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  The geometry of the triangle
  Romanova Elena,  8 class,  school 57, Kazan, Russia

       Construction of triangle and calculation its angles

       Construction of  bisectors
      
       Construction of medians
      
       Construction of altitudes


> restart:with(geometry):      

The setting of the height of the triandle and let's call it "Т"
> triangle(T,[point(A,4,6),point(B,-3,-5),point(C,-4,8)]);

                                  T

        Construction of the triangle
> draw(T,axes=normal,view=[-8..8,-8..8]);

Construction of the triangle АВС

> draw({T(color=gold,thickness=3)},printtext=true,axes=NONE);     
Calculation of the distance between heights А and В - the length of a side АВ

> d1:=distance(A,B);

                           d1 := sqrt(170)

        
        Calculation of the distance between heights В and С - the length of a side ВС
> d2:=distance(B,C);

                           d2 := sqrt(170)

       The setting of line which passes through two points А and В
> line(l1,[A,B]);

                                  l1

       Display the equation of line l1
> Equation(l1);
> x;
> y;

                         -2 + 11 x - 7 y = 0

        The setting of line which passes through two points А and С
> line(l2,[A,C]);

                                  l2

       Display the equation of line l2
> Equation(l2);
> x;
> y;

                          56 - 2 x - 8 y = 0

         The setting of line which passes through two points В and С
> line(l3,[B,C]);

                                  l3

        Display the equation of line l3
> Equation(l3);
> x;
> y;

                          -44 - 13 x - y = 0

        Check the point А lies on line l1
> IsOnLine(A,l1);

                                 true

        Check the point А lies on line l1
> IsOnLine(B,l1);

                                 true

        Calculation of the andle between lines l1 and l2
> FindAngle(l1,l2);

                              arctan(3)

        The conversion of result to degrees
> b1:=convert(arctan(97/14),degrees);

                                      97
                               arctan(--) degrees
                                      14
                     b1 := 180 ------------------
                                       Pi

        Calculation of decimal value of this angle
> b2:=evalf(b1);

                      b2 := 81.78721981 degrees

        Calculation of the andle between lines l1 and l3
> FindAngle(l1,l3);

                             arctan(3/4)

       The conversion of result to degrees
> b3:=convert(arctan(97/99),degrees);

                                      97
                               arctan(--) degrees
                                      99
                     b3 := 180 ------------------
                                       Pi

        Calculation of decimal value of this angle
> b4:=evalf(b3);

                      b4 := 44.41536947 degrees

       Calculation of the angle between lines l2 and l3
> FindAngle(l2,l3);

                              arctan(3)

       The conversion of  result to degrees
> b5:=convert(arctan(97/71),degrees);

                                      97
                               arctan(--) degrees
                                      71
                     b5 := 180 ------------------
                                       Pi

        Calculation of decimal value of  this angle
> b6:=evalf(b5);

                      b6 := 53.79741070 degrees

        Check the sum of all the angles of the triangle
> b2+b4+b6;

                         180.0000000 degrees

        Analytical information about the point А
> detail(A);
   name of the object: A
   form of the object: point2d
   coordinates of the point: [4, 6]
          Analytical information about the point В
> detail(B);
   name of the object: B
   form of the object: point2d
   coordinates of the point: [-3, -5]
          Analytical information about the point С
> detail(C);
   name of the object: C
   form of the object: point2d
   coordinates of the point: [-4, 8]

   The setting of heights of the triangle points A,B,C and let's call it "Т"

   with(geometry):
> triangle(ABC, [point(A,7,8), point(B,6,-7), point(C,-6,7)]):
        The setting of the bisector of angle А in triandle АВС
> bisector(bA, A, ABC);

                                  bA

        Analytical information about the bisector of angle А in the triandle
> detail(bA);
   name of the object: bA
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: (15*170^(1/2)+226^(1/2))*_x+(-13*226^(1/2)-170^(1/2))*_y+97*226^(1/2)-97*170^(1/2) = 0

