Determine position in front of 3D plane (cartesian math)

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+3 Q:

Determine position in front of 3D plane (cartesian math)

Hi there,

I have a planar element in 3D space that I've rotated on the x, y and z axis. I'm positioning a 3D camera in front of the planar but I need to find a way to calculate the x, y, z of the camera.

I'm trying to figure out how to place the camera at x distance from the planes face. There's obviously a bit of trig involved but for the life of me can't figure it out. Gah.

Dave

+1 A:

The relation between distance of a point to a plane is

distance = (Aa + Bb + Cc + D) / sqrt(A^2+B^2+C^2)

for distance to plane Ax + By + Cz + D = 0 from point (a,b,c)

Might need to multiply by -1 to get distance positive.

The equations for the line through point (a,b,c) perpendicular to the same plane is

x = a + At; y = b + Bt; z = c + Ct

So if you have a point in the plane, you can find the equation of the line perpendicular to that plane. Then you can use the distance constraint to solve for two points a distance along that line - one above the plane, and one below.

John at CashCommons 2009-09-22 20:34:14

Sorry, that's confused me dramatically. Excuse my ill-math skills but if you could simplify/clarify for me it would be helpful.Lets say the planar is at A (493, 543, 234) - these x, y, z values are referencing from the centre of the plane - and I want the camera positions at 500 units from that point. The planars rotation is x = 34, y = 54, z = 32.So I'd like to find out what point B (?, ?, ?) - the camera - is according to the planars rotation. My best guess was to triangulate a point from the planes four vertices (which I for this case will specify as being 400 width).

2009-09-22 23:15:57

http://mathworld.wolfram.com/Plane.html

John at CashCommons 2009-09-22 23:29:48

I don't understand. What is A? What is "the planar?" Are x, y, and z in degrees? What is the sense of the rotation? It sounds like you have a finite sheet, not a plane (which is infinite). Is this true?

John at CashCommons 2009-09-23 00:18:07

Lets just follow this equation because it's making more sense than what I wrote in the above comment. If you can help me solve this John I'll award you the answer :-) Lets say my normal vector is N = (400, 500, 600) and my P = (405, 505, 605) I would then calculate my P' by doing this: Camera coords: x = Px + x * Nx y = Py + x * Ny z = Pz + x * Nz I'm I getting this right now? lol...or completely off?

2009-09-23 00:23:20

OK, we have progress. :) The point in the plane is P = (Px,Py,Pz). Your normal vector needs to have magnitude 1, so divide each component of N by sqrt(400^2 + 500^2 + 600^2) which is about 877, so N = (Nx,Ny,Nz) is about (400/877, 500/877, 600/877). You said you wanted a distance 500 from this point. So then with P' = (Px',Py',Pz'), we have Px' = Px + 500*Nx, Py' = Py + 500*Ny, Pz' = Pz + 500*Nz. Make more sense?

John at CashCommons 2009-09-23 01:30:58

Great, haha. Progress is always good.So I only need to add the magnitude to the normal vector not the secondary P value? I've coded it up and am getting very weird results. Either I need to debug it more closely or I'm getting the calculating wrong. I'll get back to you in the next 30 minutes.Thank you for your continued patience!

2009-09-23 02:33:03

No - you're starting at P (a vector), and displacing 500 units in the direction of N (adding another vector) to arrive at P'. You add 500*N to P to arrive at P'. (I really wish I could boldface letters in comments. Would make this much easier.) If you're getting weird results, are you calculating N correctly? This is where understanding the math might come in. You need the x, y, and z components of the normal vector. These might just be the directon cosines, which relate to your angles (34,54,32) (if that's what those numbers are.)

John at CashCommons 2009-09-23 17:00:14

Thanks, I'll give it a try on the weekend, I'm completely swamped with work. I see what you're saying now...ahhhhh.

2009-09-24 00:34:08

+2 A:

If you have a plane then you also have its normal vector N and some point on it P.
If you calculate P'=P+x*N you'll get the point P' which is x units infront of point P in the direction of the normal.

shoosh 2009-09-22 20:46:11

Hmmmmm. Ok, interesting. Not understanding it 100%, refer to my comment for John W. for a better explanation of what I mean. I'm pretty sure in order to solve this the equation would need to factor in the rotation of the plane to make sure it's facing it correctly.

2009-09-22 23:27:41

David, yes. Shoosh's answer is a vector equation. P', P, and N are vectors. x is a scalar. The unit vector's components are a direct result of the orientation of the plane.

John at CashCommons 2009-09-22 23:33:57

Ok, but being the non-math-focused I am I need to understand how to code these values. The Wolfram article is very helpful in explaining the process, but the math is beyond me.So how could I produce an x, y, z to position the camera from that equation? My guess would be that I have to simply replace N with the vector value.

2009-09-22 23:57:03

"being the non-math-focused" is not an excuse for lazyness. get some decent vector arithmetic book and figure out just exactly it is you want to do.

shoosh 2009-09-23 09:15:49

ansaurus

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Determine position in front of 3D plane (cartesian math)

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