Science Forums

# How far is the Horizon?

## Recommended Posts

...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. ...

if i recall, part of the substance of the program i watched relied on new found documents that affirm the size of the largest of Zheng He's ships. i'll see if i can find which channel ran it & title. they say that the ships that survived the voyages, as well as most of their voyage records, were destroyed by the Chinese not long after Zheng He's last voyage.

notably, another thrust of the program is the theory that Zheng He reached the 'new world' before Columbus. it's hotly debated if you can imagine. :lol: :phones:

• Replies 40
• Created

#### Popular Days

... 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.... :lol:

i think the 17 miles figure is for an observer on shore seeing the top of the mast.

here's a review of the show. >> People's Daily Online - Documentary on ancient Chinese explorer Zheng He aired in U.S.

programming info>> OnTV

History International

:phones:

##### Share on other sites

I visited a light house a few years back and I was told that ships could see it from 30km out - it stood around 40 m tall and was aboard a platform at sea perhaps another 5m tall. Do the math on that one :)

I once performed the calculations for a person my height standing on a featureless sphere and got much less than I expected, though I cant remember the exact number :shrug:

##### Share on other sites

I visited a light house a few years back and I was told that ships could see it from 30km out - it stood around 40 m tall and was aboard a platform at sea perhaps another 5m tall. Do the math on that one :)
The two formulae from post #14, $d = \sqrt{h^2 +2rh}$ and $h = \sqrt{r^2+d^2} - r$, can be used 2 times to calculate the distance at which 2 objects can be seen over a horizon. So, for a person in a small boat on a smooth sea, with eyes 2 meters above sea level, looking at a 40 m tall tower, the distance is $d = \sqrt{2^2 +2 \cdot 6370000 \cdot 2} + \sqrt{40^2 +2 \cdot 6370000 \cdot 40} \dot= 5048 + 22574 = 27622 \,\mbox{m}$.

To see the 40 m tower at a distance of 30000 m, you’d have to get your eyes $\sqrt{6370000^2+(30000-22574)^2}-6370000 \dot= 4.33 \,\mbox{m}$ above sea level – not difficult in a larger boat, or one with a flying bridge.

##### Share on other sites

Sounds like I got accurate information then :shrug:

##### Share on other sites

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.

I can see the moon just fine and I think it lays somewhat further than the horizon...:shrug:

##### Share on other sites

Actually Clay, I wasn't referring to a infinite space, rather the curvature of the Earth must have a limit(s) with the horizon apparently determined by height as well.

If you know what I horizon is then why are you questioning outside of the horizon which I never questioned you couldn't see it?????

Anyhow, thanks for the replies, I have a better idea now.

##### Share on other sites

I was pointing out that your opening statement could be taken in more than one way since it would seem obvious that the horizon in Death Valley is much closer than the horizon on the summitt of Mount Everest....

##### Share on other sites

Im not sure if this is what LJP07 is pointing out, but how far would the horizen be standing on a flat plane?

##### Share on other sites

That's the point, the term horizon is ambiguous as a distance....

##### Share on other sites

Im not sure if this is what LJP07 is pointing out, but how far would the horizen be standing on a flat plane?
Infinitely far away.

The only ambiguity C1ay is that there are two things, that differ for a height not tiny compred with the radius, one is the arc and the other is the segment of the tangent line.

##### Share on other sites

Infinitely far away.

The only ambiguity C1ay is that there are two things, that differ for a height not tiny compred with the radius, one is the arc and the other is the segment of the tangent line.

ok, as a suspected. Now how many degrees, given from zenith will it take to reach the horizon? this will be independant of the height of the observer will it not?

##### Share on other sites

I took LJP07 as asking how far could the human eye itself see, and not having anything to do with Earth's roundness ?..

Just to get rid of sphere-ness (as someone already did...) – you're on a flat plain of land, not sphere, and no horizon for the land itself because it's infinitely flat. So, (if you're looking straight ahead, parallel to the land) how far can you place something and still see it ? After what distance does the eye 'stop seeing' ?

Sun is averagely 1 lightyear away, so, we can see one lightyear away. Stars are further... So say if it's 'night' (or just without a local sun), and you line up stars infinitely – one on each lightyear mark , and you can't see any further then say the 8,447,357,572,861th star. That makes the eye's (event?) horizon some 8.447 trillion lightyears, right ?

Is that what you'er originally asking LJP07, or rather was it the land-horizon's distance in respect to Earth's sphericity ? :hihi:

##### Share on other sites

Just to get rid of sphere-ness (as someone already did...) – you're on a flat plain of land, not sphere, and no horizon for the land itself because it's infinitely flat.

Yep - you will see a 'line' marking the 'horison', but that line just needs further magnification to be 'more of the same'. It won't be a line behind which stuff disappears; because your line of sight runs parallel to it they never meet, you'll see essentially to infinity.

So, (if you're looking straight ahead, parallel to the land) how far can you place something and still see it ? After what distance does the eye 'stop seeing' ?

That all depends on the quality of the medium through which the light is propagated. On Earth, our vision quite regularly fails before we see the actual horison, due to air pollution and such. If in a vacuum, the air won't spoil your sight. Then how far and to what level of detail you can see depends solely on the quality of your eye, in other words, how well your lens can focus incoming light on your retina, and the concentration of rods and cones on your retina that basically makes up your eye's 'resolution '.

