I've been given the following question as a review problem for my upcoming digital image processing midterm. This question isn't going to be graded and to my knowledge, may not appear on the test, but I'm not sure how to go about solving it.
A sphere/cube is placed at a distance of M meters in front of a camera. The object's area/volume is A/V. The camera's focal length is X. The object on the imaging plane has an area of projection of a/v. How can we calculate a/v?
This problem relates to another question, where I'm asked to find the height of a projected image. I understand that one well enough. If we state that the object's height is H and its projected image height is h, the formula would look like this: h = HX/M
I know that the ratio of the object's area to its volume will be preserved, but how can I go about calculating a/v?
In the case of a cube, the side is $S=V/A$. The corresponding side of the image is $s=S\frac {X}{M}$. Thus, the image area is,
$$s^2= \left(\frac VA \frac{X}{M}\right)^2$$
In the case of a sphere, $A/V = (4\pi R^2)/(4\pi R^3/3)$. So, the radius is $R=3V/A$. The corresponding radius of the image is $r=R\frac {X}{M}$. Thus, the area of the image is,
$$\pi r^2= 9\pi \left(\frac VA\frac{X}{M}\right)^2$$.