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Geometry example

Owned by Enrique Escolano
Last updated: Jun 02, 2022
5 min read

This example consists on a simple cylinder, like the one shown on the right.
It contains points, curves of type straight lines, circumference arcs and curved NURBS with circumference shape, and surfaces of the types planar, Coon and NURBS with cylindrical form, and there is a single volume.

Note: This model could be found at: Examples\Cylinder_ASCII.gid

RAMSAN-ASCII-gid-v7.6
UNKNOWN 0
0
1 Tops 0 1 0 0 255
2 Lateral 0 1 0 255 255
0
0
1 1 1 2 3 0 0 2 0
-1.65134 -1.60324 0
1 2 1 2 3 0 0 2 0
-1.65134 -1.60324 3.76945
1 3 1 2 3 0 0 2 0
-1.80449 -3.49553 0
1 4 1 2 3 0 0 2 0
-1.80449 -3.49553 3.76945
2 1 1 2 2 0 0 2 0
1 2
2 2 1 2 2 0 0 2 0
3 4
11 3 1 2 2 0 0 2 0
1 3 5 2 2.98214
-1.65134 -1.60324 0
-2.59749 -1.52666 0
-2.67407 -2.47281 0
-2.75064 -3.41896 0
-1.80449 -3.49553 0
0 0 0 0.5 0.5 1 1 1 
1 1 0.707107 1 0.707107 1 
11 4 1 2 2 0 0 2 0
2 4 5 2 2.98214
-1.65134 -1.60324 3.76945
-2.59749 -1.52666 3.76945
-2.67407 -2.47281 3.76945
-2.75064 -3.41896 3.76945
-1.80449 -3.49553 3.76945
0 0 0 0.5 0.5 1 1 1 
1 1 0.707107 1 0.707107 1 
11 5 1 2 2 0 0 2 0
3 1 5 2 2.98213
-1.80449 -3.49553 0
-0.858344 -3.57211 0
-0.781767 -2.62596 0
-0.70519 -1.67981 0
-1.65134 -1.60324 0
0 0 0 0.5 0.5 1 1 1 
1 1 0.707107 1 0.707107 1 
3 6 1 2 2 0 0 2 0
4 2 -0.315383 0.025526 0.949244 4.63163 7.77322
1  0  0  0  
0  1  0  0  
0  0  1  0  
-1.41253  -2.57491  3.76945  1  
14 1 1 2 1 0 0 2 0
4
1 4 2 3 
0 0 1 1 
-2.67407 -2.47281 1.88473
0.996741 -0.080672 0
0 2 5 1 2
-1.65134 -1.60324 0
-1.65134 -1.60324 3.76945
-2.59749 -1.52666 0
-2.59749 -1.52666 3.76945
-2.67407 -2.47281 0
-2.67407 -2.47281 3.76945
-2.75064 -3.41896 0
-2.75064 -3.41896 3.76945
-1.80449 -3.49553 0
-1.80449 -3.49553 3.76945
0 0 1 1 
0 0 0 0.5 0.5 1 1 1 
1 1 1 0.707107 0.707107 1 1 0.707107 0.707107 1 1 
6 2 1 2 1 0 0 2 0
4
1 5 2 6 
1 1 0 0 
-0.781767 -2.62596 1.88473
-10.6459 0.861634 -0
14 3 1 2 1 0 0 1 0
2
5 3 
0 0 
-1.72792 -2.54939 0
0 0 1
1 2 2 1 1
-2.7699 -1.5074 0
-2.7699 -3.59137 0
-0.685932 -1.5074 0
-0.685932 -3.59137 0
0 0 1 1 
0 0 1 1 
0 
5 4 1 2 1 0 0 1 0
2
6 4 
1 1 
-1.41253 -2.57491 3.76945
0 0 -1
9 1 1 2 0 0 0 2 0
4
1 2 4 3 
0 0 0 0 
-1.72792 -2.54939 1.88473
0


This is the explanation of its content:


RAMSAN-ASCII-gid-v7.6
UNKNOWN 0
0


The GiD geometry ASCII file is wrote with rules of version 7.6, and without any problemtype (UNKNOWN)
The model has two layers, created with:


1 Tops 0 1 0 0 255
2 Lateral 0 1 0 255 255
0


The layer number 1 is named 'Tops', is not frozen, visible, and with RGB color R=0, G=0, B=255 (blue)
The layer number 2 is named 'Lateral', is not frozen, visible, and with RGB color R=0, G=255, B=255 (cyan)
Last 0 denotes the end of layers block


0
There is no meshing information attached to entities.


Then four points (code=1) are defined:


1 1 1 2 3 0 0 2 0
-1.65134 -1.60324 0
1 2 1 2 3 0 0 2 0
-1.65134 -1.60324 3.76945
1 3 1 2 3 0 0 2 0
-1.80449 -3.49553 0
1 4 1 2 3 0 0 2 0
-1.80449 -3.49553 3.76945


p1=(-1.65134,-1.60324,0)
p2=(-1.65134,-1.60324,3.76945)
p3=(-1.80449,-3.49553,0)
p4=(-1.80449,-3.49553,3.76945)


the meaning of
1 1 1 2 3 0 0 2 0
is
1==type_point 1=point_id 1=label_on 2=label_on_selection_off 3=higherentities 0=num_conditions 0=id_material 2=layer 'Lateral' 0=has_mesh_data
the point 1 belong to 3 curves (1, 3 and 5) then higherentities must be 3


