
[Sponsors] 
July 15, 2006, 11:22 
projection

#1 
Guest
Posts: n/a

Hallo, Can anyone point me to an an approach as to how to find the coordinates of a projected point on a plane.
I have the following details, Corodinates of the 4 vertices of the plane and coordinate of the point to be projected Ramesh 

July 16, 2006, 02:29 
Re: projection

#2 
Guest
Posts: n/a

If I understood well what your problem is, you need to make a dot product to project. Define two orthogonal vectors on your plane (A and B) of norm 1 (divide by the norm), like an origin, then your point (P), will have PdotA in the direction of A, and PdotB in the direction of B


July 16, 2006, 02:55 
Re: projection

#3 
Guest
Posts: n/a

Could you please elaborate with some mathematical formulation. I just lost a bit of track in basic school mathematics. To repeat my problem. I have coordinates of 4 vertices A B C D of plane and point P in 3D space. I need to get the orthonormal projection of point P on plane that is coordinates of projected point PX on plane. One approach, if i am right is to define orthonormal vectors and take the dot product of it but it leads to a set of non linear equations to solve. so is there any other way


July 16, 2006, 03:35 
Re: projection

#4 
Guest
Posts: n/a

Ok, this is another approach, maybe more natural: Here the "old" reference system, is the general 3D system, where you have P expressed. The "new" system, is the plane PX. The vector OO`is a vector from the origin O of the old system to the O`new system.
First, where`s your PX plane? So you need to express your new coordinates in terms of the old`s ones. Thats made by projecting with sines and cosines. The angles should be known, because you choose how the new system is oriented. Suppose you have (all things with " ` " belongs to the new system): x`=ax y`=by z`=cz Now, that`s true for the unitary vectors x^`=ax^, y^`=by^, z^`=cz^ Now, you can express the point P as: P=OO`+P*, where P* is a vector from O`to P, but in the old coordinates (draw it and it will be clear) You know P and OO`(you choose where the new system is), so make the rest, and get P*. P* is Px* x^ + Py* y^ + Pz* z^, in terms of unitary vectors. Now, replace the unitary vectors in terms of x^`, y^`, z^` (unitary vectors of the new basis). Impose that the new coordinate z (that`s P`z, or Pz* z^`) is zero, because is a orthogonal projection. 

July 20, 2006, 08:59 
Re: projection

#5 
Guest
Posts: n/a

The explanation by Alej is fully correct, but may result a bit fuzzy to you, because of the coordinate change.
First of all, a plane is defined by three points and not by four, even if D is taken in the same plane, with numerical errors the fourth point D will never be exactly in the same plane as given by ABC. So let's take only A, B and C. Let every variable that starts with v be a vector taken of a point with respect to O and let O be your original eeehm origin, yep. Take ve1=(vCvA)/Length(vCvA) ve2=(vBvA)/Length(vBvA) Now take vPnew=vPvA Next compute two dot products, notated as "*." dot1 and dot2, both scalar: dot1=vPnew*.ve1 dot2=vPnew*.ve2 then vPx=vA+dot1*ve1+dot2*ve2 I didn't check on errors, but this must work. 

Thread Tools  
Display Modes  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Projection Methods  Petrov  Main CFD Forum  2  June 10, 2011 09:50 
Galerkin projection  sara2115  Main CFD Forum  0  June 5, 2011 10:57 
BlockMeshmergePatchPairs  hjasak  OpenFOAM Native Meshers: blockMesh  11  August 15, 2008 07:36 
free surface flow and projection methods  sylvain  Main CFD Forum  2  April 26, 2002 09:45 
A SecondOrder Projection Method for Incompressibl  Ivan  Main CFD Forum  5  April 29, 2001 07:27 