Electric motors and gears: torque, rpm questions

[cgibsong002]

Hi guys, new here to the forum. I was an electrical engineering student but am now working on more of a mechanical issue so I'm a bit out of my element.

I am working with a small electric motor attached to a gearbox. The final gear opens and closes a metal valve. I am having trouble finding a combination of motor torque/rpm and gear ratio that successfully accomplishes what we need. Too much torque and the valve is damaged, too little and the valve gets pushed back open by air pressure.

Through my research and experimenting, it seems that a high gear ratio, with a low torque motor should accomplish what I need. It is my thought that, the high gear ratio requires more reverse force to push the valve back open from air alone. Through testing the higher gear sets are way more difficult to turn the valve (but is this due to gear ratio alone or also resistance on the motor gear?). But a low torqued motor should still be able to turn the gears and will not over torque the valve.

For reference, the gear sets I have tested so far are between the ratios of 400 and 1100. I am hopefully calculating these right. Each gear set has either 4 or 5 compound gears (so 8-10 in total). I am working on getting motor specs, but I am unsure if there's any way I can actually calculate torque on the motor.

Does final gear RPM*gear ratio determine motor rpm? Because final gear is very slow and easy to observe/measure.

Thanks for the help! I apologize if I'm a bit off on my thinking.

 

[SteamKing]

Motor RPM / Gear ratio = Output RPM

Motor Torque * Gear Ratio = Output Torque

 

[cgibsong002]

edit:next post.

 

[cgibsong002]

Ok, so I finally actually dug up actual specs on the gear set and motors. My calculated gear ratios were correct.

So, motor torque*ratio=output torque. According to this, it doesn't matter whether my power is coming from the motor or from a higher gear ratio, as long as the one is raised/lowered to give the same output torque.

My question is.. does this hold true in reverse? As in:

motor torque (20g*cm)*gear ratio (1000) = 20,000 output
motor torque (5) *gear ratio (4000) = 20,000 output

In one case the motor is strong with a weak ratio, the second has a weak motor with strong gearbox. In either case, when power is applied to the system, you have the same output torque. But what about when no electrical power is applied? If simply force is applied to the output (in this case our valve), will it take the same amount of force to turn each one, or now that the electrical motor is disengaged will it require more force to turn the unit with a larger gear ratio?

 

[SteamKing]

If you want to turn the output, you must supply some input torque, whether it comes from an electric motor, your hand, or a diesel engine, etc.

 

[CWatters]

But what about when no electrical power is applied? If simply force is applied to the output (in this case our valve), will it take the same amount of force to turn each one, or now that the electrical motor is disengaged will it require more force to turn the unit with a larger gear ratio?

If you want a system that can't be driven backwards when the motor is switched off look at a worm drive.

I would use limit switches to stop the valve being damaged by the motor.