How far away is the horizon??

[weeds]

We were talking the the volunteers at the double island point light house yesterday and he reckons the horizon was only 1km away....keep in mind this light house is probably >100m above sea level.....

Soooo we are now sitting at sea level, well say 2m above sea level

I reckon the horizon is 10km away.....
Raggedy Ann reckons 4-5km away

I wonder just how far off in the distance is really is

Here is our view.....

[CraigE]

It varies depending on your distance above sea level We generally work off around 20 mile or 32 kms but this is calculated on an average of 100m above sea level.
The below link is a basic calculator.
Distance to the Horizon Calculator (http://www.ringbell.co.uk/info/hdist.htm)
Cheers
CraigE

[sheerluck]

Remembering that the earth is not a perfect sphere, but an oblate spheroid instead, the distance will alter slightly depending where you are on the earth's surface too.

[weeds]

It varies depending on your distance above sea level We generally work off around 20 mile or 32 kms but this is calculated on an average of 100m above sea level.
The below link is a basic calculator.
Distance to the Horizon Calculator (http://www.ringbell.co.uk/info/hdist.htm)
Cheers
CraigE

As mentioned we are about 2m above sea level......so does mean its about 3.2k away??

[THE BOOGER]

According to the calculator its 5.1km bit more than the 1 they told you;)

PS: Raggedy ann wins by a mile:D

[Chops]

At a course I did at school, they said a mirror reflection can be seen from aprox 25 miles away, not sure what that is in Klm's, but that was to the horizon.
Does this mean I have to change my thinking?

[weeds]

According to the calculator its 5.1km bit more than the 1 they told you;)

PS: Raggedy ann wins by a mile:D

There's a calculator? We actually that somebody must have sorted a formula for working it out....

[Homestar]

The volunteers at the lighthouse need a bit more training if they think the horizon is only a KM away. According to the calculator that CraigE posted, at 100 metres the horizon is 35.7 KM away...

I was taught by my Scout leader that it was 10km away when standing up - looks like he was a bit out as well...:D

[THE BOOGER]

Craig posted a link to the cal in the 2nd post:)

[mike_ie]

Ignoring the effect of atmospheric refraction, distance to the horizon from an observer close to the Earth's surface is about

d ≈ 3.57√h

where d is in kilometres and h is height above ground level in metres.


For an observer standing on the ground with h = 1.70 metres (average eye-level height), the horizon is at a distance of 4.7 kilometres.

For an observer standing on a hill or tower of 100 metres in height, the horizon is at a distance of 39 kilometres.

[Hoges]

There's a calculator? We actually that somebody must have sorted a formula for working it out....

In the days when "real" maths was taught in school:angel::wasntme:, we learned in yr 8 ("Form 2" as it was called in those days) that as a rough approximation, the distance to the horizon in nautical miles was 1.2 x the square root of your height above the water measured in feet.... thispost reminded me that this was one of those days I actually paid attention:eek::D

so say 2m approximates to 6 feet, and the square root of 6 is 2.5, thus 2.5 x 1.2 = 3 nautical miles which multiplied by 1.8 to convert to km = approx 5.4 km

just goes to show you can learn something useful on this forum every day!! So as Billy Connelly would say...stay awake and pay attention;)

the issue of refraction and total internal reflection is really interesting because due to atmospheric conditions, light is actually internally reflected off upper layer. This phenomenon is most evident travelling the Hay plains west of Narranderah at night... while cruising at 120kmh one night some 30 yrs ago I dipped the lights for oncoming traffic: it took 10 minutes before the traffic passed. Assuming they were cruising at similar speeds, that was a closing speed of 240kmh or 4km /min...thus 40km separation at first sighting. It was a coolish night and I'm assuming the lights were being reflected and refracted between narrow layers of air/water vapour...amazing!

[bob10]

From the Oxford companion to ships and the Sea,

HORIZON from Greek horos, 'a boundary', horizo 'form a boundary', 'limit'.