2 1 1 2 2 0 0 2 0
1 2
2 2 1 2 2 0 0 2 0
3 4


This define curves 1 and 2 that are straigth lines (type==2). In this model that close a volume all curves belong to two surfaces, then its higherentity counter is 2
The curve 1 starts in the point 1 and end in the point 2


11 3 1 2 2 0 0 2 0
1 3 5 2 2.98214
-1.65134 -1.60324 0
-2.59749 -1.52666 0
-2.67407 -2.47281 0
-2.75064 -3.41896 0
-1.80449 -3.49553 0
0 0 0 0.5 0.5 1 1 1
1 1 0.707107 1 0.707107 1
11 4 1 2 2 0 0 2 0
2 4 5 2 2.98214
-1.65134 -1.60324 3.76945
-2.59749 -1.52666 3.76945
-2.67407 -2.47281 3.76945
-2.75064 -3.41896 3.76945
-1.80449 -3.49553 3.76945
0 0 0 0.5 0.5 1 1 1
1 1 0.707107 1 0.707107 1
11 5 1 2 2 0 0 2 0
3 1 5 2 2.98213
-1.80449 -3.49553 0
-0.858344 -3.57211 0
-0.781767 -2.62596 0
-0.70519 -1.67981 0
-1.65134 -1.60324 0
0 0 0 0.5 0.5 1 1 1
1 1 0.707107 1 0.707107 1


Previous text define curves 3, 4 and 5 that are NURBS (type==11)
1=start point 3=end point 5=num control points 2=degree 2.98214=length
the 5 control points coordinates are:
-1.65134 -1.60324 0
-2.59749 -1.52666 0
-2.67407 -2.47281 0
-2.75064 -3.41896 0
-1.80449 -3.49553 0
and the knots vector is:
0 0 0 0.5 0.5 1 1 1
1=is rational, weights= 1 0.707107 1 0.707107 1


3 6 1 2 2 0 0 2 0
4 2 -0.315383 0.025526 0.949244 4.63163 7.77322
1 0 0 0
0 1 0 0
0 0 1 0
-1.41253 -2.57491 3.76945 1


And curve 6 is a circumference arc (type==11)


4=start point 2=end point (-0.315383 0.025526)=2D center 0.949244=radius 4.63163=start angle 7.77322=end angle (rad)
and
1 0 0 0
0 1 0 0
0 0 1 0
-1.41253 -2.57491 3.76945 1
is a 4x4 transformation matrix that moves the 2D arc to the final 3D location


14 1 1 2 1 0 0 2 0
4
1 4 2 3
0 0 1 1
-2.67407 -2.47281 1.88473
0.996741 -0.080672 0
0 2 5 1 2
-1.65134 -1.60324 0
-1.65134 -1.60324 3.76945
-2.59749 -1.52666 0
-2.59749 -1.52666 3.76945
-2.67407 -2.47281 0
-2.67407 -2.47281 3.76945
-2.75064 -3.41896 0
-2.75064 -3.41896 3.76945
-1.80449 -3.49553 0
-1.80449 -3.49553 3.76945
0 0 1 1
0 0 0 0.5 0.5 1 1 1
1 1 1 0.707107 0.707107 1 1 0.707107 0.707107 1 1


Surface 1 is a NURBS surface (type==14), it has 4 boundary lines: 1, 4, 2, 3, with orientations same, same, diff, diff respectivelly
Its approximated center is (-2.67407 -2.47281 1.88473)
normal=(0.996741 -0.080672 0)
0=untrimmed 2=number control points u 5=number control points v 1=degree u 2=degree v
then are listed the control points, the knots u=(0 0 1 1) and knots v=(0 0 0 0.5 0.5 1 1 1)
and the weights=(1 1 0.707107 0.707107 1 1 0.707107 0.707107 1 1)


6 2 1 2 1 0 0 2 0
4
1 5 2 6
1 1 0 0
-0.781767 -2.62596 1.88473
-10.6459 0.861634 -0


Surface 1 is a Coons surface (type==6), it has 4 boundary lines: 1, 5, 2, 6, with orientations diff, diff, same, same. Then is printed its approximated center and normal.
The shape of kind of surface is defined only by its boundary.


14 3 1 2 1 0 0 1 0
2
5 3
0 0
-1.72792 -2.54939 0
0 0 1
1 2 2 1 1
-2.7699 -1.5074 0
-2.7699 -3.59137 0
-0.685932 -1.5074 0
-0.685932 -3.59137 0
0 0 1 1
0 0 1 1
0


Surface 3 is like the 1 a NURBS surface (type==14), but in this case is trimming a planar squared shape.


5 4 1 2 1 0 0 1 0
2
6 4
1 1
-1.41253 -2.57491 3.76945
0 0 -1


Surface 4 is a planar surface (type==5), that is defined by a center and normal, and the trimming boundary lines.


9 1 1 2 0 0 0 2 0
4
1 2 4 3
0 0 0 0
-1.72792 -2.54939 1.88473
0


This define the volume 1 (type==9) that is defined by 4 boundary surfaces: 1, 2, 4, 3 with orientations same, same, same, same.
and approximated center=(-1.72792 -2.54939 1.88473)


Last 0 is a NULL entity (type==0) that finish the definition of geometrical entities.


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