10.1 megapixel camera, object is 18,877.90 feet away from me, can I see it?

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Discussion Overview

The discussion revolves around the ability of a 10.1 megapixel camera to capture an object located 18,877.90 feet away. Participants explore the necessary calculations and factors influencing visibility, including object size, optical properties, and camera specifications.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asks whether the object is large enough to be seen, noting that the size of the object and the required resolution (single pixel vs. multiple pixels) are critical factors.
  • Another participant provides the dimensions of the object, specifying it as a cylinder with a height of 43 inches, width of 3 inches, and circumference of 10 inches.
  • There is a suggestion to consider optical zoom and general optics, indicating that these factors may affect visibility.
  • A participant emphasizes the need for information about the camera's optical system, stating that without it, determining visibility is impossible. They mention the importance of field of view, object size, and contrast for calculating visibility.
  • One participant raises a hypothetical scenario about seeing a 3.5-foot object from 3.5 miles away, suggesting that optimal viewing conditions should be assumed for calculations.
  • A technical explanation is provided regarding angular resolution and the Rayleigh criterion, including a formula for calculating the necessary focal length of the camera's lens based on various parameters.

Areas of Agreement / Disagreement

Participants express differing views on the necessary conditions for visibility and the calculations involved. There is no consensus on a definitive answer, as multiple factors and assumptions are highlighted.

Contextual Notes

Key limitations include the lack of information about the camera's optical system, field of view, and contrast conditions, which are essential for accurate calculations. The discussion also reflects varying interpretations of what constitutes "seeing" an object.

Max CR
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I have a 10.1 megapixel camera. If an object is 18,877.90 feet away from me, will my camera be able to see it?

This is a non-school related project that I am doing in order to be awarded scholarships. If anybody knows how I can calculate this answer please let me know.

Please include all equations and work so I know how to do this with other cameras. THank you!
 
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Is it a mountain or a mole hill?

You need to know the size of the object, and whether the requirements for "see[ing] it" are simply a single pixel or a certain number of pixels.
 
THat is true. This item is a cylinder. It has a height of 43 inches, a width of 3 inches, and a circumference of 10 inches.

Thanks
 
I assume this is a grade school project of some sort. I'd just recommend reading something like http://en.wikipedia.org/wiki/Angular_resolution it explains things as good as I could.

P.S. stating the width AND circumference of a cylinder is redundant
 
What about optical zoom, or optics in general if any?
 
Max CR said:
I have a 10.1 megapixel camera. If an object is 18,877.90 feet away from me, will my camera be able to see it?

This is a non-school related project that I am doing in order to be awarded scholarships. If anybody knows how I can calculate this answer please let me know.

Please include all equations and work so I know how to do this with other cameras. THank you!

Since you give no information about the optical system used to image the object, it's impossible to say. And in any case, "minimum resolvability" is very poorly defined.

In order to figure it out, you need at least the field of view, the size of the object, and some estimate of the contrast. From the first two you get the size of the imaged object in pixels (assuming this is a monochrome CCD and not a color CCD with a bayer filter, for example), and from the constrast estimate you can determine the signal-to-noise ratio, which will tell you if you can distinguish the object from background. For example, you can often see electrical power lines against the sky at great distances even though the line diameter is below the resolution limit of the eyeball- you detect the power line, you don't image the power line.
 
A 3.5' object visible at 3.5 miles distance??

I think we'll have to assume optimum viewing and contrast conditions. Mention them, but otherwise I'd think it reduces to a geometry problem.

Andy raises a good point. A critical piece of info missing is the field-of-view or zoom factor on the camera.
 
For your telescopic lens with a circular aperture, you need to have an angular resolution (Rayleigh criterion, diffraction limit) of 1.22 lambda / diameter = 1.22 x 3 inches/(12 x 18,888 ft) = 16 microradians = 3.3 seconds of arc, where lambda = wavelength of light (about 0.5 microns = 5 x 10-4 millimeters). Assuming your camera pixel width is .003 mm (you need to determine this), the objective lens focal length FL has to be at least
FL = (0.003 mm x 12 x 18888/ 3) mm.
 

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