[BrianG]

I agree with Squee; they will drop. Just think about it for a minute and it will make sense. At no load, the batteries will have their highest voltage. When you take off, the motor is drawing the most current so the battery voltage will drop the most. Then, as the truck reaches the speed the motor wants to be at, the current draw will decrease and the battery voltage will rise again. It won't rise to its unloaded value since there is still some current draw, but much less than take-off current. At the motor's "sweet spot", it will be the most efficient. This does not mean it will draw the least current then, it just means that most of the battery power will be on the motor rather than dissipated as useless heat on the ESC.

This is probably a bit off-topic and outside the scope of this discussion, but a battery is really a constant voltage source (meaning voltage stays constant no matter how much current is drawn) in series with a small resistor. This resistor is not an actual component, but can be represented that way for this example. When you hook a load up to it, that load is another resistor. This creates a voltage divider. Basically, all that means is that these resistances create a current. This current drops a voltage on each resistor, and this drop is proportional to the resistance value.

Example; Let's say we have a GP3300 sub-C battery. It's unloaded voltage is 1.4v. Now, let's say it has an internal resistance of .01 ohms. Doesn't sound like a lot does it? Now, let's say we are drawing 50A of current from the ESC. This 50A will drop 0.5v (vdrop=amps*resistance, or 50A*0.01ohms) across the internal resistance of the battery. This means the battery will be 0.5v less than it's unloaded value of 1.4v, or, 0.9v.

Less current draw = less voltage drop and more voltage at the battery terminal.
More current draw = more voltage drop and less voltage at the battery terminal.

I've attached a pic that visually shows what I am talking about above. Hopefully this made sense.