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

1. Jun 9, 2009

### Max CR

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!

2. Jun 9, 2009

### S_Happens

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.

3. Jun 9, 2009

### Max CR

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

4. Jun 9, 2009

### maverick_starstrider

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

5. Jun 10, 2009

### Lok

What about optical zoom, or optics in general if any?

6. Jun 10, 2009

### Andy Resnick

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.

7. Jun 10, 2009

### DaveC426913

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.

8. Jun 10, 2009

### Bob S

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.