How far can i see?

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.

"Jay@HK" wrote in message
oups.com...
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.

Google: Horizon distance

"Jay@HK" wrote in message oups.com...
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.

I can see the sun, that's 93,000,000 miles away.
I can even see stars that are much, much further.

As far as looking out to sea goes, the Earth is curved and so it depends
on how tall you are and the height of the object you want to look at. With a
telescope you might see the top of the mast of a ship but not see the hull,
so think of it as a curved hill between you and the object. If we use the sun
as the object and watch it set out to sea, and knowing the radius of the
Earth is R = 3963.19 miles, let your height be h, then you'll have a right-angle triangle
of base R and hypotenuse R+h. You'll see the sun set at a distance which is the third
side of the triangle. Now all you need do is calculate x in this diagram, I've put in a plane
for your mountain because I don't do homework questions. :-)
http://www.androcles01.pwp.blueyonder.co.uk/sunset.GIF

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))

or:
acos=reversed cosine,

angle=acos(r/(r+a))
distance=(r+a)*sine(angle)

For a eyeheight of 1.80 m: 4791 m
For 10 km: 357 km

Dont shoot me is your figures are different.

In article ,
Sjouke Burry 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 squa

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?


or:
acos=reversed cosine,

angle=acos(r/(r+a))
distance=(r+a)*sine(angle)

For a eyeheight of 1.80 m: 4791 m
For 10 km: 357 km

Dont shoot me is your figures are different.

sqrt( 2 * 6378000 * 1.80 ) = 4791.74 (rounds to 4792, not 4791 :-)
sqrt( 2 * 6378000 * 10000 ) = 357155

--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stockholm dot bostream dot se
WWW: http://stjarnhimlen.se/

"Paul Schlyter" wrote in message ...
In article ,
Sjouke Burry 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 squa

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/ba...ing_mirage.gif
which knocks all your theories into a cocked hat.
Pick a clear day to visit.

On Fri, 27 Apr 2007 15:20:01 GMT, "Androcles"
wrote:


"Paul Schlyter" wrote in message ...
In article ,
Sjouke Burry 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 squa

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/ba...ing_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

On Fri, 27 Apr 2007 15:49:36 -0400, John C. Polasek
wrote:

On Fri, 27 Apr 2007 15:20:01 GMT, "Androcles"
wrote:


"Paul Schlyter" wrote in message ...
In article ,
Sjouke Burry 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 squa

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/ba...ing_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

Oops I slipped up a bit. It's 1.23 sqrt(h ft) not .866.
Use a for angle and use 1st term of series

R + h = R sec a = R(1 + a^2/2) = R(1 +h/R)
so a^2/2 = h/R
= .5*D^2/R^2 where a = D/R
D = sqrt(2R*h)
Dmiles = 1.225sqrt(h ft)
6 ft gives 3 miles
1 foot gives 1.225 miles
532 meter mountain gives 82,340 meters. 51 miles
John Polasek

"John C. Polasek" wrote in message ...
On Fri, 27 Apr 2007 15:20:01 GMT, "Androcles"
wrote:


"Paul Schlyter" wrote in message ...
In article ,
Sjouke Burry 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 squa

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/ba...ing_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

You only need one triangle, John.
http://www.androcles01.pwp.blueyonder.co.uk/sunset.GIF

Hmm... let me check.

x = R sin(angle).
Radius of Earth R ~= 3963.19 statute miles = 20,925,643.2 feet
Height of eye above sea level ~= 5.8 feet

cos(angle) = R/(R+h)
R h angle R.sin(angle)

20925643.2 5.8 0.000744542574228069 15580.03 feet
(2.950763 miles)

20925643.2 6.0 0.000757270716591041 15846.38 feet
(3.001207 miles)

20925643.2 300.0 0.005354681242856430 112049.6 feet
21.22152 miles


Ok, Excel is better at arithmetic than I. :-)
I had it written down inverted, it should be 1/2 mile per foot.
Sloppy of me, deduct a brownie point.

So now I calculate the cliffs at Dover have to be about 300 ft above sea level
to see across the Channel to France.

Lemme check again...
Well, waddaya know, they rise to 390 ft at the roundabout at the top,
according to Google Earth.
Lat 51.144309°
Long 1.331578°

You cannot see France from a ferry in the harbour, you can from the cliff top.
And the amazing thing is, I've done it. Along the coast a little way is the Battle
of Britain War Memorial:
Lat 51.098495°
Long 1.206357°
Google Earth reckons that is 21 feet ASL, Google Earth is wrong.
What does is interpolate between nearby known points, but there is a
ruddy great cliff you can fall off. Still, give them a brownie point for
trying.

Normal visibility is about 12 miles (limited by haze) unless you live in Los Angeles,
where you have to be lucky to see more than 2 miles.

"John C. Polasek" wrote in message ...
On Fri, 27 Apr 2007 15:49:36 -0400, John C. Polasek
wrote:

On Fri, 27 Apr 2007 15:20:01 GMT, "Androcles"
wrote:


"Paul Schlyter" wrote in message ...
In article ,
Sjouke Burry 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 squa

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/ba...ing_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

Oops I slipped up a bit. It's 1.23 sqrt(h ft) not .866.
Use a for angle and use 1st term of series

R + h = R sec a = R(1 + a^2/2) = R(1 +h/R)
so a^2/2 = h/R
= .5*D^2/R^2 where a = D/R
D = sqrt(2R*h)
Dmiles = 1.225sqrt(h ft)
6 ft gives 3 miles

Agreed. About 1/2 mile per foot.



1 foot gives 1.225 miles
532 meter mountain gives 82,340 meters. 51 miles


That's pushing past the atmospheric haze envelope.
You will not see another mountain at 50 miles, you'll have to
go higher. Without atmospheric dust to scatter sunlight
the sky would not be blue, it would be black, and stars would
be visible all day.
Now, assuming that haze is not a problem but cloud is....
Nah, forget it... not worth the trouble conjecturing when the
reality is that haze DOES limit visual range. Go to radar.
You can see stars all day long in infrared.

An obituary in the 1906 March issue of National Geographic magazine credits Langley as the "first astronomer to succeed in making money for the public out of his profession." To help fund his research, he found a way that the Allegheny Observatory could generate revenue, as the Western University of Pennsylvania(now the University of Pittsburgh) was not well-endowed at this time. Langley's revenue generator also helped place the United States, and then the world, into standard time zones!

http://johnbrashear.tripod.com/

And then....

http://www.aluminumstudios.com/photo...olor/index.htm



Well, this is sci.astro...
Pity we have a ****wit like Charles Francis Ph.D. (Cambridge) acting as
censor to sci.astro.research and doesn't like being called a ****wit.
The arrogant ******* rejects anything he doesn't personally like.