Calculating size of an object based on distance So, say an object that is 10 feet tall is 100 feet away. If I hold up a ruler 3 feet away, then the object in the distance would correspond to about how many inches?

Tried using this guy: http://www.1728.org/angsize.htm
to calculate the angle, which ends up being 5.7248 degrees

Then, if I solve for size using 5.7248 degrees at a distance of 3 feet I get 0.3, or 4.8 inches.

The thing is is that that does not seem accurate to me. Perhaps my perception of distance is off, but 4.8 inches looks more like a 10 foot tall object at 50 feet to me...?

I mean, it is a simple ratio really..

x/3 feet = 10 feet/100 feet  right???
Thanks to the intercept theorem this is indeed a simple ratio:

$$\frac{x}{3\,\text{feet}}=\frac{10\,\text{feet}}{100\,\text{feet}}
\qquad\implies\qquad x=0.3\,\text{feet}$$

If you want to also involve the angles, you have

\begin{align*}
2\tan\frac\alpha2=\frac{10\,\text{feet}}{100\,\text{feet}}
\qquad&amp;\implies\qquad \alpha=2\arctan0.05\approx5.7248°
\\
2\tan\frac\alpha2=\frac{x}{3\,\text{feet}}
\qquad&amp;\implies\qquad x=3\,\text{feet}\times2\tan\frac\alpha2 = 0.3\,\text{feet}
\end{align*}

So the computations you did using that tool are correct. Anything that looks wrong is likely an optical illusion.
It seems to me that x/3 =10/100 shows that x= 0.3 right so far, but .3 of a foot is 3.6 inches not 4.8 right? So that may account for the difference in perceived size at 100' and 50'. Thanks