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A simple camera problem

by xzardaz
Tags: camera, simple
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xzardaz
#1
Dec23-12, 01:10 PM
P: 10
Hello, I have a simple(or not?) math problem:

I have equations of 3 lines in R3 trought the origin:

line l1:
1*x+β1*y+γ1*z=0
2*x+β2*y+γ2*z=0

line l2:
3*x+β2*y+γ3*z=0
4*x+β3*y+γ4*z=0

line l3:
5*x+β4*y+γ5*z=0
6*x+β5*y+γ6*z=0

I know every λ, β and γ - they are real constants.

I also have a plane δ and :

l1 intersects δ in point p1,
l2 intersects δ in point p2,
l3 intersects δ in point p3

I also know:
....the distance between p1 and p2 = h
....the distance between p2 and p3 = w
....p1, p2 and p3 form a right triangle with right angle at p2 (h2+w2=(p1p3)2)

I want to find the equation of δ in terms of λ, β, γ, h and w

_______________________________

it should be easy to find the equation of the plane if I find the points p1, p2 and p3

I think I should compose a system, containing:
....the first six equations(which are linear, so it should be easy to solve with matrices),
....the equation for right angle in R3: w2+h2=(p1p2)2,
....the two equations for distances w and h: h2=(p1x-p2x)2+(p1y-p2y)2+(p1z-p2z)2 and w2=(p2x-p3x)2+(p2y-p3y)2+(p2z-p3z)2

There are 9 equations with 9 unknowns (3 points, each with 3 coordinates) - it should be solvable.

Have you got any ideas how to solve this? If there weren't 3 quadratic equations it would be easy.


*I know that there should be 2 planes, matching the conditions - one that I've drawn(see the image) and the other is at the opposite side of the origin. I think the two roots of the quadratic equations should find them.
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chiro
#2
Dec23-12, 04:08 PM
P: 4,572
Hey xzardaz.

Try using the intersections to get the normal of the plane and use one of the points as a point on the plane.

Basically a line is represented by L(t) = a*(1-t) + (b-a)*t where the line passes through points a and b.

Now if you solve for t for all the lines you build a normal by considering n = (p2-p1) X (p3-p1) which you normalize and then use the relationship n . (r - r0) = 0 where n is what you solved, r0 = p1, and r is a general point on the plane.
xzardaz
#3
Dec25-12, 04:22 AM
P: 10
Try using the intersections to get the normal of the plane and use one of the points as a point on the plane.

Basically a line is represented by L(t) = a*(1-t) + (b-a)*t where the line passes through points a and b.

Now if you solve for t for all the lines you build a normal by considering n = (p2-p1) X (p3-p1) which you normalize and then use the relationship n . (r - r0) = 0 where n is what you solved, r0 = p1, and r is a general point on the plane.
I don't think I fully understand you. Can you give me some references ?

If I solve L(t) = a*(1-t) + (b-a)*t for all the lines, I can't get the points, because there are infinite number of points matching the equations. How do I use w and h ?

chiro
#4
Dec25-12, 05:30 PM
P: 4,572
A simple camera problem

You can solve for t by using the fact that L(t) = point on plane for some t and then re-arrange to get the value of t.
xzardaz
#5
Jan1-13, 08:03 AM
P: 10
Happy new year,

L(t)=a(1-t)+(b-a)t\\

L(t)=p1x(1-t)+(p2x-p1x)t
L(t)=p1y(1-t)+(p2y-p1y)t
L(t)=p1z(1-t)+(p2z-p1z)t

L(t)=p2x(1-t)+(p3x-p2x)t
L(t)=p2y(1-t)+(p3y-p2y)t
L(t)=p2z(1-t)+(p3z-p2z)t

Is that what you mean ? If so, how do I find p1, p2 and p3 ?
chiro
#6
Jan1-13, 08:12 PM
P: 4,572
What you have is basically L_1(t) = p1 = a1*(1-t1) + (b1-a1)*t1, L_2 = a2*(1-t2) + (b2-a2)*t2 and so on.

You extrapolate your variables in terms of the t's by relating the different forms of the line equations together.

The normal is calculated in terms of (p3-p1) X (p2-p1) and this will give the normal to the plane with point p1 on the plane.


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