Help with gear ratios

Hi, guys, I’m sorry to seem like a rookie but my friends and I are having problems understanding gearboxes. From our understanding, the first stage of most drive gearboxes (the tough box) includes two motors each with their own
gear attached to it. Both gears act on the same middle gear. We tried treating it as a planetary gear system without a ring gear but we just got more confused. How do we calculate the ratio of a gearbox with two 10 teeth “planets” that are spinning at the same speed in the same direction as a 40 tooth “sun” gear? Is it a 4:1 ratio (40 for driven/10 for the drive) or 2:1 (40 for driven/adding both 10 tooth gears up.)

I also read that no matter how many motors are added, the speed of the output won’t change but the torque will increase. For example, 2 motors mean twice the torque 3 means thrice and so on. I don’t really understand how this works, shouldn’t adding more motors increase the speed of the output as well?

Thank you so much

To calculate this ratio you take the # of teeth on the gears on the motor and then divide that by the # of teeth on the gear that you a driving with the motor and small gear so in your case this 40 tooth gear. This should give you a number of .25 which means you have a gear ration of 4:1. This means that for every time the gear on the motor turns once the large gear will make a 1/4 of a turn. This means that the large gear is spinning 4 times slower than the motor speed.

Edit: This video will probably do a better job explaining it than I can.

Theoretically adding motors will do nothing to increase the output speed of the system. The only things that will do this is changing the type of motor and changing the gearing.

Think about it this way:

You have the two motors going at, say 400 RPM. That means each of those little gears is going to turn at 400 RPM. If we have 1:1 ratio, that means the middle gear is going to go at 400 RPM as well. There’s no way for the middle gear to “out-turn” the outer gears as they are the ones pushing it. However, with two motors, you have two things pushing the middle gear instead of one - hence double the torque.

First off, most gearboxes used for drivetrain applications that you’ll encounter in FRC won’t be planetary gear systems. Please don’t treat them this way. The only common ones that are would be the VersaPlanetary from VexPro and the Banebots Planetaries from Andymark. As you can see, their names literally note the fact that they’re planetary gearboxes.

To better understand motors and gear ratios, you can look at this resource here: [1]](
It may be from VEX EDR, but the same principles and concepts transfer over to FRC. Take some time to explore beyond that unit to learn some other fundamental topics commonly used in robotics as well.

Once you understand the underlying concepts behind gear ratios and gearing, you can use the JVN Design Calculator to make better educated decisions on how you gear your robot.
For example, the Toughbox Mini used in the Kit of Parts drivetrain is a two stage, single speed gearbox. Stage means the number of reductions in the gearbox to get the desired output. The way these work is much more straightforward than a planetary gearbox.

Assuming that you’re using a standard setup for the KoP drivetrain (6" HiGrip wheels, 2 CIM motors per gearbox) and a max weight robot (154 lbs which is a 120lb robot + 10lbs of bumpers + 14lb battery), you get something that looks like this: [2]](

Note that all the salmon colored cells are spots that are customizable. When designing, you can experiment and plug in different gear ratios, motors, wheel sizes, and more to see the resulting speed and current draw.

On the left there’s 2 stages of reduction: 14:50 and then 16:48 which has an overall gear ratio of 10.71:1. A ratio of 10.71:1 using 6" wheels results a free speed of 13.02ft/s and an adjusted speed (assuming a speed loss of 81%) of 10.55ft/s, which will be approximately the speed your drivetrain will go in the real world.

Note that you have a pushing current draw per motor of [EDIT: 71.14 amps], which is pretty safe. [EDIT: It’s best to make sure that your PDP breakers can tolerate your current draw, see post below for details] Current draw is distributed among your motors, so having more than one for your gearboxes is usually pretty safe. Stay with two CIMs per gearbox (total of 4 for your drivetrain) since having three CIMs (with a total of 6 for your drivetrain) can cause your roboRIO to brownout at higher speeds.

A common rookie mistake that I want to point out with gearing is that they’ll select very large diameter wheels in order to go “fast” (usually 8" or more). However, its harder to turn larger wheels and it significantly increases the current draw on your motors when your drivetrain is stalled (sometimes they don’t even move). Stick to smaller wheels and you’ll experience much more success on the field.

