one of our mentors and i cant seem to agree on this subject. say i have a 12:1 gear ratio for a gearbox. is that for every 12 revolutions of the motor the output is 1 revolution. or for every 1 revolution of the motor its twelve revolutions as an output

Good question.

A 12:1 ratio for a gearbox is such that for every 12 revolutions of the input motor, there is 1 revolution of the output shaft.

Andy B.

thanks for the help. im sure this is an easy thing to get confused

which would give you a looow speed, but high torque:)

This depends on what motor is used, and what the application is. For example, a 12:1 reduction used with CIM motors would be fast if used for an arm rotation, but slow for a drivetrain.

As an addition to the original post:

If you ever see a reduction given in decimal format (e.g. the 2008 KOP’s were 0.0784), THAT is where the output shaft rotates 0.0784 times per 1 revolution of the input (output:input). Car transmissions usually show reductions in this manner. This particular number (0.0784) was calculated by (14/50)*(14/50), where each 14-tooth gear is the input and each 50-tooth gear is the output in their respective stages. Also note what the reciprocal of this number is ( 1 / 0.0784 = 12.75), but to flip the decimal like that you also ‘flip’ the which side is the output and which side is the input, hence (12.75 : 1) is (input : output).

Whoever lost that argument was probably headed in the right direction but confused about the nomeclature, hence the argument kept going. Durn those kinds of arguments…:eek:

When did this start? I must have missed it.

All the car transmission gear ratios I’ve seen are expressed as the number of turns of the input shaft : one turn of the output shaft. For an overdrive transmission, the overdrive ratio is expressed as a decimal because the output shaft turns faster than the input shaft.

(I read Hot Rod, not Motor Trend, maybe they do things differently?)

so which would be faster for a drivetrain a 12:1 or a 9:1

A higher gear ratio means the motor has to turn more times to get the wheels to move the same distance. So, a higher gear ratio slows down the robot more.

The lower ratio (9:1) will give a faster moving robot.

thanks.

The decimal format is just the number itself, e.g. 0.0784 and not 0.0784:1.

I’ll try to explain the reasoning so the OP doesn’t get further confused… This stems from my street racing days back in college. Some of the guys were gearbox junkies and tuned their transmissions for lower top speed with ridiculous acceleration through a set RPM range known as a ‘power band’. When they would talk about shortening a gear, this was the number they were messing with. It was always less than one, and it represented the reduction from the engine output to the shift stage rather than from one shift stage to the next (which is also sometimes seen). This allowed them to figure out if their cars were still in an ‘optimal’ power band for each gear after they made a modification to an engine component (which may have shifted the power band to a higher or lower RPM range). This tweaking is similar to the FRC electric motor curves vs the loads we put on the motors. It makes sense to me because the higher the number, the faster the output shaft spins.

I guess I just assumed this verbage can be used for all transmissions since you see that same decimal format used in almost every available FRC calculator. Maybe I was wrong.

…operating with the unwritten assumption that all other things are equal.

A robot with a 9:1 ratio and 6" wheels will be slower than one with a 9:1 ratio and 8" wheels.

I give presentations about drive train design now and again and I sometimes assume a level of understanding that’s not there yet, so I just wanted to explicitly note that gear ratio alone does not determine a robot’s final speed.

To further the point, the final speed also depends on a lot of other factors like the efficiency of the gearbox, friction and losses in the drive train, and the inertia the motor is trying to accelerate. If the rule were simply that a lower ratio gearbox = higher final speed, then we’d all be direct driving CIMs to our wheel. But, of course, a direct drive CIM is unlikely to generate enough torque to even get the robot started rolling, would take forever to accelerate up to top speed, and would probably blow a breaker attempting to get there. That’d be another trade-off you’re making with the gearbox. A lower ratio gearbox is going to take longer to get up to its top speed. It’s even going to take longer to get up to the top speed of a higher ratio, lower speed gearbox. So, in summary, simple rules of thumb are usually highly simplified and approximate.

If you want to get a quick look at how gear ratios impact speed I recommend taking a look at this infamous piece of work… http://www.chiefdelphi.com/media/papers/1469

Take long look at the drivetrain modeling sheet, especially the velocity vs time plot.

Deconstruct it, learn it, build it better.

Does anyone have some actual numbers of how fast their robot managed to go as a percentage of the max unloaded motor RPM? I mean as-built with gearbox loss, etc. I’m not interested in feet/sec. figures. I was just wondering if something like 80% was realistic.

Yeah. 9:1 will (probably, depending on the motor…) give you a higher top speed. Acceleration is a totally different ball game, and requires some really fun math that no one ever got around to teaching me. (to be totally honest, I’m not a math guy… Hence the Sociology major)

I seem to recall from my lurking there being a spreadsheet floating around that calculates acceleration as well as top speed for FRC bots… Am I mistaken? Anyone know where this tool is?

It’s pretty close. 968 did a test of actual ground speed vs calculated theoretical speed of our 2008 robot. The actual speed was approximately 80% of the no load 100% efficiency speed.

Was that calculated from the ~5500 free speed CIM RPM ? Or were you assuming some sort of amperage being pulled?

No-load = free speed. That is, the speed the motor turns at when nothing at all is connected to it. Note, that this doesn’t mean no amps are being pulled. There’s still a small amount of friction to be overcome etc. Plus, the rotor coils are constantly changing orientations and making and breaking contacts. This means the magnetic fields are constantly collapsing and reforming, which means the back emf generated by the motor will vary over time. Since the voltage supply nominally stays the same and the coils have some resistance, current is flowing whenever the back emf doesn’t exactly equal the input voltage. So all DC motors have a no-load current as well as a no-load speed.

Which brings me to another silly technical point. Since the CIM is drawing current it has some non-zero input power, even at no-load. However, since there isn’t a load output from the shaft, it has zero output power. So the motor’s efficiency at no-load is actually 0%, not 100%.

I was referring to no load in the sense of robot sitting there with wheels in the air-ie:no load on the gearbox, not no load on the motor. I was a little unclear there.

As to the second point, I meant 100% efficiency as in zero frictional losses in the drivetrain, not 100% motor efficiency.