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Contrail Horizon To Horizon? What Distance?  
User currently offlineBa757gla From United Kingdom, joined Mar 2006, 760 posts, RR: 2
Posted (8 years 4 weeks 1 hour ago) and read 2716 times:

when i see contrails how far would be overhead to horizon and horizon to horizon? i dont really understand compilcated answers.

21 replies: All unread, jump to last
 
User currently offlineMrChips From Canada, joined Mar 2005, 927 posts, RR: 0
Reply 1, posted (8 years 3 weeks 6 days 23 hours ago) and read 2693 times:

Assuming you are on featureless ground at sea level and with the aircraft at the ubiquitous cruising altitude of 37,000 feet, a contrail will be about 545 miles long if it stretches from one horizon to the other.

[Edited 2006-06-17 04:12:31]


Time...to un-pimp...ze auto!
User currently offlineBobster2 From , joined Dec 1969, posts, RR:
Reply 2, posted (8 years 3 weeks 6 days 23 hours ago) and read 2680 times:

What is the formula for figuring this out?

I'd like a general equation using the observer's and the airplane's altitudes, and of course the diameter of the earth.

Extra credit: Can you take into account the fact the airplane's apparent position on the horizon is not the true position because of light refraction in the atmosphere?


User currently offlineNewark777 From United States of America, joined Dec 2004, 9348 posts, RR: 30
Reply 3, posted (8 years 3 weeks 6 days 23 hours ago) and read 2678 times:

Quoting Bobster2 (Reply 2):
What is the formula for figuring this out?

I'd like a general equation using the observer's and the airplane's altitudes, and of course the diameter of the earth.

I've done it before, just draw a diagram and start using geometry.

Quoting Bobster2 (Reply 2):

Extra credit: Can you take into account the fact the airplane's apparent position on the horizon is not the true position because of light refraction in the atmosphere?

If you want to get into some hardcore physics.  Wink

Harry



Why grab a Heine when you can grab a Busch?
User currently offlineFlyMatt2Bermud From United States of America, joined Jan 2006, 563 posts, RR: 7
Reply 4, posted (8 years 3 weeks 6 days 23 hours ago) and read 2675 times:

I think the ultimate contrails are the Shuttle going directly into space! Better view from an aircraft. It is a sight most enthusiasts will never forget. Distance (?) to the moon!!


"When once you have tasted flight, you will forever walk the earth with your eyes turned skyward" Leonardo Da Vinci
User currently offlineBobster2 From , joined Dec 1969, posts, RR:
Reply 5, posted (8 years 3 weeks 6 days 22 hours ago) and read 2671 times:

Quoting Newark777 (Reply 3):
just draw a diagram and start using geometry

I can draw a diagram if the observer is at sea level. I can't figure out how to do it when the observer is at some arbitrary height above sea level.

In the first case it's a simple right angle triangle equation. The second case involves angles that I can't figure out.

Can you give me a hint? Or better yet, the answer?  Smile


User currently offlineBobster2 From , joined Dec 1969, posts, RR:
Reply 6, posted (8 years 3 weeks 6 days 21 hours ago) and read 2649 times:

Quoting Bobster2 (Reply 5):
Can you give me a hint?

I may have figured it out. If the observer is above sea level you have two right angle triangles, just add them together.


User currently offlineLehpron From United States of America, joined Jul 2001, 7028 posts, RR: 21
Reply 7, posted (8 years 3 weeks 6 days 16 hours ago) and read 2609 times:

Quoting Bobster2 (Reply 6):
I may have figured it out. If the observer is above sea level you have two right angle triangles, just add them together.

Are you assuming flat level ground or taking into account Earth's curvature?

You can exaggerate your diagram as long as you state all/any assumptions you plan to use with it.



The meaning of life is curiosity; we were put on this planet to explore opportunities.
User currently offlineDavid L From United Kingdom, joined May 1999, 9523 posts, RR: 42
Reply 8, posted (8 years 3 weeks 6 days 15 hours ago) and read 2597 times:

As a starter, here's how to calculate the distance to the horizon:

http://www.wolfram.demon.co.uk/rp_horizon_distance.html

To get the maximum distance to a point at the same height above the earth's surface as the observer, just double the distance to the horizon. Obviously there's more to it if the target (contrail) is at a different height to that of the observer. It's a starter, as I said.

Quoting Bobster2 (Reply 2):
Extra credit: Can you take into account the fact the airplane's apparent position on the horizon is not the true position because of light refraction in the atmosphere?

 talktothehand 


User currently offlineRedcordes From United States of America, joined Jan 2006, 245 posts, RR: 0
Reply 9, posted (8 years 3 weeks 6 days 15 hours ago) and read 2590 times:

Better check your math and atmospheric conditions. If you could see a 545 mile long contrail (and it would have to linger for an hour which would be unusual) the visibility would have to be 270 miles. No way. Assuming that you could only see a contrail to a point 10 degrees above the horizon, the angle the contrail is viewed would be 80 degrees each way from the observer. The tangent of 80 is 5.67, which gives the ratio of the distance the contrail can be seen in each direction to the distance the aircraft is above the earth. Therefore, if the aircraft is at 36,000 ft. (about 7 miles) the length of the observed contrail as described would be about 5.67 x 7 x 2, which is about 80 miles. Sound reasonable? A way to check this would be to measure the time the aircraft is in view, assuming a speed of 500 miles/hr. Should be about 10 min. Assuming 42,000 ft (8 miles) the distance would be about 90 miles and in view almost 12 min. Check it out.

