read_connect(); //$GLOBALS[ezoic_db]->read->query("use 17things"); ?>

What picture will I get with various lens sizes with DIY install cameras?

I have an allocated car parking space across the road from my house. Its approx 35 to 40 feet away from where the camera would be located. Would I get a recognisable close up from a IR illuminated day/night camera and would the illumination be OK? What kind of view would I get with a 6mm lens, a 12mm lens, a 16mm lens or a 25mm lens? Normal streetlighting and camera image captured to DVR. My current camera works OK in daylight but stops working at dusk. Best lens size please? Any advice welcome.

Related Items

2 Responses to “What picture will I get with various lens sizes with DIY install cameras?”

  1. antoni m said :

    no idea but you could try

    1. use a security camera on a movement detector switch

    2. use a fast film or big iso number on a digi, put some illumunation (security lights or such) near the subject area and rig up a camera on a movement sensor switch

    a close up is like a head shot, well you might get one if the subject co-operates, the lens you list are for wide angle, you would need more like a 300-600 (15 meters right)

    buy a car alarm?

    a

  2. devilsadvocate1728 said :

    The illumination near street lights can be around LV 3, which corresponds to an exposure of about 1 second at f/2.8 and ISO 100. The illumination at a more distant location will be correspondingly less. If you have an infrared light in a reflector aimed right at your car, you should be able to do better than this.

    The field of view you get with your lens will depend on the size of the active area of your imager as much as the lens’ focal length. Assuming that the lens is reasonably rectilinear, i.e. straight lines in the “real world” are rendered as straight on a flat imager without a lot of barrel or pincushion distortion, the camera-to-subject distance divided by the field of view will equal the lens’ focal length divided by the width of the imager’s active area.

    The “1/2 inch” trade sized imager in my first digital camera was actually 6.40 x 4.80 millimeters, which is about as big as you can expect an imager for your application to be. Mine was a still camera. Many imagers for video use are smaller than that. Assuming yours is half that size, i.e., 3.20 x 2.40 mm, and you are 36 feet away, and you want a field of view of 6 feet by 8 feet (perhaps a bit narrow, but this will get you in the ballpark, and you will want as large an image as you can manage to get the amount of detail you want) the focal length f in millimeters will be given by the formula

    f mm / 3.20 mm = 36 ft / 8 ft

    or a focal length of 14.4 mm. If this is indeed a good approximation for the size of your imager’s active area, your best bet is to start with a 12 mm lens, which will give a field of view between 10 and 11 feet wide and about 8 feet high.




Message:

[newtagclound int=0]

Subscribe

Recent Comments

Recent Posts

Archives