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# Worm gear & torque calc.

## Worm gear & torque calc.

(OP)
Simple engineering problem, but not in my field....and its been a long time since I've had to calculate this stuff.

Summary:  I need to lift a 4kg load with a 40mm lever from the axis of the motor with a half turn movement.  I'd like to accomplish this half turn in 10 sec.  Here are a few questions relating to this...

1. Find the torque needed at the motor?
2. If I drop in a worm gear set, say 30:1, what is the torque required of the motor now?
3.  Where can I find a small worm gear set for around $30? The set must be metal. Most worm gear sets I've found are >$100.

Let me know if I'm missing any parameters.

Thanks,
Jason

### RE: Worm gear & torque calc.

Assuming
-- that the lever is 40 mm from the center of the motor to the load (not 40 mm overall), and
-- that the lever's own weight does not add significantly to the load on the motor, and
-- that the local acceleration of gravity [ratio of weight (in newtons) to mass (in kg) is 9.80665 meters-per-second squared (this varies from place to place):

1. The movement will be a semicircle, 80 mm diameter, 40 mm radius.  The torque is a sine wave, starting at zero (when the arm is pointed down), peaking at 160 kilogramforce-millimeters = 160 gramforce-meters = 1.56906 newton-meters = 1569.06 millinewton-meters when the arm is horizontal, and finally returning to zero when the arm is pointed up.  The formula for torque is radius times weight times cos(angle to horizontal).  {It has nothing to do with time.  Time matters when you want to calculate power or speed.  Since you want a 1/2 turn in 10 seconds, you need to turn at a rate of 1/20 turn per second, or 3 RPM, or Pi/10 radians per second.  If you assume for simplicity that the motor rotational velocity is constant (which it is not in real life), then you need a power of 0.492936 watts.  (Note that this is output power at this speed and torque, not electricity consumed, and not power on the nameplate at a nominal [different] load or speed.)}

2. This depends on the efficiency of the gear set, which depends on how well lubricated it is.  The torque required is the torque indicated above divided by the product of the gear ratio and the efficiency.  For example, if the ratio is 30:1 and the efficiency is 70%, then the torque required is 1569.06 millinewton-meters / 30 / 0.7 = 74.7173 millinewton-meters.  (The power required is the power indicated above divided by the efficiency, for example, 0.492936 watts / 0.7 = 0.704194 watts.  The gear ratio does not affect the power requirement.)

3. Find a working worm gear set in something that someone else is discarding because something other than the warm gear set is broken.  Remove the worm gear set from it.

Failing that, look on eBay.

### RE: Worm gear & torque calc.

(OP)
Thanks!
I was right on track with the calculations, just didn't know what to do with them.  So, barring any losses due to friction, motor accelleration and any other misc. assumptions made...
Power required = 0.4929 Watts = 0.0006607 hp

Without a gear set, would I specify a 1/1500 hp motor to drive this?  Maybe double that when considering friction and everything else?

I'm building a couple hundred of these...so I will need to find a better source than eBay.

Jason

### RE: Worm gear & torque calc.

I read in "Machine Design" mag to never assume that a worm gear drive to be over 50% energy efficiency (unless you diffinitely know otherwise) - considering the small size of this unit maybe even considerably less efficiency might be in order.

### RE: Worm gear & torque calc.

aloha,

I read in a mechanical engineering handbook that a rule of thumb for efficiency could be

100-worm gear ratio=efficiency of worm gear set in percent (half loaded systems)
0.5(100-worm gear ratio)= efficiency of worm gear set in percent (fully loaded systems)

However, I have found a couple equations for calculating the worm gear and wheel efficiency. I think the above rule of thumb is kind of old news.

Here is a link to a page containing a worm efficiency equation.
http://www.sdp-si.com/D220/HTML/D220T120.htm

The approximation equations are pretty good.

The exact efficiency equation when the worm is the driving member is wrong. I have the correct equation in front of me but I got it from a mechanical engineering handbook so I can't give you a link. My comparing the one I have with the one on the link the only thing you will need to change is change the bottom to "cos(n) + f cot(l)" then your calculations should be correct.

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