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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???

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  • $\begingroup$ Yes, it is just a simple ratio. $\endgroup$ – JimmyK4542 Jul 8 '14 at 4:22
  • $\begingroup$ Alright, that is what I thought. Thanks. $\endgroup$ – CryptoCommander Jul 8 '14 at 5:25
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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&\implies\qquad \alpha=2\arctan0.05\approx5.7248° \\ 2\tan\frac\alpha2=\frac{x}{3\,\text{feet}} \qquad&\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.

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    $\begingroup$ Sounds great, thanks for the help! I spent a while thinking about the problem and trying to determine "if an object is 10 feet and I"m 10 feet away from it how big does it appear (what is the effective/visual size)?" It took me a while to figure out that you couldn't just say "oh a 10 foot object 100 feet away will appear to be 4 inches tall" because it all mattered how far the 4 inches away was first.. something can't appear to be 4 inches tall from 0 feet away because it takes a foot or two for most people's vision to focus on an object.. $\endgroup$ – CryptoCommander Jul 8 '14 at 12:59
  • $\begingroup$ edit: Thanks for the help! I spent a while thinking about the problem and trying to determine "how big an object is at 100 feet" but you can't have something be, for example, 4 inches at a distance of 0 feet because it all mattered how far the 4 inches away was first.. it takes a foot or two for most people's vision to focus on an object so it is in fact a point of reference as well. I was curious about this because I wanted to know if a video game was using a different angular size(visual size) than real life in order to make it harder to use something like a bow and arrow beyond 100 feet. $\endgroup$ – CryptoCommander Jul 8 '14 at 13:06
  • $\begingroup$ for gameplay reasons, etc. $\endgroup$ – CryptoCommander Jul 8 '14 at 13:06
  • $\begingroup$ @JavaNovice14085: You are welcome. The established way to express gratitude is by upvoting and/or accepting an answer. The former needs some reputation, but the latter can be done once you are stisfied with an answer to one of your questions. You can still switch that check mark if a better answer turns up. $\endgroup$ – MvG Jul 8 '14 at 13:08
  • $\begingroup$ @JavaNovice14085: For video games it depends a lot on how large your monitor is, and how far away from it you are. They don't have a built-in angular size per se, and the developers might have different ideas of your placement than yourself. But for most games they likely face a compromise: allow users to look around a bit, even though a monitor covers only a rather small fraction of the visual field. $\endgroup$ – MvG Jul 8 '14 at 13:12
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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

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