|September 25, 2012, 21:16||
Find where force is applied
I am using simpleFoam (of v2.1.1). I am trying to find where the forces are applied and have been outputing the forces and moments using the following in the controlDict:
functionsIt worked great and I got all the outputs in the forces\0\forces.dat file:
# Time forces( pressure, viscous ) moment( pressure, viscous )
997 ( ( ( -25.6521 1575.07 -2489 ) ( 0.0440792 73.7995 8.74441 ) ) ( ( -5215.72 15.0189 86.2846 ) ( -15.0971 -0.0198099 -0.124274 ) ) )
998 ( ( ( -25.6488 1575.07 -2488.99 ) ( 0.0442406 73.8003 8.74387 ) ) ( ( -5215.69 15.0191 86.2908 ) ( -15.0984 -0.0198765 -0.124476 ) ) )
999 ( ( ( -25.6506 1575.07 -2489.01 ) ( 0.0442343 73.8004 8.74362 ) ) ( ( -5215.72 15.0171 86.289 ) ( -15.0978 -0.02023 -0.124946 ) ) )
My problem is that I can't work out how to derive the x,y,z coordinates where the forces are applied to obtain the moment. I have been struggling for a couple of hours and have to admit that geometry is one of my many weak points... Any help would be appreciated.
Thanks in advance
|September 26, 2012, 03:53||
Join Date: Aug 2010
Location: Groningen, The Netherlands
Posts: 216Rep Power: 7
I hope I got your question right.
But actually you posted your answer already:
used for the calculations of the moments
So in your case you chose the origin.
I hope I could contribute
|September 26, 2012, 07:46||
Thanks a lot for your answer. I probably need to clarify a bit. I am trying to know the distance - or vector - that define where the force is applied to generate the moment - I am just after the distance in the y-direction.
Let me know if it is unclear.
|September 28, 2012, 09:38||
Join Date: Jul 2010
Location: Always on the move.
Posts: 306Rep Power: 6
you have the moments referred to a defined point, in this case 0 0 0.
moment is force time distance.
to find the distance from your reference ponint do moment/force
Naval architecture and CFD consultancy
|September 30, 2012, 21:53||
Thanks a lot for your answer and comments.
My understanding is that:
- Forces are the integral over surface of p*Face_normal;
- Moment are the integral over the surface of p*Vector(CofR,Centroid)^Face_normal.
Based on the above, I was trying to inverse a system to get the y coordinate:
Mx = yFz - zFy
My = zFx - xFz
Mz = xFy - yFx
Just doing y = Mx/Fz was neglecting the effect of drag on the x-component of moment. I am unsure if the above is correct as it implies that MxFx + MyFy + MzFz = 0 (ie the moment is to be perpendicular to the Force), but when I calculate the expression based on the output from openFoam this is not true. So I am assuming that my interpretation is incorrect.
I am using a "work-around" using ParaView (Calculating Lift and Drag in Paraview (paraFoam)) where I construct two expressions:
1) The forces: surface integral of p*Normals
2) The x moment of the z force: surface integral of coordsY*Forces_Z
The center of application of the z force is calculated as (2)/(1).
I noticed some differences in the forces calculated (order of 0.2% so negligeable). This seems to be doing what I need.
Julien de Charentenay
Hibou Scientific Software Pty Ltd, home of
Khamsin, a SketchUp plugin for CFD Modeling, http://www.hibouscientificsoftware.com.au/khamsin
Last edited by julien.decharentenay; September 30, 2012 at 22:20. Reason: Just wanted to clarify.
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