Where is an error? Help please!
(One more same problem What is not correct in this optic problem?)
I know other simple solution this optic task (see please bellow), but I would like use this:
Simple solution - symbolic (by WolframAlpha.com site) and numeric (with nanotechnology help
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ACCEPTED SOLUTION
Accepted Solutions
Seems to work for negative values of x only.
Not sure and I had not tried (Prime is crashing all the time with your sheet) but maybe you run into the same problems with tan as I did in my solution above (y' gets infinte when x=0)
WE
What do you mean by this equation
and where does it stem from?
And why do you think that the argument of y´ should be a length (x(alpha)) while the argument of y is an angle?
If you write y´ here it has in any case to be the derivative with respect to alpha.
WE
You may be able to keep alpha as the independend variable, but you can't use y'(x(apha)).
What you mean is
which Mathcad can't unserstand and you will have to replace it by
If you do so, other errors occur, though.
But I see you already rewrote your equations to accommodate for x as the independent variable, anyway.
WE
From unnecessary ballast freed system but I don't understand what the error message would like to tell us!?
Looks like Mathcad does not like both derivaties to be in one equation.
Here we go.
It was necessary to introduce a dummy function and to redefine tan(). Not convincing, but at least it works.
Werner
I think you need to do it like the following Valery:
Note that the gradient is infinite at x = -F, which leads to difficulties for the integration routine. Hence the definition of the gradient function above.
Alan
Thanks, Alan and Werner.
The Prime 3 sheet in attach is without an error but has a false tan not sin in the last equation.
I feel the error but...
Seems to work for negative values of x only.
Not sure and I had not tried (Prime is crashing all the time with your sheet) but maybe you run into the same problems with tan as I did in my solution above (y' gets infinte when x=0)
WE
Why are you trying this the hard way?
Numerically it is possible to solve this, but it can also be done symbolically:
Here is an animation
Note that the solution also works when F is negative, then the mirror is convex instead of concave.
Luc
Those embedded videos don't work (as so often).
Better simply attach the original avi-file.
-> Better qualitiy, more convenient to watch and more reliable
Here is what I see:
OK, here is really a strange effect.
Why does this small change of the end value has such a big impact on the overall shape of the solution??
I could understand if the algorithm goes havoc near x=0, but not at the start at x=-F.
Or is it working from x.end back to the start? Don't think so as it has to use the initial conditions for x=-F first, I guess.
Werner
I noticed it wasn't visible. I even got to see my animation attachment efforts as separate items in the topics list:
but with NO access allowed. When I try to select one, Iget:
Here's a zip file with the AVi's.
Enjoy!
Luc
LucMeekes написал(а):
Numerically it is possible to solve this, but it can also be done symbolically:
Symbolically it is possible to solve this, but it can also be done numerically:
LucMeekes написал(а):
Why are you trying this the hard way?
I would like to solve this problem!
Do you know material of this mirror surface?
