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September 25, 2012, 20:16 
Find where force is applied

#1 
Senior Member

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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 Julien
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 Julien de Charentenay 

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September 26, 2012, 02:53 

#2  
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Dear Julien,
I hope I got your question right. But actually you posted your answer already: Quote:
used for the calculations of the moments So in your case you chose the origin. I hope I could contribute regards Colin 

September 26, 2012, 06:46 

#3 
Senior Member

Hi Colin,
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 ydirection. Let me know if it is unclear. julien
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 Julien de Charentenay 

September 28, 2012, 08:38 

#4 
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Vieri Abolaffio
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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 

September 30, 2012, 20:53 

#5 
Senior Member

Hi,
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 xcomponent 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 "workaround" using ParaView (http://www.cfdonline.com/Forums/ope...parafoam.html) 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. Kind regards, Julien
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 Julien de Charentenay Last edited by julien.decharentenay; September 30, 2012 at 21:20. Reason: Just wanted to clarify. 

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