gear ratio strength

VEX V5 General Discussion
1

So I’ve been working on a robotic arm for the past few months and I need to know the strength of some gear ratios.
here’s the rundown:

  • need to know how much weight a 7:1 gear ratio can handle before stalling. (I know that the vex website says 14.76 in-lbs for stall torque, but idk what that means)
  • How much power can two motors generate without the need for gearing.
  • any way to override the motors overheating with code. (I’m asking because it turns out its controlled by a PID or something)
  • how to replace turntable gears. (my first prototype used them for the shoulder, now they’re slipping!!!)

any information about this would help a lot. Thanks!

(PS I already know that a 7:1 gear ratio improves torque by 700%)

2
  1. How far away is the weight from the center of rotation?
  2. How far away is the weight from the center of rotation?
  3. I do not believe that you can do that.
  4. I do not really know, but a picture would help
3

Just a quick bit of information. First, there is no way to use code to over ride the thermal protection of the motor. That protection is achieved via a Positive Thermal Coefficient (PTC) device located inside the motor. When the PTC heats up with the motor, it chokes off current to the motor. Over riding the PTC (which is something you would have to do manually) is totally illegal.

To perhaps help with your torque questions, consider the torque-speed curve, etc. of the 393 motor here:
https://vexforum.com/index.php/attachment/56464d7f91a13_torque_speed_393.jpg

More discussion was provided by jpearman here: https://vexforum.com/t/motor-torque-speed-curves-rev2/21868/1

It is best to never stall a motor. It will suck up a ton of current and the PTC will shut it down.

4

How much strain you are putting on the motors, extremely simply put, is equal to the weight at the end of the arm times the distance the weight is from the pivot… that product has to be less than 14.76 inch pounds (inch from the distance to the pivot, pounds from the weight on the end of the arm). However there are tons of other factors like inefficiency, friction, imperfect motors, battery level, and the weight of the arm (not the weight on the arm) that will make mathematically figuring out exactly how much you can lift almost impossible without college level physics and math…

5

Out of curiosity, if we are talking about the weight the distance is from the pivot, does that mean the weight’s center of mass? We’re talking about big objects here.

6

600% you mean?

7

Yes, in physics, objects are treated as point-masses.

As for motors and gearing, it’s really tough to predict. Build your arm, then test it and compare different gear ratios.

Side note about power: it’s speed times torque. Gearing doesn’t change the power (in theory, but friction causes < 100% efficiency), but it slows down/speeds up motors.

I’ve seen gear ratios written as both input:output and output:input. Does anyone know which is correct? It’s starting to make me concerned I learned it wrong.

8

I don’t think there’s a right and wrong way. Just be specific and consistent I usually say “21:1 for speed,” for example.

9

It’s the center of mass’s distance from the pivot point times the total weight of the arm and stars or whatever you’re lifting… trebuchets go from your train of thought… here’s a nice website about it… http://formulas.tutorvista.com/physics/torque-formula.html?view=simple also if you want to be more complex the stress on the pivot point/motor is equal to the distance between the center of mass times the weight TIMES sin of the arm’s angle in relation to the ground… so the simplest formula only says the stress on the pivot when the arm is parallel to the ground… the less parallel to the ground the less torque on the pivot is going against your motor until your arm is perpendicular to the ground where the arm doesn’t have any torque… the funny thing is this is still extremely simplified…

10

There was some discussion about this a while ago:

11

This robot isn’t being used for competition. Its for a senior final project.

12

So basically for every inch there can be 14.76 pounds before stalling. Or are you saying that’s 14.76 inches is the maximum distance to lift 1 pound?

13

The center of rotation is at the shoulder. Right now the arm detached from the shoulder but I measured 27.5 inches (.6985 meters). Also the weight of the arm is around 3 pounds 8 ounces. Again this is an estimation since the arm is detached from the shoulder.

These are the best pictures I took before I disassembled it. Hope they help.

IMG_0505.JPG
IMG_0505.JPG4608×3456 3.15 MB

IMG_0508.JPG
IMG_0508.JPG4608×3456 3.8 MB

14

In that case, I suppose you could bypass/alter the PTC by taking apart a motor. At tournaments, some robot inspectors employ a method to determine if the PTC has been tinkered with. I don’t know how a bypass is best accomplished but I’m guessing the PTC can be replaced by a normal power resistor. If you are going to do something like that, then I suppose you could also perform other things on the motor to help keep it cool. I can’t vouch for any of these ideas but kids have asked about drilling air cooling holes, adding heat sinking materials inside the motor housing, etc. all of which are illegal in a tournament. Illegal or not, just keep in mind that the PTC is placed there to protect the motor and to prevent it from burning up. There are also PTCs in the Cortex to prevent it from providing too much current on a set of output pins, so if you are doing this non-competition experiment with more than one motor using a Cortex, you might run into the problem of those PTCs tripping out, too.

