TheMindGarage

Clarification on adder mechanisms - surely I can't be creating free energy?

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I have always believed that an adder sums torque and averages rotational speed. However, I seem to have stumbled across a problem in my understanding. For example, take this hypothetical case (I'm using SI units for this example, not RPM and Ncm which are commonly used for LEGO motors):

Motor A rotates at 2Hz and has 4Nm torque - power is 8W.

Motor B rotates at 4Hz and has 2Nm torque - power is 8W. Total power of the two input motors is 16W.

Adder output rotates at 3Hz and has 6Nm torque - power is 18W.

Where the heck did those extra 2 watts come from? If I was on Youtube, I'd stop here and claim that I've created "free energy", solved the world's problems and defeated the laws of physics. But clearly that isn't the case. So what's going on here? The only thing I think that could explain it is that the torque is not summed and that the output torque is only 5.67Nm instead of 6Nm. Please could someone with more experience in this field clarify?

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I think your calculations are correct in theory, the torques sum up and the speeds are averaged BUT in practice the final speed will be lower than average because of the inevitable friction and energy loss in the mechanism of the adder itself. The adder will slow down the motors by simply working and being driven by them, while the torque should remain unaffected.

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Just now, Sariel said:

I think your calculations are correct in theory, the torques sum up and the speeds are averaged BUT in practice the final speed will be lower than average because of the inevitable friction and energy loss in the mechanism of the adder itself. The adder will slow down the motors by simply working and being driven by them, while the torque should remain unaffected.

I'd have expected the speed to stay equal to the average of the two input speeds (although of course any resistance or friction will slow down both the output speed and the input speeds). Perhaps someone should test this out using more extreme differences in speed and torque.

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Isn't it speed that is averaged rather than frequency? Thus, you'd average 1/2 and 1/4 to get 3/8, or 8/3 Hz. At 6Nm of torque, that'd be 8/3*6 = 16W as expected.

(It'll be less due to friction etc).

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you can't sum everything upward. The system is still based on the inputs - you've got 16W going in - you can't manufacture more. You are losing torque in the gearbox, but in a calculation sense, you're missing the drop in reacted torque through the gearbox. The supplied torque of a motor is dependent on its speed, which will not stay constant at 2 or 4Hz for the same power output.

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4 hours ago, JonathanM said:

Isn't it speed that is averaged rather than frequency? Thus, you'd average 1/2 and 1/4 to get 3/8, or 8/3 Hz. At 6Nm of torque, that'd be 8/3*6 = 16W as expected.

(It'll be less due to friction etc).

+1

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5 hours ago, JonathanM said:

Isn't it speed that is averaged rather than frequency? Thus, you'd average 1/2 and 1/4 to get 3/8, or 8/3 Hz. At 6Nm of torque, that'd be 8/3*6 = 16W as expected.

(It'll be less due to friction etc).

+1

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6 hours ago, JonathanM said:

Isn't it speed that is averaged rather than frequency? Thus, you'd average 1/2 and 1/4 to get 3/8, or 8/3 Hz. At 6Nm of torque, that'd be 8/3*6 = 16W as expected.

(It'll be less due to friction etc).

Speed is frequency. 1Hz = 1 revolution per second. I'm just using SI units to make the power conversion easier.

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1 hour ago, TheMindGarage said:

Speed is frequency. 1Hz = 1 revolution per second. I'm just using SI units to make the power conversion easier.

Speed required for this calculation is the angular velocity and the frequency is the inverse of angular velocity. So angular velocity would be more like 1/Hz.

You have a fundamental error in your calculation. JonathanM is right.

There is something fishy here. Trollphysics an answer??

Most likely answer is that the speed and torque are not summed as is assumed. Probably power and torque are summed and speed is achieved from there.

Edited by Ultimario
Should properly check the problem first.

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35 minutes ago, Ultimario said:

Speed required for this calculation is the angular velocity and the frequency is the inverse of angular velocity. So angular velocity would be more like 1/Hz.

You have a fundamental error in your calculation. JonathanM is right.

But surely revolutions per second is the same as rotational speed? 2Hz is 2 revolutions per second since both have units s^-1. Unless I've done something horribly wrong...

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My version:
An adder is basically two inputs connected to the same output via two sets of gears. As we know, in a gear train torque is transferred inversely proportional to the angular speeds of the gears: T2/T1=W1/W2. The output speed of the adder is obviously an average of the input speeds: W=(W1+W2)/2=3 Hz. The first input transfers the following torque to the output: Ta=T1*W1/W=4*2/3=8/3 Nm. The second input transfers the following torque to the output: Tb=T2*W2/W=2*4/3=8/3 Nm. The output torque is a sum of the torques received by the output from the inputs: T=Ta+Tb=(T1*W1+T2*W2)/W=16/3 Nm=5.33 Nm. The output power is 16 W.

Edited by sheo

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Pretty much all asumptions in the therad are wrong :classic: (in one way or another). In the system, there's equllibrium. You can't have 4 Nm on one input and have 2 Nm on the other input at the same time. They will be equal (assuming 1:1 relation between the input shafts gearing to the differential). So, if you have a 6 Nm load* on the output, you'll have 3+3 on the inputs. 3Nm x 4Hz + 3Nm x 2Hz = 6Nm x 3 Hz. Which is true.

