Distance to horizon at sea level
Ron Dwelle
distance to horizon in miles = square root of the height of your eye (in
feet) above sea level. In excellent visibility, you might get another 15%.
Cheers,
On 23 Nov 1995, don reinberg wrote:
> Would someone please remind me how far one is able to see towards the
> ocean's horizon standing at a beach at sea level.
don reinberg
ocean's horizon standing at a beach at sea level.
Don
rein...@hooked.net
Captain Ron Jensen
rein...@hooked.net (don reinberg) wrote:
Don,
It all depends upon how tall the individual is. The Geographic Range
Table in the USCG Light List publication starts at 5 [five] feet off the
water and ranges to 1000 feet. At 5 feet, the eye is able to see to the
horizon 2.6 nautical miles [depending upon clarity of the air]. At 10
feet, the eye can see 3.7 nm. Obviously it is possible to view objects
much further if those objects are elevated.
Ron
--
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E-Mail: jen...@chelsea.ios.com Phone:714 / 965-3047
P.Bennett
>Would someone please remind me how far one is able to see towards the
>ocean's horizon standing at a beach at sea level.
>Many thanks.
>Don
>rein...@hooked.net
distance to horizon (in nautical miles) = 1.17 times square root of height of
eye (in feet)
(or maybe 1.14 times...)
Peter Bennett VE7CEI | Vessels shall be deemed to be in sight
Internet: ben...@triumf.ca | of one another only when one can be
Packet: ve7cei@ve7kit.#vanc.bc.ca | observed visually from the other
TRIUMF, Vancouver, B.C., Canada | ColRegs 3(k)
GPS and NMEA info and programs: ftp://sundae.triumf.ca/pub/peter/index.html
P.Bennett
>The horizon at sea level is 11 miles if viewed with good
>vision! Be sure to compensate for your height if standing.
>--
Huh??
The distance to the horizon (nautical miles) = 1.17 times the square root of
the height of eye (in feet). (or 2.1 root of height in metres)
To get a distance to horizon of 11 miles, you need a height of eye of 90 ft!
(according to the esteemed Mr. Bowditch)
P.Bennett
>Gentlemen... What an interesting collection of answers.
Yes, quite...
>From The
>Annapolis Book of Seamanship, page 206: Geographical range (the distance
>to the horizion fron the top of a light house) in nautical miles is 1.144
>times the square root of the height of light or the eye. For 6 feet it
>is 2.8 nm, for 15 ft it is 4.4, you would have to be at 95 feet to get
>11.2 nm.
I think the earth has grown slightly - older references used 1.144, but the
1981 edition of Bowditch uses 1.17...
>The color of the light does make a difference, red and green
>have a range 25% less than white light. Hope this helps. Cheers! PHIL
>
No - the distance to the horizon is the same for all colours, and determines
whether it is _possible_ to see something, given perfect visibility.
The nominal range (the distance at which a light can be seen, assuming nothing
gets in the way) of a coloured light is less than that of a white light, but
that has no effect on the distance to the horizon.
Will Tory
vision! Be sure to compensate for your height if standing.
--
Will Tory Glendale, AZ
Internet =
re...@primenet.com
Phil Lenihan
to the horizion fron the top of a light house) in nautical miles is 1.144
times the square root of the height of light or the eye. For 6 feet it
is 2.8 nm, for 15 ft it is 4.4, you would have to be at 95 feet to get
Lyle Lichty
> In article <496er4$b...@ixnews5.ix.netcom.com>, Phil Lenihan <plen...@ix.netcom.com> writes...
>
There is even more here than meets the eye!
> >From The
> >Annapolis Book of Seamanship, page 206: Geographical range (the distance
> >to the horizion fron the top of a light house) in nautical miles is 1.144
> >times the square root of the height of light or the eye.
>
> I think the earth has grown slightly - older references used 1.144, but the
> 1981 edition of Bowditch uses 1.17...
>
I'm not really a sailor and I do not have a copy of Bowditch, but its numbers
seem a bit off to me. A quick analysis shows that the distance to the horizon in
nautical miles is 1.065 times the square root of eye height in feet. (If you are
interested in the derivation: set up a right triangle with one side as the
earth's radius, one as the distance to the horizon, and one as the earth's radius
plus eye height. Use the pythagorian theorem and the binomial expansion to find
the distance.) The interesting question is why Bowditch has a different
number. Do you know if they derived it or determined it experimentally?
