# Center of Pressure

 Register Blogs Members List Search Today's Posts Mark Forums Read

 July 27, 2008, 23:31 Center of Pressure #1 Yakn Guest   Posts: n/a 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 #2 Glenn Horrocks Guest   Posts: n/a 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 #3 Yakn Guest   Posts: n/a Thanks for the reply. The following are the equations being used. This is for head-on 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 #4 Yakn Guest   Posts: n/a Any comments anyone? Thanks....

 July 30, 2008, 22:10 Re: Center of Pressure #5 Glenn Horrocks Guest   Posts: n/a 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 #6 Yakn Guest   Posts: n/a 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 x-axis? Thanks, Yakn.

 August 3, 2008, 18:49 Re: Center of Pressure #7 Glenn Horrocks Guest   Posts: n/a 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 #8 Yakn Guest   Posts: n/a 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 skew-symmetric 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
#9
New Member

Michael Donovan Almighty
Join Date: Aug 2009
Posts: 2
Rep Power: 0
Quote:
 Originally Posted by Yakn ;89261 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 skew-symmetric 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.
so the only way is to calculate by your self. There is not any feature in CFX in order to obtain the center of pressures. In my case I hace calculated the flow around a rocket and the next step is locating the Cp, but I don't know how to obtain it by CFX... Is very sad use the tipical ms-dos software to obtain the Cp with a basic sketch of my rocket... Can anyone help me?

thank you very much!!

 August 9, 2009, 19:22 #10 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 11,284 Rep Power: 88 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
#11
New Member

Michael Donovan Almighty
Join Date: Aug 2009
Posts: 2
Rep Power: 0
Quote:
 Originally Posted by ghorrocks 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
I have calculated the center of pressures obtaining previously the force and the torque pressure net over the surface of my rocket. After that I have made a change of center of reduction of the moment: Mo=Mp+OPxF, where Mp is 0 beacuse P is the center of pressures and I only have had to solve a linear system of equations in order to obtain the coordinates of the center of pressures (OP). In my case "O" is the coordinates origin son P is directly the coordinates of the center of pressures.

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post emueller CFX 0 May 5, 2009 11:08 praveenreddy FLUENT 2 December 3, 2007 12:33 Virag Mishra CFX 0 August 27, 2007 10:27 Astrid CFX 0 December 21, 2006 05:07 giorgio FLUENT 0 June 29, 2004 12:36

All times are GMT -4. The time now is 05:53.