View Full Version : Earth Curvature
8th May 2003, 14:07
I am in need of some trigonometric assistance as I am mathematically impaired :(
I am looking for a number, for each 60 miles that the earth's curvature causes a drop in elevation relative to an 'absolute straight' line.
The reason, you ask? Our aircraft is equipped with an HGS (heads-up guidance system, for the mnemonically dis-advanteged) and I was suggesting to the other pilot that the clearance of a weather cell could be determined by lowering the HGS combiner and referencing whether the cell is above or below the horizon line displayed on the combiner.
My 'theory' is based on the idea that each degree that the cell top is below the horizon, at 60 nautical miles, is a clearance value of approximately 6000'. The old 'rule of 60' states that "at 60 NM, each degree equals about 1 NM (6000') of lateral distance" or something to that effect.......
His response was that, because the earth is curved, the cloud will be cleared by a different value........I say the the amount of curvature of the earth's surface is neglible at the distances we are talking about (up to 300NM) and that the idea is plausible....
Anybody able to provide a little help here? I need a basic rule of thumb for some number (say 60 or a 100 NM) that the earth drops away from an 'absolute' straight line.....
Thanks in advance.........
8th May 2003, 14:17
At 300Nm the Horizon is ~5 degrees lower than the local horizon (tangent to the earths surface).
I get roughly half Bevan's figure.
I make it 2.6 degrees at 300nm.
The solution is a bit long-winded though - so bound to be a mistake somewhere!
If you draw a tangent to the Earth's surface, and make the line 300nm long.
Join both ends of the tangent to the centre of the Earth.
You've got a right angle triangle.
The angle subtended at the centre of the Earth is atan(300/3443) = 5 degrees ( where 3443 is Earth radius approx.)
If you then calculate the length of the longer line from the Earth centre to the end of the tangent from 300/sin(5 degrees) =3456 nm.
The difference from the centre of the Earth to either end of the tangent being 3456 - 3443 = 13 nm.
And 13nm in 300 = 2.6 degrees.
The Earth's surface drops away from the tangent at roughly 0.5nm in 60 = 0.5 degrees.
8th May 2003, 20:54
Agree with Mukka.
If you draw a tangent at the earth's surface, and travel along it 60nm, you will find there is approximately a 0.5nm 'drop'.
R = radius of Earth (nm)
Tangent is 60nm long
Therfore for a right angled triangle,
Hypoteneuse (H) = root(R^2 + 60^2)
The 'drop' is H - R = ~0.5nm.
8th May 2003, 23:31
But where is the horizon on that ???? Is the horizon at zero degrees or lower ?????
If you stood on the beach and looked straight out to the horizon, then travelled 60nm along that line, you would find that the sea was 0.5nm (3000') below you.
If you don't mind me asking - are you a member of the Flat Earth society?
9th May 2003, 00:39
If l is the distance along the tangent and R is the radius of the earth then the drop of the surface of the earth increases in a non linear way with distance and is found by the formula
d = R-(Rcos(asin(l/R)))Herewith a convenient table
Distance Angle Drop
30 0.499243 0.130702
60 0.998525 0.52284
90 1.497882 1.176501
120 1.997353 2.091835
150 2.496976 3.269051
180 2.996789 4.708418
210 3.49683 6.410266
240 3.997138 8.374984
270 4.497752 10.60302
300 4.99871 13.0949
330 5.500051 15.85119
360 6.001816 18.87252
390 6.504042 22.15961
420 7.006771 25.71321
450 7.510043 29.53417
480 8.013898 33.62337
510 8.518378 37.98179
540 9.023524 42.61046
570 9.529378 47.51049
600 10.03598 52.68305
9th May 2003, 00:45
Slightly off topic, but in flight its useful to remember that the distance to the horizon (in nautical miles) is 1.17 times the root of the altitude.
i.e. from 10000' horizon is 117nm away. from 30K its 202nm etc.
9th May 2003, 00:57
mukka, nah, nearly joined, but someone convinced me that we are at the bottom of a goldfish bowl..... cant wait till a cat puts its paw in trying to grap something.....
Maybe i'm mad ??:zzz: :sad:
9th May 2003, 08:54
Here's a somewhat related question:
What is the "speed" of the sun as it traverses the earth's surface and how does one correct for increasing altitude above the earth. We have all noticed how long it takes for the sun to set when westbound at the higher latitudes, say 55N. It is around 1000 kts at the equator, but anyone have the exact numbers and formulas?
9th May 2003, 16:28
keendog, the cosine of the arcsine can be simplified.
