One differential has two motors on one side, one on the other. Both sides are geared the same. Will any speed problems be seen? At stall? I have no idea how this will turn out and I can’t prototype for a few days.

It will only have an output torque of one motor.

Why’s that? Where did the extra torque go?

If you have a differential, one side connected to a motor, one side connected to nothing, the output will just drive the side that is connected to nothing.

In this case, the output will just drive the side with only one motor because it is easier, therefore it will only have the torque of one motor.

I realize I phrased that poorly. (Even if you understand it I’m restating it)

Two sides of a differential have motors going into them. One side has one, the other has two. The output is defined as the casing of the differential. When the differential is locked, the output spins. In this scenario, how will the torques stack up.

I’m fairly sure that if there is one motor on each side, the overall torque is of two motors (on the spinning casing) because the locked differential acts just like a single axle with a gear (the casing) on it.

So, in a way, I would think the 2-1 scenario would have torque of three because, when locked, the situation basically looks like this:

M–|-----M

M–|

Although, I do have a prototype nearly finished. (w/o motors) so I should be able to mess around during Thanksgiving break.

Well, it is a bit more complicated and, surprisingly, you both, @Mr_L_on_Yoshi and @tmwilliamlin168, are correct to some degree.

Lets look at this analogy:

If you assume that person on the left could provide a unit of lifting force and a person on the right could provide twice as much, then together they could lift a weight corresponding to the two units of force. Left person will struggle at 100% of his/her max, while the person on the right will be underutilized at 50%.

Now, imagine you add some extra weight. The person of the left couldn’t handle this anymore and will start dropping the sofa, while the person on the right will keep lifting it up. If you assume that left side drops as fast as the right side is being lifted, then the center of gravity (of the sofa) will stay in the same spot. In case of differential, when the resistance at the output is too high for one of the inputs to handle, the single motor side will be backdriven by the side with two motors, .

So far we assumed that max torque out of the motor is a fixed number. However, electric motors deliver almost no toque at the idle speed (just enough to compensate for internal friction) and, as you start adding load to their output, they will be slowing down to generate additional torque:

https://vexforum.com/index.php/attachment/56464d7f91a13_torque_speed_393.jpg

(I assume everybody has motor-torque-speed-curves diagram in the eng. notebook)

Lets look at our unlocked differential. As you increase the load on its output, the side with one motor will slow down twice as fast as the side with the two motors, such that torque coming from left side is equal to the torque coming from the right side. At some point single motor will overheat its PTC, “drop the sofa” and will be backdriven by the other side, which still have plenty of power margin.

I don’t know what @Mr_L_on_Yoshi has in mind, and assume it is not a simple unlocked asymmetric differential which would be a waste of a motor.

If there is some sort of a locking mechanism, such as brake, ratchet, or worm gear(s) then you could create very interesting system by preventing the backdriving of the weaker side. You just have to be careful not to make it too complex, because then it may introduce extra friction and reduce overall reliability.

Ah, I should have specified. This is for a drivetrain, where two sets of differentials are on the drivetrain. Each set, one on each side, has two differentials. Now, these differentials have one (or 1-2) motor/s going into each side. However, the second differential is connected by chain on one end and gears on the other. The configuration looks a but like this, where the “|” are gears and the “{” are sprockets/chain:

M-|–[DIF]–{-M

///|–[DIF]–{-

/ = whitespace (the fourm edits it out)

In this configuration, when the motors are acting together at the same speed, the upper casing spins, when the lower remains still. When the motors act against eachother, the upper casing stays still and the lower spins. I can confirm that this concept works; my prototype will attest. (just ironing out kinks now)

So, for a drivetrain, I believe this would be sound. What I got out of @technik3k 's explanation is that the torque scenario applies once at the very peak of stall and when back driving. **

@Mr_L_on_Yoshi - it is a clever way to have two “independent” outputs without employing multiple ratchets: {D1=M1+M2} and {D2=M1-M2}, but I still see no reason to have an extra motor on one side.

It feels VEX differentials are not very strong - with higher torques the plastic casing will start bending introducing extra friction. However, if you want to experiment with the differentials, here is an idea:

…|-M1…M2-|…

L–[DIF]–M3–[DIF]–R

Where L and R are the wheels, M1 and M2 are the motors driving differential’s casings and M3 is a motor driving connected inputs of both left and right side differentials.

Let say each motor delivers a unit of torque at a unit of angular velocity with the total available power P = 3 * torque * angVel = 3 (for all three motors).

Then if you drive M1 and M2 forward and M3 backward, you will get 1/2 unit of torque (going forward) on each of the wheels at {3 x Speed} thus utilizing all available motor power.

If you are to drive all three motors forward you would get 1 x Speed on both L and R, but only up to 1 x torque for each, with a total power output = 2 power units, thus underutilizing your three motors at 66.6%. M3’s PTC will be the limiting factor of the entire system. If max torque is the most important characteristic for this drivetrain scenario, you may just as well be driving each wheel directly by M1 and M2 and save an extra motor for somewhere else.

In the third scenario if you drive M1 and M2 forward, but break M3 with a small forward current (that is not large enough to trigger PTC) you would get the 1/2 x torque and 2 x speed on each of the wheels for a total output of 2 units of power.

In the real life applications they will likely use specialized devices for more efficient braking, but you have only VEX motors available, so you would need to experiment. If you are lucky a single motor may be able to “break” four driving motors without tripping PTC.

Also, if you add some clever ratcheting you could build a drivetrain that goes full torque forward and reduced torque but 1.5 times faster in backward directions (no pistons or shifting required). If you need even higher (i.e. 2x or 3x) speed difference you could do it too, but then you would get much weaker torque in backward direction, so it all depends on the driving needs vs how many motors you could allocate.

Wait… what about putting this system on its head? Power the two casings with motors and have the geared differential outputs as the outputs?

…|M1

D-|–[DIF]–{-S

…|–[DIF]–{-

…|M2

The above differentials could be described by the following formulas:

( D + S ) / 2 = M1

(-D + S ) / 2 = M2

This is essentially how differentials are defined. Then solving those for D and S gives:

S = M1 + M2

D = M1 - M2

If D and S are the sides of the tank drive this would be very similar to the tank transmissions that are used in the tracked vehicles.

You would need to run M1 most of the time while braking M2 for forward or backward motion and apply power to M2 for the steering.

I got the math (figured that out earlier :p) but D and S are not the wheels perse. D, for “drive” corresponds to one wheel where S, “strafe” corresponds to another. There would be two whole sets of these, connected partially together. A full diagram would look like this:

…|M1…M3|

L-|–[DIF]–{----S----{–[DIF]–]-R

…|–[DIF]–{…{–[DIF]–]

…|M2…|M4

So, when running M1 and M2 both in the same direction, L spins. Similarly, when M3 and M4 spin together, R spins. This creates a functioning drive. However, when M1 and M2, as well as M3 and M4 are opposing eachother, S spins. (A third set could be added here for added functionality)

I get that this concept here would work. Im just curious as to if the torques here would stack up properly. (Still strafe with 4 units of power)

Also, thanks for explaining, @technik3k

This is actually a very clever design!

If you could cut your friction losses to minimum and match the motors … this could give mecanum wheels some run for their money!

Ha! Finally a working design! XD

Say, would this final design work with asymmetrical torque too? Say, M1 and M3 really have 2 motors each? Would the outputs, when spun, have the “proper” amounts of power? I’ve done a bit of testing on my prototype and it feels like that’s the case… but I’m very unsure.