high-strength motors and gears

Does anyone know what the gear ratio is for:

  1. a high-strength motor geared for speed
  2. a high-strength motor geared for torque


  1. how many teeth are on the larged advanced (high-strength) gears. The really big ones that you need to screw together, not the 60 tooth ones.

Gear ratio as in the ratio of how much the speed of the really tiny internal motor is reduced before output? Or do you just want the RPM and stall torque for each configuration?

The largest high strength gears come in 60 tooth. The largest normal gears come in 84 tooth, which can be screwed together to achieve the same thickness as a high strength gear, but without the option for a metal hub.

Your questions don’t make a lot of sense. What is the gear ratio of a motor?
But here are some answers

  1. 160rpm (1.6x speed of 3 wire motor, with same torque)
  2. 100rpm (same as 3 wire motor, with more torque)
  3. 84 teeth on the largest available gear, which looks like high strength if you put two of them back to back.

See also downloadable Inventor’s guide pages on the product order pages.

That information is here on the vexrobotics.com Website. The motor has two RPM ranges (100 and 160). VEX does not publish the internal gear ratio, but if the motor in the housing is running at 25,000 rpm, it’s about 250:1 at 100 rpm.

So is this. The largest “regular” gear has 84 teeth.

Looks like a triple play…

what i mean for the motor is the interior gearing. what are the ratios that it can be at. Like, for instance, I know the interior gears can be switched out so it can have more speed, and the gears can be switched out to give it more torque.

I think your question is answered above – you can gear it for 100 rpm (more torque) or 160 rpm (more speed). Does that answer your question?

i think so. so let me see if i get it. if i have a motor with interior gearing for torque and then have it run into a 3:1 gear ratio for torque on my robot, then the entire ratio would be 750:1?

Why do you want to know the ratio relative to the internal motor?

For all intents and purposes, the Vex motors are closed systems. The output of the Vex motor module is “the motor’s output” - for the sake of discussion between Vex’s aspiring engineers, ratios do not account for the motor’s internal gearing.

You don’t need to worry about the actual rpm of the motor itself. All the VEX motors are gearhead designs, so you only need to worry about the rpm at the shaft. For the 393 motors, this will either be 100rpm or 160rpm nominal top speed. Since you will rarely see maximum rpm, you should either test the configuration you wish to use, or just use some reduction factor for testing. I usually figure about 60% of theoretical maximum for planning purposes.

Once you know what you want to do with your mechanism, you can figure out the gearing and mechanism you need.

Let’s take one simple example. You have an arm 10" long and you want to raise it from the floor to about a foot high in three seconds. You can figure the pivot height yourself, but it doesn’t make much difference in this example.

If the tip of the arm starts at the floor, the tip of the arm will move a little more than 12" when raised. To simplify things, let’s just assume that the arm tip goes up 12". The total circumference of the arm tip (if it could swing 360 degrees) is pi x (radius squared), or 3.14 x 100, or 31.4". You want to move the arm tip 12", which is 12/31 or just about .4 of one revolution.

To move .4 revolutions in 3 seconds is the same as .4 revolutions in 3/60 of a minute, or .05 minutes.

That is N r/min x .05 min = .4 r, or r = 8 revolutions per minute. Your target gearing is 8rpm. If you estimate that the effective speed of the motor is 60 rpm, you could gear your arm at 7:1 (the 12 tooth gear driving the 84 tooth gear) to move your arm up a foot in about 3 seconds.

It’s almost like some engineer thought all this out before the parts were designed. :slight_smile:

The only reason i am asking all of this is because i have two high-stregth motors for driving, set for speed, and a 1:1 gear ratio using chain. the speed for the driving is perfect, but I want to use these two motors for something else and wanted to know if there was some sort of specific gear ratio inside of the motor as compared to a regular two-wire motor so as to gear from the two-wire motor so I would still have the same speed and with the high-strength motors.

The HSM set to speed outputs 160 RPM. The internal ratio does not matter because the internal motors between the 393 and the 269 are different; you’re comparing apples and oranges anyway.

In order to get a 269, which normally outputs 100 RPM to output 160 RPM, you need to overgear the motor 1:1.6 (that’s a sprocket with 1.6x more teeth on the motor as on the output). Luckily, Vex sells two high strength sprockets that give a 1:1.67 ratio.That’s the 30 tooth sprocket for the motor and the 18 tooth for the output.

Keep in mind that by using a lower torque motor, your system will have overall less acceleration and more current draw unless you use multiple lower torque motors to make up for the difference. Whether or not this is necessary is something you can experimentally determine pretty easily.

so, in short, i can replace a HSM with a 269 motor with a 30:18 gear ratio and should still have the same acceleration?

You will have the same (similar) top speed, but you will not have the same acceleration. If you want more acceleration at the same top speed, you’ll want to put multiple 269 motors into your drivetrain.

ok thanx i think i got it from here.

And then when you can’t find the “massless arm” in the vex catalog,
and want to lift some tubes at the end of the arm,
you’ll also need the wikipedia page for Power_(physics)

  • To lift a mass M against gravity field g to height h is to do Work=mgh
  • In English system, mg = weight, so 1lb weight already includes g factor.
  • Power = Work/time = torque(in-lbs) * angular velocity (rpm)

The HS 393 motor datasheet as shipped lists stall torque at 13.5in-lb;
Since both speed and torque are mostly linear in a DC motor w gearhead,
60% of speed should be 40% of torque = 5.4in-lbs at 60rpm = 1rps;
and this is the max power point to aim for.
With Rick’s example 7:1 ratio, makes this torque 7*5.4 = 37.8 in-lbs,
but at the end of a 10 inch arm its only 3.78in-lbs,
so one HS 393 motor should be able to lift 3.78 lbs up 1 foot in 3 seconds.
However, the 7:1 gear train efficiency may be only 85%,
which would limit the work output to lift just 3.2 lbs up 1 foot in 3 seconds.
Does that theory match practice? (and/or where did I mess up?)

Current should be linear with torque; 3.6A stall current * 40% torque = 1.44 amps,
so you’d better watch the power budget as well, when lifting the full load.

Even though everybody agrees the internal gear ratios of the motors don’t really matter, it did make me curious. So, I cracked open a 393 and started counting teeth. I’ve posted my results over on the wiki (here).

The short answer is that the DC motor runs at around 16,000 RPM, and that the two reduction options are approximately 160:1 (high-torque) and 100:1 (high-speed).


  • Dean

A VEX engineer told me that the free speed of the motor in the 393 is 15,680 rpm, and the 269 is 28,613 rpm. More or less. :slight_smile:

That perfectly fits the gear ratios - I’ll update the wiki page.


  • Dean