 

*there's no input torque without a load tourque on the output.

 

EDIT: or I'm wrong? No, see next post.

I think the misleading thing that tricks our minds is that we think in terms of gear rations (we are using that most of the time), but it's not the case applicable here.

Anyway, physics was too long ago...

Edited by Lipko

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To support my claim, see this link:

https://www.quora.com/Transmission-mechanics-In-an-open-differential-an-engine-sends-the-same-amount-of-torque-to-both-the-wheels-The-same-amount-of-torque-would-mean-same-speed-How-does-this-differ-from-the-locking-differential

So basically in an open differential (which is the inverse of an adder), the torque is always the same on the two outpus shafts, no matter what their speed is.

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A last note:
Most free energy believers and "inventors" make the similar mistake when dealing with motors and power (for example the guy who could use a few watts motor to drive a 20 kW and clamied to have some 100000% effeciency).

A power of a motor (or generator) is not the power it gives (consumes) at any given time. It's just a rated power (a maximum value which is not damaging the system or the motor/generator). The actual power output of a motor always depends on its load. No load, no power consumption. More load than the rated value, the motor will give that power, but something will burn... Same is true for a generator. If there's no electric load on the generator, you don't need much power to rotate it (you only need power because of the friction).

Okay, enough of my ego...

Edited by Lipko

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5 hours ago, Lipko said:

So, if you have a 6 Nm load* on the output, you'll have 3+3 on the inputs. 3Nm x 4Hz + 3Nm x 2Hz = 6Nm x 3 Hz. Which is true.

Just a crazy thought: judging from the description of the initial task, motor A can be replaced with motor B (with the same power) plus a shameless 2:1 gear reduction (which gives a speed of 2Hz). If your calculations are correct, one of two identical motors has to provide only 1.5Nm torque (3Nm after gear reduction) while the other one has to provide 3Nm torque and the same speed. That's just unfair! :)

Edited by sheo

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23 hours ago, Sariel said:

I think your calculations are correct in theory, the torques sum up and the speeds are averaged BUT in practice the final speed will be lower than average because of the inevitable friction and energy loss in the mechanism of the adder itself. The adder will slow down the motors by simply working and being driven by them, while the torque should remain unaffected.

Theory couldn't be use in case practice do not confirms it:) Imagine that there is no friction and power loss. Even in that case law of conservation of energy should still be working as it is fundamental law of nature.

So, I state that initial calculations of TheMindGarage are not correct.

 

21 hours ago, JonathanM said:

Isn't it speed that is averaged rather than frequency? Thus, you'd average 1/2 and 1/4 to get 3/8, or 8/3 Hz. At 6Nm of torque, that'd be 8/3*6 = 16W as expected.

(It'll be less due to friction etc).

Calculations of JonathanM are not correct, too.  What kind of speed do you assume?
Proof is simple: try to calculate the same thing but assume that motor A rotates at 2Hz and has 4Nm torque (power is 8W) and motor B rotates at 8Hz and has 2Nm torque (power is 16W). Total power of the two input motors in that case is 24W. According to your calculation we need to average 1/2 and 1/8 and we get 5/16, or 16/5. At 6Nm that'd be 16/5 * 6 = 19.2W. Where does the rest of power go? Law of conservation of energy is still working and in case of ideal mechanism without friction and power loss there is nowhere for power to go. Yes, of course, in real mechanism there are always power loss because of friction, but they can be calculated theoretically, too.

Here is the explanation: frequency is the speed. 

14 hours ago, TheMindGarage said:

But surely revolutions per second is the same as rotational speed? 2Hz is 2 revolutions per second since both have units s^-1.

In that statement TheMindGarage is right.
Linear speed of the outer point of the gear can be calculated by following equation: V=ω*R, where is linear speed, ω is angular velocity and is the distance between the center point of the gear and the outer point (the radius).
Angular velocity can be calculated by simple equation: ω=2*π*ν, where π is Pi, mathematical constant, and ν is frequency or number of rotations that occur during 1 second. 
So there is no point on averaging inverse values.
 

11 hours ago, Lipko said:

Pretty much all asumptions in the therad are wrong :classic: (in one way or another). In the system, there's equllibrium. You can't have 4 Nm on one input and have 2 Nm on the other input at the same time. They will be equal (assuming 1:1 relation between the input shafts gearing to the differential). So, if you have a 6 Nm load* on the output, you'll have 3+3 on the inputs. 3Nm x 4Hz + 3Nm x 2Hz = 6Nm x 3 Hz. Which is true.

This statement is wrong, too. Why should be there any type of equilibrium in the system? By what rule?
When there is 6Nm * 3Hz = 18W power at the output the law of conservation of energy is broken. There is no point to believe that the law of conservation of energy is wrong or not working:)

11 hours ago, Lipko said:

*there's no input torque without a load tourque on the output.

