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Alexey Ivanov

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Does everyone have a good idea of ​​the work of the Draghilev method? For example, in this answer https://www.mapleprimes.com/questions/235407-The-Second-Example-Of-Finding-All-Solutions#answer291268
( https://www.mapleprimes.com/questions/235407-The-Second-Example-Of-Finding-All-Solutions )
there was a very successful attempt by Rouben Rostamian to calculate the line of intersection of surfaces without applying the Draghilev method.
Let now not 3d, but 8d. And how will the solve command work in this case? Imagine that aij ((i=1..7,j=1..8)) are partial derivatives, and xj (,j=1..8) are derivatives, as in the above example. f8 is responsible for the parametrization condition.

 

restart;
 f1 := a11*x1+a12*x2+a13*x3+a14*x4+a15*x5+a16*x6+a17*x7+a18*x8; 
 f2 := a21*x1+a22*x2+a23*x3+a24*x4+a25*x5+a26*x6+a27*x7+a28*x8; 
 f3 := a31*x1+a32*x2+a33*x3+a34*x4+a35*x5+a36*x6+a37*x7+a38*x8; 
 f4 := a41*x1+a42*x2+a43*x3+a44*x4+a45*x5+a46*x6+a47*x7+a48*x8;
 f5 := a51*x1+a52*x2+a53*x3+a54*x4+a55*x5+a56*x6+a57*x7+a58*x8; 
 f6 := a61*x1+a62*x2+a63*x3+a64*x4+a65*x5+a66*x6+a67*x7+a68*x8; 
 f7 := a71*x1+a72*x2+a73*x3+a74*x4+a75*x5+a76*x6+a77*x7+a78*x8;
 f8 := x1^2+x2^2+x3^2+x4^2+x5^2+x6^2+x7^2+x8^2-1; 
allvalues(solve({f1, f2, f3, f4, f5, f6, f7, f8}, {x1, x2, x3, x4, x5, x6, x7, x8}))


And this is how the Draghilev method works in this case.
 

restart; with(LinearAlgebra):
f1 := a11*x1+a12*x2+a13*x3+a14*x4+a15*x5+a16*x6+a17*x7+a18*x8; 
f2 := a21*x1+a22*x2+a23*x3+a24*x4+a25*x5+a26*x6+a27*x7+a28*x8; 
f3 := a31*x1+a32*x2+a33*x3+a34*x4+a35*x5+a36*x6+a37*x7+a38*x8; 
f4 := a41*x1+a42*x2+a43*x3+a44*x4+a45*x5+a46*x6+a47*x7+a48*x8; 
f5 := a51*x1+a52*x2+a53*x3+a54*x4+a55*x5+a56*x6+a57*x7+a58*x8;
f6 := a61*x1+a62*x2+a63*x3+a64*x4+a65*x5+a66*x6+a67*x7+a68*x8; 
f7 := a71*x1+a72*x2+a73*x3+a74*x4+a75*x5+a76*x6+a77*x7+a78*x8;

n := 7; 
x := seq(eval(cat('x', i)), i = 1 .. n+1):
 F := [seq(eval(cat('f', i)), i = 1 .. n)]: 
A := Matrix(nops(F), nops(F)+1):
 for j to nops(F) do for i to nops(F)+1 do A[j, i] := op(1, op(i, op(j, F))) 
end do:
           end do: 

# b[i] and b[n+1] are solutions of a linear homogeneous system and at the 
# same time they serve as the right-hand sides of an autonomous ODE.
for i to n do

 b[i] := Determinant(DeleteColumn(ColumnOperation(A, [i, n+1]), n+1)) 
                                                                     end do:
 b[n+1] := -Determinant(DeleteColumn(A, n+1)):


Only the original seven linear homogeneous equations with eight variables are needed. We solve them according to Cramer's rule, and in order to have uniqueness when solving the ODE, we use a point on the curve (according to the theory). (This point is sought in any convenient way.)
If we want to get a parameterization, then additionally, directly in dsolve, we can add the following:
 

for i to n do 
b[i] := simplify(Determinant(DeleteColumn(ColumnOperation(A, [i, n+1]), n+1))) 
end do:
b[n+1] := simplify(-Determinant(DeleteColumn(A, n+1))); 
deqs := seq(diff(x[i](s), s) = b[i]/(b[1]^2+b[2]^2+b[3]^2+b[4]^2+b[5]^2+b[6]^2+b[7]^2+b[8]^2)^.5, i = 1 .. n+1):

 

An example of uniform motion along a generalized coordinate using the Draghilev method. (This post was inspired by school example in one of the forums.)
The equations used in the program are very simple and, I think, do not require any special comments. DM is a procedure that implements the Draghilev method with "partial parameterization".

DM_V.mw

When K = 1, parameterization is carried out by changing the angle of rotation of the wheel. That is, uniform rolling is carried out.

For K = 4, the coordinate corresponding to the position of the slider is parametrized.

 

When K = 6, the slider moves with acceleration, according to a given equation. Hence, we have carried out the parameterization with respect to “time”.



With the help of such techniques, we can obtain the calculation of the kinematics of both lever mechanisms and various types of manipulators.

It seems to me that Draghilev's method can be applied quite successfully to the solution of Diophantine equations. Here is a simple example where we find two solutions at the intersection line of two ellipsoids:
  x1^2-x1*x2+x2^2+x2*x3+x3^2-961=0;
  (x1-3)^2+10*x2^2+x3^2-900=0;

Solutions: (11, -4, -26) and (10, 1, 29).

 


Based on the text of the program, it is possible to solve various examples with Diophantine equations.
3d_1.mw

Explanations.
f3 is an auxiliary equation for finding the starting point, NPar is a procedure that implements the Draghilev method, the red color of the text is the place where the integer values of the points on the integral curve are filtered.

 Can be compared with the solution of the
isolve function
 

 restart:
  f1 := x1^2-x1*x2+x2^2+x2*x3+x3^2-961;
  f2 := (x1-3)^2+10*x2^2+x3^2-900;
  isolve({f1, f2})

 

For a long time I could not understand how to make a kinematic analysis of this device based on the coupling equations (like here). The equations were drawn up relative to the ends of the grips (horns), that is, relative to the coordinates of 4 points. That's 12 equations. But then only a finite number of mechanism positions take place (as shown by RootFinding[Isolate]). It turns out that there is no continuous transition between these positions. Then it is natural to assume that the movement can be obtained with the help of a small deformation. It seemed that if we discard the condition of a constant distance between the midpoints of the horns (this is the f7 equation at the very beginning), then this will allow us to obtain minimal distortion during movement. In fact, the maximum distortion during the movement was in the second decimal place.
(It seemed that these guys came to a similar result, but analytically "Configuration analysis of the Schatz linkage" C-C Lee and J S Dai
Department of Tool and Die-Making Engineering, National Kaohsiung University of Applied Sciences, Kaohsiung, Taiwan Department of Mechanical Engineering, King’s College, University of London, UK)

The left racks are input.


The first text is used to calculate the trajectory, and the other two show design options based on the data received.
For data transfer, a disk called E is used.
Schatz_mechanism_2_0.mw
OF_experimental_1_part_3_Barrel.mw
OF_experimental_1_part_3.mw

One way to show all solutions of a polynomial in one variable.
The root is the intersection of curves representing the imaginary part of the equation (red) and the real part (blue). These equations are obtained after representing the variable as the sum of its imaginary and real parts. The circle limits the area where all the roots are located (according to theory).
Example      -15*x^7+x^2+I*x+2=0;
polynomial_roots_graph.mw

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