Torque and RPM Graph for Castle Creation products ?

[Pdelcast]

Originally Posted by VintageMA

I know this is an odd follow-up to this graph, and I understand the multiplication of the torque at the motor through the gearing to the wheels, but I have a follow-up as to how we can apply this to real setups?

Is there a way to calculate how much torque an application will need for a certain amount of acceleration or top speed? Obviously you would want to pick a motor where the torque your application needs doesn't fall off the top of the efficiency scale where it starts to drop back down from max eff%.

Also - I am confused with the oz/in ratings when comparing to servos. Picking a servo for an application we see ratings in the 125oz/in all the way up to 400oz/in. I know this already takes into account the reduction gearing in the servo and there is also much the throw of the servo compared with a spinning motor. But, when I look at that graph and see ratings for 25-30oz/in of torque for these huge motors is just seems very small in relation to a servo putting out 5-20 times more torque.

Does where I am going make sense? Can anyone help explain the relation a bit more?

Power is torque times RPM -- so to get horsepower, you actually take torque in oz/in, and multiply by the RPM. If a servo can make 120 degrees in .2 seconds (for example) and maintain 200oz in of torque (which servos CANNOT do -- they measure torque at stall (0 RPM)) -- Then you would get 100 RPM at 200 oz/in.

To find horsepower, multiply the RPM by the oz/in and divide by 1 million. So for a servo, that would be 100RPM * 200ozin / 1000000 -- or 1/50th horsepower. However, the best servos right now can't sustain that kind of output for very long.

Now, take a motor like the one in the graph -- it's turning 33,500RPM at 30oz-in of torque. 33,500 * 30 / 1000000 = 1 horsepower, or 50 times more power than the servo's motor is outputting. NOW, that said, that same motor can also output 240oz/in at about 15,000 RPM (240 * 15000 / 1000000 = 3.6 horsepower) on 16V.

Now to illustrate why RPM is important -- Take that motor (at 30ozin/33,500RPM) and put it in a servo. If we add gearing to bring the RPM down to about 100 RPM (similar to a good servo) we would multiply the torque by the same ratio as we divide the RPM. So, 33,500 / 100 = 335 (gear reduction.) Torque would then be 30ozin * 335, or about 10,000 ozin.

So, in the same application (RC Servo) a 1515/1Y motor would generate about 10,000 ozin of torque at 100RPM, and about 160,000 ozin of torque (about 800ft/lbs) at stall.


Make sense?