Gear Ratio help....plZ..someone???

[Matt Adams]

Given:
- 2 Drill Motors
- 2 12” and 4 10” wheels will be powered
- The final robot weight is roughly about (or will be) 130 lbs
- Speed we would like to achieve is around 6-7 ft/sec.
-Only 4 wheels will have contact with the surface at one time, the 12" wheels centered and "rock" between the two 10"
- The 12" wheels are pneumatic and the 10"s are hard rubbered

Find: optimal Gear ratio.

Assume:
Drill Motor Characteristics, Low Gear:
RPM: 450
Stall Torque: 319.95 in lbs
Stall Current: 127.0 A
Breaker Limit: 40 A
Torque at Breaker Limit: 103.35 in-lbs

Solution:
Okay, so you have 130 lbs of robot that needs to be pushed having 4 wheels of different diameters in contact with the ground at the same time. It's assumed that you don't want to get into a pushing match and trip the circuit breakers that are on the drill motors.

What makes this a tricky situation is that you have different wheel diamters and presumably an uneven weight distribution since the robot will be rocking. For the sake of assumptions, I'll have to assume that there is more weight on the center 12" wheels, say 60% of the robot's weight, and the remaining percentage will be rocked back and forth.
Assuming the robot has even weight side to side, we'll break down the overall robot to each motor.

Now, we're dealing with 65 lbs, and 60% of it is on the center wheel, that's 39 lbs. We'll assume that you get a descent coefficient of friction of about 1.2 since you have those nice pneumatic wheels, which means that you need to be able to exibit a torque equal to:

12 in / 2 * 39 * 1.2 = 280 in-lbs.

On the back wheel

10 in / 2 * 65 lbs * 40% * 1.2 = 156 in-lbs.

What is very unfortunate is that as you can see, you'll need to have more torque on your middle wheel than on your outtermost wheel, but you need to hold a gear ratio of 10:12 between those wheels, or you'll have a situation where one wheel will be forced to slip on the carpet. That's bad!

So what you're forced to do is actually design based on the conservative wheel, the larger middle one.

The total torque required for the drive train is 436 in-lbs, which means you'll need to have a 4.2:1 ratio between your drill motor and your primary drive shaft, which could presumabley be the center drive wheel. From here, you can place any sprocket ratio of 5:6 from the 12 inch to the 10 inch wheel.

This is the required ratio to make sure you don't trip the 40 A breakers on your drills. In my opinion, this is how you must design a drive train.

Your overall speed would be ((450 RPM / 60 s )/ 4.2) * 1 ft dia * 3.1416 = 5.61 ft / s

The short story is that you really can't ever do much better than this with just the drills if you don't want to run the risk of tripping your breakers. If you wanted to adjust the ratios a little bit, you could edge yourself up to 6 or 7 feet per second, but you'd have to understand that you can't get in pushing matches.

Also note that I entirely nelglected mechanical efficientcy losses beetween sprockets, which could amount to perhaps as high as 15% overall.
I hope that his has been useful for everyone, if you have any questions comments or corrections, please let me know!

Good luck!

Matt