Motor Calculations

[John Gutmann]

I was just looking at some things for the small CIMs and felt like doing a few calculations.

I wanted a output speed of 10 fps on a 6 inch wheel.

So the circumfrence of the wheel is about 1.57 feet this means it needs to rotate 6.36 times. This comes to 381.60 rpms.

I then looked at that data sheet and found the most efficient load and speed. which is 45.00 oz-in and 4614.00 rpms.

Ok so now you have to gear it down from the 4614 rpms to 381.6 rpms. this means you need a gear ratio of 12.09. So now I multiplied 45 oz-in of torque by the ratio, which gives me 544.05 oz-in of torque.

544.05 oz-in = 34 lb-in = 11.3 lbs is the max weight of this robot

This doesn't make much sense to me seeing as how teams use 2 CIMs the go faster then 10 fps and carry way more weight. this means from what I got to beable to carry the 133 pounds the max speed would be 10 in/sec.

From this I am thinking I went terribly wrong somewhere, If somebody could help me out with this it would be great.

[RogerR]

...544.05 oz-in = 34 lb-in = 11.3 lbs is the max weight of this robot..

this number is not the max weight of the robot, but rather the force the robot will create at the wheels (at this speed). to find the max acceleration at this point on the torque curve, divide this number by the mass (not the weight), which in this case is approx. 4.13 slugs. this will give you an acceleration of 2.73 ft/s^2.

[ConKbot of Doom]

no, that seems about right... that's the amount of 'thrust' that will be at the wheel. So, at the max efficiency, your robot will have 11.3 pounds of acceleration pushing it forward, and that is while you are at 10 fps, when it is slower, you will have more acceleration then that.

[Jack Jones]

Read this (http://www.chiefdelphi.com/media/papers/1713)