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ReplyTo : The Nitwit
ReplyFrom : Nitpicker Timothy
Hi Steve, I'm sorry to have frustrated you.
The reason I'm doing that is because you are always finding these wonderful
solutions and I can't stand that. :) No, not really of course, but it is
really difficult, as we all know by now, to find a method to harnas these
amounts of energy. I guess, if there was a easy and cheap way we would have
found it by now. We all are hoping to find some ingenious way to get around
all the problems at once, so all solutions are welcome. But also there is a
great possibility that these solutions will fail.
The only thing I can write is, let the new ideas come and try to explain why
you think they work.
Before you read any further I have to warn you :)
TAKE SOME VALIUM TABLETS AND TRY TO RELAX...
>see my web page
>for an idea that is consistant with Steve and Timothy's objections.
I almost dare not saying it but, how accurate is your calculator?
Let me recalculate the values of the various components:
a=55.47 m/s^2 (purple) b=27.73 m/s^2 (red)
Purple x-comp = a sin( 21) = 19.88 (you got 51.62)
Purple y-comp = a cos( 21) = 51.79 (you got 20.3023)
Red x-comp = b cos(-48) = -18.55 (you got -18.9)
Red y-comp = b sin(-48) = -20.61 (you got -20.3021)
I checked all cos and sin formulas in the drawing and all seem to be OK, the
only things that are wrong are the final decimal numbers.
If you now sum the several components:
y=51.79-20.61=31.18 (This value is too high to compensate)
(You wrote the formulas where you calculated the accelerations wrong but the
answers seemed right)
If you think other angles would make your system work, I still am certain it
If you want to give it a try here are more general formulas:
t is the angle for which you took 21
a=2b (reflected momentum is twice the absorbed momentum)
remember t is between 0 and 90
x=Px+Rx=2b sin(t) + b cos(-2t)
y=Py+Ry=2b cos(t) + b sin(-2t)
only t=90 will make y equal to zero, but t=90 means the light goes straight
on, whithout touching the mirror or absorber.