1. The line which limits an observers view of the surface of the Earth and of the visible heavens. In astronomical navigation three meanings must be distinguished:
a. The visible horizon, that which is actually seen.This however is affected by the dip of the horizon which depends on the refraction of light by the atmosphere and the observers height above above the sea.
b. The sensible horizon, the true horizon at sea level at the observers position on the Earths surface, corrected for dip; it is the projection on the celestial sphere of a plane tangential to the Earths surface at that point.
c. The rational horizon, this is the projection on the celestial sphere of a plane parallel to the sensible horizon but passing through the centre of the Earth instead of tangential to its surface. In measuring the altitude of a heavenly body considered as infinately distant , the radius of the Earth is insignificant, and normally the sensible and rational horizons coincide. For some some purposes, however, they must be distinguished.

Simple really. However your friends may have been close, for the layman, the visible horizon depends on the height it is observed from. and the refraction of light. Next time you are at the beachfront, look to Moreton, we know how far it is away, then check the horizon at different times of the day, on clear days and sunny, and then cloudy days. Look to the gap between Moreton & Redcliffe. Bob

[Xtreme]

All I know is that it must be a loooooooooooooooooog loooooooooooooooog way because no matter how far or how long I drive towards it, I never seem to get there. :wasntme:

[UncleHo]

Somewhere way back in my primary school days,we were told that the horizon across water at a height of 6 feet was 14 miles,also that at WW11 crows nest height it was 40 n/miles.

[Tombie]

Cape vessels anchored up here are 5-7km out.
From the beach at the waterline they are visible to the water itself.

[slug_burner]

Lots of equations around to approximate but that calculator that was linked earlier is as good as any. The horizon at sea is going to be as close to any of these equations. The topography of the ground will be the limiting factor unless out on a flat plane.

d= square root of 17 x h, where d is in km and h in meters is what i was taught, that I remember, in some course somewhere in my educational journey. For the 2 metres it works out to about 5.8 km which is close to the web link answer.

[Bushie]

Relates to a puzzle we used to ask new starters at the airport.

When you can see 3 or 4 aircraft on approach to the runway (particularly at night) which one was the closest?
Invariably they picked the one lowest in the sky, when it generally was the one highest in the sky. The others being higher above the ground but lower on the local horizon.


Martyn

[Tank]

Standing on the beach with the water lapping at your feet, a person 6' tall can see 3 miles, regards Frank.

[stallie]

Agree with Hoges,

I always used 1.17*sqrt of your height in feet above sea level to give the answer in nm (used it mainly when flying).

Never considered it from sea level, but working it out..

1.17*sqrt6=2.866nm
*1.852nm = 5.3 km.

So a smidge over 5km for a 6 foot bloke on the waterline.

edit - Tanks and I were posting at the same time and looks like we agree.... :BigThumb:

[tangus89]

The earth drops away at a rate

h = d^2 / 2R

d = distance from origin
R = radius of earth ~6375 km (near enough for the bush) depends on topography and bearing you are looking at.

so

d h
100 m 1 mm
500 m 20 mm
1 km 80 mm
5 km 2 m
50 km 200 m

this agrees with what was said before that you can see ~5km with your eyes 2m above sea level.

[Saitch]

Isn't it true that you can see "over" the horizon for a short distance as light rays are affected by the Earths gravity & therefore bend? Probably different on "Discworld"

[Bushie]

Isn't it true that you can see "over" the horizon for a short distance as light rays are affected by the Earths gravity & therefore bend? Probably different on "Discworld"

Not affected by gravity, but affected by refraction through the atmosphere. If I remember right when the apparent sun is just about set the actual sun is already below the horizon, and only visible due to the refraction.
When I was surveying we had to correct observations to stars for refraction, and the amount varies with temp and pressure (these affect atmospheric density)


Martyn

[Saitch]

Martyn
I could be horribly wrong here but I was of the opinion & training that light is affected by gravity. I'm in the surveying proffesion & in the 70's when using G8 Geodimeters, which could measure 80 + kilometres, I'm sure there was a formula to calculate curvature of the LASER caused by gravity. Surely light must be affected by gravity otherwise there'd be no black holes.

[stallie]

Martyn and Saitch,

You are both right. Both atmospheric refraction and gravity can influence the final path of light.