Hope this helps!

This is some bad advice. 80A per motor is waaaaay too high. Sure the CIMs themselves can probably handle it, but the 40A max breakers cannot. According to this datasheet, they will trip at 200% in ~1 second, which means you are more than likely to trip breakers when geared that high. You also need to pay attention to the total current draw relative to the 120A main breaker. In this case, 4 motors at 80A is a total current draw of 320A, so it will trip in ~.5 seconds. You are also likely to brownout the roboRIO, considering you’re going to pull more total current than a 6 CIM drive geared to the same speed because the motors are running at a much less efficient spot on the curve.

If you’re gearing conservatively, you’re going to want to stick below 40A per motor. If you’re willing to push it, you can go up to ~50A since the breaker can technically hold 135% of rated current indefinitely (according to the datasheet). You’ll still want to watch your total current draw for your main breaker and avoiding brownouts.

So to find the gear ratio itself, it’s just the driven gear teeth/ the drive gear teeth, but to find the torque, it’s the radius of the driven/the radius of the drive *2

I just want to make sure I understand completely thank you so much.

In this case, the torque from each motor is multiplied by two because there are two motors. If there were 3 motors it would be *3. If there was only one motor it would be *1.

Basically, the equations are:
ratio = driven/drive
speed = input_speed/ratio
torque = input_torque*ratio

In this case, the input torque is really the input torque per motor times the number of motors (which is where you get the *2 from).

Omg, thank you so much that clears everything up for me.

I think I was getting confused by mixing torque and speed together but this really explained everything.

If only there was a Chief Delphi equivalent of Reddit gold I could give to you. Thank you so much

Glad I could help. It’s not quite reddit gold, but ChiefDelphi does have a reputation system (like a karma system on other forums). If you look in the top right hand corner of each post, there are three symbols. If you click on the scales, you can add to someone’s reputation. Normally, you can give them positive rep (upvote), negative rep (downvote), or neutral rep. That changes the number of green dots that appears in their signature, and it serves as a way to tell who is known for being helpful and giving good advice. Only the person you’re giving the rep to (and admins) can see what you write. As a new user you can only give neutral rep for now, but after a few more posts you’ll be able to give positive and negative rep too. I’d suggest reading ChiefDelphi’s FAQ on the reputation system here (the whole FAQ is very helpful as a new user if you have some time to read through it).

Actually if you were to accurately measure the output speed of a transmission with one motor and again with two of the same model motor you will see an increase in output shaft speed.

If you had a motor that had a true free speed of say 5000 rpm and then connected it to a 5 to 1 transmission you will not see the calculated output of 1000 rpm. Why? Friction. That friction is due to both the bearings and the gear to gear interaction. When you add a second motor you do add another friction point with it’s gear to gear interface but you don’t add any more friction to the bearings. (in fact proper placement of the second motor can reduce the friction on the driven gear shaft) That means that while the total friction in the gear box increases with the additional motor it does not double.

So back to our 5000 rpm motor on a 5 to 1 transmission. Put one motor on it and the actual output rpm may be say 950 rpm instead of the calculated 1000 rpm. Put an additional motor on that transmission and the actual output speed will rise. No it still won’t rise to that theoretical 1000 rpm but it could rise to say 970 rpm.

^^ This, mostly. In 2015, we designed for 40A peak draw per CIM. The past two years, we have designed based on a peak draw of 50A per CIM for up to four CIMs at at time. We never tripped a breaker, nor had a brownout in the first twenty minutes with a battery, excepting only the incident where we totally bunged up the programming and smoked our rope-climbing CIM by ignoring a limit switch and stalling it at full voltage in reverse when we weren’t even climbing*. As such, I consider 50A peak loading per CIM for 4 CIMs to be pretty conservative.

  • Ratchet wrenches sometimes have unintended, but useful consequences. Play the breaks when you get them.

This is at best a second order effect, where the fraction of energy dissipated by friction is reduced a bit by greatly increasing the amount of energy on the drive shaft. Adding a second motor to a gearbox essentially doubles its torque (and acceleration, if the wheels don’t slip on the carpet), but increases the top speed by a few percent at best.