[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.