        Construction of the triangle
> draw(ABC,axes=normal,view=[-8..8,-8..8]);

 Construction of the triangle ABC

> draw({ABC(color=gold,thickness=3)},printtext=true,axes=NONE);     

 Construction of the bisector of angle А

> draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3)},printtext=true,axes=NONE);    

The setting of the bisector of angle В in the triangle АВС

> bisector(bB, B, ABC);

                                  bB

       Analytical information about the bisector of angle B in the triandle
> detail(bB);
   name of the object: bB
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: (-15*340^(1/2)-14*226^(1/2))*_x+(-12*226^(1/2)+340^(1/2))*_y+97*340^(1/2) = 0

         Construction of the bisector of angle В
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3)},printtext=true,axes=NONE);    



    The setting of the bisector of angle С in the triangle АВС

> bisector(bC, C, ABC);

                                  bC

        Analytical information about the bisector of angle С in the triangle
> detail(bC);
   name of the object: bC
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: (14*170^(1/2)-340^(1/2))*_x+(13*340^(1/2)+12*170^(1/2))*_y-97*340^(1/2) = 0

        Construction of the bisector of angle С
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3),bC(color=blue,thickness=3)},printtext=true,axes=NONE);  

 Calculation of the point of intersection of the bisectors and let's call it "О"

> intersection(O,bA,bB,bC);coordinates(O);

                                  O


     7 sqrt(85) - 3 sqrt(2) sqrt(113) + 3 sqrt(85) sqrt(2)
  [2 -----------------------------------------------------,
       sqrt(85) sqrt(2) + sqrt(2) sqrt(113) + 2 sqrt(85)

          -16 sqrt(85) - 7 sqrt(2) sqrt(113) + 7 sqrt(85) sqrt(2)
        - -------------------------------------------------------]
             sqrt(85) sqrt(2) + sqrt(2) sqrt(113) + 2 sqrt(85)

       Construction of the bisectors and  marking of the point of intersection  "О" in the triandle
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3),bC(color=blue,thickness=3),O},printtext=true,axes=NONE);
> restart:
> with(geometry):
       The setting of the heights of the triangle points A,B,C and let's call it "Т"
> point(A,7,8),point(B,6,-7),point(C,-6,7);

                               A, B, C

        Let's call "Т1"
> triangle(T1,[A,B,C]);

                                  T1

        Construction of "Т1"
> draw(T1(color=gold,thickness=3),axes=NONE,printtext=true);
  The setting of the median from the point В in the trianglemedian(mB,B,T1,B1);
> median(mb,B,T1);

                                  mB


                                  mb

        Construction of the median from the point В
> draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mb},printtext=true,axes=NONE);

The setting of the median from the point А in the trianglemedian(mA,A,T1,A1);
> median(ma,A,T1);

                                  mA


                                  ma

        Construction of the median from the point А
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),ma},printtext=true,axes=NONE);
The setting of the median from the point С in the trianglemedian(mC,C,T1,C1);
> median(mc,C,T1);

                                  mC


                                  mc

        Costruction of the median from the point С
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),mA,mC(color=maroon,thickness=3)},printtext=true,axes=NONE);




Calculation of the point of  intersection of the median and let's call it "О"

>intersection(O,ma,mb,mC);coordinates(O);

                                  O


                              [7/3, 8/3]

        Construction of medians and marking of the point of  intersection "О" in the triangle
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),mA,mC(color=violet,thickness=3),O},printtext=true,axes=NONE);
> restart:with(geometry):
> _EnvHorizontalName:=x:_EnvVerticalName=y:       The setting of the heights of the triangle points A, B, C  and let's call it "Т"
> triangle(T,[point(A,7,8),point(B,6,-7),point(C,-6,7)]);

                                  T

       Construction of the triangle
> draw(T,axes=normal,view=[-8..8,-8..8]);