Sun is averagely 1 lightyear away, so, we can see one lightyear away.

Round about 8 lightminutes away, actually. But, yes. You can see at least eight lightminutes away. But you can't resolve any features on the sun's surface (although you can sometimes see sunspots late in the afternoon when the sun is very much reddened and dim just before sunset). You basically see the sun simply because it's very big and very bright, not because you've got good eyes. Even almost-blind people can see the sun, but they can't resolve things at arm's length.

Stars are further... So say if it's 'night' (or just without a local sun), and you line up stars infinitely – one on each lightyear mark , and you can't see any further then say the 8,447,357,572,861th star. That makes the eye's (event?) horizon some 8.447 trillion lightyears, right ?

The eye's 'event' horizon depends solely on the 'brightness' of the object in question. Resolution is another matter entirely, and depends on the actual physical construct of the retina, much like dpi on a printed page, where the 'dots-per-inch' quality of the image is produced by the number, quality, and density of the cones and rods that make your eye sensitive to light.

##### Share on other sites

Now how many degrees, given from zenith will it take to reach the horizon? this will be independant of the height of the observer will it not?
Sure.

$\alpha=\arccos\frac{R}{R+h}$

Of course it stands to reason that if you multiply the sine of this angle by (R + h) you should get the same length as Craig's equation, indeed they are equivalent ways of computing this length and it isn't to difficult to show this purely mathematically.

##### Share on other sites

(I just looked up lightyear, and as I read it, it makes no sence to how I applied it. So: 8 'lightminutes'.)

I think I see... So, if one's standing on a flat plain in a vacuum with light, it basically depends on their eye's 'perfection' to how well they see – if they can't see details on things further than five feet, and need correction for anything further, their 'horison' will be closer than that of a person with perfect eyesight because the clarity in which perfect-vision sees things is better, and more detail can be seen of an object further away compared to one who needs correction and is uncorrected...

If we're using perfect eyes for an example, with all its rods and cones and able to focus light perfectly, is the horison still infinitely far away ?

In the image I made, Observer 1 has bad vision, and 2 perfect vision. They're far enough apart to where Obs1 is in 2's vision, but can 2 see past Observer 1 ? Won't each's views, at least from the sides' perspective, 'close in' from the left and right and anything past that 'triangle' won't be visible ?, and since 1 has such bad eyesight, 1 won't be able too see much/anything past his horison, and that means 1 cannot see 2, right ?

##### Share on other sites

I took LJP07 as asking how far could the human eye itself see, and not having anything to do with Earth's roundness ?..

Just to get rid of sphere-ness (as someone already did...) – you're on a flat plain of land, not sphere, and no horizon for the land itself because it's infinitely flat. So, (if you're looking straight ahead, parallel to the land) how far can you place something and still see it ? After what distance does the eye 'stop seeing' ?

There are at least 2 question imbedded in here, only one of which has been addressed so far in this thread.

The first is “what is the maximum distance that an eye can detect a point of light?” As noted in post #14 and post #31, this is a function of the power of the light source, and the sensitivity of the eye, so you can’t assign a maximum value considering just the eye. For example, if I experimentally determine that I can detect when an LED flashlight is switched on at 300 m, but not at 301 m. If the brightness of the light is increased by a factor of 4 – for example, if it uses 4 LEDs instead of 1 – I would be able to detect it at twice the distance, 600 m. (assuming no obscuring atmospheric effects)

So say if it's 'night' (or just without a local sun), and you line up stars infinitely – one on each lightyear mark , and you can't see any further then say the 8,447,357,572,861th star. That makes the eye's (event?) horizon some 8.447 trillion lightyears, right ?
The “standard” limit of light detection for the human eye is 6 on the standard astronomical magnitude scale. Some peoples eyes are slightly better, some worse, and some viewing locations better and worse, and variable according to weather and other factors, but for a “standard” human eye looking from a standard place on a standard night, a web search turns up suggestions that the most distant normal (eg: not exploding) star visible with a naked human eye is about 16,000 light-years distant (Hipparcos 5926 (V762) in Cassiopeia). For exploding stars the maximum distance is much greater - a type 1a supernova (absolute magnitude of -19.5) would be visible up to about 4,000,000 light years. Whole galaxies of stars are (obviously) much brighter than individual stars, and can be seen as single points of light at up to about 3,000,000 light (eg: the M33 galaxy)

The second question is “how far away can an eye distinguish 2 light sources from one?” Like the first question, this depends not only on the eye, but on the object being viewed – obviously, we can see, and distinguish detailed features, of large objects at a greater distance than small ones.

The answer to this question depends primarily on 3 factors: the wavelength of the light; the size of the lens focusing it (about .01 m for a typical human eye), and the distance between the objects to be distinguished. It’s customary to represent this as the eye’s (or any other instrument’s) minimum [wiki]angular resolution[/url], usually in minutes and seconds of arc. A typical human eye with “perfect” eyesight can resolve about 2-1’ (arcminutes, or 1/10800 to 1/21600 of a full circle) using typical “white” light. Taking all this into account, a human being can tell that two light placed 1 meter apart are two, not 1, up to between about 10 to 20 km.

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.