Register to reply 
A simple camera problem 
Share this thread: 
#1
Dec2312, 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 l_{1}: λ_{1}*x+β_{1}*y+γ_{1}*z=0 λ_{2}*x+β_{2}*y+γ_{2}*z=0 line l_{2}: λ_{3}*x+β_{2}*y+γ_{3}*z=0 λ_{4}*x+β_{3}*y+γ_{4}*z=0 line l_{3}: λ_{5}*x+β_{4}*y+γ_{5}*z=0 λ_{6}*x+β_{5}*y+γ_{6}*z=0 I know every λ, β and γ  they are real constants. I also have a plane δ and : l_{1} intersects δ in point p_{1}, l_{2} intersects δ in point p_{2}, l_{3} intersects δ in point p_{3} I also know: ....the distance between p_{1} and p_{2} = h ....the distance between p_{2} and p_{3} = w ....p_{1}, p_{2} and p_{3} form a right triangle with right angle at p_{2} (h^{2}+w^{2}=(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 p_{1}, p_{2} and p_{3} 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: w^{2}+h^{2}=(p_{1}p_{2})^{2}, ....the two equations for distances w and h: h^{2}=(p_{1x}p_{2x})^{2}+(p_{1y}p_{2y})^{2}+(p_{1z}p_{2z})^{2} and w^{2}=(p_{2x}p_{3x})^{2}+(p_{2y}p_{3y})^{2}+(p_{2z}p_{3z})^{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. 


#2
Dec2312, 04:08 PM

P: 4,575

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*(1t) + (ba)*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 = (p2p1) X (p3p1) 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. 


#3
Dec2512, 04:22 AM

P: 10

If I solve L(t) = a*(1t) + (ba)*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 ? 


#4
Dec2512, 05:30 PM

P: 4,575

A simple camera problem
You can solve for t by using the fact that L(t) = point on plane for some t and then rearrange to get the value of t.



#5
Jan113, 08:03 AM

P: 10

Happy new year,
L(t)=a(1t)+(ba)t\\ L(t)=p_{1x}(1t)+(p_{2x}p_{1x})t L(t)=p_{1y}(1t)+(p_{2y}p_{1y})t L(t)=p_{1z}(1t)+(p_{2z}p_{1z})t L(t)=p_{2x}(1t)+(p_{3x}p_{2x})t L(t)=p_{2y}(1t)+(p_{3y}p_{2y})t L(t)=p_{2z}(1t)+(p_{3z}p_{2z})t Is that what you mean ? If so, how do I find p_{1}, p_{2} and p_{3} ? 


#6
Jan113, 08:12 PM

P: 4,575

What you have is basically L_1(t) = p1 = a1*(1t1) + (b1a1)*t1, L_2 = a2*(1t2) + (b2a2)*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 (p3p1) X (p2p1) and this will give the normal to the plane with point p1 on the plane. 


Register to reply 
Related Discussions  
Trillion FPS camera developed at MIT camera can 'watch' the movement of light  General Physics  21  
Simple lens question (camera)  Introductory Physics Homework  8  
Simple Lens Diagram for a Camera  Introductory Physics Homework  3  
Question from simple lens diagram (camera)  Introductory Physics Homework  6  
Making a simple hispeed camera  Computing & Technology  7 