one man

Alexey Ivanov

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@David1 one (first) line, if it is rotated, can cross another fixed line in an infinite number of points. Because the first line has two degrees of freedom, and the intersection point of the lines has one degree of freedom. Then your system of equations must have two free variables.
The task is to calculate the manipulator?

You acknowledge that your variables are the angles of rotation, these angles are only three, then the variable is not 15, but 9.
Look at the definition of what is called a polynomial system of equations. It does not depend on our desire.

I think it would be great to show colleagues your whole task.

@fereydoon_shekofte  Sorry, I know English worse than I know Maple.

A small addition to the text, and we have a tangent cube.
One face of the cube belongs to the tangent plane.

Thank you, Yuri Nikolaevich, r: = sqrt (add (d [i] ^ 2, i = 1..nops (d))) is already available to my understanding.

r: = sqrt (`+` (`~` [`^`] (d, 2) [])); Instead of r: = sqrt (add (`~` [`^`] (d, 2))); ?
(I'll never understand this.)
Many thanks, now everything works.


I do not know how to use the "proc" function. I wanted, at least formally, to master this function also for another task.
But your  text is reproduced with an error. (My max available version is Maple 17)
Look, please, what's wrong:

> K := NPar([f1, f2], X0, 9.6, 100);

Error, (in NPar) invalid input: add expects 2 arguments, but received 1



@vv  I understand what feelings arise for professionals when they see my texts. Yes, I met Maple, but I did not learn to use it. I borrowed pieces of text, looked help. There was only a desire that the idea be confirmed and the algorithm will work.
But I will try to behave more civilized.
Thank you for specific assistance and constant attention.

@Ronan  Thank you for your interest to this topic.
Want to draw your attention that for my part there is practically no reaction to the replicas of a certain person (I will not mention his name). For myself, I have long concluded that it is useless to communicate with him for many reasons, or rather, even harmful to one's own reputation and health.

@fereydoon_shekofte  Thank you for your kind words.
When I had a job, I worked on developing algorithms, but I was not able to implement many ideas. A few years ago I met Maple, and Maple inspired me to return to those tasks.
I would be happy if this was useful for someone.

Do you understand that the graphs depict the equations of the first derivatives of f (\ epsilon, \ phi)?
You do not show me the f (\ epsilon, \ phi), and I myself integrated the partial derivatives and looked at the plot of
the f (\ epsilon, \ phi) with help of plot3d. And you yourself can see that the minimum of f (\ epsilon, \ phi) in this definition area is an infinite flat set.


Sorry, I still do not see your original function f (\ epsilon, \ phi).
As for your last equations, I took them from fsolve, and denoted x and y.
On an implicit graph x, it is clear that \ epsilon is practically 0, and the graph of y is empty.

I think you still have these options:
1) to look for min ((f (\ epsilon, \ phi)) ^ 2). It's working with the function itself without derivatives.
2) scale the variables "\ epsilon" and "\ phi" so that the range "\ phi" is much wider for fsolve.

@kuwait1 Ah, here's the thing. And where is your f (\ epsilon, \ phi)?

By the way, it may well be that your min has an infinite number of solutions.

@kuwait1  You really have only one equation: either x or y. There are many ways to solve one equation with several variables. First try the easiest way: set the desired values of the epsilon and  then solve one equation for another variable using fsolve.

The graphs show that your equations are equivalent:
with(plots, implicitplot):
implicitplot(x, `ε` = -5 .. 5, phi = -5 .. 5, numpoints = 20000, color = red);
implicitplot(y, `ε` = -5 .. 5, phi = -5 .. 5, numpoints = 20000, color = blue);
implicitplot([x, y], `ε` = -5 .. 5, phi = -5 .. 5, numpoints = 20000, color = [red,blue]);



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