# Center of Pressure on a body

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 May 8, 2011, 23:27 Center of Pressure on a body #1 Member   Greg Givogue Join Date: Aug 2010 Location: Ottawa Canada Posts: 57 Rep Power: 14 Hi CFDers! I have a general question regarding center of pressure (CP) on a body. How do CFD codes determine CP? Center of pressure is defined as the location where the resultant force acts on a body. The software I'm using provides forces and moments about an arbitrary point in space. If I try to use these results to determine CP (x y z) I get the following equations; Mx=-Fy*z+Fz*y My=Fx*z-Fz*x Mz=-Fx*y+Fy*x (bounded by the surface coordinates of x y z) The problem is that these equations form a singular matrix and hence 1 or more equations are not independent. Any thoughts? Essentially we'd like to take our results and use them for some basic structural analysis - to do so we the location of the resultant force. Thanks Everyone!!

 May 24, 2011, 17:02 #2 Member   james nathman Join Date: May 2011 Posts: 59 Rep Power: 14 The equations you state have many solutions. Given a point which has zero moment, then any point along the line of action of the total force also has zero moment. You must apply a constraint. Discussions about center of pressure for an airfoil apply (without explicitly saying it) that the c.p. lies on the chord line. The constraint is your choice: specified x, y, or z or planar combination of; closest point to the point about which the moment and force are given; etc. The constraint must not be parallel to the line of action of the force. C of P is not, in general, useful. In many cases, a fluid generates a moment along the line of action of the force. So, the best you can do is find a point which minimizes the moment, not zeros it. Your algorithm should remove the moment parallel to the total force before trying to find CofP. Your structural analysis will require the application of moment, so you might as well apply them in the original system. Think of a rotating propeller producing thrust. There's no place where the thrust can act that will generate the torque required to spin the propeller. caio.df likes this.

 May 25, 2011, 12:06 #3 Member   Greg Givogue Join Date: Aug 2010 Location: Ottawa Canada Posts: 57 Rep Power: 14 Hi James, Thanks for the reply and the suggestions. I recognized that CP was actually a line when the equations formed a singular matrix. I have tried constraining one of the variables (z) and using the fact the line must be the same direction as the total force, however I get a line of action that does not cross the body. The points I chose for z are contained within the limits of the body. The body is essentially the shape of a fuselage and it is fully encapsulated. Can you explain what you mean by "the constraint must not parallel the line of action of the force"? and "your algorithm should remove the moment parallel to the total force before trying to find CP" ? Thanks for the help, Greg

 May 25, 2011, 13:50 #4 Member   james nathman Join Date: May 2011 Posts: 59 Rep Power: 14 remove the moment parallel to the total force: Your equations imply the moment is produced by the force. In general, the fluid forces on a body include a pure couple. For example, a propeller (rotating about the X axis) has a nonzero thrust (Fx) and a nonzero torque, Mx. The other force and moment components are zero. Your first equation is Mx = 0. which has no solution because it doesn't recognize that the propeller torque is a pure couple, not produced by the thrust. The pure couple is M (dot) f where f is the unit vector in the direction of the total force. If (Mx,My,Mz) is the moment at (x,y,z)=(0,0,0) and (Fx,Fy,Fz) is the total force, the equation for Mx should be Mx - x_component_of [M(dot)f]= -Fy*z+Fz*y now the propeller gives 0=0 for the first equation and a solution is possible after replacing one of the force equations with the constraint equation. constraint must not parallel the line of action of the force: For the propeller case, if the constraint is Z=0, or Y=0, these define planes parallel to the thrust, Fx, and multiple solutions (any x) still exist. A generally useful constraint is 0 = Fx*x + Fy*y+ Fz*z this constrains the cp to be in a plane normal to the total force. In other words, this constraint identifies the point closest to the origin that is a CofP.

 May 26, 2011, 09:06 #5 Member   Greg Givogue Join Date: Aug 2010 Location: Ottawa Canada Posts: 57 Rep Power: 14 Hi James, In your example you're ignoring that the torque on a propellor (Mx) is created by the drag on each blade. So there should be components of Fy and Fz that contribute to Mx (Mx does not equal 0 from my 1st equation). You've probably thought of this but it seems wrong to me. Thanks, Greg --------------------------------------------------------------------------------------------------- Ignore my last comment - I understand what you're saying now - the resultant force will not have components of Fy and Fz.

