# How can I estimate the spatial resolution of consumer grade electronics?

I have been looking at stereo vision applications, and one working example involves using two USB web cams, mounted together and calibrated.

Is there a way to estimate the theoretical spatial resolution of such a set up? The goal is to avoid purchasing and testing blindly but rather to be relatively confident ahead of time.

I do understand that the EXACT resolution will probably require a test setup due to lensing, etc.

What information would be needed to make this determination? Would the minimum focal length and the maximum resolution be sufficient? Pixel density obviously plays some role in this, and is advertised on most digital cameras (which have a much higher maximum resolution), if I was attempting to compute an estimate, how should I incorporate this?

As an example consider the following setup:

• a 4" minimum focal length (MS HD Webcam focal length)
• a maximum resolution of 1920x1080
• object is on the order of a couple of centimeters
• details on the order of tenths of millimeters
• the distance between cameras and objects should be smallish, I was considering something like 15 cm to 25 cm, flexible based on the answer here
• the distance between cameras, the number of cameras can be flexible but should stay in the smallish range ... if I need $10,000 in cameras, it's a bad solution • What do you mean with "resolution"? The distance by which two points have to be separated to be distinguishable? The amount of "real" space covered by your pixels? – Jonas Aug 17 '11 at 22:23 • The distance by which two points have to be separated to be distinguishable. – Stephen Aug 17 '11 at 22:47 ## 1 Answer Consider the following figure: The cameras are separated by baseline 2b - say 10cm, so b=5cm. Your focal length of 4" sounds looong (about 100mm) for a webcam, but hey, let's go with it. f=100mm (or also 10cm!). From$xr-xl = \frac {2bf}{z}$an object at 10m away will have a pixel difference between the two cameras ($xr-xl$) of 0.1*0.1/10 = 0.001m. So, if your pixels are 5 micron square$xr-xl\$ is about 200 pixels.

Which is a lot - large disparities are expensive to search for.

With a more normal focal length of say 5mm, your disparity is more like 10 pixels. 11m is about 9 pixels, so your distance resolution is about 1m @ 10m.

If you have a specific camera in mind you can run the numbers for your particular geometry.

• z is the separation between the camera and the viewed object? I'm thinking a much smaller value, like 15 cm, making the denominator fractional. In your example then ( xr - xl ) = 0.06...m, at 5 microns/pixel ~ 13,334 pixles. Is this an order of magnitude better or worse? – Stephen Aug 17 '11 at 21:25
• Ah, at those sorts of ranges you are probably going to need cameras which "look-in" on one another so the imagers are not in the same plane. That changes the geometry significantly. I'll have to see if my book has some easy equations... – Martin Thompson Aug 18 '11 at 8:42
• Did you find any further information that might help me understand the setup better, or mark this as answered, or can I provide better detail to help you? – Stephen Sep 7 '11 at 1:38