Gear ratio RPM

So for a project we need to pull 10 pounds as far as possible in 10 seconds. there is only one rule:

you can only use two motors

So what I need to know is what the best gear ratio would be for this? I was thinking of using a 2:1 or 4:1 ratio (possibly a 5:1). However what would the speed of those ratios be. All I know is the rpm of the standard motor. Any information would be helpful.


On a complete side note, if you only have to go straight, make an elastic powered drive that you wind up before doing the challenge.

Ignoring the speed lost due to the motors’ torque-speed curve, a 2:1 gear ratio is 2*motor output, or 200 rpm. 4:1 is 400 rpm, and 5:1 is 500 rpm. However, you will also need some torque, and 2 motors geared up may not have enough torque to pull 10 pounds, especially if the weight is dragged, rather than sitting on wheels. An elastic powered drive is actually a really good idea.

I know that but how far would it go in inches? Also, what do you mean by elastic powered drive?

You know those little toy cars that some of us use to play with as kids where you put them on the ground then pull them backwards and then release them. they use a spring to store that winded up energy and the car went forward. I was thinking something like that but on a way bigger scale. If optimized to your size limit, you could make it go straight for probably 50ft.

A torsion spring made out of a stack of rubber bands as a drive. You may do it with 0 motors really.

Kind of like a wind up glider you may have had as a kid (until the balsa wood broke). Twist the rubber band over and over again. Let loose and the propeller spins.

Do this but for the robot on a larger scale. Have a brake to hold the wheels at the start and let er rip! It would take a lot of rubber bands to make a torsion spring strong enough to start a 10+ pound robot.

I’m not sure if i can use rubber bands in this?

Yeah I can’t use rubber bands in this.

The only constraints are that I can only use two motors. I need to maximize their effectiveness somehow. any loopholes that I could use?

I guess it all comes down to how you interface with the weight. Can you build a stationary winch system? Or does it have to be a drive base with wheels. I would think stationary winch would he a lot better because you wouldn’t have to move the weight of the robot.

If it is a drive base can the weight be put on the robot, (carried) or does it have to be pulled?

It has to be pulled by a sled. if you search “vex basebot” you will be able to see the robot I have to work with. However I can modify it.

So if you can’t carry the weight it becomes an interesting question. What material is the weight and what surface is it on. Does the robot have to pull a 10 pound weight on wheels or is it rubbery on carpet?

Its on a tile floor and the weight is one similar to what you would use in a gym (steel circle with a hole in the middle). Also we are pulling on 2.75 wheels but we may change the size of them

So on a tangent (original question actually) the gear ratios will increase torque but decrease speed proportionally. If there is a 1:2 reduction on a vex motor (100 rpm) it will have double the torque (3.34 nm instead of 1.67) but run at 50 rpm.

My rule of thumb on the 393 motors is 7 in-lb of torque per motor for good running. You may be up against that in this challenge.

  1. do you have to use Vex motors or can you get some other kind? That could get you more torque.

  2. Is this a Vex competition legal inspection or you just have Vex parts laying about to use?

  3. pneumatic kick starter

Other motors could be more powerful than Vex but you would have to buy them and integrate them with different power and H-Bridges. Lots of work, lots of expense.

Making these motors non-competition legal for Vex is an option too. Short circuit the PTC with a little wire or solder. That would get you some more operational range but may have catastrophic damage if you get it wrong.

You can also use a different battery and charger to run more volts through the Cortex. I’ve heard stories of 10 or 11 volts but do this at your own risk. Get away with overcharging and no PTC like it’s Sack Attack or Gateway! (Don’t get me wrong. I am in favor of the current rule set for competitions. This is just not that. :slight_smile: )

You could make a pneumatic kick starter to give you an initial push. The getting rolling is the hard part for these poor little Vex motors. A push start can help. Was it 62 that did that or someone else?

