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July 27, 2008, 23:31 
Center of Pressure

#1 
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Hi,
I know this has been posted previously; however, I did not find the posted information helpful and I'm hoping there are people out there who might be able to help me with the issue. I am trying to find the center of pressure for a vehicle in CFX. I tried both CEL Expressions and the force/torque equations method suggested in previous posts. The first gives me numbers that seem unreasonable. The latter doesn't work either...multiple solutions essentially. Any other suggestions or any ideas why the CEL Expressions aren't working? I can post up what I used if there are any questions regarding that. Thanks, Ykan. 

July 28, 2008, 01:57 
Re: Center of Pressure

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Hi,
I have done it before, it works fine. Please post your CEL and we will have a look. Glenn Horrocks 

July 28, 2008, 09:20 
Re: Center of Pressure

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Thanks for the reply. The following are the equations being used. This is for headon flow (ie. a velocity component only in the x direction).
CPX=((areaInt_x(X*Pressure)@car+areaInt(X*Wall Shear X)@car)) /((areaInt_x(Pressure)@car+areaInt(Wall Shear X)@car)) CPY=((areaInt_y(Y*Pressure)@car+areaInt(Y*Wall Shear Y)@car)) /((areaInt_y(Pressure)@car+areaInt(Wall Shear Y)@car)) CPZ=((areaInt_z(Z*Pressure)@car+areaInt(Z*Wall Shear Z)@car)) /((areaInt_z(Pressure)@car+areaInt(Wall Shear Z)@car)) 

July 30, 2008, 01:42 
Re: Center of Pressure

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Any comments anyone?
Thanks.... 

July 30, 2008, 22:10 
Re: Center of Pressure

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hi,
Use the force_x and torque_x functions. Something like: CPX = torque_x()@car / force_x()@car Glenn Horrocks 

August 1, 2008, 14:43 
Re: Center of Pressure

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Hi,
Thanks for the response. I tried that but I'm getting really small values. They don't make sense..... Also, in terms of torques and forces, shouldn't the forces in the z direction and y direction contribute to the torque about the xaxis? Thanks, Yakn. 

August 3, 2008, 18:49 
Re: Center of Pressure

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Hi,
Sorry, yes you are right, you will need to fix the vector directions up. Glenn Horrocks 

August 4, 2008, 02:15 
Re: Center of Pressure

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Ok. However, if I try to set up a matrix of the force and torque relations relative to the center of pressure, I get a skewsymmetric matrix which cannot be inverted to solve for the center of pressure.
Is there a way to do this without using the torques and forces method? Thanks, Yakn. 

August 9, 2009, 18:57 

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Michael Donovan Almighty
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thank you very much!! 

August 9, 2009, 19:22 

#10 
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Glenn Horrocks
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My original post was incorrect. Yes, you have to calculate it but it is not that hard. From the definition of Torque = r x F, where r is the displacement vector to the point the force acts on and F is the force vector. You can get the torque vector using the torque_x, torque_y and torque_z commands and similarly the forces using force_x, force_y and force_z. It is just then a matter of expanding out the cross product and solving for the displacement vector.
Glenn Horrocks 

August 11, 2009, 12:39 

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