Atmospheric refraction as the light travels though different air densities can cause it to bend - hold a pencil in a glass of water and you'll see what I mean. Similarly, spear fishermen standing in the shallows never aim at the fish, they alway aim slightly short of the fish.

In Antarctica I've watched mountains hundreds of miles away that are not possible to see due to the curvature of the earth appear and disappear as I've been standing on the ground. This is due to light being refracted in layers of cold air near the surface. (I've also seen Fisher Massif (a large blocky mountain at the end of the Lambert Glacier, Antarctica) reflected upside down in the sky due to total internal reflection but I digress...)

As for gravity though, light will not not be bent by gravity per se. According to Einstein's theory, space is "bent" by gravity, thus anything travelling in a straight line (light) will appear "bend" relative to a distant observer (but not to the object moving) as it travels through the bent space.

This theory was proved early last century when during a total solar eclipse, light from a star which should have been obscured by the sun's disc (when measured by the apparent distance to other nearby stars) was visible just beside the sun. The light, travelling in a straight line through space bent by the sun appears to be moved.
https://www.aulro.com/afvb/images/imported/2012/12/165.jpg

More recently, Hubble took a fabulous image of the same effect on a massive scale - light from a quasar and a galazy behind a massive black hole object (black holes have gravity so strong it sucks space in on itself) bending space so that the light heading out in slightly different directions is focussed by the "bent" space producing five separate images of them are apparent. Mind blowing stuff.

NASA - Hubble Captures A "Five-Star" Rated Gravitational Lens (http://www.nasa.gov/multimedia/imagegallery/image_feature_575.html)

[JDNSW]

Yes, light is affected by gravity as in the prvious post. But the effect is too small to be observed on earth and would not be a factor in surveying, except possibly for some star shots involving light rays passing close to the sun.

John

[stallie]

Good point JD, I am wondering whether even with the best most accurate laser you could still couldn't measure it as both the emitter and the sender are in the same patch of "bent" space-time - i.e. in my diagram above imagine you were on the surface of the sun surveying with sub pico-mm accuracy. the laser beam would still be the same distance from the surface.

I've just read that there was to be a satellite launched to measure the effect from the earth as there's a theory called frame dragging, where as the body rotates, it pulls the curved space-time with it.

My brain hurts.

[bob10]

With all of our equations, theories, and navigational extracts, I defy anyone to swim to it. Or sail to it, for that matter, Bob:)

[Bushie]

Yes, light is affected by gravity as in the prvious post. But the effect is too small to be observed on earth and would not be a factor in surveying, except possibly for some star shots involving light rays passing close to the sun.

John

Not much of a problem, most stars are too hard to see during the day. :D:D


Martyn

[Sparksdisco]

With all of our equations, theories, and navigational extracts, I defy anyone to swim to it. Or sail to it, for that matter, Bob:)


Quite right,
By the time i swam to the horizion i think there would not be an earth left;)

[CraigE]

The main factor is m above true sea level. Standing at the waters edge is not necessarily true sea level. Dose get very confusing unless you know exactly how far above sea level you are.
For searching we work on a few generic distances.
Inland more than 60kms, usually you are over 100m above sea level, so provided you can see the horizon to start with we work on 20 mile.
If on a hill, you can see significantly more.
In a plane this increases vastly.
However for us the horizon is only a reference point and does not have a lot of values apart from looking for signals such as smoke or mirror refraction.
At the beach standing at the high water mark we generally work at a height above sea level of around 5 metres. Then things like swell also come into play as well as clarity, sea mist, cloud, haze, glare etc.
Maximum visible distance to the horizon is calculated on an ideal day.

[slug_burner]

The main factor is m above true sea level. Standing at the waters edge is not necessarily true sea level. Dose get very confusing unless you know exactly how far above sea level you are.
For searching we work on a few generic distances.
Inland more than 60kms, usually you are over 100m above sea level, so provided you can see the horizon to start with we work on 20 mile.
If on a hill, you can see significantly more.
In a plane this increases vastly.
However for us the horizon is only a reference point and does not have a lot of values apart from looking for signals such as smoke or mirror refraction.
At the beach standing at the high water mark we generally work at a height above sea level of around 5 metres. Then things like swell also come into play as well as clarity, sea mist, cloud, haze, glare etc.
Maximum visible distance to the horizon is calculated on an ideal day.