If you wanted to handle the PTC motor limitations in a legal way, you might consider having a look at the Smart Motor Library. Smart Motor Library - Technical Discussion - VEX Forum

15

Well I’m using my vex teams parts for this so I can’t physically modify the motors.

16

Let me restate: I need to know how much weight two vex motors can lift with an 7:1 gear ratio. the distance is 27.5 inches.

17

The power of a motor is related to roughly the product of its torque times the speed at which you operate it. To run the motor at its max output power, you want to consult the torque-speed curve (linked in my post above) and look at the Power curve on that graph. You’ll see that you get the most power out of a motor when you are running it at about 7 inch-pounds of torque. That gives you a motor speed of roughly 60 rpm, or roughly 1 rev per second. (Keep in mind these are all ballpark numbers because of friction, battery condition, etc.) . At max power, one motor with your 7:1 ratio should give you about 49 inch-pounds of torque but, because it is geared down, you will lose speed to about 1/7 rev per second. Two motors will provide about 98 inch-pounds of torque at that ratio. 98 in-lb divided by 27.5 inches will give you a rough idea of how many pounds you can lift at that distance. But friction can play an important role and you can do fancy things with rubber bands to assist your lift so the motors aren’t strained.

18

I only skimmed the responses so I’m sure this will retread. Here’s the basics.

Torque is equal to Force times distance. To fully get this, you need to treat force and torque as vectors and do cross products, but you can simplify here. For an arm in FRC, all of the torque the arm applies to the gear reduction is the weight of the arm itself times the X component of the distance between the arm’s center of mass and the pivot. So in the worst case scenario (horizontal arm, assuming all of the load of the arm is placed at the end of the arm to be safe), the torque is the weight of the load times the length of the arm. A 100 inch-pound torque would be induced by a 10 pound load applied at the end of a 10 inch long arm at horizontal. For the motor to move the arm, it has to output a torque greater than the torque caused by the load of the arm.

DC motors output their peak torque at stall (zero speed) and their peak “free” speed at no load (zero torque). You can actually plot speed versus torque as a straight line going through the points (free speed, 0 torque) and (0 speed, stall torque). This will give you an approximation of the adjusted speed when you load your motor to any portion of its stall torque.

When using a motor at full throttle you will always be at some point on this curve. When a motor is at stall, it draws its stall current, and when the motor is at free speed, it draws its free current (usually very low relative to stall). This is important because the thermal breakers in the Vex system limit how long you can draw a sustained current, and this limit is below the stall torque of multiple motors. So keep that in mind too. I dont’ know these limitations off the top of my head.

If you had no current limitations, and you wanted to move the arm at the perfect ratio of speed to torque, you would try to gear the arm to operate at max mechanical power (or slightly below this to allow for inefficiency). This would be the speed and torque provided at half of stall torque / free speed. To be safe, maybe 70% of free speed / 30% of stall torque.

So what is this torque? If you have a 7:1 reduction, and 14.76 in-lb of torque at stall, and 100 rpm free speed (I’m not pulling these numbers from anywhere but this thread, check them yourself), then at 30% of stall torque the motor would run at 70 rpm and with 4.43 in-lb torque. With a 7:1 reduction, that results in 10 rpm for the speed and 31 in-lb of torque. We’ve got a 28 inch long arm, so the working load it can easily lift assuming all of it is applied at the very tip of the arm would be 1.1 lbs. It could lift more than this (1.845 lbs at max power, 3.69 lbs at stall torque), but when picking a gear ratio you should try to set the optimal point around here. Keep in mind I didn’t account for the weight of the arm itself in these calculations - you could assume the weight of the arm is uniform throughout the arm and subtract that weight * half the length of the arm from the torque to figure out your actual working torque.

How can you increase the torque available? Either reduce the gearing (slows the arm down, more torque), or add more motors (adds power to the system, basically doubling the stall torque), or counterbalance the arm with springs, reducing the load placed on the motors by applying force / torque in the direction opposite of gravity.

Hope this helps.