Do the torque balance equation, with the pivot at your feet. And you might as well assume that you’re horizontal, because you are for part of the pushup, and that’s where the maximum force will be. If we assume for simplicity that your arms are at one end of your body and your center of gravity is halfway down, you’re pushing half your weight. Given runner pat’s experiment, though, this is probably not a justified assumption.
This is true, but that horizontal distance is going to vary as the cosine of your angle from horizontal. Doing pushups against a slightly inclined wall (so that you are vertical at the top) would be extremely easy, as nearly all your weight would stay on your feet.
You’re right, but so is Machine Elf. By changing your angle relative to horizontal, you’re moving your center of gravity.
Standing straight up your center of gravity is, necessarily at your feet (otherwise you’d fall over). So as you get closer to vertical (heads up) more and more of your weight is carried by your feet and less by your arms. If you were to raise your feet instead, more of the weight would be carried by your arms.
It’s been a LONG time since my high-school physics class, so there’s a good chance I’m saying this totally backwards.
Another data point - 156 currently, in push up (top and bottom positions the same) 118, about 75% of my body weight. RP’s was 78% of his body weight. The difference probably related to differences in our centers’ of gravity (RP’s been disabled for a few years and does a lot of handcycling so likely has a center of gravity located higher up).
Kevbo, you’re making the same mistake I did. Tilting the floor is not the same thing as extending your arms further. The horizontal distance from your feet to your arms will vary with cos(theta), but so will the horizontal distance from feet to center of gravity, and it’s only the ratio of those two numbers that matters.