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# How far is the Horizon?

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This probably sounds a silly question but the human eye has a maximum capability in vision so it makes sense to say that there must be a certain limit we can see and the horizon seems to represent that point. No matter where you stand on the planet, you should see the same distance of horizon. You cannot see beyond this point as it's too far away. But how far is this point?

I thought that it might not be detectable because the horizon might look like a muddled load of whats ahead of it and so looks like a definitive spot but only looks like an illusion. That's one idea but then I have another idea that contradicts it.

Any definitive conclusions here?

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How far is the horizon? All depends on how tall you are. Really. Ride the elevator to the observation deck of a tall building, and the horizon is considerably farther than at ground level. It has nothing to do with the human eye's capability - rather, the shape of the globe.

How far can the human eye see? Look up at night and count the stars. That's quite a distance.:doh:

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I don't think that's true.

Envisage an endless landless ocean. Imagine two people standing at ground level, one person is 6 feet, the other is three feet, it has nothing to do with there eyesight. ( Assume eye sight is their best and equal ). I'm sure they would see to the same distance regardless of height????

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Turtle once told me you can see 17 miles outward until the curve of Earth takes it away.

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They would see the same distance if your endless landless ocean was perfectly flat. In reality, the Earth being a sphere, your horizon ends as a function of your height and the curvature of the planet. An average six-foot man standing on the beach looking over the ocean, has an horizon of around 20kms, if I can remember correctly. I'll scrounge up some links somewhere.

If you're standing on a flat plane (not spherical - don't know where in the known universe you're gonna find it) your vision will end at a distance depending on the quality and thickness of the air. If your flat plane exists in a perfect vacuum, you'll see essentially to infinity - but remember; if you look straight ahead, your line of vision will be parallel to the plane's surface, in other words, they'll never meet. But this is mere hypothetical, such a plane does not and cannot exist. Our horizon is bound by the spherical nature of Earth, and our height above it. Lie flat on the floor and see how far your horizon goes, as an aid to thought.

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Look here:

Distance to the Horizon

Usually the visibility is limited by scattered light in the lower atmosphere. Even under extremely clear conditions, it's unusual to see more than a couple of hundred kilometers.

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look here!

...Numerically, the radius of the Earth varies a little with latitude and direction; but a typical value is 6378 km (about 3963 miles). If h is in meters, that makes the distance to the geometric horizon 3.57 km times the square root of the height of the eye in meters (or about 1.23 miles times the square root of the eye height in feet).

3.57 km is not = 1.23 miles. 3.57 kilometers = 2.2182952 miles

so for a 6 foot person, their formula (using the correct miles figure) gives

sgrt6*2.2182952 miles=5.4336913388651586159761639159565 miles. call it 5 and a half miles. still not the 17 mile figuree i gave Orby in another thread, but that comes from sailors and refers to the distance a mast on a ship at sea disappears from the view of someone ashore. ;) :doh:

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Turtle, I suspect the conversion takes into account not only Km to miles, but also meters to feet in the 2nd part of the equation. Or the ratio of km to meters is 1/1000 where miles to feet is 1/5280.

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Turtle, I suspect the conversion takes into account not only Km to miles, but also meters to feet in the 2nd part of the equation. Or the ratio of km to meters is 1/1000 where miles to feet is 1/5280.

pardon? the part i quoted & highlighted is simply in error. :doh:

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Subject: Curvature of the Earth

Hidden by the curvature of the Earth

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3.57 km is not = 1.23 miles. 3.57 kilometers = 2.2182952 miles

I don't believe this is the conversion intended by the article, as pointed out by Zythryn:

Turtle, I suspect the conversion takes into account not only Km to miles, but also meters to feet in the 2nd part of the equation. Or the ratio of km to meters is 1/1000 where miles to feet is 1/5280.

pardon? the part i quoted & highlighted is simply in error. :eek2:

So, to clarify, the part of the article you quoted:

...Numerically, the radius of the Earth varies a little with latitude and direction; but a typical value is 6378 km (about 3963 miles). If h is in meters, that makes the distance to the geometric horizon 3.57 km times the square root of the height of the eye in meters (or about 1.23 miles times the square root of the eye height in feet).

No need to be prickly about it.

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pardon? the part i quoted & highlighted is simply in error.

My point was they didn't say that 3.57km = 1.23 miles.

What the quoted text said was the formula to find the horizon was

3.57km * sqrt(height of eye in meters)

1.23miles * sqrt(height of eye in feet)

Because sqrt(height of eye in meters) <> sqrt(height of eye in feet) the first part of the formula should not be a simple conversion from km to miles.

However, the 2nd formula is incorrect. Your conversion is actually closer than what they stated:)

Made the same mistake myself Turtle, forgot to convert back from miles to km to see if they were indeed equivalent.

They are, the formula works out:

3.75km * sqrt(height of eye in meters) = 1.23 * sqrt(height of eyes in feet)

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My point was they didn't say that 3.57km = 1.23 miles.

What the quoted text said was the formula to find the horizon was

3.57km * sqrt(height of eye in meters)

1.23miles * sqrt(height of eye in feet)

Because sqrt(height of eye in meters) <> sqrt(height of eye in feet) the first part of the formula should not be a simple conversion from km to miles.

roger after In's post i worked it out both ways >> 6 feet = 1.828 meters; 3.57 km * sqrt1.828 = 4.8278228136500618678129415815426km

1.23 miles * sqrt6 = 3.0128723836233090607826594118882 miles

4.8278228136500618678129415815426km = 2.99987 miles

However, the 2nd formula is incorrect. Your conversion is actually closer than what they stated:)

and yet the 3 mile figure agrees with the corrections in the article Doug gave. (scroll to end of page)>> Hidden by the curvature of the Earth

:eek2: no where near the 17 mile anecdotal figure i gave, but as i say it refers to seeing the top of the mast disappear over the horizon. but then, isn't the horizon just a horizon?

the bear went over the mountain,

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This probably sounds a silly question but the human eye has a maximum capability in vision so it makes sense to say that there must be a certain limit we can see and the horizon seems to represent that point.
As several have pointed out already in this thread, the distance to the horizon is a geometric effect having to do with the spherical shape of the earth. Except on very featureless parts of the earth, such as far offshore and to a lesser extent in great plains and hardpan (duneless) deserts, local terrain is much more significant than the Earth’s large-scale geometry – you can see as far as the tops of the nearest higher-elevation terrain.

