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Theoretically, what is the minimum distance between the camera and an object, where the object begins to not be visible ? Suppose that a camera has f as focal distance and the sensor size 2/3" and the object is a square of 2m x 2m and we are filming in the daylight.

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This is not a practical answer (it is not a practical question), but it is a precise answer.

Let's define "not visible". If in an image, I will offer a description of "not visible" that the object is not more than one pixel size in the image, which certainly will not be considered visible (probably 5 or 10 pixels works as well ...), but "it depends", on the overall image size (pixels) and the enlargement at which it will be viewed, and of course, the size of the object, and what we mean by "can be seen".

Then the calculator at http://www.scantips.com/lights/subjectdistance.html can be used to fill in the other details, such as sensor size and image size and focal length. All of the details there will affect it.

Then if you specify "size of object in image" to be 1 pixel (or 5 or 10 pixels), then it will compute the necessary distance to the object (in the same units as was specified for the "real size of the object, perhaps feet?) A one pixel dot cannot be recognizable as anything.

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    I know the question says "shooting in daylight", but lots and lots of things in astronomical photography are visible despite having a diameter less than one pixel. Visibility depends on contrast as well as brightness. – hobbs May 16 '16 at 2:03
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I will just give you a glimpse of what are you asking, so you can do your own math.

We need to take in account:

The object

1) What is the color. Diferent colours have diferent wavelengths, so this affects on the sensor reception, difraction, atmospheric absortion, etc.

2) The contrast with the background. This is pretty obvious, a white board on a white surroundings is less visible than a bright softbox flash on a dark surrounding.

The atmospheric conditions

This is obvious too, haze day, desert, refraction of the hot-cold air, water droplets.

The sensor

Sensor size, resolution, capacity, quantum characteristics...

Exposure time

Probably this has less effect on a daytime shoot than on astronomy, but this affects too.

Lens

Lens absorption, sharpness, aberration, difraction.

Focal length

Are you using your Iphone or did you rented Hubble to take a shoot of a daytime of the moon?

The main factors, on ideal conditions are focal length versus sensor resolution.

Study this: https://en.wikipedia.org/wiki/Angular_resolution

and this: https://en.wikipedia.org/wiki/Angle_of_view

After that we need to take into account the resolution of the sensor.

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    I think the word "theoretical" encourages plugging-in optimal values for each variable. White object/Black b/g; O% haze, O density atmosphere; diffraction limited lenses, blue collimated monochromatic lighting (to determine maximum sensor circular area); etc. etc. etc. – Stan May 15 '16 at 23:35
  • Add aperture to the list - due to it's effect on diffraction. Also, precision of focusing and motion blur will play an important role. – MirekE May 16 '16 at 2:15
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Under bright sunlight conditions, a young person with 20/20 vision can resolve an object that is approximately 3000 diameters distant. A 2 meter square object has a diagonal measure of 2.8 meters. This object if viewed from 2.8 X 3000 = 8,400 meters, will appear to be a point without discernable dimension.

The 3000 times its diameter rule of thumb is too stringent for photographic purposes. This is because all lenses have unresolved aberrations. Also, the contrast of the photographic system adds difficulty to resolving objects. The accepted limit for viewing a photographic image is: The image of an object will be seen as a point if it sustains 3.4 minutes of arc. This works out to an image viewed from 1000 diameters distant. See C.B. Neblette The Photographic Lens Library of Congress 64-20637.

Doing the math: Object diameter 2.8 meters Lens focal length 50mm Object distance 100 meters Image size 1.4mm Viewing distance 1.4 X 1000 = 1,400mm (seen as a point)

Object diameter 2.8 meters Lens focal length 50mm Object distance 280 meters Image size 0.5mm Viewing distance 0.5 X 1000 = 500mm (seen as a point standard reading distance).

Object diameter 2.8 meters Lens focal length 100mm Object distance 560 meters Image size 0.5mm Viewing distance 0.5 X 1000 = 500mm (seen as a point standard reading distance).

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    How about a simpler approach? An object will lose its identity if its image is a point with no dimension. The max size is 0.5mm viewed from 20 inches. The lens traces out of cone of light. The length f the cone is the focal length. The width of the image is 0.5mm. If a 50mm lens is mounted, the ratio is 50 ÷ 0.05 = 100. This is the ratio distance to image size inside the camera. The same ratio relates to outside the camera. If the object has a diameter of 2.8 meters, it will image as a point if it is 2.8 X 100 = 280 meters downstream of the camera. – Alan Marcus May 16 '16 at 2:26
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If both the subject and the camera are located on Earth (or any planet), the distance to the horizon is an upper bound. As explained in the link, with the camera at 1.7 meters above the ground and assuming a spherical planet, the curvature of the Earth limits sight to about 2.9 miles not counting any refraction caused by temperature changes in the air column.

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There is no "one answer", it depends from too many variables. It depends on the size of the object, the focal lengh of the lens mounted on the camera, the resolving capability of the lens (its optical resolution, so to say), the resolution of the sensor...and the definition of "not visible" for who is watching the final image. Uh, and the type of light hitting your object.

Maybe someone could theoretically come up with some kind of formula to put all those things together, but I seriously doubt.

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