One other thing I thought of. Does adding wheels make the toque needs of the drive wheels lesser? I forget if the weight on that wheel causes the torque to go down or not. So 4 or 6 omnis in front to carry the load should help distribute the weight. It is only 10 pounds and people have heavy robots out there.

Doubling up the drive wheels width wise can increase traction too.

Lastly, clean the wheels prior to your run. The dirt on them from a school tile floor can cause reduced traction. And you need every bit you can.

I was thinking of using a 2:1 or 4:1 ratio (possibly a 5:1).
How do you come up with those numbers, and what do you mean by them?
Showing you have done some work is a good way to get better answers.

Is this your task?

As is common in classroom experiments, it looks like the default setup works (starts and moves).
You can consider that as iteration zero, and scale your wheel speed up or down with gears or wheel sizes to get motor speed to be closer to the max power point, which is 50% of free wheel speed.

Going off this file here:

Traction is the name of the game here, and that depends on how much torque your motors can effectively convert to traction. So is speed.

For a given motor power output, there is an inverse relationship between speed and torque, as given from one of the mathmetical definitions for power:

P(rot) = torque * angular speed
(derived for rotational motion from P = Force * speed)

What that means is, you can’t have both speed and torque. It either has to be geared for speed or for torque, and you need to find that perfect balance.

The motors also need to be able to torque against the weight of the robot and the weight. The surface is also a factor, because a rougher surface will make it harder for your robot to pull your robot. However, since you and your classmates are competing on the same surface, this wouldn’t affect the relative results. Another thing is that the hardest time for your motors is the start up, where they have to both provide torque to get the robot rolling up to speed, and to keep it rolling. It’s not as bad once it gets up to speed because the robot’s weight will naturally carry itself on. Your robot’s no good if the motors can’t survive the start-up. Don’t burn out the motors in the start.

Per your project rules, you can’t augment the motors with rubberbands, so it really boils down to three variables:
-The gear ratio
-The weight of the robot, the weight, and the latter’s rig.
-The size of your wheels (Bigger size = more speed. Smaller size = more torque)

So what I suggest is that you first try to remove as much weight as possible from your robot, then find that optimal gear ratio and wheel size. Also, see if distributing your weight across multiple wheels will help increase speed. The point here is to get your motors to run the wheels as fast as possible without tripping the internal PTC (burning them out).

I really can’t tell you how far it’ll go without knowing specifics about your robot, but my perspective is that 10 pounds is a lot for two motors.

@TerminatorTech ;
Please let us know

  1. what you ended up doing,
  2. what was the best solution from your classroom (gear ratio and distance)
  3. What was distance of the default direct drive example .

My prediction is that best case is about 6 feet * gear_for_speed_multiplier
and that you should gear up if the default distance was > 6 feet, and gear down if < 6 feet.

look at the 393 motor curve. You want the motors running at peak torque and efficiency, not necessarily peak rpm geared down. 393 peak power is at around 50 rpm, roughly power level 60. Gear down from there to get all of the necessary torque.

  • peak torque (0 rpm @ 13 in-lbs torque = zero power) or
  • peak efficiency(85 rpm @ 2in-lbs torque which is only 50% of max power.
  • I don’t understand what you mean by “power level 60”, surely the best results come from full power setting of 127 on the motors.
    Are you mixing concepts of set_motor(60) = 50rpm @ no load, with full 127 power loaded down to 50rpm?
    If not, what do you mean by “power level 60”?

My 6’ calculation is based on max power point 50RPM x 10 seconds x wheel circumference at 1:1 ratio,
and the gear up or gear down based on initial RPM (or initial distance) should make sense from there.
There is my “spherical cow in a vacuum” math, not including friction losses that would show up if you actually used gears or chains.
Anyone have different math? Or a different prediction? Or actual test data?

That is a good reminder to put one motor on port 1…5, and the other motor on port 6…10, even though the max power point of 393 is less than 2A per motor, so if you are tripping the PTC, you are doing it wrong.

If the COF of delrin on school floor is 0.1, my perspective is that it is tiny. Once we have any real data, we can make a guess at the COF.