> >The color of the light does make a difference, red and green
> >have a range 25% less than white light. Hope this helps. Cheers! PHIL
> >
>
> No - the distance to the horizon is the same for all colours, and determines
> whether it is _possible_ to see something, given perfect visibility.
Perhaps not. The only reason I can think of for Bowditch's number is that a
difference in air density or humidity between the lighthouse height and sea
level causes the light to refract (bend) down. In such a situation one could see
farther than the straight-line distance. This might conceivably apply to light
from a lighthouse, but I doubt if any difference in density between my eyes and
sea level would have an effect.
Sailing provides some wonderful applications of physics. Here is another neat
one, if indeed the bending of light due to differences in air density really does
allow a lighthouse light to be seen farther than the distance to the horizon.
Lyle Lichty
lic...@cornell-iowa.edu
Jens M Andreasen
On 27 Nov 1995, Lyle Lichty wrote:
> Sailing provides some wonderful applications of physics. Here is another neat
> one, if indeed the bending of light due to differences in air density really does
> allow a lighthouse light to be seen farther than the distance to the horizon.
lighthouse
\ ____ person
\/~~~~\/
/~~~~~~\
Allthough the foot of the lighthouse is behind the horizon, the top isn't
and can be seen at a greater distance, being a function of the hight of the
lighthouse as well as the hight of eye.
// Jens M Andreasen
Ron Dwelle
than the square-root of height (in feet). On rare occasions, such as
mirage on the Great Lakes, it is sometimes much more. From the USCG
Light List and charts, I just calculated half a dozen "visibility distances"
based on stated height, and the Coast Guard seems to use ~1.14 to ~1.17 as a
multiplier (at least I can't find a consistent number). On the other
hand, I don't know what "height" refers to exactly--the center of the
light? the top of the lighthouse?
Cheers
kdi...@xmission.com
Here's a loose description of how distance to horizon works. Maybe it will be
helpful, it certainly isn't rigorous.
The basic formula is
distance to horizon in miles =
k * square root (height above ground in inches)
Obviously there have been a few simplifications here to cause the units to
work out so conveniently, but it's close.
k is a constant that describes the curve of the signal (light, microwave,
radio, etc) compared to the curve of the earth considered as a sphere. If
there were no atmosphere, k would equal 1 and that would be that.
(k is the ratio of the apparent earth radius to the true earth radius)
Because the atmosphere refracts (bends) light and electromagnetic radiation, k
is normally greater than 1 and the horizon is further than simple spherical
geometry would predict. Under certain conditions, k can be less than 1 and the
horizon moves closer.
Working with VHF radio, k is commonly assumed to be 4/3. With visible light
over water, k typically becomes 1.14 to 1.17. Realize, however, that k is a
real measurable quantity that varies continually depending on temperature,
layering, humidity, and other factors.
The bottom line is that the square root of the height of your eyeball above
the water multiplied by 1.14 is an approximation. The actual k value would
need to be known to calculate the true horizon distance.
I do believe I've achieved my goal of putting everyone to sleep. <g>
>Cheers
Keith Diehl -- Salt Lake City
http://www.xmission.com/~kdiehl
kdi...@xmission.com
>The basic formula is
> distance to horizon in miles =
> k * square root (height above ground in inches)
Please excuse my carelessness and substitute feet for inches in my previous
posting. My apologies for the error.
Brian Smith
I actually know the *definitive* answer but I am not letting on ;-)
B.S!
Andy Champ
: Whoopee! A technical flame fest in the making!
: I actually know the *definitive* answer but I am not letting on ;-)
I had a good think and I think one of the existing ones is correct given two
things:
(1) you are actually measuring straight distance from the eye to the horizon,
rather than the ground distance
(2) (much more important) the height of the observer is small compared
to the diameter of the Earth.
Can't swear to the constants though.
Andy
kdi...@xmission.com
HIGH NERD FACTOR AHEAD!!
her...@cats.UCSC.EDU (William R. Ward) wrote:
>You need to adjust for the units. One length is in miles, the other
>in inches. I believe that the constant k of 1.14-1.17 includes some
>compensation for converting units as well as the aforementioned
>radius.
>--Bill.
You're right. Lets do this once mathematically and end the thread.
Picture a circle with two lines originating at it's center. To create a pretty
picture for this example, the angle formed by the lines is about 30 degrees.
