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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

 

 

Hello,

I use for the 3D visualization the component CAD geometry with STL files.
My STL files are created from CATIA with parts mesured in mm.
In MapleSim, in order to keep mm, I have, of course, to set "mm" in the inspector tab of the components "CAD geometry".
But, I have also to put the scale factor to 0.001.
I don't understand why I should to set it to 0.001 because :
- the CADs from CATIA are in mm.
- and the option in the inspector tab of the components "CAD geometry" is also in mm.
Would you have some precisions about the scale factor for the "CAD geometry" element ?

Thank you for your help.

Using the Plot Component it is possible to obtain
an interactive worksheet for drawing visually plane geometry
elements, e.g. select a triangle and draw its incircle.
Do you know if such worksheets are available somewhere?

Hello,

I am trying to use Apollonius procedure from geometry package. Here is an example:

restart;
with(geometry):
circle( c1, [ point( c1c, 0 , 0 ), 5 ] );
circle( c2, [ point( c2c, 5 , 4 ), 2 ] );
circle( c3, [ point( c3c, 13 , 0 ), 3 ] );
A := Apollonius( c1, c2, c3 );

Unfortunatelly the Apollonius method does not give any result. The only message is:

intersection: there is no point of intersection

 

Anyone know what is wrong in my code?

Best regards

Rafał Nowak

Let ABC be a triangle with A(3,0,0),B(0,6,0),C(0,0,4) and let H(x,y,z) be its orthocenter .Find the coordinates of point H ?

https://en.wikipedia.org/wiki/Surgery_theory

glue two geometric object

i use maple 17, 

will future version of maple do this?

or which software can do this?

Hello,

Concerning the 3D visualization of my multibody systems, in the visualization windows, i can see both :
- the display of geomtry of the elements which has been defined as simple forms (as cylindrical geometry)
- the display of the geometry of the elements where the display of the geometry has been defined with CAD.

However, concerning the 3D animation, i have only see the components where the display of the geometry is defined as simple forms (as cylindrical geometry).

Have you some ideas why I can not see the elements which has been defined with CAD ?

For your information, the CAD geometries have been defined with STL files and, in the CAD geometry component, I let the box "Transparent" empty.

Thank you for your help

I have some triangles ABC with vertices

1) A(-13,-5,5), B(-5,11,-11), C(-3,-9,15) has centre of out circle is (3, 3, 3), orthocentre (-27, -9, 3) and centroid (-7, -1, -3). 

2) A(-6,6,-1), B(-5,-1,-3), C(2,10,7) has centre of out circle is (1, 2, 3), orthocentre (-11, 11, -3) and centroid (-3, 5, 1). 

How can I write a program to find a triangle with integer coordinates of vertices, centroid, orthocenter and center of the triangle in geometry 3D? 

How to convert an  algebraic equation to a parametrized one ?

 

How to sketch the curve of a circular cylinder , x^2+y^3^=a^2 with the parametric equations lying on it ?

x=a*cos't) , y=a*sin(t) , z=b*t , if t varies from 0 to 2*Pi, the point P starts at(a,0,0) and moves upward.

How to show that the parametric equations are indeed lying on the cylinder ?

How to plot the whole things ?

 

Hi,

I'm using my own Maple package to make questions in Maple T.A.. I'm working with finite automata so I'd like to draw graphs in the text of the questions. I have a function that draws automata, it uses plots:-display as it's output. Is there a way that I can use plots:-display in plotmaple, like this? (This is example doesn't work, I'd like to know why and how I have to modify it.)

$circ=maple("geometry:-circle('Circ', [geometry:-point('PP',0,0),3]);
DRA:=geometry:-draw(Circ);
DISP:=plots:-display(DRA);
DRA;");
$plo=plotmaple("plots:-display($circ),plotdevice='gif', plotoptions='height=250, width=250'");

Thank you for the answers.

How to get the equations of circles A, B, C, such as circle A with center (1,1) is drawn in the first quadrant.

Circle B with radius 2 and circle C are placed so that each circle is tangent to the other 2 circles and the x-axis.

THe 3 circles are on the first quadrant.

1) do it with Maple

2) do it by hand

3) draw the figure

 

Regards

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