Alexey Ivanov

## 1220 Reputation

12 years, 97 days

## Example on the surface...

x1^2+2*x2^2+x3^2-x2*x3 -100=0;

## Another example...

Another example, but from 2d .
x2^3+x1^2-x1*x2-16=0;
In some seconds, the Method finds the following integer solutions:
(-33284, -1024), (-86, -18), (-60, -14), (-6, -2), (-4, 0), (-2, 2), (4, 0), (4, -2), (46, -14), (68, -18), (32260, -1024).

I think this is not bad for a numerical approach to Diophantine equations that works without enumeration of the coordinate grid.

@acer Thank you.

## ......

@acer  Sorry, but I have problems with English and with online translation. I didn't understand what you wanted to say. Just in case, I'll tell you what I did. I built graphs and saw that there are intersection points for all ns. Then I tried the NextZero function. From point 0, NextZero skips the solution for the first value of ns, so I moved the starting point to the right. This point, for example, can be either 1.*10^(-14) and 4.*10^(-10) ...

## ......

@C_R
To be honest, I don't know much about joints. I think, for this scheme, the joints can be any suitable. And the L5 itself can be made of variable length due to the splined connection. It's just a mixer.
And it also seemed to me that in the pictures from the Internet, the "horns" holding the container (first picture) are made of springy plates that can compensate for small deformations.

## Purely for clarification...

@C_R
The percentage of change in the length of L5 in a full cycle during the operation of the original program.

## It's all right...

@C_R
But I think it's all right. MapleSim works analytically and therefore cannot have deformation. In the proposed simulation, the mechanism is calculated numerically. If geometrically, then in one case the surfaces touch each other (when link lengths = a/sqrt(3)), and in the other case they intersect "shallowly".
(Visually L5 deforms a lot in your example. In the table (in the program text OF_experimental_1_part_3_Barrel.mw) these are hundredths of the length.)

## Yes...

@C_R
Yes, only my the lower distance is not 2, but 2.4.
If the distance is 2 and L1=2/sqrt(3), L5=1.35, then Maple also works fine, but the distortion is much stronger. Printed in program OF_experimental_1_part_3_Barrel.mw
If the distance is 2, and L1= 2/sqrt(3) and L5= 2/sqrt(3), then Maple throws an error and shows that the system of equations has no solutions. It is necessary to study the system with these data. It is possible that it has an infinite number of solutions and this will be the best option.

## Thanks...

```L1 := 1.6; L2 := 1.5; L5 := 1.35;
f1 := (x7-g1)^2+(x8-g2)^2+(x9-g3)^2-L1^2;
f2 := (x10-g1)^2+(x11-g2)^2+(x12-g3)^2-L1^2;
f3 := (x1-CD1)^2+(x2-CD2)^2+(x3-CD3)^2-L1^2;
f4 := (x4-CD1)^2+(x5-CD2)^2+(x6-CD3)^2-L1^2;
f5 := (x4-x1)^2+(x5-x2)^2+(x6-x3)^2-L2^2;
f6 := (x7-x10)^2+(x8-x11)^2+(x9-x12)^2-L2^2;
f7 := ((x1+x4)*(1/2)-(x7+x10)*(1/2))^2+((x2+x5)*(1/2)-(x8+x11)*(1/2))^2+((x3+x6)*(1/2)-(x9+x12)*(1/2))^2-L5^2;
f8 := (x10-x7)*(g2-x8)-(x11-x8)*(g1-x7);
f12 := x4*(CD2-x2)-x5*(CD1-x1);
f9 := (x1-x4)*(x7-x10)+(x2-x5)*(x8-x11)+(x3-x6)*(x9-x12);
f10 := ((x7+x10)*(1/2)-(x1+x4)*(1/2))*(x4-x1)+((x8+x11)*(1/2)-(x2+x5)*(1/2))*(x5-x2)+((x9+x12)*(1/2)-(x3+x6)*(1/2))*(x6-x3);
f11 := ((x7+x10)*(1/2)-(x1+x4)*(1/2))*(x10-x7)+((x8+x11)*(1/2)-(x2+x5)*(1/2))*(x11-x8)+((x9+x12)*(1/2)-(x3+x6)*(1/2))*(x12-x9);
T := Isolate([f1, f2, f3, f4, f5, f6, f6, f7, f8, f9, f10, f11, f12], [x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12])```

L1 := 1.6 is the length of the levers; L2 := 1.5 is the width of the horns; L5 := 1.35 is the distance between the midpoints of the horns. And some of these parameters are printed (after calculations) in one of the given programs (in particular, L5).
The system of equations itself gives 32 solutions, but it seems that there are only 4 fundamental provisions themselves. In other words, the system of equations does not contain free variables, and, accordingly, we do not have degrees of freedom of mechanism. Of course, we are talking about a specific mathematical model.But discarding f7, we formally get one degree of freedom, and with it a slight deformation of the "yellow" lever (in our case, L5)
To be honest, I made this mechanism out of curiosity. I was interested in testing exactly this approach. And this is where my knowledge ends.
If we talk about inverse kinematics, then I do not quite understand what "switching" is. But, I think that we can always replace the right construction (rhombus) with the left one. It is necessary to add equations corresponding to a rhombus to the mathematical model of the left device, and then the left device will work like a rhombus. That is, we virtually remove unnecessary degrees of freedom. This, of course, if I understand your question correctly.
I often look at the forum, but did not see your message, because the "flag" for me was not red. For this it is necessary. to have addressed to me. How do you now have a red flag.

## ​​​​​​​Yes...

@C_R
Yes, this method is equally suitable for both sequential and parallel manipulators, including the Stewart platform. We take the trajectory we need and solve the inverse problem.
(By the way, here on the forum at the end of the topic in the comments there is an example.)

## This is not an answer...

I won’t say that I understood what the text of the message is about, but as for solving the inverse problem of the kinematics of manipulators, there is one simple and unambiguous way to solve the inverse problem – a way to reduce this problem to the kinematics of lever mechanisms. This is achieved by mathematically reducing the number of degrees of freedom to 1 (that is, by the introduction of additional restrictions using mathematical equations).
Maple examples:
https://www.mapleprimes.com/posts/214084-One-More-Way-Of-Inverse-Kinematics
https://www.mapleprimes.com/posts/215233-Another-Example-Of-A-Real-Manipulator-Model

## Zeros of the derivative...

Why an optimization package? The zeros of the derivative of the Gamma function indicate its extreme values. Having found the zeros of the derivative in the desired range, you will select the amplitudes you need.
Or did I not understand something?

## I wish......

@vv
I wish I could see all these pictures at school. In due time, of course.
I wanted to see what the area of polynomial roots looks like and how to graphically interpret the presence of complex and real roots in the same area. Well, with a reasonable dimension, I think it is possible to somehow understand what's what, using various examples.
By the way, going around the circle, you can find all the branches leading to solutions, and go to them. (As an alternative solution.)

## Just find out how many solutions?...

Have you tried using the nops procedure?
For example:
SN:=(solve(..));
nops(SN);

## @mmcdara   It's okay. This...

@mmcdara   It's okay. This is done intentionally - this can be seen from the text of the program: look at the display, where graphs of different types are located in one direction and in the opposite direction.
(This is if I understood you correctly, because I use an online translator.)

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