Calculating 180 and 360 Camera Coverage
Hello John,
For typical high ceiling room 9 feet and AoV 360 with most comment camera resolution 1080 the width of scene = 6.28 x 9 = 56.5 and PPF =34. Also same result of Aov 180 when we increase the resolution and same data input just the PPF will increase
The question what is the dimension of each wall in the room (square, rectangular). For example, we have room with dimension of 12L x 12W x 9H it is covered.
Hi John,
Is it possible to update the calculator to differentiate between regular fisheye and Immervision Panamorph (the latter is assumed to produce more resolution at the edges...).
Also camera positioning information should be used in the calculation : mounting height and orientation (ceiling or wall mount).
We are looking at how to best incorporate camera height. This is especially important for 360 cameras where a camera mounted 30' high, is minimally 20 something feet away from people/cars, which will ensure poor quality video even if an object is directly underhead.
Filip, as for the Immervsion vs Fisheye differentiation, Immervision claims better quality at the edge. If there were any benefits, it would be a slightly different / wider effective FoV. We have tested Immervision and Fisheye lenses head to head (see here) and Immervision is better than some but worst than other fisheye, when testing regular, non edge areas. As such, it would not change the calculation / equation, only the potential area it was used over.
Great article; if I have a 180 20MP camera with 4 5MP imagers I would use the 5MP in the caculator?
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Help! I'm totally confused...
What does the field width refer to in 360 mode?
In regular mode the width meant:
the width of the object that would exactly fill the FOV horizontally at the specified distance given the specified AOV.
What does it relate to now?
Width in 360 mode is the circumference of a circle, so any given distance, the width of the fov is 6.28 x that distance.
Thanks!
Is the PPF number you give an average for the whole width, or at the edge, or center? It seems like it would vary greatly between the center and the edge of object.
Sorry, but still having trouble. Here is a practical problem:
Assuming 360 mode and using your fisheye round picture of the parking lot, how would I go about determining the PPF of where the bushes are on the left?
Would I measure the distance to the bushes disgonally down or just the horizontal component of just the vertical? Thanks.
I was using your parking lot picture as an example, it looks to me like maybe 15 feet high? It doesn't let me change the picture in 360 mode.
But what do you want to do in reality? What's your actual application you are working on?
Btw, it's not 15 feet high, it's ~10, which means that diagonal vs horizontal won't make much of a difference for monitoring.
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