pic: Plywood 8WD Concept

[Deke]

Theoretically you can triple load a pinion gear as the drivetrain rocks on one set of wheels or the other as a worse case scenario. If both sides of the gearbox are loaded evenly, its more of a double loaded worse case scenario.

[Adrian Clark]

Originally Posted by Infinity2718

Theoretically you can triple load a pinion gear as the drivetrain rocks on one set of wheels or the other as a worse case scenario. If both sides of the gearbox are loaded evenly, its more of a double loaded worse case scenario.

The worst case load scenario would be during a high speed direction change in high gear. The pinion could see upwards of 6x stall torque.

-Adrian

[tcjinaz]

Wow, what responses for a summer post.
I was only thinking what wonderful designs two heads can come up with...

[Deke]

Originally Posted by tcjinaz

Wow, what responses for a summer post.
I was only thinking what wonderful designs two heads can come up with...

I enjoy these discussions, it encourages thinking, growing, learning, and corrections. Its always a humbling experience, but that's where the growing comes from.

[Deke]

Originally Posted by Adrian Clark

The worst case load scenario would be during a high speed direction change in high gear. The pinion could see upwards of 6x stall torque.

-Adrian

Yes, thanks for the clarification. So in a typical gearbox that AM or Vexpro sells, that would be 2x the stall torque. So this layout would be 6x stall torque or triple of what the gear usually sees.

[Gregor]

Originally Posted by Adrian Clark

The worst case load scenario would be during a high speed direction change in high gear. The pinion could see upwards of 6x stall torque.

-Adrian

Could you explain where the 6x number came from?

-Curious Student

[AdamHeard]

Originally Posted by Gregor

Could you explain where the 6x number came from?

-Curious Student

2x stall (as that's what occurs when you give full voltage and the motor is at free speed in the wrong direction, the back EMF of the motor is going the "wrong" way and adds to the supply voltage so you get double voltage, double current, double torque).

3 motors, as the last pinion transfers the torque of the other 2 motors.

That works out to 6x total, and really points out the danger of passing load down pinions.

[s_forbes]

Originally Posted by AdamHeard

2x stall (as that's what occurs when you give full voltage and the motor is at free speed in the wrong direction, the back EMF of the motor is going the "wrong" way and adds to the supply voltage so you get double voltage, double current, double torque).

This doesn't sound quite right... you see max current and torque at a stall condition (motor speed = 0 rad/s). Torque at free speed of a DC motor should be ~0. Are there by chance any experiments that confirm this is the case, or theory to back it up? Where's Ether?

I would expect to see very little torque on a motor pinion if the robot is going full speed and full voltage in the opposite direction was applied. Although, I've been wrong many times before...

[AdamHeard]

Originally Posted by s_forbes

This doesn't sound quite right... you see max current and torque at a stall condition (motor speed = 0 rad/s). Are there by chance any experiments that confirm this is the case, or theory to back it up? Where's Ether?

I would expect to see very little torque on a motor pinion if the robot is going full speed and full voltage in the opposite direction was applied. Although, I've been wrong many times before...

I design motors, it's right

I'll break it down a little more thoroughly. Regardless of how it got that way, if a motor is rotating at some speed, it will generate a voltage (the back EMF). So lets say you were going full speed one way, and then decide to go full voltage the other way. Right before the switch, the Back EMF opposes the battery voltage and you get a very small voltage differential to drive current (which is equivalent to the current required to generate the torque required to match the friction).

Let's say the battery is at 12V, and the back EMF is 11.8 (made up numbers), some current calculated by V=IR would flow.

When you switch the polarity, the 12V and 11.8V no longer oppose each other, you now have 23.8V across the motor. You once again find I from V=IR, and find that this I is ~ double the I that would be calculated at usual stall conditions. Since Torque is proportional to current, you will therefore see about double stall torque.

Another way to visualize this is to extend the classic torque/speed graph to include negative free speed, you'll see the torque there is 2x the stall torque.

Extending it past full speed positive shows that you need to apply external torque to get a motor going faster than free speed as well.

An important takeaway from all this is most FRC systems are capable of seeing 2x stall torque if they reach full speed one way, and negative full voltage is applied.

[s_forbes]

Originally Posted by AdamHeard

Another way to visualize this is to extend the classic torque/speed graph to include negative free speed, you'll see the torque there is 2x the stall torque.

This looks like a sensible way to demonstrate the point. I did some googling for a bit to try and find an extended motor curve that goes into the negative rpm region, but haven't turned up any good results yet. I'd be interested to see how the power and efficiency curves fit on the negative side as well.

[Deke]

I would like to crunch some numbers here to see what kind of loading the cim pinions usually take, i have not looked at the numbers before to see where they stand. It seems to me that the simple and conservative way to approach this is to use the stall torque at a steady state, but there are some dynamics involved here. Now a question would be how much of an impact do the system dynamics play into the tooth loading? When the motors reverse is there a ramp or instant loading? Do the wheels slip? I think if those parameters can be modeled, an accurate calculation can be made regarding the loading.

Does that logic make sense or am I missing something?

[Ether]

Originally Posted by s_forbes

Where's Ether?

MIA I guess :-)

I agree with Adam's analysis with one minor quibble: with 2x stall current you may not get 2x stall torque. Depending on the motor design, the magnetics may saturate before that. Does anyone have stall torque test data for CIM at 24 volts?

[AdamHeard]

Originally Posted by Ether

MIA I guess :-)

I agree with Adam's analysis with one minor quibble: with 2x stall current you may not get 2x stall torque. Depending on the motor design, the magnetics may saturate before that. Does anyone have stall torque test data for CIM at 24 volts?

Very fair point, and rather likely actually (that saturation occurs, leading to less than 2x but greater than 1x usual stall torque).

[AdamHeard]

Originally Posted by AdamHeard

Very fair point, and rather likely actually (that saturation occurs, leading to less than 2x but greater than 1x usual stall torque).

Also worth pointing out that in especially regards to drive with so many paired motors, you will have 4 or more motors trying to draw near double stall current, over 1000 Amps for 4 cims. This will cause a large voltage drop, until the current draw reaches an equilibrium with that. This will greatly reduce the Torque.

My hunch is you will still experience greater than 1x Stall torque at nominal voltage, but not the classical 2x I originally stated.

[magnets]

Originally Posted by AdamHeard

Also worth pointing out that in especially regards to drive with so many paired motors, you will have 4 or more motors trying to draw near double stall current, over 1000 Amps for 4 cims. This will cause a large voltage drop, until the current draw reaches an equilibrium with that. This will greatly reduce the Torque.

My hunch is you will still experience greater than 1x Stall torque at nominal voltage, but not the classical 2x I originally stated.

Your hunch is definitely right. A robot that doesn't have enough torque to spin its wheels when its up against a wall can break traction with the ground if you try to stop it too quickly.