 May 26, 2011, 09:56 #6 Member   Greg Givogue Join Date: Aug 2010 Location: Ottawa Canada Posts: 57 Rep Power: 14 Hi James, Here's an example of some of the numbers I'm working with. Datum for moment calcs (1.33 0 -2.08) fuselage limits x(-21.5, 11) y(-2.3, 2.3) z(-2.08,3.003) Fx =-281.2741973 Fy =-706.7381177 Fz =-274.1519911 Mx= 1649.523746 My= 12493.59808 Mz= -29011.77046 I tried applying the ideas you have suggested above but I still get CP values (lines of action) that do not lie anywhere near the body... Can you give it a shot? Thanks and I appreciate all your help, Greg

 June 28, 2011, 20:43 #7 Member   james nathman Join Date: May 2011 Posts: 59 Rep Power: 14 The CofP closest to the moment datum is at (38.0, -13, -5.6) The pure couple (moment parallel to force) is (574, 1443, 560) The line of action of the force does not pass through the fuselage limits. Fluid dynamics strikes again.

 July 13, 2011, 18:46 #8 New Member   J. Join Date: Jul 2011 Location: USA Posts: 2 Rep Power: 0 James, Could you walk us through how to set up M(dot)f, I have a similar problem to Greg's and I guess I need a refresher to find x, y, and z components of M(dot)f. Thanks, J.

 July 13, 2011, 20:03 #9 Member   james nathman Join Date: May 2011 Posts: 59 Rep Power: 14 F=(Fx,Fy,Fz) M=(Mx,My,Mz) unit vector in force direction f=(Fx,Fy,Fz)/FMOD where FMOD=sqrt(Fx*Fx+Fy*Fy+Fz*Fz) Moment parallel to F (pure couple) is (Fx,Fy,Fz)*(Mx*Fx+My*Fy+Mz*Fz)/FMOD/FMOD Pure couple is M(dot)f*f, actually. Do as I meant, not as I said.

 July 15, 2011, 12:01 #10 New Member   J. Join Date: Jul 2011 Location: USA Posts: 2 Rep Power: 0 Blackjack, Thanks for the quick reply. I follow you now, and everything checks out. J.

 July 18, 2011, 22:35 #11 Member   Greg Givogue Join Date: Aug 2010 Location: Ottawa Canada Posts: 57 Rep Power: 14 Thanks Blackjack and Desk Pilot, Basically my CFD analysis produced Forces and Moments at the cg of an aircraft represented by a fuselage with and without a large center line fuel tank. We wanted to take these results to determine the stresses at specific points around the structure of the tank in order to do some back of the envelope stress calcs. If I follow your logic and find those moments at the points of interest, shouldn't they be the same as the ones I would get from re-running the sim with the datum at the same point? Thanks again, Greg

 July 24, 2011, 16:41 #12 Member   Greg Givogue Join Date: Aug 2010 Location: Ottawa Canada Posts: 57 Rep Power: 14 Hi James, I re-ran the CP calcs using the following equations; 0=Fx*x+Fy*y+Fz*z My - (Mdotf)*fy=Fx*z - Fz*x Mz - (Mdotf)*fz=Fy*x - Fx*y and I get (-21,-7.8,-1.52) which is different than your answer for CP. Can you tell me which set of equations you used? I'm assuming you used the planar constraint? Thanks, Greg Last edited by Greg Givogue; July 24, 2011 at 16:56.

 November 29, 2011, 19:26 #13 Member   JPConway Join Date: Nov 2011 Posts: 33 Rep Power: 13 blackjack (or others), hoping you can help me out on this. I'm trying the same problem but in 2D Fx (drag) = -3.48N Fy (lift) = 60.65N Mz = 17Nm measured at origin i've got FMOD = 60.75N i've got f = (-0.0573,0.998) not sure where to go from there. I think the equation is suppose to be Mz-z_component of (Mdotf)*fz = -Fxy+Fyx but the second moment term zeros out if I use the equation described by blackjack below (Rearranged to match equation above). ((Mx*Fx+My*Fy+Mz*Fz)/FMOD)*((Fx,Fy,Fz)*/FMOD) ((0*Fx+0*Fy+Mz*0)/FMOD)*((0)*/FMOD) let me know. thanks.

