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How to get a particular plane in CFD-Post?

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Old   April 2, 2014, 23:06
Default How to get a particular plane in CFD-Post?
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Hi, all.
I want to get a circular region in CFD-Post and the equation is x^2+y^2<=1 && Z=1. I insert a plane Z=1, but this region is too big. I just want to display the circular region. How to do this ?
Thanks.
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Old   April 3, 2014, 01:09
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Quote:
Originally Posted by Raijin Thunderkeg View Post
Hi, all.
I want to get a circular region in CFD-Post and the equation is x^2+y^2<=1 && Z=1. I insert a plane Z=1, but this region is too big. I just want to display the circular region. How to do this ?
Thanks.
1. Insert plane.
2. Definition -> Method = XY Plane
3. Plane Bounds -> Type = Circular
4. Plane Bounds -> Radius = 1

Z is not radius but distance between coordinate system XY plane and your plane. Also note that by default plane type is "slice", so you'll see only intersection with your geometry. If you want to see full cyrcle then set plane type to "sample".
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Old   April 3, 2014, 01:14
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You can go to turbo mode in CFD post and intialise Rotational axis by setting axis definition then you will get Radius as variable.

With this Radius as variable you can use Surface of revolution for circular plane.
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Old   April 3, 2014, 02:19
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Quote:
Originally Posted by Antanas View Post
1. Insert plane.
2. Definition -> Method = XY Plane
3. Plane Bounds -> Type = Circular
4. Plane Bounds -> Radius = 1

Z is not radius but distance between coordinate system XY plane and your plane. Also note that by default plane type is "slice", so you'll see only intersection with your geometry. If you want to see full cyrcle then set plane type to "sample".
Thanks for your reply. If this circle is eccentric, this method doesn't work.
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Old   April 3, 2014, 02:24
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Thanks. I will try what you said.
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Old   April 3, 2014, 04:06
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Quote:
Originally Posted by Raijin Thunderkeg View Post
Thanks for your reply. If this circle is eccentric, this method doesn't work.
You then may insert "User Surface" with "method" = "offset from surface", "type" = "translational" and define direction and distance. This will give you circle which center is shifted from the origin of coordinates.
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Old   April 3, 2014, 06:30
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You can define any shape you can define analytically by defining a variable as function of x,y and z; and then doing a user surface of a contour plot of it. For instance, if you define a variable as r = sqrt(x^2+y^2), then you can draw a contour at any radius you like and make it into a user surface. Likewise you can extend this to ellipses, parabolas or anything else you can define analytically.
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Old   April 3, 2014, 21:39
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Quote:
Originally Posted by ghorrocks View Post
You can define any shape you can define analytically by defining a variable as function of x,y and z; and then doing a user surface of a contour plot of it. For instance, if you define a variable as r = sqrt(x^2+y^2), then you can draw a contour at any radius you like and make it into a user surface. Likewise you can extend this to ellipses, parabolas or anything else you can define analytically.
This is awesome. Thank you.
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Old   January 31, 2015, 14:25
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Can I select a specific Region? Can I vary the plane size? I want a very small plane at a specific location.


I have a box let say 1m x1m x10m.


I want a plane at 0.2 x 0.2 and 5m in the z axis of that box.

How is that possible?

Awaiting reply

Thanks.
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Old   January 31, 2015, 17:42
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You can define arbitrary shapes using limits on planes, or using contours. For instance the contour between x=1 and x=2 will be a strip 1m wide.
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Old   January 31, 2015, 23:58
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Hassan Iftekhar
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I am having the same issue. I could not shift the plane

Could you explain further
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Old   February 1, 2015, 15:01
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Quote:
Originally Posted by ghorrocks View Post
You can define arbitrary shapes using limits on planes, or using contours. For instance the contour between x=1 and x=2 will be a strip 1m wide.
Can you please explain. Thanks.
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Old   February 1, 2015, 18:48
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You are seeing the elements which lie wholly within the plane. That is why it is jagged.
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