one man

Alexey Ivanov

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8 years, 22 days
Russian Federation

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Yes, there are several ways to solve this problem. The fact is that in some similar examples there is only a finite number of variants of such triangles, and it seemed to me that the easiest way to decide (for fun) is to use the optimization package. 
For example, x^4+ y^4 + z^4 - 1 = 0, the maximum area of the triangles is approximately 2.08.

It seems  there is a way to exit the program when a check for the existence condition of the tetrahedron is "NO". We put the program text in a pseudo-loop where the break statement works. If the check passes, then the  print ("GO") is executed; if not, the print ("GO") is not executed. A simple check by point F: if its third coordinate is other than 0, then the program is executed, for example, point (F, 1, 1, 0.000001), if its third coordinate is 0, then the program is not executed.

restart: with(geom3d):
for j to 1 do 
point(A, 0, 0, 0), point(B, 2, 2, 0), point(C, 0, 2, 0), point(F, 1, 1, 0):   
gtetrahedron(T1, [A, B, C, F]):
i := nops(op(0, detail(T1))); if i = 1 then print("Ok, gtetrahedron") else    print("No gtetrahedron, NO"); break end if:
end do:


@tomleslie  Thanks a lot. I didn’t want to extend the text of the check, especially since I couldn’t do it as gracefully as you. I have very poor Maple proficiency. I was more interested in understanding how to stop further program execution in the case of `no_tetrahedron_possible`

I think the author of the topic simply wanted that to him be shown  how to establish uniform rotation of one of the vertical struts of the mechanism (input link of mechanism).
I don't have MapleSim either.

@Rouben Rostamian  

 You wrote: “I don't understand your code for making a cylinder. 
plots [arrow] when head_length = 0 will look like a cylinder. In this case, the cylinder diameter is width=1, and the length of the cylinder is equal to the distance between the middles of the segments with red points. The direction of the arrow is also set along the axis of the cylinder.

What about Markiyan Hirnyk  and MaplePrimes Search?

Directly by using Maple functions dsolve/numeric/BVP
managed to find only one solution for many initial data ('approxsoln').
If anyone, of course, is interesting.

Digits := 20:
dsys := {diff(x(t), t, t)+.2*(diff(x(t), t))+x(t)^3-.3*cos(t) = 0, x(0) = x(2*Pi), (D(x))(0) = (D(x))(2*Pi)}:
sol := dsolve(dsys, numeric, abserr = 1.*10^(-5), approxsoln = [x(t) = .5, (D(x))(t) = 1.5]):
plots[odeplot](sol, [diff(x(t), t), x(t)], 0 .. 2*Pi)


@Janeasefor Thank you very much, very pleased  that you liked it.

In principle, I think it is possible to reduce all “inverse kinematics” to one subroutine (for example, in Maple) for any kind of manipulators in order to obtain a solution based on a straight line segment. More precisely, I do not think, in fact, everything is checked in Maple.

I do not mind in any way. I have long been interested in finding out whether dsolve works after the “event”.  And it was appeared a reason to do it.

The word "correct" does not quite match the situation. It’s just that Anatoly Vladimirovich Draghilev himself believed that the spelling of his last name in English “Draghilev” corresponded to reading in Russian, well, he asked for it.
Once again, thank you for your attention to this issue. It seems to me that the main thing is that you did a very good job.

In the neighborhood of a solution point there will always be a continuous solution (according to the theory). But constraints on variables will have to be chosen from the found set of solutions.
Draghilev's method for your situation is well stated here
If translation from Russian is available to you, then

If there are continuous solutions of the system of m underdetermined equations (m*n, n>m), it is possible to obtain the dependence of all variables: xi, i = 1..n (where "a" is one of xi) from a certain parameter. For example, the  Draghilev's method. It is suitable?

I'm afraid you misunderstood. In its original form, your system most likely has no solutions.
It was shown that the separately numerator and denominator (that after "/") have the same solutions. These were equations without denominators in fsolve.

Your system of equations seems to have dependencies and features. In the first text I removed the denominators and found a solution with the help of fsolve. The same values of variables give 0 of denominators (second text).
Please check carefully.

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