[Edited 2006-06-17 12:48:03]

[Edited 2006-06-17 12:49:37]


"The only source of knowledge is experience." A. Einstein "Science w/o religion is lame. Religion w/o science is blind."
User currently offlineDavid L From United Kingdom, joined May 1999, 9523 posts, RR: 42
Reply 10, posted (8 years 3 weeks 6 days 13 hours ago) and read 2571 times:

Quoting David L (Reply 8):

Actually, to get the theoretical horizon to horizon distance you'd work out the distance to the horizon for an observer at height h, do the same for a point at, say, 6 NM altitude, add the two together then double the result.

OK, now we can start getting "real"...

Quoting Redcordes (Reply 9):
Assuming that you could only see a contrail to a point 10 degrees above the horizon, the angle the contrail is viewed would be 80 degrees each way from the observer.


User currently offlineSlamClick From United States of America, joined Nov 2003, 10062 posts, RR: 68
Reply 11, posted (8 years 3 weeks 6 days 12 hours ago) and read 2558 times:

Quoting Redcordes (Reply 9):
the visibility would have to be 270 miles. No way.

Ahh, you must live on the eastern seaboard.

Visiblity on that order is common out west. The limits to the visibily in the western states is almost always the distance to the horizon. I've been able to see the top of Mount Hood, near Portland Oregon from FL330 over Twin Falls Idaho. The setting sun was behind it and it was clearly visible. Far more than 270 miles.



Happiness is not seeing another trite Ste. Maarten photo all week long.
User currently offlineDavid L From United Kingdom, joined May 1999, 9523 posts, RR: 42
Reply 12, posted (8 years 3 weeks 6 days 12 hours ago) and read 2552 times:

Quoting SlamClick (Reply 11):
Visiblity on that order is common out west.

But not much chance of that where the thread starter lives.  Sad


User currently offlineSlamClick From United States of America, joined Nov 2003, 10062 posts, RR: 68
Reply 13, posted (8 years 3 weeks 6 days 12 hours ago) and read 2551 times:

Quoting Newark777 (Reply 3):
I've done it before, just draw a diagram and start using geometry.

You must keep in mind that the contrail is curved. That is, it is a constant height above a curved earth surface called sea level.

So draw two concentric circles, an inner and an outer one. The inner one is the surface, the outer one is my 'orbit' above it. Draw a straight line across them just touching the inner circle represents line-of-sight. Now you can do the geometry and it is fairly simple.

Easiest way, though is to take the accepted formula and double it. It goes like this:

Distance to the horizon in nautical miles is 1.23 times the square root of your height above the surface in feet.


So I am flying at 35000' and my calculator says the square root of 35K is 187.08

187.08 X 1.23 = 230.1 nautical miles.

So I can begin to see you 230 nautical miles away. That means you can see my contrail pop up over the horizon at that distance. It will remain in view, if I pass directly over your head until I get that same distance off in the other direction, or 460 nautical miles. That is about 529 statute miles.

Of course the only time you would have me in sight that long is if I flew in a straight line directly over the top of you. The fact is I am more likely to have a closest point of approach some miles distance from you. That will shorten the actual distance you can see my contrail. This is because the area of sky you can see is actually a circle with you at its center. If I just fly across the outer edge of that circle you will not be able to see my contrail for very long.

I'd guess that most of the times I saw this phenomenon back in the days of big recips flying at high altitude, most of the contrails were many miles distant and rarely over a hundred or two miles of actual trail seen.



Happiness is not seeing another trite Ste. Maarten photo all week long.
User currently offlineDavid L From United Kingdom, joined May 1999, 9523 posts, RR: 42
Reply 14, posted (8 years 3 weeks 6 days 12 hours ago) and read 2549 times:

Quoting SlamClick (Reply 13):
Draw a straight line across them just touching the inner circle represents line-of-sight.



Quoting SlamClick (Reply 13):
So I can begin to see you 230 nautical miles away. That means you can see my contrail pop up over the horizon at that distance.

Much better than my 3 stage effort, especially since the observer is likely to be as near as dammit to sea level. SlamClick strikes again.


User currently offlineTimz From United States of America, joined Sep 1999, 6750 posts, RR: 7
Reply 15, posted (8 years 3 weeks 6 days 3 hours ago) and read 2494 times:

Quoting SlamClick (Reply 11):
The limits to the visibily in the western states is almost always the distance to the horizon.

SlamClick didn't mean to suggest that the calculated horizon distance was the farthest one could see-- mountains can of course be seen farther, sticking up beyond the "horizon".