The setting of the altitude in the triangle from the point Сaltitude(hC1,C,T,C1);
> altitude(hC,C,T);

                                 hC1


                                  hC

        Analytical information about the altitude hC from the point С in the triangle
> detail(hC);
   name of the object: hC
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: -99+_x+15*_y = 0

        Construction of the altitude from the point С
> draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hC},printtext=true,axes=NONE);     

  The setting of the altitude in the triangle from the point Аaltitude(hA1,A,T,A1);
> altitude(hA,A,T);

                                 hA1


                                  hA

        Analytical information about the altitude hA from the point А in the triangle
> detail(hA);
   name of the object: hA
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: -28-12*_x+14*_y = 0

        Construction of the altitude from the point А
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hA1},printtext=true,axes=NONE);       The setting of the altitude from the point В

> altitude(hB1,B,T,B1);
> altitude(hB,B,T);

                                 hB1


                                  hB

        Analytical information about the altitude hB from the point В in the triangle
> detail(hB);
   name of the object: hB
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: -71+13*_x+_y = 0

        Consruction of the altitude from the point В
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hB1(color=blue,thickness=3),hB1},printtext=true,axes=NONE);     
 Calculation of the point of intersection of altitudes and let's call it "О"

>intersection(O,hB,hA,hC);coordinates(O);

                                  O


                               483  608
                              [---, ---]
                               97   97

        Construction of altitudes and marking of the point of intersection "О" in the triangle
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hB1(color=blue,thickness=3),hB1,O},printtext=true,axes=NONE);




 

 

 

 

 

 

 

 

 

 

 

 

 

Construction of standard quadrilaterals

      Muchametshina Liya,  8th class,  school № 57, Kazan, Russia


                   Square

                  Rectangle     
                  
                  Rhombus        
 
                  Parallelogram

                   Trapeze

Construction of square

> restart:
> with(plottools):
       Сoordinates (x;y) of the lower left corner of the square and the side "а"
> x:=0;y:=3;a:=6;

                                x := 0


                                y := 3


                                a := 6

      Construction of the square
> P1:=plot([[x,y],[x,y+a],[x+a,y+a],[x+a,y],[x,y]],color=green,thickness=4):
> plots[display](P1,scaling=CONSTRAINED);

The setting of the second square wich moved relative to the first on the vector (2;-3) (vector can be changed) and with side "а-1" (the length of a side can be changed)P2:=plot([[x+2,y-3],[x+2,y-3+a-1],[x+2+a-1,y-3+a-1],[x+2+a-1,y-3],[x+2,y-3]],color=black,thickness=4):
> plots[display](P1,P2,scaling=CONSTRAINED);

Construction of rectangle

> restart:
> with(plottools):
        Сoordinates (x;y) of the lower left corner of the square and the "а" and "b" sides
> x:=0;y:=2;a:=3;b:=9;
>

                                x := 0


                                y := 2


                                a := 3


                                b := 9

       The rectangle is specified by the sequence of vertices with given the lengths "a" and "b"
> l:=plot([[x,y],[x,y+a],[x+b,y+a],[x+b,y],[x,y]]):
> plots[display](l,scaling=CONSTRAINED,thickness=4);
Construction of rhombus

> restart:
> with(plottools):
      The coordinates (x;y) of the initial vertex of the rhombus and the half of the diagonals "a" and "b"
> x:=0;y:=2;a:=3;b:=4;

                                x := 0


                                y := 2


                                a := 3


                                b := 4

       Rhombus is specified by the sequence of vertices with the values "a" and "b"
> ll:=plot([[x,y],[x+a,y+b],[x+a+a,y],[x+a,y-b],[x,y]]):
> plots[display](ll,scaling=CONSTRAINED,thickness=4);

Construction of parallelogram

> restart:
> with(plottools):
      (х;у) - the starting point, (i;j) - the displacement vector of starting point, "а" - the base of the parallelogram
> x:=0;y:=0;i:=4;j:=5;a:=10;

