PDA

View Full Version : How high would would you have to go to see America East Coast in the sky from UK?



clacton0
2010-Feb-11, 12:58 PM
HI

How high would would you have to go to see America East Coast in the sky from UK?

Would you be able to see this from 12 miles up?

Also one you got to the high would you be able to pick up US TV and Radio signals from line of sight?

:)

Or the UK from the East Coast of the US.

antoniseb
2010-Feb-11, 01:22 PM
Hi clackton0, welcome to the BAUT forum.

12 miles won't do it. I'm frequently 8 miles above one or the other, and can only see one or two hundred miles. You could do this as a trig problem. I imagine the answer will be between one and two thousand miles.

You'd need a fairly large surface area or very directional antenna to pick up the line of sight signals.

neilzero
2010-Feb-11, 01:23 PM
I don't think 12 miles is high enough to see across the Atlantic, because the atmosphere is not sufficiently transparent, but TV and radio signals make it across occasionally even close to sealevel, but not reliably. Variously mechanisms such as ducting and bouncing off ionized layers make long distance communications possible. I don't think USA and UK TVs are compatible, unless the new high definition digital TV is compatible. Very high power helps, but stations on the same frequency = channel often interfere to the point of unintelligible, and unwatchable. An extremely directional antenna also helps. Neil

NEOWatcher
2010-Feb-11, 01:25 PM
Here's a quick back of the napkin test that only works at the equator.
Basically the thought is that you are looking for the tangent at the item you are looking at, and intersecting it with a line from the center of the earth to your location.
The difference in these lines is an angle equal to the difference in longitude.
The tangent becomes the base of a right triangle and is the opposite line of the angle.
The center to location line (extended) is the hypotenuse
The last side is about 4000 miles and is the adjecent angle
So the height would be 4000 miles less than the hypotenuse.
In this case 60 degrees difference in latitude.

A bit of trig (with ballpark figures), and:
hypotenuse = adj / cos(angle)
Or
Hypotenuse = 4000 / 0.5
So height above the Earth becomes 8000 - 4000.

At non-equator locations, it will be less, but you would have to figure out the angle. Not too hard at equal latitudes where you can multiply the longitude angle by cos(latitude).

Buttercup
2010-Feb-11, 01:37 PM
Here's a quick back of the napkin test that only works at the equator.

Oh sure, NEOWatcher. :lol: :lol: What a cinch! ;)

Actually I was going to answer in the exact fashion you did, but figured I'd wait and see if someone else knew the answer and would type it out. :p

Interesting question though.

[And yes I am severely mathematically challenged]

clacton0
2010-Feb-11, 02:12 PM
Hi clackton0, welcome to the BAUT forum.

12 miles won't do it. I'm frequently 8 miles above one or the other, and can only see one or two hundred miles. You could do this as a trig problem. I imagine the answer will be between one and two thousand miles.

You'd need a fairly large surface area or very directional antenna to pick up the line of sight signals.

Thanks for the welcome I have been here the last few months but just not posted much :)

novaderrik
2010-Feb-11, 07:05 PM
I don't think 12 miles is high enough to see across the Atlantic, because the atmosphere is not sufficiently transparent, but TV and radio signals make it across occasionally even close to sealevel, but not reliably. Variously mechanisms such as ducting and bouncing off ionized layers make long distance communications possible. I don't think USA and UK TVs are compatible, unless the new high definition digital TV is compatible. Very high power helps, but stations on the same frequency = channel often interfere to the point of unintelligible, and unwatchable. An extremely directional antenna also helps. Neil

the earth itself isn't "sufficiently transparent", either, and there is a lot of curved earth between England and the US.

Centaur
2010-Feb-11, 08:36 PM
How high would would you have to go to see America East Coast in the sky from UK?




The distance from the Earth’s center that one must be to view another location would be the Earth’s radius divided by the cosine of the great circle angular separation. At the latitudes for the stated problem the Earths’ radius is about 6370 km. Plymouth, England and St. John’s, Newfoundland (Canada) are separated by 31.433° on a great circle. The formula for that is somewhat complex, so I'll ask you to assume it is correct. An observer at one of those locations must be 1096 km or 681 miles above the Earth’s surface to see the other. If we substitute New York City for St. John’s, then the separation is 48.347° and an observer must be 3214 km or 1997 miles above the Earth’s surface.

Argos
2010-Feb-11, 08:43 PM
To put it in perspective, even aboard the ISS [max altitude ~400 km] it wouldn´t be possible.

