Cool physics/navigation question.

[Beady]

Thanks! Am trying to work through it. Um, where do you think you screwed it up?

D in miles = 1.221 X (sq rt of H in feet).

Therefore, assuming that you can stand on the water and that your eyes are therefore 6 feet off the water:

D = 1.221 X (sq rt of 6) = 1.221 X 2.45 = 2.99 ~ 3 miles.

The furthest spot you can see is ~3 miles away. Likewise, another person an additional ~3 miles away, or six miles from you, could see the same spot.

Now, say you and your friend are 2,000 miles apart, and you want to find the minimum altitude you both must be at to see a common point, do this:

1000 miles (remember, there are two of you) = 1.221 X (sq rt of "X")

1000/1.221 = sq rt of "X"

sq rt of (1000/1.221) = X squared

sq rt of 819 = X squared

28.6 miles = X squared

sq rt of 28.6 = X

5.35 miles altitude = X


I had a feeling something was wrong when I found myself finding a square root twice.

1.221 gives statute miles; I forget the number for km.

I'm quite sure my formula is right. It is also corroborated by Gribble's. Beady said his formula came from the Coast Guard. I suspect it's a simple approximation that works on small distances.

Actually, it comes from "The Mariner's Pocket Companion." It was an annual published by the Naval Institute Press, but seems now to be defunct. Prior years' copies appear to be available through Amazon.

BTW, my "squiggles" are the result of rounding; some of those decimals can get pretty outrageous.

 

[dogjones]

D in miles = 1.221 X (sq rt of H in feet).

Therefore, assuming that you can stand on the water and that your eyes are therefore 6 feet off the water:

D = 1.221 X (sq rt of 6) = 1.221 X 2.45 = 2.99 ~ 3 miles.

The furthest spot you can see is ~3 miles away. Likewise, another person an additional ~3 miles away, or six miles from you, could see the same spot.

Now, say you and your friend are 2,000 miles apart, and you want to find the minimum altitude you both must be at to see a common point, do this:

1000 miles (remember, there are two of you) = 1.221 X (sq rt of "X")

1000/1.221 = sq rt of "X"

sq rt of (1000/1.221) = X squared

sq rt of 819 = X squared

28.6 miles = X squared

sq rt of 28.6 = X

5.35 miles altitude = X


I had a feeling something was wrong when I found myself finding a square root twice.

1.221 gives statute miles; I forget the number for km.



Actually, it comes from "The Mariner's Pocket Companion." It was an annual published by the Naval Institute Press, but seems now to be defunct. Prior years' copies appear to be available through Amazon.

BTW, my "squiggles" are the result of rounding; some of those decimals can get pretty outrageous.

Still doesn't seem to quite work. According to Gribble's graph/calculator http://newton.ex.ac.uk/research/qsystems/people/sque/physics/horizon/ it looks like the horizon would be at 1000k at about 79k up (about 49 miles).

Still seems a pretty cool cheat though. When I was young I remember being told that the horizon is always about 7 miles away. Which your formula corroborates somewhat - most of the coastal cliffs in Bermuda are about 36 feet!

 

[dogjones]

Stellarium is a fantastic program....and you will really like the price.

Marvellous. Thanks!

 

[Beady]

I just cheated and went to Wikipedia. They have a slightly shorter formula than mine:

(can't seem to copy it directly. It's "d = sq rt 1.5h")



So, running both my scenarios through this:

1) d to horizon in miles = sq rt (1.5 * 6)

d=sq rt * 9

d = 3 miles

So far, so good.


2) 1,000 = sq rt 1.5X

1,000 squared = 1.5X

1,000,000 = 1.5X

1 million/1.5 = X

666,666 feet = X

666,666/5,280 = 126.3 miles altitude

OK, found another mistake in working my first version:

1000 miles (remember, there are two of you) = 1.221 X (sq rt of "X")

1000/1.221 = sq rt of "X"

sq rt (1000/1.221) squared = X squared

sq rt of 819670761 = X squared

670761/5280 = 127.04 miles

So, *correctly worked*, my first version agrees with the Wikipedia version.

 

[ddt]

I just cheated and went to Wikipedia. They have a slightly shorter formula than mine:

(can't seem to copy it directly. It's "d = sq rt 1.5h")

Wiki's formulae are pictures . Use LaTeX for proper formulae here, as Ziggurat suggested.

The wiki page on horizon gives your formula, and says it is an approximation. That page gives three different formulae, and it may be instructive to see how they are derived. I attached a small picture of the situation, with:
M: the center of the earth
B: Bermuda
A: a vantage point above Bermuda
C: the point on the horizon (originally the point half-way between Bermuda and Durham, UK)
R: the radius of the earth (6,378 km)

Wiki first gives a crude approximation:
[latex]$$ d \approx \sqrt{13 * h} $$[/latex]
where d is measured in kilometers and h in meters. It says this approximation is good enough for terrestrial circumstances (up to roughly 10 km). Then it gives a finer approximation:
[latex]$$ d = \sqrt{2*R*h + h^2} $$[/latex]
and finally, it gives the exact formula:
[latex]$$ s = R * \arccos \frac{R}{R+h} $$[/latex]

You may notice the change in letter: from d to s. The first formula doesn't actually measure the distance from B to C, but rather the distance from A to C, which is an approximation.

The second formula follows easily from the Theorem of Pythagoras. Look at the triangle MAC, and this gives you:
[latex]$$ (R+h)^2 = R^2 + d^2 $$[/latex]
so, solving for d, you get formula (2); this is an exact calculation of d, under the assumption that the earth is a perfect sphere.

Next, for the first formula, note that the term [latex]$h^2$[/latex] is very small in comparison to [latex]$2*R*h$[/latex], so just get rid of it and write
[latex]$$ d \approx \sqrt{2*R*h}$$[/latex]
Filling in the value of R, rounding and adjusting for h being in meters rather than kilometers gives formula (1).

The third formula is derived by observing that
[latex]$$ s = \phi * R $$[/latex]
and furthermore that
[latex]$$ \cos \phi = \frac{|CM|}{|AM|} = \frac{R}{R+h} $$[/latex]
Plugging the last into the former gives you formula (3).

As you can see from the picture, for small [latex]$\phi$[/latex] and thus small h, the distances d and s are nearly the same. However, for the kind of distances dogjones inquired about, this definitely does not hold anymore and the first two formulae are inadequate.

 

[Soapy Sam]

Starcalc is also good - easy to use, covers all the bases and free.
http://www.m31.spb.ru/StarCalc/

 

[Beady]

The wiki page on horizon gives your formula, and says it is an approximation...However, for the kind of distances dogjones inquired about, this definitely does not hold anymore and the first two formulae are inadequate.

They're good enough for back-of-the-envelope calculations and general curiosity. If I ever launch a high-altitude rocket, I'll be sure to explore the question more completely.

 

[steve s]

Stellarium is a fantastic program....and you will really like the price.

Thanks for the link. That's a neat program. It's depressing though that I had to bump up the Light Pollution setting to about 8 in order to simulate the view I have from my back yard. Oh, the curse of living in suburbia.

Steve S.