Can anyone help me out, I'm currently spec'ing a job for some CCTV cameras for a really paranoid, really rich person. He only wants a 4 camera system but wants as close to 100% coverage as possible.
I stupidly suggested drawing a diagram for him with the positions of the cameras on the house and their approx field's of view so he can see any possible blind spots.
So I need to way to workout with the size & type of lens on the camera what view that camera will have.
Anyone point me in the right direction?
one of the camera's I'm looking at has a 3.6mm lens and the other 2.8-12mm Varifocal Lens.
Cheers
Rob
I only know that a small apature gives better depth of field :) so I guess the lens which give the best light would be best.
all the cameras I use have 1/3" ccd's if that matters and use the lens I mentioned above. There must be a formula to work out the 'cone' view of the camera.
this is what I want, but with reletively accurate cones.
(http://i150.photobucket.com/albums/s81/robdav99/camerasexample.jpg)
Think I have found it.
distance x 4.8 / width of view
so if I want to see 18m away and a 21m wide picture I need a 4.1mm lens.
Good job, I've got no idea! Only lens I've got I know the field of view for is my Sigma 10-20mm which is about 110 degrees but I couldn't tell you how I know that or how to work it out either.
this looks handy
http://www.pelco.com/sites/global/en/sales-and-support/downloads-and-tools/tools/lens-calc.page
I'm guessing you've got drawing software for this, but if not I find QCADs free version works well enough for simple 2D sketches.