One line ends at the circle, the other extends slightly beyond the circle.
What we've drawn is a line from the center of the earth to the surface of the
ocean, and another line from the center of the earth to the height of your eye
above the surface of the ocean.
The length of one line is obviously r, the radius of the earth. The other line
is (r + h), the radius of the earth plus the height of your eye above the
surface of the earth.
Now we pick our approximations here. What value would you like to use for the
radius of the earth? One reference I have says that the equatorial radius is
3963 statute miles ( 3441 nautical miles) and a polar radius of 3950 statute
miles (3430 nautical miles). Let's use the polar radius.
Draw one more line from your eye at the end of (r + h), to the point where r
(the other line) intersects the circle. Call this line D, the distance to the
horizon.
This should form a right triangle with sides of r, r+h, and D.
Using the Pythagorean Theorem
r^2 + D^2 = (r + h)^2
or rearranging for our purposes
D^2 = (r + h)^2 - r^2
squaring (r + h) gives r^2 + 2rh + h^2
thus D^2 = r^2 + 2rh + h^2 - r^2 which simplifies to
D^2 = 2rh + h^2.
Because 2rh is MUCH greater than h^2, we can ignore h^2 and simplify to
D = sqrt(2rh) = sqrt(2 * 3950 mi * h ft * 1/5280)
The last term is added to convert h from feet to miles.
Solving gives sqrt(1.5h) or 1.22*sqrt(h) as the distance in statute miles to
the horizon.
Using nautical miles
D = sqrt(2rh) = sqrt(2 * 3430 nm * h ft * 1/6080)
D = 1.06 * sqrt(h)
These solutions disregard atmospheric refraction and the fact that the world
isn't round. The actual optical horizon will vary depending on the current
refractive index of the atmosphere (big changes) and the radius of the earth
at your location (little changes).
Yawn. Is anyone still awake? Does anyone still care?
Let's go sailing now <g>
William R. Ward
) Ron Dwelle <dwe...@river.it.gvsu.edu> wrote:
) Here's a loose description of how distance to horizon works. Maybe it will be
) helpful, it certainly isn't rigorous.
) The basic formula is
) distance to horizon in miles =
) k * square root (height above ground in inches)
Well feet, actually, but you already corrected that.
) Obviously there have been a few simplifications here to cause the
) units to work out so conveniently, but it's close.
) k is a constant that describes the curve of the signal (light,
) microwave, radio, etc) compared to the curve of the earth considered
) as a sphere. If there were no atmosphere, k would equal 1 and that
) would be that.
) (k is the ratio of the apparent earth radius to the true earth radius)
You need to adjust for the units. One length is in miles, the other
in inches. I believe that the constant k of 1.14-1.17 includes some
compensation for converting units as well as the aforementioned
radius.
--Bill.
--
William R Ward Bay View Consulting +1 408/479-4072
her...@bayview.com 1803 Mission St. #339 +1 408/458-8862 pgr
her...@cats.ucsc.edu Santa Cruz CA 95060 USA
COPYRIGHT(C) 1995 William Ward. Not for distribution via Microsoft Network.
Jens M Andreasen
On 29 Nov 1995, Brian Smith wrote:
> I actually know the *definitive* answer but I am not letting on ;-)
Sorry, I started to read this thread after it had been going for a while,
but: What was it that was wrong with the oldfashioned way of calculating
trigonometry ??? This would involve sinus, and no constants except for
the radius of the earth ...
// Jens M Andreasen
Peter D. Engels
M Andreasen <jens...@dsv.su.se> wrote:
> On 29 Nov 1995, Brian Smith wrote:
> Sorry, I started to read this thread after it had been going for a while,
> but: What was it that was wrong with the oldfashioned way of calculating
> trigonometry ??? This would involve sinus,
With my sinus problems, the horizon's location is always uncertain :-)
--
Pete
Jens M Andreasen
On Mon, 4 Dec 1995, Peter D. Engels wrote:
> With my sinus problems, the horizon's location is always uncertain :-)
No, the horizons location is given and you are looking at it at an excact
90 degree angle. What more is given? Oh yes the radius of earth. Make
that one side of a triangle. The second side is that same radius plus your
height of eye. What we have now is a triangle with two known sides and
one known angle ... That should be enough to calculate your position :-)
- - - +-------------/ <- viewpoint
|~~~ /
|~~~~~~~~~ / <- sealevel
|~~~~~~~~~~/~~
| / ~
| / ~
| / ~
| / ~
| / ~
Or is there something I forgot?
