They key is this:
...solving for 'd' = (k*M1*g*h - M3*.25*r^2*w^2)/(M1 + M2)*g
notice in general the (k*M1-M3)/(M1+M2) relation. Let's just
substitiue for 1kg and 5 kg weights.... (k-x)/(6) also let's
look at 2kg he used. (2k-x)/(7). Notice that if x were close
to 1kg things look bleak, also if k were .5 things look....
>If you overlook the losses, doesn't the relationship still hold?
No, infact it plays an important part of how much 'd' moves,
energy is absorbed by losses so that F1 and F2 can change
... things are not linear when scaling.
>If the force relationship is [F1/F2]^2,
>
>If impulse F1 is 1/10th the force of F2, F2 will react as if F1 was a
>mechanical force of 1/100th.
Nope..
>If impulse F1 was 10x the force of F2, F2 will react as if F1 was a
>mechanical force of 1000x.
Nope again...
>So, if he is correct, a large ratio difference (in favor of impulse F1)
>would
>supply a huge amount of power. We could happily accept the losses.
>
>Is the impulse force ratio correct? (or has he made poor assumptions?)
Poor assumptions is correct. The weights free falling will
have an elastic effect to the weight it tries to pull and thus
will bounce back up - then down - then up... waisting a lost
of energy in a damped sinusoidal response. Calculate the
velocity of M1 just before impact and make believe your
in a car going X miles an hour. You hit a brick wall (large
mass) and what happens... you bounce back somewhat
from the wall.. right? No free energy here....
v/r Ken Carrigan