Craig are you saying that on Lake Eyre at approximately 15 m (49 ft) below sea level the horizon is at a different distance to that at Lake Titicaca with a surface elevation of 3,812 m?

[Saitch]

Oooooooh!!! My brain hurts.:confused::confused::confused::confused:

[mike_ie]

Craig are you saying that on Lake Eyre at approximately 15 m (49 ft) below sea level the horizon is at a different distance to that at Lake Titicaca with a surface elevation of 3,812 m?

I would imagine it's not that straightforward. Distance to horizon equations are all based on what's known as the Tangent Secant Theorem, and therefore count on the horizon being at sea level. Both lakes being inland lakes, far from the sea, the distance to the horizon would be completely dependant on the terrain surrounding the lakes.

[Sparksdisco]

what about if your standing on the edge of a black hole looking in how far to the horizin then;):p

[stallie]

You would be stretched so much by the warped space time that the electrical impulses traveling along your optic nerve would never reach your brain....

[slug_burner]

Craig are you saying that on Lake Eyre at approximately 15 m (49 ft) below sea level the horizon is at a different distance to that at Lake Titicaca with a surface elevation of 3,812 m?


Distance to horizon equations are all based on what's known as the Tangent Secant Theorem, and therefore count on the horizon being at sea level.

So given that at best we are approximating the earth with a sphere of approximate diameter of 12756.2 km the Tangent Secant Theorem gives us a horizon for an observation point of 2m above the surface of Kati Thanda – Lake Eyre of 5050.977m and on Lake Titicaca 5051.735. For the 0.75m difference in 5km it is hardly worth the effort of doing the calculations, the error in the approximation of the earth to a sphere probably accounts for any error introduced by the altitude of the surface for which you are working out the distance to the horizon. But according to the theorem yes there is a difference:o

[CraigE]

So given that at best we are approximating the earth with a sphere of approximate diameter of 12756.2 km the Tangent Secant Theorem gives us a horizon for an observation point of 2m above the surface of Kati Thanda – Lake Eyre of 5050.977m and on Lake Titicaca 5051.735. For the 0.75m difference in 5km it is hardly worth the effort of doing the calculations, the error in the approximation of the earth to a sphere probably accounts for any error introduced by the altitude of the surface for which you are working out the distance to the horizon. But according to the theorem yes there is a difference:o

As has been said the calculations are based on visible horizon, with no interference (mountains, trees, buildings etc etc). Hence why if we ever need to see the horizon you find a high point with unobstructed views. With GPS it is fairly easy as you will have your elevation and some GPS units or weather units will even calculate your distance to horizon. Generally we will only use the horizon for locating incoming aircraft, ships - boats etc. Fairly mute now with the advent of GPS for anything other than spotting.

[mike_ie]

https://www.aulro.com/afvb/images/imported/2012/12/5.jpg

We have that F bisects AC and that CD is added to it.
Apply Square of Sum less Square:

AD⋅DC+FC^2=FD
But FC=FB and so:
AD⋅DC+FB^2=FD^2

Applying Pythagoras's Theorem, FD^2=FB^2+DB^2 therefore:

AD⋅DC+FB^2=FB^2+DB^2
AD⋅DC=DB^2
DB = sqrt (AD.DC)

Make the following substitutions:
d = DB = distance to the horizon
D = AC = diameter of the Earth
h = DC = height of the observer above sea level
D+h = AD = diameter of the Earth plus height of the observer above sea level

d = sqrt (h(D+h))

PERSON STANDING AT LAKE TITICACA:
D = 12756200m
h = 3812m

d = sqrt(48641165744)
d = 220547.42m or 22.5km


PERSON STANDING AT THE SEA SHORE
D = 12756200m
h = 2m

d = sqrt (2(12756200+2)
d = sqrt (25512404)
d = 5050.98m, or 5.05km

...so yes, there's a big difference in what a person standing at Lake Titicaca can see, vs a person standing on the beach.