Quoting SlamClick (Reply 11):
I've been able to see the top of Mount Hood, near Portland Oregon from FL330 over Twin Falls Idaho. The setting sun was behind it and it was clearly visible. Far more than 270 miles.

Like I said before, Mt Hood wouldn't stand out from there, compared to the closer peaks. It might be hidden by them.


User currently offlineSlamClick From United States of America, joined Nov 2003, 10062 posts, RR: 68
Reply 16, posted (8 years 3 weeks 5 days 22 hours ago) and read 2447 times:

Quoting Timz (Reply 15):
SlamClick didn't mean to suggest that the calculated horizon distance was the farthest one could see-- mountains can of course be seen farther, sticking up beyond the "horizon".

Not only that but on numerous occasions I have seen the moon close to the horizon in the western states.



Happiness is not seeing another trite Ste. Maarten photo all week long.
User currently offlineRedcordes From United States of America, joined Jan 2006, 245 posts, RR: 0
Reply 17, posted (8 years 3 weeks 5 days 4 hours ago) and read 2361 times:

Quoting Redcordes (Reply 9):
A way to check this would be to measure the time the aircraft is in view, assuming a speed of 500 miles/hr. Should be about 10 min. Assuming 42,000 ft (8 miles) the distance would be about 90 miles and in view almost 12 min. Check it out.


I hate to quote myself, but this method might be the best after all is said and done. This would be real and take all variables into account. I got money that says in most locals you won't see an aicraft at cruise (contrails) in view for more than 15 min. (probably less, particulary in the thread-starter's location). Slam you got any cash?



"The only source of knowledge is experience." A. Einstein "Science w/o religion is lame. Religion w/o science is blind."
User currently offlineTimz From United States of America, joined Sep 1999, 6750 posts, RR: 7
Reply 18, posted (8 years 3 weeks 5 days 1 hour ago) and read 2325 times:

Quoting Redcordes (Reply 17):
I got money that says in most locals you won't see an aicraft at cruise (contrails) in view for more than 15 min.

You mean in most locales it's never clear enough-- maybe so, but I suspect the original question assumed clear air and an unobstructed view. If binoculars are allowed there are plenty of hilltops from which you can easily spot a contrail coming toward you, 150+ miles away, on a clear day. (Won't be so easy if it's crossing your line of sight, tho.)

On a clear day here in California, we can just barely make out Mt Lassen (10000 ft high) from Mt Diablo (3800+ ft high)-- that's 180 statute miles. Think we could see a 747 contrail above Mt Lassen?


User currently offlineRedcordes From United States of America, joined Jan 2006, 245 posts, RR: 0
Reply 19, posted (8 years 3 weeks 4 days 15 hours ago) and read 2289 times:

The thread-sarter is on the ground in the U.K. He asked about his typical observations, not somewhere on a hilltop in California with binoculars. He also said he doesn't understand complicated answers. Heck, many nights here in New England the visibility is over 250,000 miles--because the moon is visible! Additionally, when the visibility is extremely good, the contrails are extremely small.


"The only source of knowledge is experience." A. Einstein "Science w/o religion is lame. Religion w/o science is blind."
User currently offlineSlamClick From United States of America, joined Nov 2003, 10062 posts, RR: 68
Reply 20, posted (8 years 3 weeks 4 days 9 hours ago) and read 2268 times:

Quoting Redcordes (Reply 19):
The thread-sarter is on the ground in the U.K. He asked about his typical observations, not somewhere on a hilltop in California with binoculars.

True, and it would only be polite of the thread starter to come back and let us know if he got an answer that satisfied his curiosity.

My answers addressed only the geometric possibilities under perfect conditions plus a couple of empirical observations.

Quoting Redcordes (Reply 17):
Slam you got any cash?

Hell no! If I'd wanted cash on hand at this point of my life I'd never have devoted my life to aviation.

I think your fifteen minutes would be on the short side but then I've spent most of my life in the western US and have that frame of reference.

I cannot give your theory a fair test for several reasons. I live above five thousand feet. This makes prime contrail country only about thirty thousand feet above me, making the theoretical distance to the horizon a mere 213 nautical miles away. Further, I live in a valley of the Great Basin. Hills and mountains rise a thousand feet above me only a mile away, and three thousand feet above me only seven miles away. This shrinks my sky considerably.



Happiness is not seeing another trite Ste. Maarten photo all week long.
User currently offlineDavid L From United Kingdom, joined May 1999, 9523 posts, RR: 42
Reply 21, posted (8 years 3 weeks 4 days 9 hours ago) and read 2266 times:

Quoting SlamClick (Reply 20):
it would only be polite of the thread starter to come back and let us know if he got an answer that satisfied his curiosity.

I would have thought so, too.

I live 30 to 40 miles away from him and, if it weren't for the clouds, the visibility would probably be close to Redcordes' "10o above the horizon" scenario today (no hills bigger than that in the area), which is about as good as it ever gets here.


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