You say goodbye to your friend who is going on a 300 mile cross country flight(in fact it is to an airport 320 miles away). You stand outside and watch him fly away on a GPS direct routing to final destination. Both airports are at the same elevation. When he touches down, how much lower than the surface you are standing on will he be compared to you and when will that number someday become potentially useful to you.

From https://www.tc.gc.ca/eng/civilaviation/ ... 1-4793.htm:

"Aerodrome elevation. The elevation of the highest point of the landing area."

So really you could either be higher or lower than him depending on:

- how far below the highest elevation the touchdown point is
- how far below the highest elevation the spot you're standing on is (assuming it is part of the "landing area")
- if you're not standing on part of the landing area, you could be above or below the airport elevation

Do you mean what is the depression of the horizon at a range of 320nm?

Following, CpnCrunch's train of thought, may I assume that my friend touches down on the landing area? Do I stand on the same spot while my friend flies the 320 nm? If I do, I hope I'm not standing on the landing area of the departure airport . But that only takes care of the first part of the question.

I don't have a clue about the second part. When is this info going to be useful to me?

I think what he is getting at is how much the curvature of the Earth affects your height above or below a reference point, over a distance of 320 nm. The best answer is it depends on your frame of reference. To make matters simple, if I choose the surface of the earth as my reference framework, then the difference is zero. I think the goal of the question is to calculate how far below the plane that is tangential to the surface of the Earth at the departure point will you be when you are 320 nm away. Are we assuming the Earth is a perfect sphere for this, and for that matter, would there be a significant difference if something like the WGS-84 ellipsoid was used? (If you wanted to base your calculations on an oblate spheroid such as the WGS-84 model, you would need to know lat/long of each point, and do a lot more math...)

Some people think that I believe the earth is flat...but lets just use a perfect sphere.

Approximate values will be fine.

Part A, I have no idea how much 'lower' than me s/he will be. I suppose that it might be useful in certain heavily loaded aeroplanes, or in an engine-out situation, where it is a well-known fact that planes only climb due to the curvature of the earth!

pelmet wrote:Some people think that I believe the earth is flat...but lets just use a perfect sphere.

Approximate values will be fine.

You're going to have to confirm that you're using a weird definition of "lower" and say what is, because by any sensible definition the answer is zero.

There should be a treadmill in this question somewhere.

14.9

North Shore wrote:Part A, I have no idea how much 'lower' than me s/he will be. I suppose that it might be useful in certain heavily loaded aeroplanes, or in an engine-out situation, where it is a well-known fact that planes only climb due to the curvature of the earth!

Or...what other aviation related item is affected by the curvature of the earth.

Answer to the first part is 65 thousand feet by the way...and no, I did not calculate that.

A couple questions for your (not yours) 65,000' calculation:

1) Are you assuming the earth is spherical? Are you doing the calculation based on WGS-84, or using the precise Lat-Lon of both departure and destination aerodromes (which you didn't give us)

2) Are you assuming they are flying in a straight line, ie: altitude is rising while they approach the destination or are you taking into account the curvature of the earth for your calculation?

3) If you are taking into account the curvature of the earth for flight path and the subsequent calculation, which value are using to average your calculation?

4) Are you assuming that 0 is a tangential plane at your departure aerodrome? And again, which model are you using to measure the shape of the earth?

At it's simplest form(Spherical earth, constant flight path [changing altitude with distance], zeroed on a plane from your departure point), it's relatively simple trigonometry, although it would be far from accurate, at it's most accurate (actual shape of the earth, given Lat-Lon of both aerodromes, altitude varying to account for curvature of the earth, zero point at R=0, not the surface, the assumption that altitude is relative to your initial point and the center of the earth...) it would be vastly more complicated and dependent on numerous variables, which you did not provide.

To be frank, although it's an interesting concept to ponder about, I don't think it has much relevance unless you take it you the more complicated level. I'm guessing you've thought about most of this, given the title of the thread...

E

I think you are over-thinking this. Just assume that the 65,000' calculation is approximately correct. I'll give you a strong hint. I am reading a manual for a certain aircraft system and this is what it said.

The Apollo guidance computer?

Archie Trammell's manual?

Assuming the earth is a sphere of radius 4000 miles, and you're only referring to it dropping away from you as someone flies 300 miles away from you, then basic trig would put your friend about 5 miles lower than the plane tangent to the surface of the earth where you're standing... Very rough back of the envelope calculation though. Still thinking about how it could be useful to know this...

Useful navigation formulae:
http://williams.best.vwh.net/avform.htm

I can't be bothered to do the calculations, but it might be useful if the GPS fails and he needs to rely on radio navigation that uses line of sight.

Well if he's at 3500 ft, by the time he's 63.2 nm away he's below the horizon, so that screws up the the traditional night XC green laser strike sendoff right? Good times...

Interestingly, the answer I got is 50% more than the 60,000 ft answer. 14.9 nm = 90,700 ft.

I started by assuming the Earth is spherical, so our starting point and end point can be considered to be points on a circle with the radius of 3440 nm (approximate average radius of the Earth). I was interested in the angular distance between these points, and using 60 nm per degree, I came up with 5.33 degrees for the 320 nm distance. I took a tangent to the circle at the starting point, and comparing that to a parallel line through the centre of the circle, and with some simple trig, I came up with this: Have I missed something here?

Chris M wrote:There should be a treadmill in this question somewhere.

Awesome!

Anyway, I have no real firm idea but can only guess it would/could be important if you were operating an aircraft with an inertial guidance system.

But honestly, I don't believe it will ever be useful to me.

AOW wrote: Have I missed something here?

No.

It's even simpler if you're prepared to make the small angle approximation (not an issue at 5°):

No offence to Pelmet, but it is a bit of a silly question. Kind of like those trick questions that TC likes to ask.

But where does he get an answer of 65,000' ?

Wouldn't the rotation of the earth change that?