Inertial Load Referral to a DC Motor Shaft 2

[logbook]

This is not a student posting even though it may sound like a homework question. I am phrasing the question in a fairly general manner because the project is very sensitive.

I have a DC motor driving what is essentially a purely inertial load. The load is minimal except when being accelerated, hence I am just interested in the inertial aspects because I want to make it accelerate quickly.

It is my understanding that if the motor drove the load through a (light ideal) gear box the inertial load would effectively be reduced by the gear ratio (not the gear ratio squared for example). In other words if the gear box made the load rotate 4 times slower the inertial load referred to the motor shaft would be reduced by a factor of four. I don’t want to use a gear box, I just want to verify my thinking.

The real application is to drive the load through an offset bearing and push-rod. The net effect is to make the load oscillate (in a rotational manner) backwards and forwards by 10 degrees in each direction about a mid-point. This motion of the load is sinusoidal because of the offset bearing, much like being driven by a cam.

It seems to me that at the ends of the offset bearing motion, the motor hardly moves the output shaft at all. In this case the speed reduction effect makes the inertial load seen by the motor essentially zero. On the other hand 90 degrees further on in the cycle the offset bearing is making the load rotate at its maximum speed and therefore giving the maximum inertial load referred to the motor shaft. Does this sound plausible? What is the inertia division factor (neglecting losses)?

It seems that the position is a sinusoidal quantity so maximum velocity occurs at the point midway between the two extremes. Unfortunately the acceleration is the second derivative of the position so the maximum acceleration seems to occur at the ends of the sweep.

 

[CESSNA1]

LOGBOOK: What are the speeds and the distances here, can you post a sketch?

If the motor shaft with the eccentric is located near the pivot point of the driven device that, in effect, is a speed increasing gear and the mass moment of inertia will be amplified. Doing something like at that at 3600 rpm will generate large inertial forces.

Regards
Dave

 

[GregLocock]

Check out the theory for Hookes joints. 20 degrees pk-pk of torsional vibration will, as Cessna said, cause some huge forces and torques at any significant speed.

Incidentally I am >99% sure that your assumption that the referred mass is proportional to n^1 is wrong.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[logbook]

This is an equivalent system drawing. The speeds, distances, masses and so forth shouldn’t make any difference to the inertia ratios. ±90 degree rotation on the motor shaft from the position shown moves the load ±9 degrees.

 

[logbook]

Greg,
Why are you so confident that the inertia "gearbox factor" is not the first power of the gear ratio, and what power would you suggest? If it were a lever the force would go with the first power of the lever ratio wouldn’t it? Force & torque, lever ratio & gear ratio, all seem similar. I know the inertia itself is the mass * distance squared but F=mA and the rotary equivalent all seem pretty linear.

 

[GregLocock]

Sorry I misread you post, I thought you were driving via shaft, not a connecting rod.

Either way n^2 is much more likely



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Canuckboy]

Actually, the inertial "load" is proportional to the square of gear ratio. The analogy to the lever is correct, but remember that you should not compare forces, but rather apparent mass. Say you have a 2:1 lever, with a 1 kg mass at the long end, and the mass being accelerated at 1 m/s^2. The force directly on that mass is then 1 N. So then the force at the short end of the 2:1 lever (the end you are pushing on) is 2 N, but that end of the lever is accelerating at only 0.5 m/s^2. Remember that F=ma, hence m=F/a where m is the apparent mass at the short end of the lever. Therefore m=2N/0.5m/s^2=4kg. And you see the apparent mass is four times (i.e. the square of the lever ratio times) the suspended mass.

There is a precise analogy with electrical transformers, where a 2:1 transformer yields twice the voltage, half the current, and 4 times the electrical impedance.

 

[logbook]

I changed the diagram because it wasn’t quite right.

Thanks canuckboy for your gearbox calculation. How about the variation of inertial load with position due to the offset bearing? I have to move the motor shaft by 1/8 turn at a time very rapidly. It is evidently harder to move it by 1/8th turn at some points in the cycle compared to others.

 

[GregLocock]

well, that's simple geometry. The position as drawn is the worst case, is it not?

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[logbook]

Yes, this should be the position of the largest amount of inertia referred to the motor shaft. If the load moves ±9 degrees for a ±90 degree movement of the motor shaft I now think this gives me a reduction of inertia at the motor shaft by (9/90)^2, that is a factor of 100 reduction as a worst case.

Anyone care to agree or disagree?

 

[solider]

This my first attempt to post an image/movie. I hope this will be sort of what you are looking for.

The movie shows the 4bar cycling, which is about equivalent to what you have in your figure. The graphs start at zero degrees with the crank at 9 o'clcok. The first graph shows the crank torque as seen by the crank at low rpm, the second at 200 and the third at 2000rpm. Units are in Nmm. I've put in arbitrary amounts for the rocker mass, CofG and radius of Gyration.

You can see that the first graph, the crank is mainly just lifting and lowering the weight of the rocker and coupler (connecting rod). You can also see that the torque is about zero, when the crank moves on by about 100 degrees (in line with the coupler at the bottom of the cycle), and again when the crank is in line with coupler at the top dead centre. As the speed increases, the crank sees more of the inertia forces of the rocker and coupler, since their accelerations are increasing.

If you would like to get me to precisely model what you have in mind then let me know...I might be persuaded!

Any comments?

 

[zekeman]

If you are interested in the startup of your system then none of the explanations or solutions offered thus far addreses that problem.
To achieve quick startup, you need more power, since that would get you to the equlibrium rpm state the fastest. However, in doing so, the motor inertia must also be accelerated so that there is a practical and economic limit to the size of the motor you need for the job.
It is easy to do this calculation for a simple linear gear driven system.There, you would use the reflected inertias (I/n^2) of the components referred to either the motor or load axis. However for your system, you would need to write the dynamic equations for each of the linkages and the torque/speed curve of your motor to achieve the transient solution. Moreover,the steady state solution does not guarantee a constant RPM because of all the inertia forces.( like an IC engine) If that is your goal you would need a flywheel to approach this .

 

[solider]

I agree with zekeman, this is a very complicated thing to optimise.

If you want to minimise the acceleration time for a constant inertial load you would choose a gear box ratio that ends up matching the inertia of the load with the inertia of the motor (N = sqrt(load inertia/motor inertia).

The problem with the mechansim you have suggested is that the inertia as seen by the drive shaft varies with position. For the purposes getting going (if you want to get going that is) I would take an average inertia load as seen by the drive shaft.

The problem then is, because the load torque is varying quite a lot, there will be a tendency for the mechanism to race at crank angle of minimal torque and slow down at maximal torque so that it has a per-cycle velocity fluctuation. To minimes this velocity fluctation problem you either put a flywheel on (seriously increases acceleration time) or maximise the gear ratio at the gearbox but making sure you can achieve the mechanism speed (again, your acceleration time is increased becasue you are having to accelerate the motor more than optimally.

 

[logbook]

Evidently the system constants are such that this thing will drive very quickly as a steady state. In my application it moves from stationary to stationary as fast as possible, having moved through a suitable angle such as 90 degree on the motor shaft. Practically it seems that a smaller motor is better for fast acceleration because its own ratio of inertia to torque gets better as it gets smaller, and the inertial load of the rest of the system is relatively small (so far).

Clearly this idea will fail at some point with smaller and smaller motors, but it has got me where I need to be to date. I have as yet to get the time to play with seeing what the maximum acceleration limit of the current mechanism is.