SimScale CAE Forum

Calculating Center Of Pressure

Is there a way to calculate the center of pressure of a car model in simscale CFD?
If yes, could anyone guide me as to how to go about this.

Hi @varkotian,

Just an idea here, you could use two result control items at two arbitrary points and use the pressure moments to calculate the centre of pressure for each axis thus giving a coordinate in 3 dimensions. Then from that point, theoretical no moments would be pressure and thus it could be said that all forces act through that point? This might need a bit of thinking about but once you have done the calculation once you could write a file or something to do it in the future much quicker.

But to answer your question explicitly, there is no automatic calculation of the centre of pressure in simscale (at least to my knowledge).

Let me know if this helps, or even doesn’t help and I’ll scratch my head some more :S


Thank you for the idea. It worked and I found out the Center Of Pressure easily.

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I tried calculating using the above method but ended up with same equations twice.
Could someone please explain the calculations used to find the center of pressure.

And I’d be interested for what type of application you’re doing this? Vehicle / building aerodynamics?

I am doing this for vehicle aerodynamics.

This is the method I had in mind:

I can’t guarantee the maths is perfect at 10.30pm but the method should be good… remember this is for 1 axis only, this has to be done 3 times for 3 ordinates. The points in which the moments are around are arbitrary points in the result control forces and moments item.

@varkotian feel free to chip in here as you successfully managed this, is this the way you did it? any shortcuts?

Kind regards,

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I understood this method but I don not think the two arbitrary points and the center of pressure will all lie on the same line. And two forces(Down force and Drag) will act on the body instead of the one force used in the above method. Because of these reasons I think this method may not be correct. Please correct me if I am wrong.

I found the points using the above method but the moments about this point is not zero. This may not be the correct way to find the center of pressure.

H @pdarda, thanks for the reply this is making for an interesting topic. Firstly is there any chance I could borrow your project? I need some real numbers to try myself.

To get working method I’m going to write out the set of steps and assumptions I made and you can tell me which ones you disagree with, hopefully, this will give us both an understanding of why this does/doesn’t work… can we do that?

  • Moments about a point can be represented by a force and a distance
  • If the force is acting perpendicular to the axis the distance can be that from the point to the force parallel to the axis.
  • (2D) if the force position of the force is unknown we can find it if we have two-moment forces measured from the same force at different points giving a position along that axis.
  • We know the distance (or can find) between the two points along the axis.
  • We know the moments at these two points.
  • We do not need to know the magnitude of the force.
  • To consider lift (or downforce) (down the y-axis) and drag (along the x) the calculation will be needed to be done twice or even a third if there are forces around the vertical axis.

So this implies that it could be found but does not have to be on the geometric line between the two points, in fact, the way I have put them on the same line in the example is miss leading. I should have drawn one higher and one lower.

So how close are your calculations to zero moments?

Hope to get your thoughts soon,


yes, you can have my project,
Here’s the link :

I think that as we have two forces here ,that are perpendicular to each other, instead of only a single force we may not be able to calculate the position even if we had two-moment forces. We will get only one equation having two unknowns(x and y co-ordinate) which cannot be solved.


I’m convinced that they can be treated as two separate pairs of equations solved for each force. I did the calculation, got close… 15N could be put down to calculation errors, 15N was the highest of the 3.

look at run 0001 and let me know what you think.

Kind regards,

I understand that we cannot get exact answer. There will be some errors. So, we have to be satisfied by the approximate values which we get from the calculations.

Could you share the calculations using two forces so that I could have a better understanding of the calculations.


@pdarda I’ve just done another example and I think I see what you mean, so I turned the complexity up and looked at moments in 3D:

syms Px Py Pz

F0=[39.39, -200.28, -159.87]%N;

P0=[Px, Py, Pz]%m;


moments1=[-111.32, -220.29, 224.98];%Nm


%hand solution



c =

2.1161   -0.0369    0.3246

The only one that looks dodgy is the x ordinate. I have just run out of core hours so maybe you could tell me how it goes?


Here is a link to a conversation very much like ours, think it boils down to being able to simplify for 2D (like we originally arrived at) or using M=r x F where x is the cross product (hopefully similar to my matlab script above)


I am confused which numbers you have used. So after simulation you are given pressure force in x,y,z direction and also for pressure moment in x,y,z. Therefore can you help me realize how should I use these numbers, I will link my simulation (Repaired FSAE Car-20m/s-run 1-complete formula).

I will be very grateful for your help.

A manual (graphical in CAD program) method for the case of two orthogonal forces (say fx and fz) acting on a point and a moment (say ymoment) about that point in an axis orthogonal to the plane of the two forces, I believe that you can determine the CP positon as follows (a general formula to the below is much more difficult due to quadrant issues of the force vector and the polarity of the moment but that should be no sweat for the programmers at simscale, and with a little further application of this method, solving for a true 3D CP position would be a great feature to add to the Forces and Moments output :slight_smile: ):

  1. Determine the force vector of summing the two forces whose force magnitude is F.
  2. Offset this force vector by a value of abs(ymoment/F). Pay attention to polarity of ymoment to make sure that you offset the force vector on the correct side.
  3. Thats it.

The below image should help out with my poor explanation. fx (drag) fz (lift) and ymoment (pitching moment) come right from your simulation force plot output as pressure values (in my case I can ignore viscous and porous values as they are so small). Once you convert the ymoment into two force vectors as described above, the purple and blue force vectors cancel each other and you are only left with the green force vector at the CP.

My ulterior motive for posting this is to get feedback so that I know I did this correctly :slight_smile: , its just that before I did the CFD on this case, may brains CP calculator would not have placed the CP as shown…


Nice one @DaleKramer!

Cheers and keep up the good work!