Azimuth: which way to go, and see

Let’s say I’m located at 0,0. I want to walk to a very tall mountain, which summit is located at 10,10. The direction (azimuth?) I need to move toward is 45° (likewise, if I start at the summit and want to return, the direction is 45+180=225:

$ echo 10 10 | gmt mapproject -Af0/0
10      10      45

Is this correct?

What If I just want the direction I have to look towards (being at 0,0) to the the summit at 10,10. Will this also be 45°?

Sometimes, when hiking, you’ll come across one of these devices with a spinnable arrow that indicates which direction buildings, objects and other POI’s are located and where to place your gaze.

Light doesn’t follow great circles. How (if any) does this affect this?

Heuristics are welcome!

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You are aware that if you start at 0,0 always following the azimuth 45, and keep swimming and walking and walking and walking (maybe swim a bit) and walking, you’ll reach the North Pole?

The altitude does no influence the azimuth between two points.

Didn’t quite get the hiking question. When hiking the distances are short so following a constant direction should get you to the destination.

Can confirm your command. Re-reading the documentation it is not entirely clear which points is previous and which is next when a point is given via -A. I think that section could need some clarification [result is correct though]

I find this confusing (wrong?).
“ Upper case F , B or O will convert from geodetic to geocentric latitudes and estimate azimuth of geodesics (assuming the current ellipsoid is not a sphere).”

So we go from geodesics to geocentric to compute a geodesic azimuth?

Thank you both. And pardon ignorance.

You are aware that if you start at 0,0 always following the azimuth 45, and keep swimming and walking and walking and walking (maybe swim a bit) and walking, you’ll reach the North Pole?
The altitude does no influence the azimuth between two points.

Yes, I do think I knew that.

Didn’t quite get the hiking question. When hiking the distances are short so following a constant direction should get you to the destination.

I’m assuming a (very) long distance. Thats why I introduced the ‘tall mountain’.

To put it another way: the photons reaching my eyes from a mountain far, far away, what will the ground track of those photons look like?

To put it another way: the photons reaching my eyes from a mountain far, far away, what will the ground track of those photons look like?

The shortest distance. That is, a (geodesic) great circle.

Thanks @Joaquim.

That track would be a loxodrome, right?

Right

So in my example, giving an answer of 45 deg.; this is valid only when situated at 0,0. Once you take the first step, this value will (ever so slightly) change? And this applies to every step of the journey?

You are dangerously getting to the point where I’ll suggest checking with

Geodesic buffers

and (and related functions)

https://www.generic-mapping-tools.org/GMTjl_doc/documentation/gdalfuns/#invgeod

Have no computer with me to check things further.

Thanks again, @Joaquim. I’ll read.

A follow-up;

Again; If I’m sitting at 0,0. If I choose a direction/azimuth to look (45 deg in this case), how do I generate a ground track of ‘what’s in the path of what I will be seeing’? It my mind this is simply creating a great circle with the given azimuth; e.g.

gmt begin test png
echo -e "0 0\n10 10" | gmt plot -Gred -Bxafg -Byafg -Sc0.5c
gmt project -C0/0 -A45 -Q -G1 -L0/2000 | gmt plot
gmt end show

What I don’t understand is why the line is a bit off, not hitting 10,10, exactly.

This is analogous to an e.g. ship plotter that projects a line in the direction the vessel is moving telling you what you would hit if you were to sail in this direction “for a long time”.

Well, we are not doing Cartesian things here; this is on a sphere. It wont go through (10,10. Here is another calculation: Given azimuth 45, go ~14 degrees in that direction:

echo 0 0 | gmt vector -Tt45/14.00194217d
10.0000000033 9.85107611971

Same result as project it seems.

Thanks @pwessel.