                                x := 0


                                y := 0


                                i := 4


                                j := 5


                               a := 10

     The parallelogram is defined by the sequence of vertices
> P1:=plot([[x,y],[x+i,y+j],[x+i+a,y+j],[x+a,y],[x,y]]):
> plots[display](P1,scaling=CONSTRAINED,thickness=4);
 If  i= 0  it turns out the rectangleget.
       If  j= а  it turns out the  square.
       If  a := sqrt(i^2+j^2) it turns out the rhombus. a:=sqrt(i^2+j^2):

Construction of trapeze

Trapeze general form
> restart:
> with(plottools):
>
        (х;у) - the starting point, (i;j) - the displacement vector of starting point, а - the larger base of the trapezoid
> x:=0;y:=2;i:=1;j:=5;a:=11;

                                x := 0


                                y := 2


                                i := 1


                                j := 5


                               a := 11

         The trapez is defined by the sequence of vertices     
> P1:=plot([[x,y],[x+i,y+j],[x+i+j,y+j],[x+i+a,y],[x,y]]):
> plots[display](P1,scaling=CONSTRAINED,thickness=4);
Rectangular trapezoid
> restsrt:
> with(plottools):
> x:=0;y:=2;i:=0;j:=6;a:=11;

                                x := 0


                                y := 2


                                i := 0


                                j := 6


                               a := 11

> P1:=plot([[x,y],[x,y+j],[x+j,y+j],[x+a,y]]):
> plots[display](P1,scaling=CONSTRAINED,thickness=4);
Isosceles trapezoid
> restart:
> with(plottools):
> x:=0;y:=2;i:=4;j:=6;a:=15;

                                x := 0


                                y := 2


                                i := 4


                                j := 6


                               a := 15

> P1:=plot([[x,y],[x+i,y+j],[x+j+i,y+j],[x+a,y],[x,y]]):
> plots[display](P1,scaling=CONSTRAINED,thickness=4);



 

 

 

Hello, when you run an error occurs:

Error, (in plots:-display) unknown plot object: circle
Error, (in plots:-display) expecting plot structure but received: P[0]
Error, (in plots:-display) expecting plot structure but received: PP

I'm not good in Maple. Please help me understand.

Programm: https://dropmefiles.com/Qnkzn

I attempted to show that two lines are parallel.  I started with a problem in Geometry for which I do not have the solution.

I tried several ways with Maple to show this to be true.  Most of the time, I ended when maple could not determine if a-b = c-d, etc.

brg_proof.txt contains a statement of the problem and my latest maple code.

Question: How should I approach the proof, by the compass and straight edge method?  Is this possible in maple?

intersection in the geometry package does not seem to recognize assume.

restart: with(geometry):

assume(p[1]<>0, p[2]<>0, p[3]<>0);
assume(q[1]<>0, q[2]<>0, q[3]<>0);
point(T,[p[1],q[1]]);
point(U,[p[2],q[2]]);
point(V,[p[3],q[3]]);
point(Op,[0,0]);

line(OT,[Op,T]);
line(OU,[Op,U]);
line(OV,[Op,V]);

point(B,2*q[2],solve(subs(x=2*q[2],Equation(OU)),y));
coordinates(B);
IsOnLine(B,OU);

PerpendicularLine(AD,B,OT);
ArePerpendicular(AD,OT);
sol:=solve({Equation(AD),Equation(OT)},{x,y});
eval(x,sol);
point(A,eval(x,sol),eval(y,sol));  ## the intersection exists
intersection(xA,AD,OT); ## fails
about(p[1]),about(q[1]);

Hi all,

I have three points in 3d space say A1=[a11, a12, a13]; A2=[a21, a22, a23] and A3=[a31, a32, a33]. I want to fill the triangle formed by these points. How can I do that?

