Line of sight

[TheStu]

Not sure how best to categorize this question, or how to phrase it, but here goes.

Assuming that we are on Earth, how would one calculate the greatest distance that someone could see (parallel to the ground, so it looking straight ahead) assuming no topological features in the way (hills, mountains, lakes) or atmospheric interference?

I think I read once that it was 125 miles before the curvature of the earth comes into play.

 

[iCyborg]

Hm, I'm not sure if I'm doing something wrong, but I'm getting a very small number: around 4,7km assuming a human whose eyes are at h=1.70m and Earth radius R=6,400km.
I can't really see any mistakes as it's seems simple:
if L is the length we're looking for, it's really the distance of the tangent to the earth from the eyes. The tangent is orthogonal to the radius at the touching point, so you have a right-angled triangle with sides L, R and hypotenuse (R+h), h is the height above ground, i.e. where the eyes are.
So L^2 = h(2R+h), and you get L=4,67km. I guess you better stand on something...

 

[SonicIce]

i wondered too
http://blogs.discovermagazine.com/badastronomy/2009/01/15/how-far-away-is-the-horizon/

 

[Ninjahedge]

You also have to define what "seeing" is.

Are you looking for an object at ground level, or just teh head and shoulders of another person the same height?

Any real reason for the curiosity on this one, or was it a passing brain hiccough?

 

[TheStu]

You also have to define what "seeing" is.

Are you looking for an object at ground level, or just teh head and shoulders of another person the same height?

Any real reason for the curiosity on this one, or was it a passing brain hiccough?

The scenario is thus;

A young man, we will call him Blark Jent, is on top of a windmill (maybe 12m) and from that vantage point can see the skyline of a city, we will call it Letropolis. They are in Kansas, a flat state, so I was wondering based on that information if a reasonable estimate of the distance between Letropolis and Rallville could be determined.

 

[bobdole369]

I can't be bothered to do you homework, but while I was on a cruise I noticed and photographed this:

http://www.flickr.com/photos/dolespics/2950457800/

http://www.ndbc.noaa.gov/station_page.php?station=SANF1

I was on deck 6 or so - so thats roughly 100 feet up too.

Given the 432mm equiv - or 12x zoom - you can imagine that the structure, which is fairly large, appeared just discernable to the naked eye. We were 3 miles out.

I've been out about 3-5 miles onboard yachts as part of my job, the skyline (condos and office buildings) are visible but not quite the beach till you are roughly 3 miles out. This is at something like 15feet up from the waterline. You get a much better view up in the pilothouse, which can be up over 45ft from the waterline.

 

[Ninjahedge]

We are also ignoring two important things.

Gravity (MINOR bending) and refraction (MAJOR bending).

We are talking about a perfect world here, right?

You would be able to calc it easily given the formula for it, the radius and the heights.

If your radius is, say, 5000' (I know it isn't) and you are up 1000', you look across and see another building at the 500' mark it all works out to simple trig.

The length from you to the tangent point is [(6000')^2-(5000')^2]^½ = 3316.6'
The length from the tangent point it is [(5500')^2-(5000')^2]^½ = 2291.3'

The distance you are viewing is 5607.9'.

Now stop giving us your trig homework!

 

[TheStu]

We are also ignoring two important things.

Gravity (MINOR bending) and refraction (MAJOR bending).

We are talking about a perfect world here, right?

You would be able to calc it easily given the formula for it, the radius and the heights.

If your radius is, say, 5000' (I know it isn't) and you are up 1000', you look across and see another building at the 500' mark it all works out to simple trig.

The length from you to the tangent point is [(6000')^2-(5000')^2]^½ = 3316.6'
The length from the tangent point it is [(5500')^2-(5000')^2]^½ = 2291.3'

The distance you are viewing is 5607.9'.

Now stop giving us your trig homework!

I haven't studied Trig in like 10 years... which I think is part of the problem.

 

[C1]

http://www.calculatoredge.com/electronics/lineofsight.htm
http://www.kagstrom.no/LineOfSight.htm

 

[maddie]

If I may.