It’s a bit deceptive to describe the human eye, or any light-sensing organ or device, as having a maximum range. While my unaided eyes can’t “see” ordinary printed text beyond a distance of a couple of meters, it can “see” the configuration of stars and clusters of stars thousands or millions of light-years away. Eyes are passive sensors – how far they can see depends on the brightness and feature size of the thing being seen, in addition to the particular eye’s light sensitivity, and, for resolving features, lens size and optical precision.

so for a 6 foot person, their formula (using the correct miles figure) gives

sgrt6*2.2182952 miles=5.4336913388651586159761639159565 miles. call it 5 and a half miles.

For those of us who like deriving (or seeing derived) our own formulas, here’s the one for the diagram in Infi’s post:

$r^2 +d^2 = (r+h)^2$, where $r$ is the sphere’s (planet) radius, $d$ the distance you can see, and $h$ the height of your eyes above the surface of the sphere.

Solving, $d = \sqrt{h^2 +2rh}$, about the same as the quoted formula, when r is much larger than h.

My eyes are about 1.65 m above the ground (don’t forget, your eyes are about .15 - .2 meters below the top of your head, not on the top of it!). Using 6370000 m for the Earth’s radius, the formula gives a distance to the horizon of 4585 m.

Solving for height given distance, $h = \sqrt{r^2+d^2} - r$.

So plugging in 27000 m (16.78 miles), we get a required height of 57.22 m – a pretty tall boat mast by 19th century, or even modern standards (the George Washington Bridge, is about 64 m above the water, AFAIK taller than any ship currentl afloat)

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I am 6'4", but I have terrible eyesight. How do I adjust the formula? :eek2:

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...Solving for height given distance, $h = \sqrt{r^2+d^2} – r$.

So plugging in 27000 m (16.78 miles), we get a required height of 57.22 m – a pretty tall boat mast by 19th century, or even modern standards (the George Washington Bridge, is about 64 m above the water, AFAIK taller than any ship currentl afloat)

i simply can't get enough of your calculations Craig; you epitomize the word 'computer'. :)

so, off to sea what is the case of the largest documented wooden sailing vessel. fortuitously i watched a 2 hour show on just this topic and find that the treasure ships of the eunich Zheng He are the largest wooden vessels ever constructed. this reference does not give the height of the masts, but if they were only a bit taller than the vessels were wide then indeed they might be seen at 17 statute miles.

"Treasure ships", used by the commander of the fleet and his deputies (nine-masted, about 126.73 metres (416 ft) long and 51.84 metres (170 ft) wide), according to later writers (no proof of the supposed great size of these ships exists, and as stated above, they are improbably large). The treasure ships purportedly weighed as much as 1,500 tons.126.73m by 51.84 m (415.780ft by 170.078ft)[10][11] By way of comparison, a modern ship of about 1,200 tons is 60 meter (200 ft) long [1], and the ships Christopher Columbus sailed to the New World in 1492 were about 70-100 tons and 17 meter (55 ft) long.[12]

I am 6'4", but I have terrible eyesight. How do I adjust the formula? :doh: :hihi:

squatting wearing glasses. :hihi:

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so, off to sea what is the case of the largest documented wooden sailing vessel. fortuitously i watched a 2 hour show on just this topic and find that the treasure ships of the eunich Zheng He are the largest wooden vessels ever constructed. this reference does not give the height of the masts, but if they were only a bit taller than the vessels were wide then indeed they might be seen at 17 statute miles.
When I think of big 19th century sailing ships, I think of the Cutty Sark: 85.4 m (280’) long, 970000 kg displacement, and 46.3 m (152’) from deck to the top of the highest of her 3 masts. Guessing at her waterline-to-deck height, I’d make the top of her mast about 50 m above the water, lightly heeled and moderately loaded. As with most ships, her crow’s nest, wouldn’t have been at the extreme top of the mast, but several meters lower, near her highest (t’gallant) spar.

A 9 (!) mast, about 140 m (!), 1500000 kg (!) ship like one of admiral Zheng He’s Treasure ships, likely had a lower height-to-length ratio that a clipper ship, but still might have been taller. These ships were 15th century oddities, though, who’s actual dimensions are the subject of debate. The biggest wooden sailing ship of unquestioned origin, I’ve just discovered, is the six-masted schooner Wyoming: 100 m (330’) long (137.6 m (450’) including her immense bowsprit), 3700000 kg, and a famous example of a wooden ship to big to be truly seaworthy, foundering and sinking with all hands 24 years after her 1910 launch.

Another really big sailing ship, this one of modern steel construction and very seaworthy, if the US Coast Guard’s Eagle, 90 m long, 1700000 kg, and 50 m tall. She’s a spectacular ship, very fast under sail, and not slow under diesel power. Formerly a WWII German transport ship, she’s said to have shot down 3 aircraft in combat!

PS: I love that ship (the Cutty Sark)! Built a plastic and fabric kit model of her when I was a pre-teen, and got to visit her in 2001. I was shocked to learn she was badly damaged by a fire a couple of months ago – see the link above for details

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