If I know an objects real size and I get its apparent size by putting a ruler 1 foot from my eye and measuring it. If the object is perpendicular to my eye is that enough information to determine it's distance from me? What would the calculation be?

  • $\begingroup$ You want the ruler and the object parallel. Then if some point on the object is a distance D and the corresponding point on the ruler is a distance d from your eye, you multiply the apparent size by D/d to get the actual size (by similar triangles). $\endgroup$ – almagest Mar 23 '16 at 14:58
  • $\begingroup$ Thanks Almagest. I think it's the same formula that Noah wrote right? Now just want to figure out if distance from eye to ruler is a big factor and needs to be part of the formula. $\endgroup$ – Steve Mar 25 '16 at 11:56

Let $m$ denote the $m$easured distance on the ruler, $l$ the $l$ength of the object, and $d$ the $d$istance from you and the object. Because of similar triangles:




  • $\begingroup$ Thanks Noah! What about the distance from my eye (or camera) to the ruler? Does it just need to be close or does that distance play into the calculation? Lets say l = 2,500,000 inches and m = 1 inch then said object is 2,500,000 inches away correct? Just thinking if I move the ruler closer or further from eye/camera then m would increase and decrease right? $\endgroup$ – Steve Mar 25 '16 at 11:52
  • $\begingroup$ Found this calculator which seems to be what I need but wondering why it doesn't have a way to put in m in the same units as l. They just have reference pixels, degrees and other units. Seems it's more for astronomy? $\endgroup$ – Steve Mar 25 '16 at 12:00
  • $\begingroup$ As for your first comment, I assumed the distance from your eye to the ruler was 1 foot. If the distance can vary, let's call it $x$ and the new equation is $d=\frac{xl}{m}$. As for the calculator, I don't have an answer for that. $\endgroup$ – Noah May Mar 26 '16 at 22:06
  • $\begingroup$ So if distance of ruler from eye is 1 foot then no need for the x (or x is = 1) but what for example x be if distance from eye was 2 feet? x would be = 2? $\endgroup$ – Steve Mar 27 '16 at 14:04
  • $\begingroup$ Yes, it would be.Don't forget that $x$, $l$, and $m$ have to be the same units. $\endgroup$ – Noah May Mar 29 '16 at 15:57

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