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?
Reply to
Tim Wescott
Loading thread data ...
Tim Wescott fired this volley in news:w snipped-for-privacy@giganews.com:
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
Reply to
Lloyd E. Sponenburgh
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.
Reply to
Tim Wescott
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?
Reply to
jim
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.
Reply to
David Billington
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
Reply to
Tim Wescott
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.
Reply to
Tim Wescott
Hi Tim
I looked up gear design in "Mechanical Engineering Design" 7 edition. It is the standard design text book for almost all upper level ME classes.
The book talks about efficiency for worm gears but doesn't talk about it for bevel, helical and spur gears.
For these max power is a function of failure modes.
I am GUESSING it is the same ether way
Dan
Reply to
d32
One more thought. Worm gears has sliding friction but properly designed spur etc experience rolling friction which is not a strong function of force loading
Reply to
d32
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.
Reply to
jim
jim fired this volley in news:4eudnQy9gfkqx2XMnZ2dnUVZ snipped-for-privacy@bright.net:
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
Reply to
Lloyd E. Sponenburgh
I vaguely remember Boston Gear as having some information on worm gear effi ciency and driving worm gears in reverse. They also had some info on lubri cation. They highly recommend synthetic gear oil. Their gear reducers com e with synthetic gear oil and I believe the warranty is void if non-synthet ic gear oil is used.
Dan EE not ME
Reply to
dcaster
I ran across at least one paper. It was mostly about the wisdom (or lack thereof) of using a worm drive so that you could depend on it being self- locking.
I was kind of hoping for information on reverse vs. forward efficiency because with the right combination of friction, inertia, and external forces, it can have a strong effect on the dynamics of the electro- mechanical system. When performance is important, the guy doing the control rules for the system needs to take those dynamics into account.
Reply to
Tim Wescott
Where does the lost power go? It must be heat.
Reply to
David Lesher
Side topic:
I have read that the Wehrmacht's tanks, esp. the later ones that were so large, suffered badly from final drive failures. So much so that a road march of them for any distance would disable say 33%. This helped me better understand Allied anti-railroad tactics...forcing them onto the roads.
[They also lacked tank retrievers of enough strength to salvage them....]
The failures were because of 2 basic reasons: they lacked sufficient chromium to fully harden the gearing, and the second was my question.....
This source (that I now can't find again...) said there was a way to better design/machine the ring gears needed, but Germany lacked the tooling/resources to use that approach, and instead used less strong methods. I inferred the better way needed more time or a better mill but beyond that, I don't know.
I'm curious about what that might have meant, and wonder if anyone can speak to gear design issues...
Reply to
David Lesher
Only in fragments. If the issue was an internal-tooth ring gear, as for a planetary gearset, that wasn't a milling issue. That was a gear-shaper issue. They also can be broached, but I don't see that as likely during the war, on big gears. Perhaps that's the tooling that Germany lacked. It requires very big pieces of high-quality tool steel, and chromium shortages would be an issue.
Germany had some very good gearmaking capability but that of the US was better at that time. Gleason was the world leader in making big, strong gears of several types.
BTW, the US had power-transmission issues on some tanks at the time, too. Caterpillar made a big 24-cylinder porcupine diesel engine that was supposed to be the end-all for our largest tanks. But it had so much torque that it twisted off driveshafts like they were swizzle sticks.
Reply to
Ed Huntress
I haven't been able to find adequate details of the problem from a military historian / machinist. Supposedly the bombing had purposely targeted the makers of gear hobbers and shapers, along with ball bearings. The stainless tanks and plumbing for the hydrogen peroxide subs consumed most of their chromium and nickel.
This may help but I can't check it with dial-up:
formatting link

The Wiki is as good as anything else I've found:
formatting link

jsw
Reply to
Jim Wilkins
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.
Reply to
Tim Wescott
Steven Ambrose has some interesting comments in "Citizen Soldier" about the Sherman tank. The allied soldiers in Europe reportedly disliked it because it was so wimpy compared to the big German tanks -- but Eisenhower could get three Shermans for every one big US tank (I can't remember what we had), and a Sherman used less gasoline. The thinking was that as soon as the allies broke out in Normandy and started for Berlin that the mobility of the Sherman tanks would outweigh the size difference.
It's hard to say who was right, but we did win using those little tanks.
Reply to
Tim Wescott
Yeah, I've always wondered how we shot up Tigers with Shermans. Did the Shermans have the high-velocity 90mm guns, our answer to the German 88s?
Reply to
Ed Huntress

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.