Visibility range of objects

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Korbl
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Visibility range of objects

Postby Korbl » Tue Jul 24, 2012 2:49 am UTC

Say you have a petrified tree, 1.6 km in diameter, 8 km in height, in a barren desert, in the middle of the day.

Assuming no obstructions, from how far away could the human eye detect the tree?

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PM 2Ring
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Re: Visibility range of objects

Postby PM 2Ring » Tue Jul 24, 2012 4:41 am UTC

Korbl wrote:Say you have a petrified tree, 1.6 km in diameter, 8 km in height, in a barren desert, in the middle of the day.

Assuming no obstructions, from how far away could the human eye detect the tree?


That's some tree! It will be visible from quite a distance. Do you want to take atmospheric distortion and the curvature of the Earth into account?
http://en.wikipedia.org/wiki/Visual_acuity#Other_measures_of_visual_acuity wrote:The smallest detectable visual angle produced by a single fine dark line against a uniformly illuminated background is also much less than foveal cone size or regular visual acuity. In this case, under optimal conditions, the limit is about 0.5 arc seconds or only about 2% of the diameter of a foveal cone. This produces a contrast of about 1% with the illumination of surrounding cones. The mechanism of detection is the ability to detect such small differences in contrast or illumination, and does not depend on the angular width of the bar, which cannot be discerned. Thus as the line gets finer, it appears to get fainter but not thinner.


According to Google Calculator, 1 / tan(0.5 arcseconds) = 412 529.612

Korbl
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Re: Visibility range of objects

Postby Korbl » Tue Jul 24, 2012 5:37 am UTC

It's for a D&D game, basically it's Yggdrasil.

I have no clue what an arc second is (I stopped at chemistry in High School, where I did poorly). Can you give me an idea of that distance in feet or miles or at least metric distances? Curvature and atmospheric distance aren't super important.

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gmalivuk
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Re: Visibility range of objects

Postby gmalivuk » Tue Jul 24, 2012 6:14 am UTC

412529.612 times the height of the tree.

(You gave the dimensions in metric, but prefer USCS answers?)
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Korbl
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Re: Visibility range of objects

Postby Korbl » Tue Jul 24, 2012 6:18 am UTC

Thanks to both of you, this helps a lot.

(yeah, I'm an american, and a culinary type at that, so I'm used to Imperial, but I figure I'd use metric since I was asking people who are probably more used to that. Does that make sense?)

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PM 2Ring
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Re: Visibility range of objects

Postby PM 2Ring » Tue Jul 24, 2012 6:30 am UTC

(I'm Australian, so I tend to use metric measures. I did learn imperial measures as a child, but I don't use them much these days.)

Korbl wrote:It's for a D&D game, basically it's Yggdrasil.

I have no clue what an arc second is (I stopped at chemistry in High School, where I did poorly). Can you give me an idea of that distance in feet or miles or at least metric distances? Curvature and atmospheric distance aren't super important.


Ok. I asked about curvature because I guessed that this might be on a flat world, or at least on a planet with a different radius to the Earth. If it's not on a flat world then at large distances some (or all) of the tree will be below the horizon, hence not visible.

An arc second is a unit of angular measure. There are 60 seconds to a minute, and 60 minutes to a degree; the terms arc-second and arc-minute are used to distinguish these units from the units of time.

A line 2 units long at a distance of 412 529.612 units subtends an angle of 1 arc-second, so it will be visible under ideal viewing conditions, according to that Wiki link. So on a flat world the 8km high tree will be visible at a distance of over 1.6 million kilometres (ignoring atmospheric distortion). However, to get any kind of detail, we'd need to use the figure of 5 minutes of arc mentioned earlier in the article, giving a distance of a shade over 5,500 km. On Earth, that's a little over half the distance from a pole to the equator, so horizon issues will definitely be significant. :) At a height of 8km, the horizon is roughly 320 km away, so the 8km high tree will be below the horizon for observers more than 320 km distant from it.

FWIW, the formula I used for relating height, distance and angular size is
(height / 2 ) / distance = tan(angle / 2)

Korbl
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Re: Visibility range of objects

Postby Korbl » Tue Jul 24, 2012 7:36 am UTC

Ok, that's better. Yeah, on a flat world the thing would be visible pretty much anywhere... which would be amusing... "You know that dark line that's always been on the horrizon? Yeah, it's the tree." but I think I just finalized my decision that the world is round.

Thank you all very much for the help.

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Re: Visibility range of objects

Postby Tass » Tue Jul 24, 2012 10:13 pm UTC

Note that the air has to be exceptionally clear and calm for anything to be visible at that distance. Especially in a desert you'd get all sorts of mirages. Under exeptional circumstances someone might see it from as far away as 500km even though it should be below the horizon. Often people would see it distorted or upside down or high in the air.

But most often it would simply be lost in the blue mist of distance unless you are within 100km or so.

Edit: Actually you specified "in the middle of the day". In the middle of the day in a desert the air will tend to be warmer near the ground, this makes light refract upwards, and it will be lost below the horizon even when much closer than 320km. On the other hand in early morning when the ground is cold the light would curve down following the curvature of the earth, possibly allowing someone to spot it as the first rays of sun hits the top, from said 500km or so.


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