Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to find out the distance between the centers of $2$ circles. Say, circle $1$ $(\theta,\phi)$ circle $2$ $(\theta,\phi)$

The radius of this circle is found using $d\tan(\theta)$ where $d$ is the range (different from radius)

$$cd=\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}$$ is the formula I'm going to use to find out the distance between the $2$ points.

But can anyone help me in defining $x$, $y$ and $z$?

I know to convert it to Cartesian coordinates if say the points are $(r,\theta)$ where $r$ is the radius [$2$ dimensional]

$cd$ is a euclidean distance between the centre of circle#1 (formed due to angles $az(t,n),el(t,n)$ in space from transmitter $t$, at trial $n$ and distance $d$ away from origin) and circle#2 (formed due to angles $az(r,n),el(r,n)$ in space from receiver $r$, at trial $n$ and distance $d$ away from origin)

share|improve this question
1  
It is difficult to understand what you are asking. –  copper.hat Dec 10 '12 at 16:44
    
the centers of the circles are defines by angles say (theta, phi) I need to find out the distance between the centers of the 2 circles using cd=sqrt((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2) but I do not know what x, y and z are –  Nidhi Dec 10 '12 at 16:45
    
but r1 and r2 are the radius. I have 2 angles as my center. will the formula still work? –  Nidhi Dec 10 '12 at 16:49
    
my problem is 3 dimensional. :( –  Nidhi Dec 10 '12 at 16:49
    
I am not sure I understand exactly what you are asking. What do you mean by saying that you have a circle given by $(\theta, \phi)$? –  Thomas Dec 10 '12 at 16:50
show 3 more comments

1 Answer

In spherical coordinates, and generally in $\mathbb R^3$, it takes three coordinates to specify a point. If you only have $\phi, \theta$ you have a ray from the origin. You need the distance from the origin to get a point. Then, given $(r,\theta,\phi)$ for each point you can convert to Cartesian coordinates with $x=r \sin \theta \cos \phi, y=\sin \theta \sin \phi, z=r \cos \theta$

Added, based on your comment: The easiest is to erect a coordinate system at each transmitter. From transmitter 1, you know the point is at $(r_1,\theta_1,\phi_1)$, which you can convert to $(x_1,y_1,z_1)$. Then from transmitter 2, you know the vector is $(r'_2,\theta'_2,\phi'_2)$ where the primes indicate that the coordinate system is centered on transmitter 2. Hopefully your axes are aligned, that makes it easier. Convert this to $(x'_2,y'_2,z'_2)$, still referenced to transmitter 2. Then if transmitter 2 is at $(t2_x,t2_y,t2_z)$ in the transmitter 1 frame, the second point is at $(x'2+t2_x,y'_2+t2_y,z'_2+t2_z)$. Now you can use the usual Cartesian distance function.

share|improve this answer
    
No I have tried this but the problem is will r be the radius or the range. I have 2 different parameters. I'll try to attach a figure –  Nidhi Dec 10 '12 at 16:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.