Below sea level values throw the formula into disarray, due to having to get the square root of a negative value, but it applies to all values above zero.

Of course, this is the theoretical, and assumes that the horizon you are looking at is at sea level. Easiest real world scenario to prove effects of terrain - stand on a beach that has a mountain range behind you. Looking out to sea, the horizon id about 5km away. Looking behind you, the horizon i wherever the top of the mountain is, probably 6-700m away....

[bob10]

https://www.aulro.com/afvb/images/imported/2012/12/5.jpg

We have that F bisects AC and that CD is added to it.
Apply Square of Sum less Square:

AD⋅DC+FC^2=FD
But FC=FB and so:
AD⋅DC+FB^2=FD^2

Applying Pythagoras's Theorem, FD^2=FB^2+DB^2 therefore:

AD⋅DC+FB^2=FB^2+DB^2
AD⋅DC=DB^2
DB = sqrt (AD.DC)

Make the following substitutions:
d = DB = distance to the horizon
D = AC = diameter of the Earth
h = DC = height of the observer above sea level
D+h = AD = diameter of the Earth plus height of the observer above sea level

d = sqrt (h(D+h))

PERSON STANDING AT LAKE TITICACA:
D = 12756200m
h = 3812m

d = sqrt(48641165744)
d = 220547.42m or 22.5km


PERSON STANDING AT THE SEA SHORE
D = 12756200m
h = 2m

d = sqrt (2(12756200+2)
d = sqrt (25512404)
d = 5050.98m, or 5.05km


Below sea level values throw the formula into disarray, due to having to get the square root of a negative value, but it applies to all values above zero.

Of course, this is the theoretical, and assumes that the horizon you are looking at is at sea level. Easiest real world scenario to prove effects of terrain - stand on a beach that has a mountain range behind you. Looking out to sea, the horizon id about 5km away. Looking behind you, the horizon i wherever the top of the mountain is, probably 6-700m away....
Simple, really, cant work out how those hopeless Mayans got it all wrong, actually. Bob

[slug_burner]

https://www.aulro.com/afvb/images/imported/2012/12/5.jpg

We have that F bisects AC and that CD is added to it.
Apply Square of Sum less Square:

AD⋅DC+FC^2=FD
But FC=FB and so:
AD⋅DC+FB^2=FD^2

Applying Pythagoras's Theorem, FD^2=FB^2+DB^2 therefore:

AD⋅DC+FB^2=FB^2+DB^2
AD⋅DC=DB^2
DB = sqrt (AD.DC)

Make the following substitutions:
d = DB = distance to the horizon
D = AC = diameter of the Earth
h = DC = height of the observer above sea level
D+h = AD = diameter of the Earth plus height of the observer above sea level

d = sqrt (h(D+h))

PERSON STANDING AT LAKE TITICACA:
D = 12756200m
h = 3812m

d = sqrt(48641165744)
d = 220547.42m or 22.5km


PERSON STANDING AT THE SEA SHORE
D = 12756200m
h = 2m

d = sqrt (2(12756200+2)
d = sqrt (25512404)
d = 5050.98m, or 5.05km

...so yes, there's a big difference in what a person standing at Lake Titicaca can see, vs a person standing on the beach.

Below sea level values throw the formula into disarray, due to having to get the square root of a negative value, but it applies to all values above zero.

Of course, this is the theoretical, and assumes that the horizon you are looking at is at sea level. Easiest real world scenario to prove effects of terrain - stand on a beach that has a mountain range behind you. Looking out to sea, the horizon id about 5km away. Looking behind you, the horizon i wherever the top of the mountain is, probably 6-700m away....

Mike,

I think you have made an error. On a plane or lake some distance above or below sea level both point B and C are raised or lowered by the altitude (3812m for Titicaca and -15 for Kati Thanda – Lake Eyre and DC is only 2m (not 3812+2 or -15+2, otherwise the water in the lake would all run down one end real quick).

Above or below sea level does not matter. You have to calculate a new value of AC allowing for the altitude. Just use DB^2=AD.DC.

