Hey, Im kinda stuck on finding the distance from the origin to the plane. What I know so far is I need to use trigonometry and find the distance between the origin and the point on the plane that the planes nonrmal points to. I think I narrowed down the equation to: A = B * cos(alpha) (Heres a text image describing what the vars are)
but if alpha is the angle, how to I find it? I'm just real confused, so basically, how do I find the distance from the origin to a plane? Thanks EDIT: Added source tags around the text image so it would look right [edited by - MattCarpenter on December 17, 2002 11:29:51 PM]

What information do you have about the plane? Do you have the equation for it? It is quite easy to find the plane normal from the equation. Then you can find the normal that intersects with the origin and find the point on the plane which touches that normal. Then find the distance between the origin and that point.

Firebird Entertainment

The only information that I know about the plane is the three verticies from a triangle that forms the plan, and its normal :-/

I guess I don''t have the equation for it then, since its being made just from a triangle

hey matt...
Isnt the distance to the origin just D in the plane equation if the normal is nomralized?
Ie
Ax + By + Cz = D

C ya,

Simon

A novel way to get the distance to a plane is to use the doppler effect. If you know the base frequency that the plane produces, and the speed at which the plane is traveling, you can regress the hyperbola of frequency versus time of the observed sound to find the distance to the flight path.

EDIT: Oh, and airplanes don't have "normals". Only stewardesses.

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Don't listen to me. I've had too much coffee.

[edited by - sneftel on December 18, 2002 1:26:35 AM]

all you need to specify a plane is a single point on it, and the normal. You definately have that, and you could derive the plane equation if you really wanted to.

the point on the plane that the planes nonrmal points to.

The normal doesn''t point anywhere on the plane, it is just a vector which is perpendicular to the plane. Vectors don''t point anywhere - they just specify a direction. If you are looking up in Bratislava, Slovakia then you are looking at something different to someone looking up in Brisbane, Austalia.

If you were wanting to find the shortest distance from the origin to the plane, then take any point on the plane, call it P. Then take the dot product of the vector from the origin to that point, call it OP, with the normal.

pff... u advanced algebra person

lol

quote:

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Original post by boxdot3
hey matt...
Isnt the distance to the origin just D in the plane equation if the normal is nomralized?
Ie
Ax + By + Cz = D

C ya,

Simon

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Where you have the equation Ax+By+Cz=D and store the coefficients in a vector you have [[N],N.P] (normal, normal dot any point on the plane)

It is true that normal.point is the same for all points on the plane

The point Q on the plane nearest the origin is a vector from the origin perpendicular to the plane. N.Q=N.P by definition. Since Q is perpr to the plane, Q=kN

N.(kN)=N.P
k(N.N)=N.P
k=(N.P)/(N.N)

assuming you have a 4D vertex structure xyzw where N is in xyz and N.P is in w,

k = vertex.w / ([vertex.xyz] DOT [vertex.xyz])

here my brain fails, either k is the distance or the distance squared. try a few examples to see if you need to square root.

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...your plane: Ax + By + Cz + D = 0;

distance from the origin:

d = |D| / sqrt(A^2 + B^2 + C^2);

You can use any of the three vertices (P, for example) and the normal N to calculate the plane equation:

(Q - P) dotProduct N = 0

After getting the plane equation of Ax + By + Cz + D = 0, the formula in the previous poster can be adapted to calculate the distance.

Kidd