A general solution for varying heights of observers, objects and sizes of spheres

Height of observer eyes --- h
Radius of Planet (Earth) --- R
Height of object --- H

Angle between viewer (eyes) and tangent of planet surface is

arccosine (R/(R + h)) = Z degrees

Distance is measured on the surface

2piR x 360/Z = d

Do it on the opposite side of the tangential point of contact for the object of height H

Angle

arccosine (R/(R + H)) = Y degrees

Distance

2piR x 360/Y = D

Total distance to object = d + D

 

[iCyborg]

I haven't studied Trig in like 10 years... which I think is part of the problem.

You don't need trig. All you need is Pythagorean theorem which is something taught in elementary school.

My formula is for seeing the ground. It's easy to generalize: if you're looking from a windmill at h1(=12m), and want to see skyline at h2 (need to specify this), then the max distance is:
sqrt(h1(2R+h1)) + sqrt(h2(2R+h2))
R is Earth radius. It's just two right-angled triangles, each with two sides given, and you want the third. Each one has sides of the form D, R, R+h, so D^2 = (R+h)^2 - R^2 = 2Rh+h^2 = h(2R+h)

 

[TheStu]

Given the following values;
h1 = 12m
h2 = 400m
R = 6400000m

I got 83,948m or just under 84km as the answer.

Thanks for reminding me of pythagoras, I just assumed that the problem would be more difficult due to the curved surface.

 

[Modelworks]

You get into this a lot with microwave communications and point to point links. Many people think they should be getting a perfect signal because there is nothing between point A and B, they forget the earth isn't flat and don't add in the fresnel zones.
http://www.afar.net/fresnel-zone-calculator/

 

[maddie]

You don't need trig. All you need is Pythagorean theorem which is something taught in elementary school.

My formula is for seeing the ground. It's easy to generalize: if you're looking from a windmill at h1(=12m), and want to see skyline at h2 (need to specify this), then the max distance is:
sqrt(h1(2R+h1)) + sqrt(h2(2R+h2))
R is Earth radius. It's just two right-angled triangles, each with two sides given, and you want the third. Each one has sides of the form D, R, R+h, so D^2 = (R+h)^2 - R^2 = 2Rh+h^2 = h(2R+h)

Actually you do need trig.

According to your method, the distance calculated is the straight distance between two points in a 3 dimensional coordinate system. Distances on a sphere, such as a planet, is calculated on the surface.

Do you think the empire state building and the Eiffel tower are further apart at the top than at the base on a map. Your straight Pythagoras method will give different values.

 

[canis]

Actually you do need trig.

According to your method, the distance calculated is the straight distance between two points in a 3 dimensional coordinate system. Distances on a sphere, such as a planet, is calculated on the surface.

Do you think the empire state building and the Eiffel tower are further apart at the top than at the base on a map. Your straight Pythagoras method will give different values.

Read OP's title.

If I may.


A general solution for varying heights of observers, objects and sizes of spheres

Height of observer eyes --- h
Radius of Planet (Earth) --- R
Height of object --- H

Angle between viewer (eyes) and tangent of planet surface is

arccosine (R/(R + h)) = Z degrees

Distance is measured on the surface

2piR x 360/Z = d

Do it on the opposite side of the tangential point of contact for the object of height H

Angle

arccosine (R/(R + H)) = Y degrees

Distance

2piR x 360/Y = D

Total distance to object = d + D

It is trivial to calculate arc length. But the equations you wrote are wrong. It should be Z/360 instead of 360/Z and Y/360 instead of 360/Y.

 

[Net]

great circle distance <= maxDistance

notation for less than or equal to: <=

 

[ZMX]

Net's diagram has it right.

The number you heard about 120 miles is for maximum ideal atmospheric visibility. This means regardless of line of sight and how big an object is, if it's 120 miles away, it isn't visible.

...Unless it's emitting its own light source powerful enough to penetrate 120 miles of atmosphere.

 

[Ninjahedge]

Net, that is what I was doing.

The tangental point is perpendicular to the LOS. You now have two right triangles.

I had to sketch it out too before I did the forehead smack and realized it was that simple....

 

[canis]

..

 

[canis]

..

 

[Pulsar]

A 6 foot person would be able to see around 3 miles.

 

[maddie]

Read OP's title.



It is trivial to calculate arc length. But the equations you wrote are wrong. It should be Z/360 instead of 360/Z and Y/360 instead of 360/Y.

You're correct.