# Feynman's argument on machine limits: how are the machines reset?

Feynman makes an argument in his Lectures that loses me on one point. His argument (abbreviated):

We suppose that there is a reversible machine—which lowers one unit of weight by one unit of distance, and at the same time lifts a three-unit weight. Call this Machine A. Suppose this particular reversible machine lifts the three-unit weight a distance X. Then suppose we have another machine, Machine B, which is not necessarily reversible, which also lowers a unit weight a unit distance, but which lifts three units a distance Y. We can now prove that Y is not higher than X. Let us see why. Let us suppose that Y were higher than X. We take a one-unit weight and lower it one unit height with Machine B, and that lifts the three-unit weight up a distance Y. Then we could lower the weight from Y to X, and use the reversible Machine A, running backwards, to lower the three-unit weight a distance X and lift the one-unit weight by one unit height. This will put the one-unit weight back where it was before, and leave both machines ready to be used again!

But how are the machines ready to be used again? Doesn't that imply they're back to the states they were in at the start of the problem? Sure, the one-unit weight can now ride Machine A again, but to make his point about perpetual motion, Machine B has to be reset.

Are we to assume that when the machines don't move weights, they can be set to any state "for free"?

The 1kg and 3kg weights start at height zero. We connect them to machine B and it lowers the 1kg weight to $h=-1$ and raises the 3kg weight to $h=Y$. We now swap the weights onto machine A, and it lowers the 3kg weight to height $h=Y-X$ and raises the 1kg weight back to $h=0$.
So the 1kg weight is back where it started and the 3kg weight is at $h=Y-X$. Feynmann's point is that if $Y\gt X$ then the 3kg weight is higher than it started so we have gained potential energy from nothing i.e. we have a perpetual motion machine.