 November 29, 2011, 20:12 #14 Member   james nathman Join Date: May 2011 Posts: 59 Rep Power: 14 Yes, Mz=-Fx *y + Fy * x Now add a constraint by replacing either of the other two equations with, e.g. (the usual airfoil constraint) y=0. Then x=Mz/Fy which is the usual airfoil value The other moment equation tells you z=0. Again, the location of the center of pressure depends on the constraint, but all valid centers lie along the line of action of the force.

 November 29, 2011, 20:31 #15 Member   JPConway Join Date: Nov 2011 Posts: 33 Rep Power: 13 y=0 seems like a pretty abirtary constraint are you making that assumption based on the origin being in a certain location? for example at the leading edge? or somewhere on the chord? in this case the origin is simply the origin of the CAD model which was chosen arbitrarily. that being said, the location based on that constraint seems reasonable. i also tried your approach with 0 = Fx*x + Fy*y + Fz*z = Fx*x + Fy*y but the value that is obtained isn't reasonable. I'm still having trouble wrapping my head around there being multiple centers of pressure. Are you speaking numerically or physically? Could you elaborate on this statement: "Again, the location of the center of pressure depends on the constraint, but all valid centers lie along the line of action of the force." Thanks.

 November 30, 2011, 15:01 #16 Member   james nathman Join Date: May 2011 Posts: 59 Rep Power: 14 Dommasch says the center of pressure is the location of the resultant force acting on the airfoil. Beer & Johnston say that a force can be replaced by another at a different location provided the two forces have the same line of action and are of the same magnitude. Physically and mathematically there are an infinite number of points where the moment produced by a 2-D body is zero. When you tried 0=Fx*x+Fy*y, what was the moment about the calculated point? If it was zero, then it was a center of pressure according to the above definition. If the moment wasn't zero, there was a mistake in the calculation (could be mine, but my procedure has, so far, always produced a zero moment in 2-D). Try different constraints: y=0.1, y=0.2, etc. Plot x vs. y. The points will all lie along the line of action of the force. Any one of them could be chosen. y=0 is an arbitrary constraint because there's not sufficient information to define a single point. What's your definition of center of pressure? What makes the result you got previously unreasonable?

 November 30, 2011, 15:43 #17 Member   JPConway Join Date: Nov 2011 Posts: 33 Rep Power: 13 center of pressure is as defined here (http://en.wikipedia.org/wiki/Center_of_pressure) The center of pressure is the point on a body where the total sum of a pressure field acts, causing a force and no moment about that point. I must be missing something.... using the equations: 0=Fx*x+Fy*y and Mz=-Fx *y + Fy * x we can get down to: 0=Fx*x+Fy*y y=(Fx*x)/Fy Mz=-Fx *y + Fy * x Mz=-Fx*(Fx*x)/Fy) + Fy*x Mz=(-Fx^2/Fy)*x + Fy*x x=Mz/(-Fx^2/Fy + Fy) plugging in the values measured at the origin I get these numbers: x=0.28m y=-0.016m the numbers are reasonable with respect to the geometry and the location of the center of mass of the 2d object. however the calculated moment at that point is equal to 17Nm which is what was measured at the origin. since as you mentioned this is suppose to be the center of pressure where Mz=0, I must still be missing something. thanks.

 November 30, 2011, 18:24 #18 Member   james nathman Join Date: May 2011 Posts: 59 Rep Power: 14 Yes, I get x=.28. However, I get y=.016 (sign disagrees) More importantly, the moment at that location is 1 micro Nm. (zero in my book). I think you just didn't transfer the moment correctly to the CofP. You're almost there.

 November 30, 2011, 18:44 #19 Member   JPConway Join Date: Nov 2011 Posts: 33 Rep Power: 13 y=(Fx*x)/Fyy=(-3.48N)(0.28)/60.65 which is where i get the negative from moment wise, I'm not following i'm just plugging and chugging into: Mz=-Fx *y + Fy * x Mz=-(-3.48) *(-0.016) + 60.65 * (0.28) Mz=16.926 (note: using +0.016 gets to 17.03N so it is entirely possible that +0.016 is correct but that would imply we should use the positive force which puts us back at 16.926)

 December 1, 2011, 00:22 #20 Member   james nathman Join Date: May 2011 Posts: 59 Rep Power: 14 0=Fx*x+Fy*y yields y=-Fx*x/Fy So y is +.016 The equations used to calculate the CofP location cannot be used to calculate the moment at the CofP. The moments in those equations are the moments about the origin.