Shane
Kenneth Carrigan wrote:
> Thanks Jer - for sending this again
> Shane,
> I think I answered this sort of in previous post...
>
> 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