out}} + T_{in\ 4}\frac{\omega_{in\ 4}}{\omega_{out}}$$
Where ##T## is the torque and ##\omega## is the angular velocity (or RPM).

Instead of the mechanical power (##T\omega##), you could even use the sum of electrical power going in (voltage X current of an electric motor for example) or the "combustion" power (mass flow rate X heat of combustion for a specific fuel) as the input power - or even a mix of all of those. Of course, the power losses (usually through heat) have to be included in the equation; and the closer you are to the "true" input (e.g. the quantity of fuel or electricity going into a motor), the more losses you may have.

It's the principle of conservation of energy.

• Lnewqban
For any arrangement of motors and transmissions you can think of, this will be true about power ##P##:
$$P_{out} = P_{in}$$
Having 4 motors as inputs and a single drive axle, you get:
$$P_{out} = P_{in\ 1} + P_{in\ 2} + P_{in\ 3} + P_{in\ 4}$$
$$T_{out}\omega_{out} = T_{in\ 1}\omega_{in\ 1} + T_{in\ 2}\omega_{in\ 2} + T_{in\ 3}\omega_{in\ 3} + T_{in\ 4}\omega_{in\ 4}$$
$$T_{out} = T_{in\ 1}\frac{\omega_{in\ 1}}{\omega_{out}} + T_{in\ 2}\frac{\omega_{in\ 2}}{\omega_{out}} + T_{in\ 3}\frac{\omega_{in\ 3}}{\omega_{out}} + T_{in\ 4}\frac{\omega_{in\ 4}}{\omega_{out}}$$
Where ##T## is the torque and ##\omega## is the angular velocity (or RPM).

Instead of the mechanical power (##T\omega##), you could even use the sum of electrical power going in (voltage X current of an electric motor for example) or the "combustion" power (mass flow rate X heat of combustion for a specific fuel) as the input power - or even a mix of all of those. Of course, the power losses (usually through heat) have to be included in the equation; and the closer you are to the "true" input (e.g. the quantity of fuel or electricity going into a motor), the more losses you may have.

It's the principle of conservation of energy.
Thanks, jack action. You have been most helpful and thorough in your explanation. Gold star !!!!

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• berkeman
I believe that your ideal guess is correct.

Because of that arrangement, all the motors and pinions and central shaft will be forced to turn at the same rotational speed as of any one of the single motors.

Each motor will be consuming the same amount of electricity than when working alone; therefore, the input work will be four times greater than if using a single motor.

The output work will need to be similar to that of a single motor times four.
The only way to have that for the same rotational speed is by having increased the output torque four times.

Energy = work = force x distance = torque x rotated angle

Power = work/time = force x velocity = torque x rotational velocity

https://en.wikipedia.org/wiki/Work_(physics)#Units
Thank you sir! Your input is greatly appreciated.

• Lnewqban
No offense, berkman, but I can always count on you to be the Negative Nancy when I ask any sort of question. That being said, there are several differences; consequently, the most notable of which is this particular question dealt with overall power, not efficiency. It also did not utilize any sort of gearbox or power transfer medium other than the bevel gears. This is also a theoretical system not a real live system. Need I continue?

• weirdoguy and russ_watters
Sir, I realize my questions may seem rather remedial to you guys, but there is a reason why I am asking them. More often than not, the question is a rhetorical question and I just need validation of my presumption.

I value y'all's opinion tremendously. That's why I continue to post questions to this site. Even amidst all the sarcasm and shots being fired, I have thick enough skin to understand that a lot of this seems elementary to you guys, and you really mean no harm in your comments.

Thanks and keep up the good work.

• weirdoguy
Sir, I realize my questions may seem rather remedial to you guys, but there is a reason why I am asking them. More often than not, the question is a rhetorical question and I just need validation of my presumption.

I value y'all's opinion tremendously. That's why I continue to post questions to this site. Even amidst all the sarcasm and shots being fired, I have thick enough skin to understand that a lot of this seems elementary to you guys, and you really mean no harm in your comments.

Thanks and keep up the good work.
Just keep in mind that we don't allow multiple thread starts on the same question. I'll look again at your past thread(s) to see if they should be merged with this new thread start...

Sir, I realize my questions may seem rather remedial to you guys, but there is a reason why I am asking them. More often than not, the question is a rhetorical question and I just need validation of my presumption.

I value y'all's opinion tremendously. That's why I continue to post questions to this site. Even amidst all the sarcasm and shots being fired, I have thick enough skin to understand that a lot of this seems elementary to you guys, and you really mean no harm in your comments.

Thanks and keep up the good work.
This concerns me. You clearly are not giving us the full story while are asking questions which at face value have trivially obvious answers. To me this makes the questions pointless and possibly feeding some sort of false premise (which, again, you aren't telling us). This makes us uncomfortable and probably doesn't actually help you - you just think it does.

Your prior thread was just plain bad/wrong. This one is trivially obvious/pointless. Thinking they are guiding you in a positive direction is worrisome.

• weirdoguy, jrmichler, berkeman and 2 others