
Re: Model rocket center of pressure/location of aerodynamic forces
Shivani Patel Jun 16, 2015 5:47 PM (in response to Jason Meyers)Is your main concern to create a plot for Center of Pressure vs angle of attack? To determine stability?
This is what SOLIDWORKS has to say on the subject:
It is impossible to determine the point of application of a force from its Force and Moment components. All that can be obtained in the line (not point) of application of the force: the lien of application is the infinite line passing through the actual point of application and parallel to the force vector.
Extra information from another source has to be used to determine the point of application on the line.
The Line of application of a force can be determined by using the torque (M) definition: M=[r*F]. Here r is a vector to force application point from the origin O of the coordinate system, F – force vector, M – torque vector. In obtained calculation results (Surface Parameters for instance) you get components of force and torque. So, you can solve the equation regarding r. The equation can be rewritten regarding coordinates as following:
Mx = y * Fz  z * Fy
My = z * Fx  x * Fz
Mz = x * Fy  y * Fx
Here (Mx, My, Mz) – torque components, (Fx, Fy, Fz) – force components, (x, y, z,) – coordinates of force application point. So, we have three equations and three unknown values (x, y, z) which can normally be found from these equations.
In fact, one of the system of equations is undeterminate and cannot be solved (it is as if you had 3 unknowns and only 2 equations).
To determine the line, you can for instance assume two values for z (0 and 1 for instance) and calculate the corresponding values for x and Y. this will give you two points and the line of application pases through these points.
To determine the point of application from the line requires extra information which may not be possible to obtain.
In some special cases, it will be possible. For instance if you are trying to to determine the point of application of a force on a flat face. In that case, the point of application will be the intersection of the flat face with the line of application.
In the case of the force on a whole body with nothing but flat faces, you can calculate the point of application the force on each face separately, and then find the equivalent force and point of application for the entire body.

Re: Model rocket center of pressure/location of aerodynamic forces
Andrei Popov Jun 16, 2015 7:05 PM (in response to Jason Meyers)I have never done this before but this is how I would do it:
1. I would define an axis or a point on an axis with a dimension distance from CG or from a fixed point ( rocket nose) and I would use this point as a reference for the further calculations.
2. In flow simulation you can set a surface goal the total force on the rocket surfaces, this force would produce a torque about the above axis. Also you can set a torque goal, I am not sure if you can select the axis to be the reference of the torque but you can try and see.
3. Then you can set a parametric study where the control parameter is the position of the axis w.r.t. the CG, you can select the dimension that defines the position of the axis. The goal of the study should be the torque goal = 0, Solidworks will make several iterations by changing the axis position until the torque is zero.
Position of the center of aerodynamic forces
SolidWorks Quick Tip  Parametric Studies in SolidWorks Flow Simulation