When you do so, the formula d = R-(Rcos(asin(L/R)))
becomes d = R - sqrt(R^2-L^2) (using L instead of
your l to avoid confusion with the number 1).
Your number d is not the distance along a radial
drop (the way gravity pulls us). It is the distance
to the Earth along the line that is perpendicular to
In principle there is a value of the distance to Earth
whatever L may be. But the formula for d cannot be
evaluated for L > R because the arcsine makes no
sense for such values of L/R. The geometric explanation is
that the drop along the perpendicular line misses the Earth
when L > R.
For small values of L the difference between
your d and Fat Dog's exact radial drop distance,
sqrt(R^2+L^2)-R, is not too great. For example, if
L = 60 nm then the inaccuracy shows up only in the fourth
decimal place. However, at 600 nm the error is about 1.5%.
10th May 2003, 06:50
Several years ago I saw a table giving depression of the natural horizon with altitude - ie the higher you go, the more "down" you have to look to see the intersection of sea & sky. So at FL300 the natural horizon is a poor reference to use when trying to determine whether something is above or below your altitude. Is this what your question refers to?
But in the context of the original post, I suppose the HUD will give the pilot an actual (tangential to surface of earth) horizon mark. So you could say "Well that cloud is 2.5° below the actual horizon (or 0° pitch reference) and is 300nm away so we will just clear it" or words to that effect.
Disclaimer - never used a HUD, don't have the table of horizon depression against altitude with me. So this is all just speculation eh? But I do think that 2.5° at 300nm is correct... :p
Sun moves at 15° per hour east to west. 1° = 60nm times cosine latitude.
Thus, at 60° latitude, 15°/hr = 15 x 60 x cos(60) = 450 knots.
You could beat the sun by walking around the earth if you pick the right place...
11th May 2003, 09:29
I think the original questioner is still scratching his head. If so, try this.
The earth isn't really spherical, but we're going to pretend it is. So no use claiming to be super accurate. I guess a sphere with 3440 nm radius is about as good as any. We're also ignoring refraction.
Also we're ignoring the earth's spin-- so a vertical line points to the center of the sphere. And a horizontal line is defined as a line perpendicular to a vertical line. So a horizontal line doesn't point to the horizon-- right?
If a horizontal line just touches the earth's surface, and you are on that line X nautical miles from the point where it touches, how far above the earth's surface are you? If you want the answer in feet, it's 0.883 times X squared. That's correct to within 0.2% beyond 200 nm.
11th May 2003, 11:55
I need a basic rule of thumb for some number (say 60 or a 100 NM) that the earth drops away from an 'absolute' straight line.....
The question seems to be answered above. But remember that the aircraft itself will be dropping away from an absolute straight line as it travels around the earth. Yes, this is negligible when compared to the earth's radius, but is not negligible at all when compared to a small change in altitude. A calculation error of 3000 feet is not negligible when it comes to flying around or through storm cloud tops!
Therefore, if you want to know if you will fly above or through a cloud seen some distance away, you must take earth curvature into account.
This is difficult without a diagram. Draw a section of the earth, with radius r and angle subtended at the earth's core of 1°. Distance across the earth's surface will be 60nm of course. Put the aircraft above the earth's surface at one radial line, and draw a tangential line (your "absolute straight line") out from it. Draw a cloud above the surface at the other radial line at the same height above the surface as the aircraft model.
The cloud is well below the tangent line - but the aircraft will still fly through it. Draw a straight line from the aircraft to the cloud. This makes an isosceles triangle with the two earth radial lines, and the angle between the tangent line and the aircraft-cloud line is half the angle subtended at the earth's centre - 0.5° in this case.
To answer your question: The earth's curvature does matter. For each 60nm you are from cloud, the cloud must be at least half a degree below your horizontal in order for you to fly above it.
My apologies if I've taught you to suck eggs here - I think this answers the spirit of your question, even if it has not given you the magic number you were after.
11th May 2003, 13:38
OK, a third formula now. If anybody is still keeping score, then Fat Dog's
formula remains exact (modulo all the simplifying assumptions Tim Zukas
enumerates), keendog's formula is a good approximation, and
Tim Zukas's approximation very nearly splits the difference between
the two dogs' answers.
Since 6080/6886 is about 0.883 the conversion to nm is X^2/6886.
Now if one expands the answers of the two dogs in a Maclaurin series (http://ascc.artsci.wustl.edu/~bblank/distanceToEarth.jpg)
(which in this case is just Newton's Binomial Formula)
one finds the simple quadratic approximation X^2/6886 to be
nearly midway between the two earlier answers. The
shows that all three answers are very nearly the same for X < 200 nm.
(Just thought I'd make all this clear.)