That statement is correct in some way. But we can assume that motor, for example, rotates at 2Hz and has maximum of 4Nm torque on that rotational speed.

I think, it is quite obviously that the output will be rotating with 3Hz. If anyone doubts it, let me know, I will provide some drawings and more explanation.

So the question is what is the maximum torque, that we can use on the output on that rotational speed?

9 hours ago, sheo said:

Just a crazy thought: judging from the description of the initial task, motor A can be replaced with motor B (with the same power) plus a shameless 2:1 gear reduction (which gives a speed of 2Hz). If your calculations are correct, one of two identical motors has to provide only 1.5Nm torque (3Nm after gear reduction) while the other one has to provide 3Nm torque and the same speed. That's just unfair! :)

This is another proof that Lipko's statement is wrong.
 

13 hours ago, sheo said:

My version:
An adder is basically two inputs connected to the same output via two sets of gears. As we know, in a gear train torque is transferred inversely proportional to the angular speeds of the gears: T2/T1=W1/W2. The output speed of the adder is obviously an average of the input speeds: W=(W1+W2)/2=3 Hz. The first input transfers the following torque to the output: Ta=T1*W1/W=4*2/3=8/3 Nm. The second input transfers the following torque to the output: Tb=T2*W2/W=2*4/3=8/3 Nm. The output torque is a sum of the torques received by the output from the inputs: T=Ta+Tb=(T1*W1+T2*W2)/W=16/3 Nm=5.33 Nm. The output power is 16 W.

I totally agree with this explanation. It is the implication of the law of conservation of energy.

I have the LEGO Speed computer. Some day I will try to provide an experiment that will prove my and sheo's statement:)

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@Limga

If the speed of the system is constant (no acceleration) then there has to be equilibrium. By system I mean the adder with its inputs and output.

I don't see where I implied that the law of conservation of energy is wrong. I only said that the power that goes into the system is dictated by the load and this power can be higher than the rated power of an input* (obviously the power of the input comes from the electric system driving the motor - no conservation of energy broke anywhere, the electric system is capable to supply energy, but you'll probably burn your motor). I also said that the output rotates at 3 Hz, that's the most obvious thing.

 

*okay, that's obviously not entirely true, only to some extent. I have to think about this part... The other things still holds (that there are equal torques on the inputs in a non-accelerating adder)

Edited by Lipko

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1 hour ago, Lipko said:

@Limga

If the speed of the system is constant (no acceleration) then there has to be equilibrium. By system I mean the adder with its inputs and output.

I don't see where I implied that the law of conservation of energy is wrong. I only said that the power that goes into the system is dictated by the load and this power can be higher than the rated power of an input* (obviously the power of the input comes from the electric system driving the motor - no conservation of energy broke anywhere, the electric system is capable to supply energy, but you'll probably burn your motor). I also said that the output rotates at 3 Hz, that's the most obvious thing.

 

*okay, that's obviously not entirely true, only to some extent. I have to think about this part... The other things still holds (that there are equal torques on the inputs in a non-accelerating adder)

Why there should be an equilibrium of torque? By what law of physics?
Imagine the following situation: we use only one motor to power the inputs. One input is connected directly to the motor and another with gear reduction 1:2 from the same motor. In that case the torque on one input will always be two times higher independently of the load on output. The speed of the system will remain constant anyway, even in that case.

I agree, that the power that is produced by the motors is only dictated by the load, but it couldn't be any higher than the maximum power of an input. As usual DC-motor has linear RPM/Torque curve, they deliver maximum of power at half of their maximum rotational speed. 
They just cannot provide more power than this at certain voltage level, no matter how much current the source can provide. So if motors, for example, can provide maximum 8W each the system just can't have 18W on the output. No matter how high the load is.

Edited by Limga

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2 hours ago, Limga said:

Why there should be an equilibrium of torque? By what law of physics?
Imagine the following situation: we use only one motor to power the inputs. One input is connected directly to the motor and another with gear reduction 1:2 from the same motor. In that case the torque on one input will always be two times higher independently of the load on output. The speed of the system will remain constant anyway, even in that case.

I agree, that the power that is produced by the motors is only dictated by the load, but it couldn't be any higher than the maximum power of an input. As usual DC-motor has linear RPM/Torque curve, they deliver maximum of power at half of their maximum rotational speed. 
They just cannot provide more power than this at certain voltage level, no matter how much current the source can provide. So if motors, for example, can provide maximum 8W each the system just can't have 18W on the output. No matter how high the load is.

You are right of course. My physics knowledge is rather rusty.

But anyway: my equilibrium statement still holds on this situation, because the original poster implies 1:1 ratio between the inputs (since he uses simple average to calculate the output speed). The law here is https://en.wikipedia.org/wiki/Mechanical_equilibrium. Rotational equilibrium applies on single objects. It's easy to imagine that the torque on the inputs is the same if you fix the crown gear. It's also easy to imagine that the torque that's needed to fix the crown gear is the sum of the two equal input torques (if we assume 1:1 ratio between the crown gear and the output' gear, like the OP implied). It's no different if everything rotates at a constant speed.

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