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September 3, 2014, 06:56 

#21  
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Meimei Wang
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Quote:
I'm also looking for the way to extract COP location in CFX. I run exactly the same case by Fluent and CFX. The other values obtained (e.g. lift&drag) are almost exactly the same. Fluent can export COP directly. I use your CEL to obtain COP in CFX. Fluent and CFX give very different value! So I guess your CEL formula shall have problem. I can't figure out what the problem is. Anyone could help on that?
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September 3, 2014, 11:18 

#22 
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Albert Mathews
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Anna,
thx for looking into this. question, in Fluent, is the COP calculation integrating pressure and viscous forces or just pressure? regards 

September 3, 2014, 19:30 

#23 
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Glenn Horrocks
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There is no need to use functions like "areaInt(Pressure+Wall Shear)@CarBody". The function force()@CarBody is the one to use.
Also, the basic derivation of centre of pressure is T=d cross F where T is the torque vector defined around your origin point, F is the force vector and d is the displacement of the force to give the torque. This means your COP is at d. When you evaluate this it gives you a set of simultaneous equations to solve to get the value of d from T and F. T is evaluated from torque_x/y/z() and F is evaluated from force_x/y/z. In your case you may be able to simplify this because you know soemthing about the COP. For instance you know it is on the plane in a 2D simulation. 

September 3, 2014, 21:41 

#24  
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Meimei Wang
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Quote:
Thanks. I'll calculate in this way. Btw, what's wrong with the formula given as below? areaInt(y*(Pressure+Wall Shear))@CarBody/areaInt(Pressure+Wall Shear)@CarBody
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September 3, 2014, 23:44 

#25 
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Glenn Horrocks
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areaInt(Pressure+Wall Shear)@CarBody will give you the magnitude of the force vector. I ahev not checked your equation in detail, but for your equation to work you will need to have assumptions about the position on the COP.
But my main problem is why use these functions when a direct function [force()] already exists? It is a little pedantic, but I prefer to keep my stuff simple as possible. 

September 4, 2014, 10:20 

#26 
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Edmund Singer P.E.
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Glenn:
The method you describe provides a nonunique solution. The force vector acts in a line and you can have many solutions to the set of equations along that line. That method works well if you are looking for the center of pressure acting on a surface or a plane (or one where you know one of the points). This allows you to fix one of the points and then the other 2 points become unique. It also can work for an airfoil. You can assume that the center of pressure acts along the cord and then the other 2 points become unique. Alberts equations are, by definition, the center of pressure. 

September 4, 2014, 10:32 

#27 
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Glenn Horrocks
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You are correct, thanks for the correction and clarification.


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