// Jens M Andreasen
Colin and Alida
> nautical miles is 1.065 times the square root of eye height in feet.
> interested in the derivation: set up a right triangle with one side as the
> earth's radius, one as the distance to the horizon, and one as the
earth's radius
> plus eye height. Use the pythagorian theorem and the binomial expansion
to find
> the distance.) The interesting question is why Bowditch has a different
> number.
The formula given for distance of the VISIBLE horizon is 1.17*sq. root
of ht. of eye in feet, this is stated as an approximate measure. The
visible horizon is more distant than the GEOMETRICAL horizon due to
terrestrial refraction. This difference is taken into account for
"normal" conditions, and is tabulated in table 8 of Bowd. V2 (1981 ed.)
The distance of the GEOMETRICAL horizon is what you have derived, and is
stated in Bowditch as 1.06 * sq. root of ht. of eye in feet.
> Perhaps not. The only reason I can think of for Bowditch's number is that a
> difference in air density or humidity between the lighthouse height and sea
> level causes the light to refract (bend) down. In such a situation one
could > see farther than the straight-line distance.
You are so right.
> Sailing provides some wonderful applications of physics.
And mathematics - I never would have guessed that spherical trig would
have been as exiting as figuring the courses and distance to sail a
great-circle voyage, being a merchant seaman first, and mathemetician much
after, but it's always fun to get the answer.
Colin Dewey
Colin and Alida
Dwelle <dwe...@river.it.gvsu.edu> wrote:
> From experience, it is clear that distance to the horizon is often less
> than the square-root of height (in feet). On rare occasions, such as
> mirage on the Great Lakes, it is sometimes much more. From the USCG
> Light List and charts, I just calculated half a dozen "visibility distances"
> based on stated height, and the Coast Guard seems to use ~1.14 to ~1.17 as a
> multiplier (at least I can't find a consistent number). On the other
> hand, I don't know what "height" refers to exactly--the center of the
> light? the top of the lighthouse?
>
> Cheers
The height listed for a light will be the height of the light itself, not
the structure. Charted objects on land, buildings, hills, and such, are
usually (in the U.S.) given in feet above Mean High Water, which is the
average of all high waters over a 19 year period
Colin and Alida
Smith) wrote:
> Whoopee! A technical flame fest in the making!
>
> I actually know the *definitive* answer but I am not letting on ;-)
> B.S!
No one's been called an idiot yet, so it's still discussion (however idle)
not like the "knots-per-hour" mess.
CD
Brian Smith
: Sorry, I started to read this thread after it had been going for a while,
: but: What was it that was wrong with the oldfashioned way of calculating
: trigonometry ??? This would involve sinus, and no constants except for
^^^^^
Right! Not something to be sneezed at (^_^)
B.S!
Motu Iti
We look for something on the horizon - like another boat. So the question
should be; How far away is that boat on the horizon? Now lets assume it's
a power boat with a super structure of 12 feet above the waterline. How
far is it when it drops from sight. Or better yet a sailboat with a mast
clearance of 50 feet and it's anchor light on. How far away is when it
drops from sight?
P.Bennett
Simple - take your distance to horizon, plus the distant object's distance to
horizon, and add them together:
total distance = 1.17 root (my height) + 1.17 root (your height)
(heights in feet - for heights in metres, use 2.07)
Jens M Andreasen
Sorry, sinus was the danish word, should of course have been *sine* and
not be confused with sinusitis.
// Jens M Andreasen
Keith Diehl
>Distance to Horizon: How many of scan the horizon to look at the horizon?
>We look for something on the horizon - like another boat. So the question
>should be; How far away is that boat on the horizon? Now lets assume it's
>a power boat with a super structure of 12 feet above the waterline. How
>far is it when it drops from sight. Or better yet a sailboat with a mast
>clearance of 50 feet and it's anchor light on. How far away is when it
>drops from sight?
Figure your distance to the horizon. Then figure the distance to the horizon
as if you were at the highest point of the hypothetical other boat. Add them
together. That's about where the highest point of the other boat will be lost
from sight, disregarding waves and refraction.
(use your favorite computational method for distance to horizon) <g>