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Unread 28-12-2013, 00:04
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Re: Gearbox friction model

Reverse the first two equations and you have it. Speed loss is related to but not the same as efficiency. People typically use another constant for speed loss.
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Unread 28-12-2013, 00:33
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Re: Gearbox friction model

So,

Code:
OutputFreeSpeed = inputFreeSpeed / ratio * speedEfficiency^numStages;
OutputTorque = inputTorque * ratio * torqueEfficiency^numStages
?

What's a good estimate for these constants? .85 for both of them?
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Unread 28-12-2013, 01:02
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Re: Gearbox friction model

Quote:
Originally Posted by Bryce Paputa View Post
So,

Code:
OutputFreeSpeed = inputFreeSpeed / ratio * speedEfficiency^numStages;
OutputTorque = inputTorque * ratio * torqueEfficiency^numStages
?

What's a good estimate for these constants? .85 for both of them?
Depends on motor piloting, number of stages (yes, the efficiency per stage varies per number of stages), which motor you use, amount of stress on output shaft, heat, etc. Best thing to do is to test it out in the field.
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Unread 28-12-2013, 02:09
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Re: Gearbox friction model

I usually use these (highly inaccurate) estimates, based a little on research, and a little on experience:

Belt reductions: about .98 efficient
Gear reductions: about .95 efficient
Chain reductions: about .90 efficient
Planetary reductions: about .8 efficient
Single lead worm reductions about .6 efficient

Of course, as T^2 said, these depend heavily on a lot of factors. I've noticed (although not in the least bit empirically) that for reductions that are less efficient to start off with, misalignment, lack of lubrication, high speed, etc makes a much bigger difference in terms of efficiency. For example, it's much worse to misalign a worm gearset than a spur gearset in terms of efficiency, and belt reductions are much happier at high speeds than chain reductions. Of course, I don't have any numbers to back these assumptions up, so this may be useless to others.

The process I usually use is to assume that these efficiencies are just multiplied by the free speed of the motor to get the adjusted free speed, however, that isn't the real way you do it. I'd be very interested to hear what the actual meaning and use of efficiency is.

Really, what you want to do is to use some reasonable numbers to estimate how your system will perform, and leave enough flexibility in the system that you can regear if necessary. Maybe you won't need to redo it on all systems, but every year we've needed to regear some stuff, and not because we didn't do our math. Efficiency (like friction) is something that's very hard to analyze.
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Unread 28-12-2013, 08:39
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Re: Gearbox friction model

Quote:
Originally Posted by DampRobot View Post
I usually use these (highly inaccurate) estimates, based a little on research, and a little on experience:

Belt reductions: about .98 efficient
Gear reductions: about .95 efficient
Chain reductions: about .90 efficient
Planetary reductions: about .8 efficient
Single lead worm reductions about .6 efficient
These are great numbers for design, and I use about the same numbers when I do design estimates. However, if you'd like to get more accurate, you'll need to actually build the thing and do some testing. You'll notice some strange stuff. We saw that we were most inefficient with low loads, and most efficient with medium loads. Lubrication and accurate machining and placement of gears, belts, and pulleys really makes a difference. Another thing that we found interesting was that the efficiency changes a lot with time. When we started our shooter (it used a "hack" 1:1 banebots p60), it was around 2800 rpm. After 1 minute, it was up to 3400 rpm, and after three more minutes it went to 3100 rpm.
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Unread 28-12-2013, 09:34
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Re: Gearbox friction model

Yeah those numbers look accurate enough for my estimates (either .81 for everything or the separate ones), I'm not doing any rocket science.
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Unread 28-12-2013, 02:18
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Re: Gearbox friction model

Quote:
Originally Posted by Bryce Paputa View Post
So,

Code:
OutputFreeSpeed = inputFreeSpeed / ratio * speedEfficiency^numStages;
OutputTorque = inputTorque * ratio * torqueEfficiency^numStages
?

What's a good estimate for these constants? .85 for both of them?
In my experience you can generally just use one number for speed loss for the whole system regardless of gear ratio. "Speed loss" is really just the lowered free speed of the system caused by the negligible loading of by overcoming kinetic friction. For well greased gearboxes it shouldn't change *too* much.

I've always used 81% as a "speed loss constant", allegedly the result of experimental data collected by 229, but I wouldn't be surprised if it was just some arbitrary number somebody came up with.
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