
Geometry Geometry Math Forum 
 LinkBack  Thread Tools  Display Modes 
October 15th, 2007, 03:21 AM  #1 
Newbie Joined: Oct 2007 Posts: 2 Thanks: 0  Spherical Geometry, please help....
I was wondering if anyone knew how to solve this problem. I have a sphere which is divided into bins (sections). The sphere is divided using the angle of elevation and the azimuth (horizontal) angle. Each bin is represented by the feature vector (j,k). Where j is the angle of elevation and k is the azimuth angle. What i want to know is given a point how can i find out which bin it is in? I have the following information about the sphere:  radius, centre point, north pole, angle of elevation and azimuth angle divisions. Any help on this would be greatly appreciated. 
October 16th, 2007, 10:34 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,243 Thanks: 1439 
How are all the points you know and the angle divisions given?

October 17th, 2007, 02:42 PM  #3 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 
The section it belongs in is dependent on how the sphere is divided, but: The way the sphere is represented, you're looking at it as 2 circles intersecting at a right angle (on different planes, of course). Remember, in 3 dimensions there are 2 ways for something to be perpendicular. Let's first look only at what you have as the second circle (azimuth, k). This circle is on the x,y plane. That is, it is flat. Fortunately, you can use all of your planar trig on this one. So, you follow the circle around like normal until you reach k. The next thing you do is hard to explain without a picture. The angle of elevation (j), is going to follow a circle which is perpendicular to the first circle along the line formed between the center of the sphere and k. Now, what you have is 2 perpendicular circles, one of which is flat, the other is standing up and down, and is offset from one plane (i.e. xz or yz, depending on the representation) by an angle of k. You can demonstrate this with 2 large rings that are the same size. Put them together so they form a sort of "+" from one cross section, then you can rotate one along the other. Now that you've used k to find where to "place" this circle, you can forget it. Now, simply move along this vertical circle until you reach angle j, and you have the your position. Each section has some upper and lower bounds, based on the (j,k) you are given. If you have a point in (x,y,z), convert it to spherical coordinates: (ρ,θ,φ) where ρ is the magnitude (x²+y²+z²), θ is equivalent to k (the angle from x along the xy plane), and φ is equivalent to j (The angle from z along θ). The only important part (as long as the point is within the sphere) is (θ,φ). If (θ,φ) is between the upper (k,j) [notice my notation is backwards from yours] and the lower (k,j) of a section, it falls within that section. If you need clarification (I'm always better my second time explaining something), or I missed the question, let me know. Cheers, Cory 
October 24th, 2007, 07:08 AM  #4 
Newbie Joined: Oct 2007 Posts: 2 Thanks: 0 
Thanks, Its ok I have figured out how to do it in a simpler way. Im going to translate and rotate the coordinate system so the centre point will be the origin then from there i can use trig to define the angle of elevation and the azimuth angle. Thanks for the help anyway 

Tags 
geometry, spherical 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Calculus 3Spherical  maluita659  Calculus  2  March 7th, 2014 12:51 PM 
Spherical raindrop?  Shamieh  Calculus  17  September 25th, 2013 06:39 PM 
Spherical Trigonometry  westborn7  Trigonometry  0  August 4th, 2013 07:14 AM 
Spherical mirrors, spherical?  proglote  Physics  7  August 18th, 2011 12:29 AM 
Glide Reflection in Spherical Geometry...  TTB3  Applied Math  0  March 3rd, 2009 04:54 PM 