Re: How far can i see?

On Fri, 27 Apr 2007 15:20:01 GMT, "Androcles"
<Engineer@xxxxxxxxxxxxxxxxxxxxxx> wrote:


"Paul Schlyter" <pausch@xxxxxxx> wrote in message news:f0pg5b$s3k$1@xxxxxxxxxxxxxxxxx
In article <46300a03$0$2026$ba620dc5@xxxxxxxxxxxxxxxxxxx>,
Sjouke Burry <burrynulnulfour@xxxxxxxxxxxxxxxxxx> wrote:

Jay@HK wrote:
If I stand at the coast assume it is sea level, how far if the end of
the sea I can reach?
Further, If I stand at a 532m mountain, How far can I see?
Please let me know the quation, Thanks a lot.

for a=altitude(in meters)
and r=earth radius(in meters 6378000):

Distance=sqrt((a+r)*(a+r) -(r*r))

Why not expand and simplify that expression a bit?

Distance = sqrt((a+r)*(a+r) -(r*r))

Expand the square:

Distance = sqrt( a*a + r*r + 2*r*a - r*r )

Now, r*R - r*r will cancel out, giving you:

Distance = sqrt( a*a + 2*r*a )

Factor out a:

Distance = sqrt( a*(a + 2*r) )

Finally, close to the Earth's surface a << r and therefore
we can neglect a in one place, giving us:

Distance = sqrt( 2*r*a )

That's a bit simpler, isn't it?

2 miles horizontal per foot vertical is close enough.
Five foot high, 10 miles to the horizon.
Six foot high, 12 miles.

Greater than 13 feet above sea level and you can see the coast of France
from the famous white cliffs of Dover, except for the haze
http://www.xes.cx/pics/haze-4.JPG
and the mirage effect
http://nsidc.org/arcticmet/images/basics/phenomena/towering_mirage.gif
which knocks all your theories into a cocked hat.
Pick a clear day to visit.

Draw a couple of triangles and you'll get D/R = h/D or D = sqrt(h*R)
In short, miles per foot is given by
D = .866*sqrt(ht in ft)
which can be further simplified since .866 = sqrt(3)/2
D = sqrt(3*h)/2

For 6 feet, D = sqrt(18)/2 = 2.121 miles
John Polasek
.