What is the relationship between gear reduction and torque?
MarkS
Posts: 342
Let's say that I have a motor with a stall torque rating of 300 g/cm. I connect it to a gear box with a 2:1 gear ratio. How does this effect the output torque? Would the output double? Quadruple? What is the formula?
Thanks,
Mark
Thanks,
Mark
Comments
Incidentally, that's gm cm, not gm/cm.· That's because you multiply grams by centimeters to get torque -- you don't divide, as the slant bar would imply.· "Gm/cm" implies grams divided by centimeters.
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· -- Carl, nn5i@arrl.net
Huh... Most places list the torque rating as g/cm or in/lbs, etc. I think the slash is meant (at least it is what I meant) to denote relationship (replacing the word 'per') rather than a mathematical equation. Whether or not that is the correct usage is another matter...
Thank you.
Post Edited (MarkS) : 4/22/2009 1:48:53 AM GMT
Relationships such as this are mathmatical, they are based on, and representable by a formula of some sort.
Catalog writers, and others without physics background, often make the mistake of using a slash where it's not correct.· Engineers and physicists never do.
It actually can be very·important when you use those quantities to calculate other quantities, for example the energy required to turn that shaft through some angle against that torque.· Eventually you're going to multiply something by that torque of (3 gm cm), and if you have a slash in there you are likely to divide when you should be multiplying, and you will get an answer that is wildly wrong.
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· -- Carl, nn5i@arrl.net
-phar
That's OK for such a purpose, I suppose, but it is outside any standard set of units and unsuitable for calculation since its value depends upon the value of local gravity, which varies at different locations.·· This kind of "gram" is about 980.665 dynes most places on Earth.· Imagine trying to use it when calculating the path of a celestial body!· It's equally awkward when designing machinery.
The newton is part of the MKS (meter-kilogram-second) system of units, and the centimeter is part of the CGS (centimeter-gram-second) system; thus they ought not to be mixed, to avoid the necessity for using·conversion constants.· The MKS unit of moment (torque) is the newton meter, and the CGS unit is the dyne centimeter.
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· -- Carl, nn5i@arrl.net
-Phil
Don't bet your project on that - I knew a lot of engineers during my 36 years as an engineer at a major aerospace company and a few of them were real screw-ups!
Jerry
Back when I was a principal engineer designing medical electronics, the company (Coulter Electronics) hired a guy with a master's in electrical engineering.· The guy seemed OK at first, but when he accidentally broke a glass diode (perhaps a 1N34) in two, he glued it back together and proceeded to solder it into the circuit.
He was not an engineer.·
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· -- Carl, nn5i@arrl.net
Some of the best electronics guys I worked with were former military technicians with a lot of experience that had gone back to school to get a degree. In fact we started seeking those guys out. I have been out of the business for quite a while so I don't know how things are today but I doubt that the situation has gotten a lot better.
Jerry
If it were moment=force/length, the moment (torque) would decrease with increased length of the lever arm).
Actually it's moment=force*length and the moment increases with·increasing length.
And the work done by turning the shaft is angle*moment, where the angle is measured in radians.
Thus·work=angle*moment=angle*force*length.
If we screw up and write moment=force/length, then we get work=angle*moment=angle*force/length which gives a wildly wrong answer because we divided when we ought to have multiplied.
Dimensional analysis is crucially important.
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· -- Carl, nn5i@arrl.net