3d orthogonal projection on a plane

3dgeometrymathprojection

I have a point in 3d P(x,y,z) and a plane of view Ax+By+Cz+d=0 . A point in plane is E.Now i want to project that 3d point to that plane and get 2d coordinates of the projected point relative to the point E.

P(x,y,z) = 3d point which i want to project on the plane.
Plane Ax + By + Cz  + d  = 0 , so normal n = (A,B,C)
E(ex,ey,ez) = A point in plane ( eye pos of camera )

What i am doing right now is to get nearest point in plane from point P.then i subtract that point to E.I suspect that this is right ???

please help me.Thanks.

Best Answer

The closest point is along the normal to the plane. So define a point Q that is offset from P along that normal.

Q = P - n*t

Then solve for t that puts Q in the plane:

dot(Q,n) + d = 0
dot(P-n*t,n) + d = 0
dot(P,n) - t*dot(n,n) = -d
t = (dot(P,n)+d)/dot(n,n)

Where dot((x1,y1,z1),(x2,y2,z2)) = x1*x2 + y1*y2 + z1*z2

Related Solutions

Reverse-projection 2D points into 3D

It is not possible to give an unambiguous solution to this problem. However, here's how I would extract the different solutions:

1) Solve for the camera position and direction using the P3P (Perspective-3-Point) algorithm from the original RANSAC paper, which give up to four possible feasible solutions (with the points in front of the camera).

2) Project a ray with the camera position as origin having (X,Y) as projection in the camera and calculate its intersection with the plane.

3d to 2d Projection Matrix

This gives you two sets, each of three equations in 3 variables:

a*x0+b*y0+c*z0 = x0'
a*x1+b*y1+c*z1 = x1'
a*x2+b*y2+c*z2 = x2'

d*x0+e*y0+f*z0 = y0'
d*x1+e*y1+f*z1 = y1'
d*x2+e*y2+f*z2 = y2'

Just use whatever method of solving simultaneous equations is easiest in your situation (it isn't even hard to solve these "by hand"). Then your transformation matrix is just ((a,b,c)(d,e,f)).

...

Actually, that is over-simplified and assumes a camera pointed at the origin of your 3D coordinate system and no perspective.

For perspective, the transformation matrix works more like:

               ( a, b, c, d )   ( xt )
( x, y, z, 1 ) ( e, f, g, h ) = ( yt )
               ( i, j, k, l )   ( zt )

( xv, yv ) = ( xc+s*xt/zt, yc+s*yt/zt ) if md < zt;

but the 4x3 matrix is more constrained than 12 degrees of freedom since we should have

a*a+b*b+c*c = e*e+f*f+g*g = i*i+j*j+k*k = 1
a*a+e*e+i*i = b*b+f*f+j*j = c*c+g*g+k*k = 1

So you should probably have 4 points to get 8 equations to cover the 6 variables for camera position and angle and 1 more for scaling of the 2-D view points since we'll be able to eliminate the "center" coordinates (xc,yc).

So if you have 4 points and transform your 2-D view points to be relative to the center of your display, then you can get 14 simultaneous equations in 13 variables and solve.

Unfortunately, six of the equations are not linear equations. Fortunately, all of the variables in those equations are restricted to the values between -1 and 1 so it is still probably feasible to solve the equations.

Best Answer

Related Solutions

Reverse-projection 2D points into 3D

3d to 2d Projection Matrix

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