Thanks is advance.

I can draw a triangle using the geomtry package:

point(A, 0, 0), point(B, 2, 1), point(C, 2, 0); triangle(T, [A, B, C])
d := draw(T, axes = none, color = black); t := textplot([[.227, 0.60e-1, typeset(theta)], [1.1, 0, typeset(k__2)]])
display(t,d)

The result is as below.

 

I would like to place the label k2 below the line but negative numbers do not work. I would like to put a label k1 to the right of
the opposite side but there is no room,  and finally I would like to put sqrt(k1^2+k2^2) above the hypotenuse slanted to be
parallel with the hypotenuse if possible.

  Hi, there

How can I draw the excircles, incircles,circumcircle and their centers of a triangle simultaneously with maple13 in a geometric plot? please specify the commands.

many thanks for your help

M.R.Yegan

of a (concrete/general) triangle, making use of Maple tools in an efficient way?  Mathematica applies the barycentric coordinates and the Dirichlet distribution to this end. More generally, how to efficiently choose a random point in a given bounded region?

The docs for the package geometry say:

 

The command with(geometry,distance) allows the use of the abbreviated form of this command.

I am new to Maple. What is the abbreviated form of the distance command?

Dear All

Is there anybody who is working on contruction of optimal Lie algebra using Maple packages like DifferentialGeometry and LieAlgebra, I tried to find commands for constructing algebra in these packages but could not find such commands. I am sure these are only package that might help me. Following is Lie algebra whose optimal system is required:

 

with(PDEtools, SymmetryCommutator, InfinitesimalGenerator):

S[1], S[2], S[3], S[4], S[5], S[6], S[7], S[8], S[9], S[10], S[11] := [_xi[x] = 0, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = 1], [_xi[x] = 0, _xi[y] = t, _xi[t] = 0, _eta[u] = 0, _eta[v] = x], [_xi[x] = 0, _xi[y] = y, _xi[t] = 2*t, _eta[u] = -2*u, _eta[v] = -v], [_xi[x] = 0, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = t], [_xi[x] = 0, _xi[y] = 0, _xi[t] = 0, _eta[u] = 1, _eta[v] = y], [_xi[x] = 0, _xi[y] = 0, _xi[t] = 1, _eta[u] = 0, _eta[v] = 0], [_xi[x] = 1, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = 0], [_xi[x] = t, _xi[y] = 0, _xi[t] = 0, _eta[u] = 1, _eta[v] = 0], [_xi[x] = y, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = 2*x], [_xi[x] = x, _xi[y] = 0, _xi[t] = -t, _eta[u] = 2*u, _eta[v] = 2*v], [_xi[x] = 0, _xi[y] = 1, _xi[t] = 0, _eta[u] = 0, _eta[v] = 0]

[_xi[x] = 0, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = 1], [_xi[x] = 0, _xi[y] = t, _xi[t] = 0, _eta[u] = 0, _eta[v] = x], [_xi[x] = 0, _xi[y] = y, _xi[t] = 2*t, _eta[u] = -2*u, _eta[v] = -v], [_xi[x] = 0, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = t], [_xi[x] = 0, _xi[y] = 0, _xi[t] = 0, _eta[u] = 1, _eta[v] = y], [_xi[x] = 0, _xi[y] = 0, _xi[t] = 1, _eta[u] = 0, _eta[v] = 0], [_xi[x] = 1, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = 0], [_xi[x] = t, _xi[y] = 0, _xi[t] = 0, _eta[u] = 1, _eta[v] = 0], [_xi[x] = y, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = 2*x], [_xi[x] = x, _xi[y] = 0, _xi[t] = -t, _eta[u] = 2*u, _eta[v] = 2*v], [_xi[x] = 0, _xi[y] = 1, _xi[t] = 0, _eta[u] = 0, _eta[v] = 0]

(1)

