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Author Topic: Gearbox?  (Read 1596 times)

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Offline modeonTopic starter

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Gearbox?
« on: January 27, 2011, 08:14:07 AM »
How do I choose the right gearbox for the motor? I have 2 Ampflow E-150 Motors in possession. (http://www.robotmarketplace.com/products/MAG-E150.html) Anyone willing to share any advice?

Offline mstacho

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Re: Gearbox?
« Reply #1 on: January 27, 2011, 10:18:34 AM »
Really it depends on the kind of torque or RPM you need at the output shaft.  At the moment, the motor runs at 4700RPM, so if you wanted, say, only 47RPM at the output shaft, you'd need a 100:1 gear box.  That'd also boost your torque.

It also depends on what kind of gears you want (and what is available).  Planetary gears are great in that they don't have much backlash (IIRC), but they can be more expensive, while simpler gearboxes might be cheaper but they also may be larger and have more backlash.

I hate to keep saying it depends on the application, but it does :-)

MIKE
Current project: tactile sensing systems for multifingered robot hands

Offline modeonTopic starter

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Re: Gearbox?
« Reply #2 on: January 30, 2011, 05:03:02 PM »
I definitely want high torque. I've been reading the tutorials and I am confused with my calculations. Can you proofread my calculations?

Velocity = 2*pi*r*rps
           =2*pi*(1/6)*78.3=81.99 ft/s? ( I converted the wheels which are 4" in diameter to ft and divided by 2 to get the radius. The 78.3 is rps as it was originally 4700 rpm. ) That seems way too large....

For calculations, should torque be in lb/ft, lb/in, or oz/in? Each webpage uses different units. I wanted to calculate force but I had some difficulty.
2.24 (lb/ft) / (1/6) ft = 13.44 lb.

We're facing other robots that could be up to 50 lb. Ours will probably around 30 lb.

Offline mstacho

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Re: Gearbox?
« Reply #3 on: January 30, 2011, 05:25:41 PM »
Your calculation is technically fine, but you need to remember that 4700RPM is a maximum speed.  Under load the speed will be much less, and how much it varies depends on the motor and the load. 

That's why your gearbox will help here: with a higher torque, the speed will also lower, but it won't drop as much when you put it under moderate load.  So if you have a 100:1 gearbox, your RPM will be only 47 as a max, but under load it might not drop to much lower than, say, 40 (again, this depends on the motor).  So, assuming instead that you have a 100:1 gearbox, you'd actually get only about 0.82 feet/second.  Keep in mind that 100:1 is a bit extreme (although I *am* building a robot with a 228:1, so it's not like it's way out of the ballpark).

Your torque calculations, it doens't matter which units you use, since they can all be converted into each other.  Standard units in scientific circles are N-m (Newton-Meters), and in the USA people usually use ft-lb.  Keep in mind, the unit isn't lb/ft, its lb*f, although your calculation is fine.  Here's why:

torque = lb*ft

Therefore the force, in lb, is given by:

torque/ft = lb

Your motors are pretty big, so 13.44 pounds at the wheels doesn't sound too unreasonable (although you really should be doing everything in Newtons, because Newton's laws were designed to work in newtons, meters, and seconds :-) ).  I'm not 100% sure how force at the wheels is transferred to the body of the car, but it's probably just direct. 

Keep in mind that a gearbox will in theory considerably increase the amount of force you have at the wheels, but in reality you won't get the super acceleration you're expecting.  Here's why:

Technically speaking, if you had a 100:1 gearbox, you'd have 224lb-ft of torque, for a force at the wheels of 1344lb.  Wow.  *However*, you also have to consider the speed reduction.  In other words, 224lb-ft is a *maximum* possible, and therefore 1344lb is also a maximum.  It'll only ever happen under an extreme load (the "stall load" of the motor).  What it means is that, theoretically, your robot could push 1344 pounds before the motors couldn't give any more force.  It also means that if somehow your motors were spinning at 47RPM (that's a 100:1 reduction in speed, remember) you wouldn't have 1344lb of force at the motor.  How much force you actually get depends on the torque/velocity curve of the motor, which is something I'm still looking into :-)

I hope that hasn't confused you all that much.  Simple answer here is as follows:

With a gear reduction, you now know the maximum possible torque of the motor, and the most likely speed your robot will go without any major load (although 30 pounds is quite a bit...).  The two are related: as the speed goes up (and necessarily the load it's pushing must go down if you're already putting the max. amount of current into the motor to do that) the torque goes down.  But when the robot is stuck against another robot and pushing hard, it'll have a huge amount of bite to it :-)

MIKE

Current project: tactile sensing systems for multifingered robot hands

Offline waltr

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Re: Gearbox?
« Reply #4 on: January 30, 2011, 05:37:29 PM »
Just to add:
I agree to do all of the calculations in the metric system, much easier and a lot less confusing.

Then say from torque specs, use an on-line converter to go from N-m to what ever imperial units the motor seller uses.

 


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