Okay, I follow that.
Here's my follow up question to anybody who can answer it:
Consider a bouncy ball dropped from above the treadmill. The higher above the treadmill it is dropped from, the longer it falls, the faster it hits the belt at.
My first assumption: The faster the ball hits the belt, the less time it spends in contact with the belt. I'm not sure that this is true, but it seems likely to me for some reason.
If this is true, the faster the ball hits the treadmill, the less time the treadmill will be able to exert force on the ball, the less it will cause acceleration in the rear-facing vector (toward the back of the treadmill). So, the faster the ball is travelling when it hits the ball, the less rearward force will be imparted. This is obviously not a linear relationship to the angle the ball will leave the treadmill or how far down the treadmill (or off it) the ball will bounce, but less it seems like the effect would be similar to a golf ball bouncing down the cart path vs. the fairway vs. the ruff (less friction in golf being compared with less treadmill contact time for the ball).
If there is an analogy in elasticity and impact in runners tendons/legs, then it seems that those runners with a more effective tendon/ligament elasticity and those with shorter ground contact time would have an easier time running on a treadmill. The reason would be that the treadmill would spend less time imparting rearward force to the runners body through the legs. In other words, assuming two runners apply the same amount of horizontal vector force per footstrike, but a different amount of vertical force per footstrike (caused by better elasticity, for instance), the runner with the higher vertical force is going to be better at treadmill running relative to normal ground running. The reason would be that the ground gives just a normal force and the treadmill actually applies a force in a specific direction?
I'm thinking that there is a logical error somewhere in this thought, but I can't tell where. My guess is that there is a corresponding frictional normal force on regular ground, or that I'm still conflating the two different inertial reference points in my ball analogy, and so it only carries over to the first few steps on a treadmill for a runner. If anyone can sort through this and give me an answer, I'll be very happy.