Arnold Rimmer
2010-Feb-12, 12:02 AM
Fun fact:

If the earth was a perfect sphere and the surface totally flat, and you were my height (1.90m), so your eyes were at 1.80m altitude, the furthest spot on the ground visible to you would be ca 5 km away.

Jeff Root
2010-Feb-12, 12:38 AM
If you have a globe of the Earth, you can place a straightedge tangent to
the surface of the globe at the point you'd like to be able to see, and then
measure how high above your location the straightedge is.

Every home should have a globe of at least one planet.

-- Jeff, in Minneapolis

joema
2010-Feb-12, 01:33 PM
HI

How high would would you have to go to see America East Coast in the sky from UK?

Would you be able to see this from 12 miles up?

Also one you got to the high would you be able to pick up US TV and Radio signals from line of sight?...
This is the "radio horizon", which is the locus of points at which direct rays from an antenna are tangential to the surface of the Earth. If the Earth were a perfect sphere and there were no atmospheric anomalies, the radio horizon would be a circle. To compute the radius of the circle drawn on the earth in such a case use the formula:

Radio Horizon = 1.23 * sqrt(Antenna Height), where:

Radio Horizon = distance in miles from antenna
Antenna height = height in feet

Using this formula an antenna 12 miles (or 63,360 feet) high would have a radio horizon of 309.6 miles. So it wouldn't reach the U.S. from the U.K.

To see (either visually or via line-of-sight radio frequences) the U.S. from the U.K. would require an altitude of about 600,000 feet (113.6 miles, or 182.88 km).

Jeff Root
2010-Feb-12, 02:18 PM
To see (either visually or via line-of-sight radio frequences) the
U.S. from the U.K. would require an altitude of about 600,000 feet
(113.6 miles, or 182.88 km).
Which U.S. and which U.K. did you have in mind, there?

-- Jeff, in Minneapolis

01101001
2010-Feb-12, 03:05 PM
If you have a globe of the Earth, you can place a straightedge tangent to
the surface of the globe at the point you'd like to be able to see, and then
measure how high above your location the straightedge is.

And put analog computer programmer on your cv/resume!

joema
2010-Feb-13, 12:29 AM
Which U.S. and which U.K. did you have in mind, there?

-- Jeff, in Minneapolis
The same ones mentioned in the 1st post of this thread.

Jeff Root
2010-Feb-13, 05:42 AM
Joema,

Think about the altitude you computed (113.6 miles, or 182.88 km), and think
whether that altitude makes any sense.

Do you think maybe the numbers are way, way off?

If you have a globe, do as I suggested above: Put a straightedge on
the globe so that it is tangent to the surface at the point in the U.S.
that you want to see, and then estimate how far above your location
in the U.K. the straightedge passes.

-- Jeff, in Minneapolis

Locomotion
2010-Feb-13, 06:38 AM
Do you think maybe the numbers are way, way off?

Yea, you need the British Virgin Islands for those to work.

joema
2010-Feb-13, 08:43 AM
...Do you think maybe the numbers are way, way off?

If you have a globe, do as I suggested above: Put a straightedge on
the globe so that it is tangent to the surface at the point in the U.S.
that you want to see, and then estimate how far above your location
in the U.K. the straightedge passes.
You are right, the numbers are off. Thanks for spotting that. Your method of using a globe illustrates the importance of sanity-checking calculations, which I didn't do.

The error stemmed from using a "rule of thumb" equation which gives an approximation that's correct for normal-height radio antennas. For extreme altitudes it becomes inaccurate.

For that case we need to use an equation derived from the Pythagorean theorem: http://en.wikipedia.org/wiki/Pythagorean_theorem, which is:

d = SQRT (2 * h1 * R + h1^2) + SQRT (2 * h2 * R + h2^2), where:

d = line of sight distance in meters
h1 = height of antenna 1 in meters
h2 = height of antenna 2 in meters
R = earth's radius in meters (6.38E6 m)

Using 50 meters for the height above the ocean of U.S, we find that an altitude above the U.K. of about 1.7E6 meters or 1,056 miles is required to see the U.S. from the U.K, either visually or via line-of-sight radio frequencies.

Check that on your globe and see if it's more accurate.

Tom Servo
2010-Feb-13, 09:05 AM
I wonder if you could find out using google earth. If its view angle is correct then just look at the altitude reader at the point were you can see the two contries.