Gearbox efficiency while back-driving

I suppose if there's a "sci.engr.mechanical" group I should post it there -- but r.c.m has some smart mechanical types, and s.e.c may still have
some lurkers who might come out of the woodwork for this one.
So here's a question for the mechanical engineers in the group(s).
I was giving a seminar on control systems last week, and had the embarrassment of not only having a huge mathematical error in one of my slides, but had one of the smarter audience members question my underlying assumptions -- and I didn't have answers for either problem while standing there.
The basic problem is this:
If I'm putting a gearbox into a control system, and I have a data sheet (or measurements) for the gearbox that tell me it's efficiency in the designed direction of power transfer (usually when it's gearing down), I would like to know what its efficiency is in the backward direction.
In other words, if I have a gearbox with a gear-down ratio of K:1 and 100% efficiency, then when I drive a torque into the thing I should get K * torque out. But what I really get is K * torque * efficiency.
I know both from experimentation with one sample, and from working out the math, that if I have a single-stage worm gear that I try to back- drive, its efficiency in the backward direction is pretty close to
h_b = h_f / (1 - 2 * h_f), where h_f is the efficiency in the forward direction.
But I don't know if this is general to all gearboxes, or even to all worm gear trains -- it's quite possible that I messed up my calculation and then lucked out on the one sample that I experimented on.
So -- anyone know? Are there any mechanical engineering texts that I should buy to check up on this?
--

Tim Wescott
Wescott Design Services
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I don't know the numbers, Tim, but I know that unless that worm has an awfully steep pitch, it's back-drive efficiency is almost zero. There's probably a critical pitch, combined with lubricant selection and materials selection where the efficiency suddenly and exponentially approaches zero.
Lloyd
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On Tue, 30 Jul 2013 12:09:06 -0500, Lloyd E. Sponenburgh wrote:

That's certainly my experience. I used to work on a product that had a worm drive, and this honkin' big flywheel on the motor that drove it. The worm drive was selected because its compact and because you can't back-drive the thing with power off. The flywheel on the motor was added because with the power on, when you commanded a reduction in speed the gearbox would seize, then the inertia of the driven mechanism would immediately break the output shaft of the gearbox.
I've heard of worm-drive gearboxes being changed from "can't back-drive" to "back-drives just fine" with a lubricant change. This from a fellow engineer that I trust. Apparently the lubricant in question was one of those "more expensive than gold" aerospace things that engineers sometimes use to patch over mistakes in design that aren't discovered until it's too late to fix them in a more reasonable way.
I've actually got a gearmotor in my office that uses a worm drive that can be back-driven. It's about 70% efficient in the forward direction, and about 57% efficient in the reverse direction, which is consistent with the formula that I cooked up. I suspect that it's got a two-, three- or four-lead worm, though.
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Tim Wescott
Wescott Design Services
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On 30/07/13 18:29, Tim Wescott wrote:

Vibration can also make the difference between a gearbox that won't back drive to one that will so the operating environment can make a difference.
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On Tue, 30 Jul 2013 20:31:39 +0100, David Billington wrote:

I know. It's a complex subject. That's why I was thinking that maybe the "real" answers would be buried in some mechanical engineering text.
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Tim Wescott
Control system and signal processing consulting
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Tim Wescott wrote:

That can't be right. It makes no sense. If h_f was 50% h_b would be infinite. If h_f was between 50% and 100% then H_b would be between minus infinity and -1. for example: If h_f were 80% then h_b would be -133%. What does an efficiency of -133% mean?

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On Tue, 30 Jul 2013 13:40:12 -0500, jim wrote:

AAAAAGH! I got it backwards AGAIN!!!! That's what's on the erroneous slide! Crap! Crap! Crap! Proofreading's a bitch.
It should read
{ (2 * h_f - 1) / h_f if h_f > 1/2 h_b = { { 0 if h_f <= 1/2
where h_b = 0 means that the gear locks when you back-drive it.
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Tim Wescott
Control system and signal processing consulting
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Tim Wescott wrote:

OK that formula produces numbers that are at least possible.
My wild guess is that there is no standard relationship for the efficiency forward and back on a worm gear.
Your formula has another problem. The efficiency in reverse is never really zero. If you get a worm gear that is theoretically not reversible started going in reverse it can usually be driven backwards to keep it going. It is just under static friction that it jams.

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Nooo... that's not even close. Try that with a 100:1 worm drive. You couldn't reverse-drive such a box with all the horsepower and "pre- starting" in the world.
We're not talking about "theoretical mechanical advantage" here. There are niggly little things like _actual_ mechanical friction at work, also.
Lloyd
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Where does the lost power go? It must be heat.
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On Fri, 02 Aug 2013 15:11:01 +0000, David Lesher wrote:

Yup. One of the reasons to select for a more efficient gearbox (assuming you don't care about motor size or input power used) is overheating.
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Tim Wescott
Wescott Design Services
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