Help with Gear Ratio Calculations

[kklokreach]

Hello everyone, sorry I know this is a very rookie question but I've searched everywhere and can not find the answer. How exactly do you calculate the gear ratio for a gearbox by hand?? I understand that it is Input gear/Output gear but I don't understand which constants to use and what to do when converting because when I do the math I get FPS and Torque output to be in the thousands.
Thanks in advance!

Nick

[DonRotolo]

Can you show us your work so we can try to spot the error?

[MattC9]

Now are you wanting to do it by hand? Because there are gearbox calculators out there for FRC applications that I would be happy to get you a link.

[Rauhul Varma]

Originally Posted by kklokreach

How exactly do you calculate the gear ratio for a gearbox by hand?

The gear ratio of a gearbox is usually expressed in turns of the input to turns of the output. To find this you divide the number of teeth on the bull gear (the gear on the output side) by the number of teeth on the pinion gear (the gear on the input side). For each additional stage you multiply the ratios of each stage together. So if your first stage was 12:48 (input : output) and the second stage was 20:40, your overall gear ratio would be (48/12)*(40/20):1 which is equal to 8:1.

For gears in series like this:


You only need to use the number of teeth on the first gear and the number on the last gear as the intermediary gear(s) do(es) not affect the ratio. For example if the above picture was a 12 tooth gear connected to a 40 tooth gear connected to a 72 tooth gear the ratio would be (72/12) = 6:1 which is the same as if you calculated each pair's ratios and multiplied them together; (40/12)*(72/40) = 6:1

Originally Posted by kklokreach

I understand that it is Input gear/Output gear but I don't understand which constants to use and what to do when converting because when I do the math I get FPS and Torque output to be in the thousands.

First lets define some variables:
tau(i) = your motors' stall torque
tau(o) = your gearbox's max torque output
omega(i) = your motor's free speed (unloaded rpm)
omega(o) = your gearbox's max rpm output
r = you gearbox's overall gear ratio
R = your wheel radius (in feet)
v(o) = your robot's maximum translational velocity

To get the maximum torque the gearbox will output you simply multiply your motor's stall torque by the gearbox's gear ratio:

tau(o) = tau(i) * r

To get your robot's maximum speed you first need your max rpm which is:

omega(o) = omega(i) * (1/r)

Then to get the max speed you need to convert the rpm (angular velocity) to translational velocity, which is:

v(o) = omega(o)*(2piR)/60

(note you're robot will never be traveling at that speed as you will have energy loss due to friction and other sources and your motors will always have some amount of load on them as well which will make them rotate more slowly than they do at their free speed)


The simplest way to do all of these calculations would be to just use a premade excel spread sheet like JVN’s design calculator which can be found here:

http://www.chiefdelphi.com/media/papers/2755

[Monochron]

More shameless self-promotion!!

http://www.chiefdelphi.com/forums/sh...d.php?t=128531
I made this handy motor / gearbox reference sheet that has most of the equations you need for gearbox calculations.


If you want more specific numbers (factoring in things like ideal voltage vs. actual voltage, COF, etc.) you will need to find some of the other reference sheets here on CD. If that is what you are after trying searching for "Motor Data" or "MCALC". If you can't find the right thing, I can probably link them for you.

[kklokreach]

Thanks this helps a lot, i finally got the numbers to come out right. Though now I have another question, how does the number of motors affect the speed and torque of the drivetrain? Of course this means more torque and speed due to an extra motor, but how does this translate to the fps and torque mathematically?
-Nick

[Tyler2517]

I don't think adding a motor directly increases your speed. It does increase acceleration as you are adding more power but the max rpm of your drive will not be increased. It does add pushing power but in most frc scenarios drive trains are traction limited and not limited by the motors but the do help. Most teams use the extra acceleration to reach higher speeds. Correct me if i am wrong.

Originally Posted by kklokreach

Thanks this helps a lot, i finally got the numbers to come out right. Though now I have another question, how does the number of motors affect the speed and torque of the drivetrain? Of course this means more torque and speed due to an extra motor, but how does this translate to the fps and torque mathematically?
-Nick

You can add as many motors as you want* and your final speed isn't going to change. Motors increase torque, but not output velocity (if you have one motor at a thousand rpm, and you add another motor at a thousand rpm, the first motor isn't suddenly going two thousand rpm). Torque is calculated linearly. Double the amount of motors, double the amount of torque.

*This is assuming all motors have the same or similar output speeds. Of course adding a motor that's twice as fast as the original motor is going to speed things up a bit, but that's a situation you should rarely come across in FRC. CIMs and Mini CIMs, for example, are about the same speed.

[kklokreach]

Originally Posted by Andrew Lawrence

You can add as many motors as you want* and your final speed isn't going to change. Motors increase torque, but not output velocity (if you have one motor at a thousand rpm, and you add another motor at a thousand rpm, the first motor isn't suddenly going two thousand rpm). Torque is calculated linearly. Double the amount of motors, double the amount of torque.

*This is assuming all motors have the same or similar output speeds. Of course adding a motor that's twice as fast as the original motor is going to speed things up a bit, but that's a situation you should rarely come across in FRC. CIMs and Mini CIMs, for example, are about the same speed.

Oh ok, so if you add M amount of cim motors then your torque increases by xM? Like having two CIMs means double the amount of torque and three CIMs triples the amount of torque?
-Nick

[DonRotolo]

Originally Posted by kklokreach

Oh ok, so if you add M amount of cim motors then your torque increases by xM? Like having two CIMs means double the amount of torque and three CIMs triples the amount of torque?
-Nick

Yes, almost exactly.

In the real world, additional loss due to friction is a consequence of additional motors, but compared to the motor output it is very small and can safely be ignored in most cases.

[Chris Fultz]

the on line calculators are great - but it is good to take some time to do some hand calculations so you understand what is happening in the gearing system. after you understand, then the calculator tools are great for making faster decisions and playing "what if'.

[highlander]

Originally Posted by Chris Fultz

the on line calculators are great - but it is good to take some time to do some hand calculations so you understand what is happening in the gearing system. after you understand, then the calculator tools are great for making faster decisions and playing "what if'.

Can't agree more.