Victors Non-linear!!!

[KVermilion]

Originally Posted by Manoel

Interesting data, just be aware when comparing your results with Rickertsen2's that you used a resistor and he used a motor (resistor + voltage source + inductor model). Of course, your tests are better, because this resistor setup will not interfere with the Victor's output waveform.
With your data, we can say for sure that the Victor's are linear (something the IFI guys already told us, for the matter), I'm just a little puzzled about the current "crop" at the graphic's extremes, were you using a battery to feed the Victor or maybe a different power supply with a maximum output current?

Well, the "crop" (where the current levels flat?) must be due to the victor having a (rather large) dead area of the ends at the extreames. I used the standard "EX18-12" SLA battery from exide (the one from the kit) This battery has put out over 65 amps, so I don't think that was an issue.

Yes the victors are linear, but the motors are not... power curves.

[Gdeaver]

In the current test, was the Victor operating with the default IFI calibration? If so it would be interesting to see if calibrating the victors under program control would remove the flat part of the curve . First gave us the means last year with the inclusion of the current sensor to implement a constant current motor control . Did anybody try it? If so did it help?

[Al Skierkiewicz]

Wow, there is a lot of stuff to answer in this post.
First to Manoel, hey you passed the test. The correct answer is total resistance is .008 ohms. When I answer questions I usually use the original question as part of the answer. It helps lower the confusion level. 8 milliohms doesn't sound like much until you try and draw stall current through it. V=I*R=130*.008=1.04 volts. I like using the wire analogy since most everyone understands that more wire is a problem. .008 ohms=8 ft. of #10 or two conductors of #10, 4 ft. long.
Andy, in most cases we don't worry about the resistances since they will add up to whatever they add up to. If the robot runs OK ignore them. If it doesn't, we need to know that there are losses and approximately what they are for troubleshooting. If you measurre 10 volts at the motor in stall, that seems OK. (We can discuss all circuit losses on another thread if you like.) Measure 4 volts and you know there is some really big problem somewhere.
K, nice owrk on the graphs. As to the non linearity at the ends of the graph and the center offset, I would you say you did a nice job of demonstrating why I always calibrate the Victors. Theoretically, a calibration would have fixed both ends and the center at the same time. Can you calibrate and regraph? The pulse drawing is a great visualization of a 75% duty cycle. Since the Victor switches at 120 Hz, 1/120=8.3333 mSec as you have drawn.
As to the other incongruities, you have all read and seen the graphs of current vs. time for an inductor above, but you do need to remember there are many more variables in a simple robot system to consider. If your robot runs near what your design speed was calculated to be, and is controllable, then practice like mad. If you can't controll it or it doesn't run straight, then start to look for problems.
Finally, Andy, the math on this doesn't simplify for voltage as you thought. The voltage term in this equation is a time dependent function because you are solving for current vs. time. So very simply you have I(t)=V(t)/R. You can still manipulate that equation as for Ohm's law if you keep the time dependent reference in. i.e. V(t)=I(t)*R. Theoretically the internal resistance of the battery, all the wire and connector losses, and the series resistance of the Victor should be included in the calculation for current to be accurate. For simple analysis in this discussion, those losses were ignored.

[KVermilion]

Originally Posted by Al Skierkiewicz

K, nice owrk on the graphs. As to the non linearity at the ends of the graph and the center offset, I would you say you did a nice job of demonstrating why I always calibrate the Victors. Theoretically, a calibration would have fixed both ends and the center at the same time. Can you calibrate and regraph?

Yeah, I'll have to figure out how to actually calibrate it, first

As long as about 5-3 ticks each way of deadzone can be kept, I don't see why not.