Messing with the motor outs (A and B)

Just a thought:

(lets call the four motor outputs A1, A2, B1 and B2)

What would happened if I connected a GM10 (with the original spring still attached) between B2 and A2? Like in the illustration below

I was hoping to achieve the following:

 Motor A Motor B GM10 position Stop Stop Center Stop Forward Rightmost Stop Backwards Leftmost Forward Stop Leftmost Forward Forward Center Forward Backwards Leftmost Backwards Stop Rightmost Backwards Forward Rightmost Backwards Backwards Cente

Or would I just fry the L293D?

Strangely I think it might
Strangely I think it might work. The motors will work more slowly than normal, since they are sharing in some cases. Might be enough were some motors stop since the other is having less resistance, taking more current.

I completed your logic table for two GM10’s.

 cmd A cmd B 4 = A1 5 = A2 6 = B1 7 = B2 A1 - B1 A2 - B2 halt halt sink sink sink sink center center halt backward sink sink sink source center leftmost halt forward sink sink source sink leftmost center forward halt sink source sink sink center rightmost forward backward sink source sink source center center forward forward sink source source sink leftmost rightmost backward halt source sink sink sink rightmost center backward backward source sink sink source rightmost leftmost backward forward source sink source sink center center

The bold columns are new. I also included the electrical states for four pins on the picaxe (28x1) that correspond with A1-B2.

The real question for me is how the resistances in the motors compare to each other. And if it matters at all. Consider the fourth line in the table.

 cmd A cmd B 4 = A1 5 = A2 6 = B1 7 = B2 A1 - B1 A2 - B2
 forward halt sink source sink sink center rightmost

Motor A will run forwards because lead A1 is sinking current wich is sourced by lead A2. The GM10 connected to A1-B1 sees two sinking leads and therefor will not get any voltage or current. The other GM10 (A2-B2) sees lead A2 sourcing and the other lead, B2 sinking. But lead A2 is now sourcing current for two motors. Which motor shall receive the brunt of it?

If Motor A (unspecified type) draws all the current because it is some heavy duty, no resistance to speak of, kinda supermotor, the current will choose that path. The GM10 will not move.

If the situation is the other way around, the GM10 will draw all the juice and Motor A will remain still.

And even if all motors are nicely balanced (say they are all the same make/model/type/age), you would see some differences under varying motor loads. A stalling motor would draw away all current from other motor(s) that happen to be connected in parallel at that moment.

Still, this would be an awesome hack. Maybe a diode or two would make it work nicely. Even if it would reduce the number of permutations in the logic table. Or would it?

Stalling, yes, that might be

Stalling, yes, that might be an issue, since the GM10 stalls when it reaches the end of its range (max left and max right).

But, if your theory is that I wont fry the L293D, Ill give it a shot.

Thank you Robologist and rik

To stay on the cautious side, youd could breadboard the whole cricuit, send pwm signals to the L293 inputs, measure current flowing throught V+ into the chip, maybe even put in a fuse at say 1 or 2 A.

Program your Picaxe to give off a modest mark/space rastio at first, slowy crank it up while cycling through the permutations in the table.

Oh man, what a nerd I am!