algorithm - Mapping coordinates from plane given by normal vector to XY plane

Question

So, I have this algorithm to calculate cross-section of 3D shape with plane given with normal vector.

However, my current problem is, that the cross-section is set of 3D points (all lying on that given plane) and to display it I need to map this coordinates to XY plane.

This works perfect if the plane normal is something like (0,0,c) - I just copy x and y coordinates discarding z.

And here is my question: Since I have no idea how to convert any other plain could anybody give me any hint as to what should I do now?

Answer 1

Your pane is defined by a normal vector

n=(xn,yn,zn)

For coordination transformation we need 2 base vectors and a zero point for the pane

Base vectors

We chose those "naturally" fitting to the x/y pane (see later for edge case):

b1=(1,0,zb1)
b2=(0,1,zb2)

And we want

b1 x b2 = n*c (c const scalar)

to make sure these two are really bases

Now solve this:

b1 x b2= (0*zb2-zb1*1,zb1*0-1*zb2,1*1-0*0) = (zb1,zb2,1)
zb1*c=xn
zb2*c=yn
1*c=zn

c=zn,
zb2=yn/c=yn/zn
zb1=xn/c=xn/zn

b1=(1,0,yn/zn)
b2=(0,1,xn/zn)

and normalize it

bv1=(1,0,yn/zn)*sqrt(1+(yn/zn*yn/zn))
bv2=(0,1,yn/zn)*sqrt(1+(xn/zn*xn/zn))

An edge case is, when zn=0: In this case the normal vector is parallel to the x/y pane and no natural base vectors exist, ind this case you have to chose base b1 and b2 vectors by an esthetic POV and go through the same solution process or just chose bv1 and bv2.

Zero point

you spoke of no anchor point for your pane in the OQ, but it is necessary to differentiate your pane from the infinite family of parallel panes.

If your anchor point is (0,0,0) this is a perfect anchor point for the coordinate transformation and your pane has

x*xn+y*yn+z*zn=0,
(y0,y0,z0)=(0,0,0)

If not, I assume you have an anchor point of (xa,ya,za) and your pane has

x*xn+y*yn+z*zn=d

with d const scalar. A natural fit would be the point of the pane, that is defined by normal projection of the original zero point onto the pane:

P0=(x0,y0,z0)

with

(x0, y0, z0) = c * (xn,yn,zn)

Solving this against

x*xn+y*yn+z*zn=d

gives

c*xn*xn+c*yn*yn+c*zn*zn=d

and

c=d/(xn*xn+yn*yn+zn*zn)

thus

P0=(x0,y0,z0)=c*(xn,yn,zn)

is found.

Final transformation

is achieved by representing every point of your pane (i.e. those points you want to show) as

P0+x'*bv1+y'*bv2

with x' and y' being the new coordinates. Since we know P0, bv1 and bv2 this is quite trivial. If we are not on the edge case, we have zeroes in bv1.y and bv2.x further reducing the problem.

x' and y' are the new coordinates you want.