No, electrodacus. You're letting your preconceptions override your search for the facts.
This is the complete model of the situation:
Let
t be the surface velocity of the treadmill, positive right.
Let
c be the velocity of the center of the gears, positive left.
Let
R be the radius of the treadmill gear (blueish), and
r the ground gear (pale yellow-orange).
Finally, let
φ be the angular velocity of the gears.
For the ground gear to not slip, we need
φ=
c/
r.
For the treadmill gear to not slip, we need
φ=(
c+
t)/
R.
Combining the two rules, we have an equation that holds whenever the gears do not slip:
c/
r=(
c+
t)/
R. Solving this for
c yields
c=
t*
r/(
R-
r).
The only impossible situation is when the two gears are the exact same size. This is the locked-up gearbox case, where non-slip motion is not possible.
In the case where
r<
R,
c will be positive, and the center of the gears will move in the opposite direction compared to the surface of the treadmill.
Circlotron observed the situation when
R<
r, about
R≃0.867
r. Then,
c=
t*(-7.5). Because
c is negative, it travels in the same direction the surface of the treadmill travels (or equivalently the wire is pulled). Because the magnitude of the ratio is so large, the forces are such that you need spool to have excellent traction (stiction, static friction) to ground: a heavy soldering wire spool is an excellent test case.
Then, when you pull the wire towards yourself, the spool will also rotate towards yourself but much faster. When you pull the wire 20mm, the spool will travel 20mm×7.5 ≃ 150mm, just as Circlotron described. (I wonder if circlotron agrees that the diameter of amount of solder in their spool was about 0.867 of the outer diameter of the spool? In any case, if you happen to have a heavy spool yourself, you can easily check the math here.)
Although I called
t and
c and
φ velocities, the math stands exactly the same if you consider them displacements instead. That is, when the treadmill surface moves right by
t (left if negative), the angle of the axis of the gears changes by
φ, and the spool/axis of the gears moves left by
c (right if negative).
There is no energy storage needed. You can start from a standstill, move the treadmill surface by a fixed amount, and measure how far the spool/axis of the gears moved. If the spool/gears were heavy enough with enough friction/traction/stiction, so that there was no slippage, the above formulae will hold.
There is no strangeness related to energy conservation either. If you do the heavy almost-full spool test, you'll find that it is quite hard to pull the wire. In other words, it is the treadmill that provides all the energy here, at every instant in time. It will all be spent in the friction/traction/stiction, if you do the test from standstill to standstill. All pure mechanics, no slapstick, no aether, no fancy theories. Plain ol' classical mechanics here.
Sizes and ratios do matter in practice, though. For example, if your heavy spool is 99% full (meaning, the diameter or radius of the wire in it is 99% of the diameter or radius of its outer edges), the ratio is 1/(0.99-1) = -100. This means that every millimeter you manage to pull the wire, the spool will travel 100 mm. It is unlikely that there is enough friction to see this happen; instead, the spool will slip. So, to see the phenomena better, use a spool with somewhat less wire.
I so wish BigClive would try this. He's got good cameras, nice bench setup, and suitable spools at the top of his shelves. Or maybe Dave would?
Me and cameras don't mix too well. It does look funky, and is a perfect example of how our intuition can lead us astray, which is the reason I answered to this thread in the first place.