Max Coverage Camera Placement

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Baris
Baris on 3 May 2012
Hi guys, at the moment i can use line of sight algorithm to my map which has altitude datas .(For instamce 400*400 meters)My cameras can view 300 meters on land. i choose cameras placement.But I want to divide this map to 4 equal pieces (4 x 200*200 meters) and to learn which points give the maximum coverage for each subdivision and for neighbors also. I mean a point -best for first subdivision - also for its neighbors such as second one. Can anyone help me?
Thanks in advance. Baris

Accepted Answer

Walter Roberson
Walter Roberson on 3 May 2012
You will need to explain how "coverage" is determined.
Presumably a camera placed in a low area would not be able to see "through" a high area, but by the same measure a camera placed in a high area would not necessarily be able to see into a dip, depending on the angle. Are the two directions to be weighted evenly?
What about degraded quality of image for viewing a curved surface? What about degraded quality of image according to distance?
Should the algorithm used be to ray-trace from each proposed camera location to calculate the angular area of the projection of the surface as seen at the proposed location? But then one should really include direction of the lighting and reflectivity and diffusion. And if the distances are high enough, refraction as well. Or, for that matter, if the distances are small enough, refraction of the light rays near the surface could be important. Frequency [mix] of the illumination should be included.
With the area divided into sections, is the view of each camera to be cut off at the boundary of the section? If one camera is at a high point in section 1 that has a good view of portions of section 2, then potentially the best position for the camera in section 2 would be to capture some additional local detail rather than trying to get the best view of just that one section.
Oh yes, nearly forgot, camera lens off-axis distortion needs to be taken into account as well: a look directly at an object is, in a lensed world, less distorted than further away from the lens axis. You are going to have different results if you have a fish-eye lens, for example.
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Baris
Baris on 3 May 2012
Thanks for reply,let me explain in details.
I have an UAV and there is a cameras under it.With this cameras i want to scan some terrrain.İ scan terrain with viewshed function in matlab.In Viewshed i choose my position and according to this position(x,y,z) scanning is possible like a circle.Of course if terrain altitude is upper then uav then it can not scan some parts.Here coverage means the points which uav can scan according to line of sight and viewshed.My cameras is a basic cameras and lenses.No need to zoom or other thing.
I want to find points both enables best coverage for self division and neighbor division.
Walter Roberson
Walter Roberson on 3 May 2012
Interesting function, viewshed(); I had not encountered that one before. http://www.mathworks.com/help/toolbox/map/ref/viewshed.html

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