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March 19, 2007, 13:54 
center of pressure

#1 
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Hello, could anyone tell me how to find the center of pressure of an object?


March 19, 2007, 17:03 
Re: center of pressure

#2 
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Hi,
You can calculate the total force and moment on a surface in either CFXPost (using the function calculator) or a monitor point during solution. The moment is calculated about the coordinate axis by default. You can work it out from there. Glenn Horrocks 

March 20, 2007, 08:41 
Re: center of pressure

#3 
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Hello I know how to calculate the Total force on the Post, but not the moment...
Pedro 

March 20, 2007, 16:56 
Re: center of pressure

#4 
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Hi,
There is a torque function in the function calculator in CFX Post. Glenn Horrocks 

March 21, 2007, 13:49 
Re: center of pressure

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Keep us posted on how this goes. I've tried to do this before, but have not been successfull. You know that the torque at the COP is zero, so you can construct 3 equations for torque about each of the three axis and get those torque values from CFXPost. You end up with 3 equations and 3 unknowns (the unknowns are the x,y,z coordinates of the COP)  sounds great! But solving those equations leads to 0 = 0. In other words you don't really have 3 unique equations. I think you have to introduce a fourth equation that places some geometric constraints on the location of the COP (maybe constraining it to a line or a plane). Anyways, let me know if you figure it out.


March 21, 2007, 19:03 
Re: center of pressure

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Hi,
I don't think you are calculating it the correct way Stumpy. The x coordinate of the COP = torque_x(about origin)/force_x, and repeat for Y and Z. No simultaneous equations are needed. Glenn Horrocks 

March 26, 2007, 11:12 
Re: center of pressure

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Concerning other thing, Do you know why it happens that when calculating the flow around a thin surface, when i want to plot for example the pressure distribution in one of the surfaces, the mesh appears as well ? If i do it on a sphere for example, the plot of the pressure on the sphere surface looks good, but on a thin surface i cannot erase the mesh... Thanks


March 28, 2007, 09:05 
Re: center of pressure

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Actually,
I'm having the same problems as Stumpy. It seems to me that the moment about Z axis has nothing to do with the Z location of COP. The three equations that I think Stumpy was mentioning are: Torque(z) = Fx ycop + Fy xcop Torque(x) = Fz ycop  Fy zcop Torque(y) = Fz xcop + Fx zcop Please tell me why these equations are incorrect! If they are correct, I think I'm stuck like Stumpy. 

March 29, 2007, 06:09 
Re: center of pressure

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Hi,
I have just shown you how to do it in this post! If you need the exact CEL function, try: COPx = torque_x()@Surface / force_x()@Surface COPy = torque_y()@Surface / force_y()@Surface COPz = torque_z()@Surface / force_z()@Surface Glenn Horrocks 

March 29, 2007, 22:33 
Re: center of pressure

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Glenn,
Thanks. I did see your post, but I wasn't convinced that it is correct. If you read my post, you would see my challenge, which I will restate now. I assume that torque_x() is torque about the xaxis. If that is the case, then the force in the x direction contributes nothing to the torque about xaxis. I listed 3 valid equations which Stumpy mentioned. I know for a fact they are valid. But I also know for a fact that they form a singular matrix. Forgetting CFX for a second. If someone could shed light on the mathematics of this, that would be helpful. Thanks for your help, but I think I'm raising valid questions here. Samer 

March 30, 2007, 14:21 
Re: center of pressure

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I tried Glenn's method on an example and it didn't seem to give sensible results. Take the AxialIni_001.res file that everyone has in the examples dir in their installation. Using the boundary S1 Blade and Glenn's method I calculated the COP to be (2.473[m], 0.775[m], 0.226[m]). If you plot this point it is way outside the domain.
Those equations Samer posted are the 3 I was refering to. Any aero guys out there? I think they regularly calculate COP on 2D airfoils, but they provide a geometric constraint by constraining the location to the cord line. Don't know if you can extend that to 3D? 

March 30, 2007, 16:38 
Re: center of pressure

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You cant get the center of pressure from the force moment calculations as described. The 3 equations form an equation for a line or vector (which is what the force acts in).
These equations do not have a unique solution, for any x, another y and z will solve these equations. In order to find the center of pressure is easy if you know one location that the pressure acts upon (say a z force acting on an x,y plane). You can set the z=0 and get the right x and y locations. Complications arise when you have an arbitrary surface with a force on it and can't reasonably remove one of the coordinates. Suffice it to say, I don't know what you are using your COP for, but it might be easier just to put a coordinate system at a convinent location for any other calcs you may need, and use a force and moment at that point, instead of a force and 0 moment at the COP. 

March 31, 2007, 09:44 
Re: center of pressure

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Thanks Stumpy and Edmund,
Stumpy, I am an Aero Engineer and I have done COP for 2D airfoil many times. As you said, it forms a line which we intersect with the chord line to get a unique COP point, if you like. Thanks Edmund. I understand now that there is not a unique point solution in 3D. The reason I needed a point was for a structures guy to add a point source to represent a blade in ANSYS. So he wants my forces to generate the moments at that point. What I ended up doing was assuming that the torque in the disk rotation direction is only caused by one of the forces, since the blade is at Top Dead Center. Once I had that COP I used the other 2 equations to get the remaining COPs. Any alternative ideas? 

April 2, 2007, 02:39 
Re: center of pressure

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Hi,
You are correct, my equation was incorrect. My apologies for misleading you. Glenn Horrocks 

April 2, 2007, 08:16 
Re: center of pressure

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Samer:
Your structures guy should easily be able to use a force and a moment that you give him. Just specify the location of the coordinates that the torques are around, and let him move it to where he wants. A forcemoment combination can be moved anywhere. If he doesnt know how to do that, you got problems.... 

April 2, 2007, 16:45 
Re: center of pressure

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Hi guys,
This is not so difficult........... What about a center of pressure with coordinates (CPX,CPY,CPZ) for a body with the outersurface called wing: CPX=areaInt(X*Pressure)@wing/areaInt(Pressure)@wing CPY=areaInt(Y*Pressure)@wing/areaInt(Pressure)@wing CPZ=areaInt(Z*Pressure)@wing/areaInt(Pressure)@wing This worked well for pressure forces. You can add viscous forces as well. ............ Astrid 

April 16, 2007, 18:44 
Re: center of pressure

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Hey Astrid, i have done it in a simple flow around a sphere and it should give right in the midle of it, but it does not, it goes well away from it. Do you know what is wrong?


April 17, 2007, 12:40 
Re: center of pressure

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Is the pressure around the sphere constant? Do you not have high pressure on the upstream side and low pressure on the downstream side?


April 17, 2007, 14:28 
Re: center of pressure

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no, the pressure around the sphere is not constant, upstream it is higher


January 13, 2014, 18:09 

#20 
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Albert Mathews
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Although this thread is old, I found it very useful. For those like me looking for more details, I worked on it a bit and found that the following equations work to find the centre of force on an object in a flow.
If I have made any mistakes, please correct me since, in that case, I would like to know how to do it correctly. In CFX Post, Expressions tab, right click, create a new expression, I chose COFy as name. For a 2D surface called CarBody, the following calculates the height of the drag force applied to the object. areaInt(y*(Pressure+Wall Shear))@CarBody/areaInt(Pressure+Wall Shear)@CarBodyLikewise for for lift/downforce, I chose COFz. areaInt(Z*(Pressure+Wall Shear))@CarBody/areaInt(Pressure+Wall Shear)@CarBody 

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