G[1] := InfinitesimalGenerator(S[1], [u(x, y, t), v(x, y, t)]); 1; G[2] := InfinitesimalGenerator(S[2], [u(x, y, t), v(x, y, t)]); 1; G[3] := InfinitesimalGenerator(S[3], [u(x, y, t), v(x, y, t)]); 1; G[4] := InfinitesimalGenerator(S[4], [u(x, y, t), v(x, y, t)]); 1; G[5] := InfinitesimalGenerator(S[5], [u(x, y, t), v(x, y, t)]); 1; G[6] := InfinitesimalGenerator(S[6], [u(x, y, t), v(x, y, t)]); 1; G[7] := InfinitesimalGenerator(S[7], [u(x, y, t), v(x, y, t)]); 1; G[8] := InfinitesimalGenerator(S[8], [u(x, y, t), v(x, y, t)]); 1; G[9] := InfinitesimalGenerator(S[9], [u(x, y, t), v(x, y, t)]); 1; G[10] := InfinitesimalGenerator(S[10], [u(x, y, t), v(x, y, t)]); 1; G[11] := InfinitesimalGenerator(S[11], [u(x, y, t), v(x, y, t)])

proc (f) options operator, arrow; diff(f, v) end proc

 

proc (f) options operator, arrow; t*(diff(f, y))+x*(diff(f, v)) end proc

 

proc (f) options operator, arrow; y*(diff(f, y))+2*t*(diff(f, t))-2*u*(diff(f, u))-v*(diff(f, v)) end proc

 

proc (f) options operator, arrow; t*(diff(f, v)) end proc

 

proc (f) options operator, arrow; diff(f, u)+y*(diff(f, v)) end proc

 

proc (f) options operator, arrow; diff(f, t) end proc

 

proc (f) options operator, arrow; diff(f, x) end proc

 

proc (f) options operator, arrow; t*(diff(f, x))+diff(f, u) end proc

 

proc (f) options operator, arrow; y*(diff(f, x))+2*x*(diff(f, v)) end proc

 

proc (f) options operator, arrow; x*(diff(f, x))-t*(diff(f, t))+2*u*(diff(f, u))+2*v*(diff(f, v)) end proc

 

proc (f) options operator, arrow; diff(f, y) end proc

(2)

``

 

Download Lie_Algebra_Classification.mwLie_Algebra_Classification.mw

   
 

 

 

   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(2)
 

``

Regards

To check the point on the belonging to the segment I use the algorithm shown in the example. This is an example of intersection of the two segments in 2d. (We not check for parallelism.) We find the point of intersection of the corresponding lines and solve the equation f1 with respect to t and f2 with respect to tt. If 0 <= t <= 1, then the point belongs to the first segment, and if 0 <= tt <= 1 then the point belongs to the second segment.
(Similarly we can check point on the belonging to the segment in 3d.)
In the example point belongs to the second segment, but not the first. These segments do not intersect.
Question: Is there a function in Maple to find the intersection of the segments or to check on the belonging segment point, to make shorter?

segments_intersection.mw 

 

Hello everyone!

i want to make a triangle which height's is a. Sorry, my English is not very good!

And I writed it by C language, but It is not right in maple. and I don't know. Can you help me? thank you very much! triangle.mw

 

I would like to define all the geometry parameters in a multibody model in MapleSim in mm.

Consequently, i define the mm unit for a parameter in 3 places :
1)  in the Parameters area of the subsystem 

2) in the Tab Inspector of the subsystem

 

3) in the Tab Inspector of the rigid body frame of the subsystem

 

The problem is that I receive the warning "Possible double conversion unit on a parameter" and the dimension seems to be as 10^-3 *mm soit too much reduced.

I see that in some examples of the library of Maplesim the parameters of the parameters area of a subsystem are defined as real and consequently, it doesn't cause this kind of problem.

However, how can I do if I want to define my parameters in the parameters area of a subsystem as a position in mm ?

Thanks a lot for your help

 

 

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