Re: CCTV lens calculator

[G. Morgan <alarmpro@xxxxxxxxxxx>]

On Thu, 09 Nov 2006 06:28:21 GMT, "A.J." <aj@xxxxxxxx> wrote:

>Ooops, read the message again and realised that the camera is mounted from
>the ceiling looking down.
>
>Distance to object     10 feet
>Lens    1.6 mm
>Field of View    30' x 22.5 '
>
>According to the online calculator at supercircuits.com.  Looks like you
>will need two cameras back to back to provide the desire coverage. They will
>be both looking down but covering a different 15' in different direction.
>but looks like you still need to get down to 3mm lens just to cover 16'
>across.
>


Actually it's going to mounted on a wall facing the door entrance that
has a vertical ht. of 20' so I can mount it higher if I can get away
with just one cam.

Basically, I'm looking for a 140° field,  mounted at least 10' high up
to 20' (hope it don't have to be that high) with person recognition in
the center of the door entry and still see any shenanigans 15' on
either side of the door.

Kinda like a half fish-eye?

--

-Graham
(remove the double e's to email)