Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am reviewing some old material I learned years ago and am having trouble figuring this one out. Can somebody confirm that I have done my math correctly, and tell me how I can recover the field of view from these parameters?

I have a $3.6$(height) by $4.8$(width) mm sensor, which displays as $640\ by\ 480$ pixels. I take an image of some item that I want to be 50px large in my image. It is $10 m$ away, and $10 cm$ tall in real life.

I compute the size of a pixel on the sensor as $\frac{3.6}{640}\ by\ \frac{4.8}{480} mm$ to get $0.006\ by\ 0.01 mm$.

I compute the magnification by $\frac{image\_size}{actual\_size}$, so I compute that the image size of $50px$ with a pixel height of $0.006mm$ takes up $0.3 mm$ of the sensor. Thus, magnification is $\frac{0.3}{100} = 0.003 mm$.

Magnification can also be viewed as $m = \frac{f}{u - f}$, so $0.003 = \frac{f}{100 mm - f}$, which will give us $f = 0.299 mm$.

From here, I know that it's possible to calculate the field of view, but how can I do it? I imagine the equation will use the image sensor size and the focal length.

Thanks!

share|improve this question
1  
Do you really have higher resolution in the height of the sensor? Resolution like 640:480 is usually given as width:height. Switching it around gives you more reasonably, square pixels: $3.6/480 = 4.8/640 = 0.0075$ mm. –  Mikael Öhman Dec 16 '11 at 17:48
add comment

1 Answer

up vote 1 down vote accepted

It's easy to check if your calculations are right from this (rather crude) drawing:

enter image description here

There you see that the size of an object is related to the size of the image according to $$h_o=h_i\frac{D}{f}$$ where $h_o$ and $h_i$ are the sizes of the object and image. $f$ is not a focal distance, but the distance from the pinhole to the sensor or film.

So, if you have a 10cm object at 10m from the camera, and you want that to be 0.5mm (50px width), you need to place the sensor 0.5mm*10m/10cm=5cm behind the pinhole. If you do that, the angle your camera will see is $\tan(\alpha)=h_\mathrm{CCD}/(2f)$. The field of view is defined as twice this angle ($\mathrm{FOV}=2\alpha$). With the above numbers, the vertical FOV is 4.1º and the horizontal FOV is 5.5º.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.