Do you know how to tell the distance from boat you are sailing on to the horizon or shore with out GPS or other devises? I heard ones that with good visibility the horizon is in about 7 miles. Is that right?
-- Adam Bartos
Nacra 5.0
SolCat 18 (sold)
Lake Zurich, IL --
Depends on how far above sea level your eyes are.
If there are mountains on shore you can see them from 30 miles out.
If you are looking for your bright orange floating VHF radio from your catamaran you can see about 200 yards in calm water.
Wikipedia:
Ignoring the effect of atmospheric refraction, distance to the horizon from an observer close to the Earth's surface is about d≈3.57√h
where d is in kilometres and h is height above ground level in metres.
Examples:
For an observer standing on the ground with h = 1.70 metres (5 ft 7 in) (average eye-level height), the horizon is at a distance of 4.7 kilometres (2.9 mi).
For an observer standing on the ground with h = 2 metres (6 ft 7 in), the horizon is at a distance of 5 kilometres (3.1 mi).
Clear as mud?
-- Sheet In!
Bob
_/)_____/)_/)____/)____/)_____/)/)__________/)__
Prindle 18-2 #244 "Wakizashi"
Prindle 16 #3690 "Pegasus" Sold (sigh)
AZ Multihull Fleet 42 member
(Way) Past Commodore of Prindle Fleet 14
Arizona, USA --
The formula was simply for a straight-line distance to the "horizon" from eyes a certain height above the ground. The horizon is the ground/water separation of the earth and sky. Buildings and bridges stick up above the horizon and are of course visible from much further. Just like if,instead of being 6' 7 and seeing 3.1 miles I was 67 feet tall I could see much further, works on either end.
Some people might refer to a "true horizon" as above vs a "visible horizon" which would be what would you would see if the true horizon was obscured by a solid line of trees or buildings. That visible horizon might be way further than the the true horizon.
I've always found bridges and sets of tall condos very misleading on the water as far as judging distance, they look so "close" but can really be very far.
I realize most of you aren't still in school like I am, so let me give you a little mathematical recap.
Pythagorean theorem a formula used to find the length of any side in a right triangle when you already know 2 sides.
a^2 + b^2 = c^2
a and b are legs while c is the hypotenuse.
A in this horizon case becomes a constant. 20,925,524.9 ft.
now C is going to be the hypotenuse where we stand so C = A+X. X being the height of the vantage point. once we have 2 of the three we solve for B the distance you can see.
EXAMPLE:
find distance man whose eyes are six feet above surface can view to horizon.
Love math nerds, the original beachcats mail list back in the 1990's was heavily composed of .edu and NASA email addresses and the inanely complicated theoretical mathematics discussions that ensued were hilarious.
I wish I could find a copy of the whitepaper done back then titled "The Butt Cheek Differential".
-- Sheet In!
Bob
_/)_____/)_/)____/)____/)_____/)/)__________/)__
Prindle 18-2 #244 "Wakizashi"
Prindle 16 #3690 "Pegasus" Sold (sigh)
AZ Multihull Fleet 42 member
(Way) Past Commodore of Prindle Fleet 14
Arizona, USA --
Yes I guess judging the distance to an object on the water would have a direct relation to how long you have been sailing and therefore depend on how many drinks you have had. This would also depend on how much wind there was as there is direct relationship to the number of drinks had to how much wind there is. Where the distance is relatively equal to the inverse of the wind times the number of drinks divided by pi.
-- Dustin Finlinson • Magna, UT
Member: Utah Sailing Association 1982Prindle 18 1986 Hobie 17
1982 Prindle 16
1980 Prindle 16(mostly)
1976 Prindle 16(mostly)
Check out "Prindle Sailors" on Facebook. --
-- ~ Vietnam Vet 69-71~ 17 Hobie w/big jib, ~18 Hobie mag,~DN Ice sailor,
and other toys.......
~~ I live in NY state on the north shore of Oneida lake in
Bernhards Bay. ~~~~~~ --