I think what you calculated when you got 22.5km to the horizon is the equivalent of the horizon for when you are standing on a 3812m mountain at the beach looking out over the sea.

Hope I am not keeping you from getting your defender going again!

[mike_ie]

Above or below sea level does not matter. You have to calculate a new value of AC allowing for the altitude. Just use DB^2=AD.DC.

No. The equation is based on a line tangental to a circle, i.e. the Tangent Secant Theorem, and won't work for negative values. Place the point D somewhere between C and F and you will see what I mean when you try to draw a line to the horizon point.


I think what you calculated when you got 22.5km to the horizon is the equivalent of the horizon for when you are standing on a 3812m mountain at the beach looking out over the sea.

Exactly right. All distance to horizon equations HAVE to work on the premise that the horizon is at sea level. You can try to factor in the height above sea level that the horizon line will be at, and add that to the radius of the earth in your equation, and simultaneously deducting it from your observers point above sea level, i.e.

d = sqrt (h-x((D+x)+h))
d = sqrt (h-x(D+x+h))

where x is the height of the horizon line above sea level.

however for an accurate answer that means you need to know the height above sea level of your horizon line to factor it into the equation, i.e. you need to know the answer before you can calculate the answer :) Which obviously, no equation can do. That's why these equations are only really used to describe a standing-in-a-lighthouse-staring-out-to-sea type problem.


Hope I am not keeping you from getting your defender going again!

Nope. Still waiting for the head gasket. Till then, plenty of time for equations :)

[slug_burner]

No. The equation is based on a line tangental to a circle, i.e. the Tangent Secant Theorem, and won't work for negative values. Place the point D somewhere between C and F and you will see what I mean when you try to draw a line to the horizon point. ...........

That is why you need to calculate a new diameter to work with, that way both C and B stay on the circle and you can still work with a tangent.


.......

however for an accurate answer that means you need to know the height above sea level of your horizon line to factor it into the equation, i.e. you need to know the answer before you can calculate the answer :) Which obviously, no equation can do. That's why these equations are only really used to describe a standing-in-a-lighthouse-staring-out-to-sea type problem.



...........

For a lake it is simple to determine the altitude of any other point on the lake.

What I was trying to point out was that a couple of thousand m in 12576km makes very little difference. The approximation of the distance to the horizon at sea level is more or less the same distance over any body of water irrespective of altitude, particularly when you start with the assumption that the earth is a sphere.

[mike_ie]

That is why you need to calculate a new diameter to work with, that way both C and B stay on the circle and you can still work with a tangent.



For a lake it is simple to determine the altitude of any other point on the lake.

What I was trying to point out was that a couple of thousand m in 12576km makes very little difference. The approximation of the distance to the horizon at sea level is more or less the same distance over any body of water irrespective of altitude, particularly when you start with the assumption that the earth is a sphere.


ASSUMING.... that the lake you're looking out over is longer in that direction than the horizon point is far, then yes :) Which is a pretty safe assumption if you're looking out over the lake and all you can see is water meeting sky.

PERSON STANDING AT THE SEA SHORE

d = sqrt (h(D+h))
D = 12756200m (avg diameter of earth)
h = 2m (observer height ASL)

d = sqrt (2(12756200+2)
d = sqrt (2(12756202))
d = sqrt(25512404)
d = 5050.98m or 5.05km



PERSON STANDING AT LAKE TITICACA:

d = sqrt (h-x(D+x+h))
D = 12756200m (avg diameter of earth)
h = 3814m (observer height ASL)
x = 3812m (lake height ASL)

d = sqrt (3814-3812(12756200+3812+2)
d = sqrt (2(12760014)
d = sqrt (25520028)
d = 5051.73m or 5.05km

Difference is negligible....

[Rohan]

Wow, I've learned a few valuable things here:

1, the maths teaching system (smile), that we had in the UK in the late 80s was woefully inadequate!

3, my brain hurts.

Thanks anyway, made interesting reading :)

Rohan.

[Hay Ewe]

interesting, will go back and read all this again when I have some time in a while