Aces High Bulletin Board
General Forums => Aces High General Discussion => Topic started by: GlassJaw on April 20, 2008, 04:46:08 PM
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In clear weather how far out would you expect to be able to see a carrier from the air? Isn't there also very long trail behind a large ship as it travels that remains visible for a while?
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Not a Navy guy, but how far you can see depends on your altitude.
I have been told that if you are on the deck, the horizon is 19 miles.
As for the second part of your question, in the book "Wings Of Gold", there is a good description of a Navy pilot who tracked a fleet by following an oil slick.
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On flat ground a man on foot sees the horizon at 2.2 miles away.
Man on horseback at 4.4 miles.
From tractor cab on a clear day you can see tree's or buildings 6 - 8 miles easy.
So yes I can believe that around 500 feet you should be able to spot a task group at 15 or so miles.
Depending of course on wave and weather conditions.
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From my old radar days in the Navy; the distance to the horizon is the square root of 2 times the hight in feet and the answer is in miles. This is the theoretical distance, and works if the earth was a perfect sphere (which it's not) and does not take into account the tidal changes to sea level. However it's a VERY good estimate. As an ET we had to figure out all kinds of fun and exciting formulas.
For example at 18 feet the horizon is 6 miles away, 2x18 = 36 and the square root of 36 is 6.
13 Years later and I still can't get this stuff out of head! :rofl
:salute
Baumer
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From my old radar days in the Navy; the distance to the horizon is the square root of 2 times the hight in feet and the answer is in miles. This is the theoretical distance, and works if the earth was a perfect sphere (which it's not) and does not take into account the tidal changes to sea level. However it's a VERY good estimate. As an ET we had to figure out all kinds of fun and exciting formulas.
For example at 18 feet the horizon is 6 miles away, 2x18 = 36 and the square root of 36 is 6.
13 Years later and I still can't get this stuff out of head! :rofl
:salute
Baumer
Cool formula.
Thanks
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I read a book about the USS Midway and a pilot in there said a CV looked the same size as a dollar bill at yer feet from 10k.
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On a flight to Orlando, FL last year I just happened to be looking out my window and spotted a carrier and support group heading towards Jacksonville. We were already descending but I estimate we were still above 20,000 ft and the distance to the carrier at 12-15 miles. Very visible, very long wake.
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From tractor cab on a clear day you can see tree's or buildings 6 - 8 miles easy.
What adjustments have to be made to the formula if the tractor doesn`t have a cab?
:devil
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What adjustments have to be made to the formula if the tractor doesn`t have a cab?
:devil
dont forget such things as.. proper inflation of tires.. tire wear.. and soil density. oh and type of tractor..
(http://www.tractorjackets.co.uk/images/TractorJack2.jpg)
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My dad was in the Merchant Marines after the Infintry and he always told me visibility was 9 miles at sea. I can't vouch for the accuracy of that. I just took his word on it.
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dont forget such things as.. proper inflation of tires.. tire wear.. and soil density. oh and type of tractor..
(http://www.tractorjackets.co.uk/images/TractorJack2.jpg)
Oh it just keeps getting deeper and deeper.
Your picture reminded me that the height of the driver would also have to be factored in.
Now...where did I put my slide rule?
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You can see the wakes of ships from orbit. Seriousy.
Ok, so they have to be making way, but the bow wave is a whole lot bigger than you might think. A CV making turns is hard to hide if you're looking down.
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From my old radar days in the Navy; the distance to the horizon is the square root of 2 times the hight in feet and the answer is in miles. This is the theoretical distance, and works if the earth was a perfect sphere (which it's not) and does not take into account the tidal changes to sea level. However it's a VERY good estimate. As an ET we had to figure out all kinds of fun and exciting formulas.
For example at 18 feet the horizon is 6 miles away, 2x18 = 36 and the square root of 36 is 6.
13 Years later and I still can't get this stuff out of head! :rofl
:salute
Baumer
You also have to remember the ocean has peaks and valleys on the surface. I learned electrons before I learned neutrons.
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No military experience to call on, but I remember reading a lot about the effect of phosphorescence (in the tropics at least). Because a ship's wake disturbs millions of plankton, they end up giving off a self generated greenish glow for as long as an hour. PT captains around Guad worried about leaving a trail of crumbs that would make them eminently trackable.
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From my old radar days in the Navy; the distance to the horizon is the square root of 2 times the hight in feet and the answer is in miles. This is the theoretical distance, and works if the earth was a perfect sphere (which it's not) and does not take into account the tidal changes to sea level. However it's a VERY good estimate. As an ET we had to figure out all kinds of fun and exciting formulas.
For example at 18 feet the horizon is 6 miles away, 2x18 = 36 and the square root of 36 is 6.
13 Years later and I still can't get this stuff out of head! :rofl
:salute
Baumer
do you ever wake up and wonder where the guy who you were supposed to relieve for the 04-08 watch is and why he didn't wake you?
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do you ever wake up and wonder where the guy who you were supposed to relieve for the 04-08 watch is and why he didn't wake you?
No, but I was the 2-8 SRO on 6 and 6s and had to do maintenance from 8 to 2. Woke up at 0630, EDO, EDPO had both signed my logs and someone had entered the 0600 data (I suspect either the EDO or the SEO). Never found out, though.
Once in a while everything works out.