# How far would you need to displace your eyes to get meaningful depth perception of the stars?

The question follows from xkcd cartoon "Depth Perception (941)". I've isolated the frames that describe the concept here.

In words, one could theoretically point two cameras at the sky, and displace them so that, if viewed as components of a projected 3D image, the starfield of the night sky would have a perceivable depth. That is, Sirius or Alpha Centauri would appear closer than, say, Betelgeuse.

The idea sounds interesting, but I was wondering whether it's actually possible. That is, how large would the displacement of the cameras need to be to create a perceivable depth? To quantify, let's say we're trying to reduce the scale from 5 light-years to 100 m. Would this require a displacement of 5 light-years / 100 m$\times$(separation of eyes)$\approx1/400$ light-years $\approx136$ AU or is it more complicated than that?

I guess the maximum we could achieve is by taking two images of the same field of stars, one year apart and combining them, to give a separation of the "eyes" of about 2 AU. I don't know enough astrometry myself to be sure.

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I did think twice about asking this, but I don't see it as open ended and it isn't purely hypothetical. Apologies if it isn't suitable for Astro.SE. – Warrick Aug 23 '11 at 8:24
I fear the answer might lie more in the realm of physiology then physics. I don't really know how to quantify 'depth perception'. – Nic Aug 23 '11 at 9:23
The rest of the comic does refer to clouds. I am more concerned with the question if the reading glasses would actually help. – Phira Aug 24 '11 at 10:31

Parallax is linearly proportional to separation, so to get meaningful depth perception to even one star, your eyes would have to be (present eye separation)*(distance to Proxima Centauri)/(longest distance at which we naturally have meaningful depth perception). Having been to Meteor Crater, I can tell you that the last quantity is definitely under a half mile, i.e. distance from rim to center, but for a conservative estimate we'll call it that.

Our formula, then, is sep=3 inches * 4.2 l-yr / 2640 feet = .0004 l-yr = 25 au. That is past Uranus, almost to Neptune.

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Aha. So my guess at the relationship was right, I just accidentally used m instead of cm for the separation of the eyes... Now fixed. – Warrick Aug 23 '11 at 11:23
If your interpupillary distance (term of art!) was "just" enough to converge on Proxima Centauri, you still would gain very little, as that would be just enough to pull it forward from an essentially black background. To pull off the experience of the local stars being at different depths, you'd have to go further... – Larry OBrien Aug 23 '11 at 21:31
For what its worth, the calculation can be done on Wolfram|Alpha. – Warrick Aug 25 '11 at 7:54
Nice! I will have to remember that the next time I need to do a unit conversion and/or proportional reasoning calculation. – Andrew Aug 25 '11 at 19:00
Great, so we may begin producing 3D images of the sky, now that the voyager probes reach the frontiers of our sole system. – artistoex Jun 10 '12 at 13:34

Apparently the highest resolution a human eye can perceive is around 30 arcseconds. The largest 1 AU baseline parallax for any object outside our Solar System is for alpha/proxima centauri, which is 0.75 arcseconds. It is the closest star to our Sun at a mere 4.37 light-years away. In order to achieve a level of perceptible parallax to the human eye you would need a baseline at least as long as 38 AU, which is about out to the orbit of Uranus. So if you had two spacecraft orbiting on opposite sides of the Sun out near Uranus then you could maybe achieve a barely perceptible degree of depth perception for a single star.

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I own a "Stereo-Telemeter" made by Wild Heerbrugg in the 40ties, used by Swiss Army to measure distance of aircraft. This Telemeter has a basis of 1.25 meters and the manual says that the stereoscopic "seeing" with the instrument will extend to about 10 000 meters in best case. This ratio is 10 000 / 1.25 = 8000

With 1 AU = 8 Lightminutes and maximal possible basis is diameter of earth orbit which is 2 AUs, we get 2AU = 16 lightminutes times 8000 = 128 000 lightminutes as maximal "reach" of stereoscopic seeing. This is about 88.8 lightdays or about a quarter of a ly. So, nothing to write home about :=(

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Just to give you an idea, this effect is actually well known, and used to find the distance to the nearest stars. The Wikipedia article Stellar parallax goes into great detail, but suffice it to say that the primary method of measuring parallax in stars is via "Annual Parallax", which is to use the Earth's position exactly six months apart as the baseline for determining the distance. Using extremely high precision, we are able to detect slight differences in stars positions that are closer than a few hundred light years with non-space measurements, and up to 1600 using space measurements.

But, the problem is that this is a very small distance. Even the closest star only has a parallax of 0.7687 ± 0.0003 arcsec. As Wikipedia quotes, "This angle is approximately that subtended by an object 2 centimeters in diameter located 5.3 kilometers away."

The bottom line is, if you had a very powerful telescope and a baseline of the Earth's orbital diameter, then you could start to see some 3D effects. But a webcam across a few hundred feet, or even miles, will only allow you to see 3D clouds, and possibly the Moon. The basic premise is true, it's just that you would need far more than a thousand times to see any 3D sky effects.

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OP didn't mention it, but the comic was about looking at clouds; it compared them to stars in terms of not normally perceiving any parallax. A few hundred feet is probably enough for that. – villageidiot Aug 23 '11 at 21:21
@user: I did say "But a webcam across a few hundred feet, or even miles, will only allow you to see 3-d clouds, and possibly the moon." – PearsonArtPhoto Aug 23 '11 at 22:12