An easy way to see it, maybe. But it’s still just lines on a screen. I don’t think that would lead to understanding at all.
I would love to see those equations.
Aside: My grandfather was a ham operator, and had a sufficient rig to be able to reach intercontinental distances (at least, when conditions were right). He had a globe that he had re-mounted so that Cleveland was at one of the poles (and the point antipodal to Cleveland at the opposite pole, of course) to make it easier to find the great circle direction to any given point on the Earth. In other words, he did define a different direction to be north.
It didn’t take all that much. On the 15 meter band with a 40m dipole I fairly regularly could talk to Japan. IIRC the transmitter was about 100 watts. If the direction mattered to him, he probably had a beam antenna.
Simplest way to imagine it… Imagine a slice of orange or watemelon. Unless the line of the slice goes directly through the centerpoint of the fruit (i.e exactly cut using lines of longitude) then the shortest path between the two points along the rind is the center line directly between the points. Going off center - ie. following the cut edges - is more distance to get to the same place. A piece cut through the center of the fruit/globe is essentially a rotation of a great circle line between two poles.
As I understood it, those radio signals bounced off the ioinsphere? Something I heard was that they tended to trave best following roughly the terminator line between night and day, goive or take an hour or two? (Although they could sometimes travel long distance at all times.)
Oh, sure, it’s not hard, if you try for it. But I imagine that there are a lot more hams who don’t have a 40m antenna than who do.
I’ll see about putting them all together soon
I’ll go into more detail later, but here’s an outline.
I wanted to figure out the horizontal centrifugal forces due to moving on a non-rotating sphere at a specific bearing at a specific latitude (longitude is arbitrary). So I developed the equations on a sphere for a great circle achieving a maximum or minimum latitude at at longitude zero (the free parameters were speed and extreme latitude). For that general equation, I calculated the bearing of the object’s motion (the dot product of the velocity vector and the north vector), which specified the starting latitude of the great circle and the time at which the object is at the desired latitude and bearing.
I calculated the second derivative of the great circle’s motion evaluated at that time.
Then I figured out the equations of motion for a loxodrome starting at the right latitude and traveling at the right bearing, and calculated the second derivative of that motion, too. The difference between those two second derivatives is the horizontal centrifugal force that an object instantaneously moving on the loxodrome will feel.
This was close to sixty years ago. I have no idea what most hams are into these days.
A nursing home?
Sorry, couldn’t resist. 
In the recent thread on hobbies dying for lack of new folks, ham radio figured prominently, helped along by a large side order of technological obsolescence.
But if
But of course we can reset the arbitrarily chosen North pole on any location on the map (say your current location) and then traveling in any direction from that location will be along a new version of “directly south” If you then created a Mercator projection map with your position as the north pole, all of the great circles would just be straight lines. (ignoring the 0.3% non-sphericalilty of the earth)
All of the great circles through your current position. Other great circles would still be curves on that map.
On the other hand, on a polar-projection map, all straight lines are great circles, and vice-versa. The main drawback of a polar projection is that it can only show less than half of the Earth (as the size of the map approaches infinity, the region covered approaches a hemisphere).
What I ment is that if you travel to go to any point along a great circle from your current position to any point on the map it would be a straight due south line on the Mercator projection map
Back when I took an interest in amateur radio (which was sometime in the Jurassic) you could get printed maps that allowed you to determine the appropriate direction for your antenna. They were an azimuthal projection, with the entire globe centred on your location and great circles radiating outwards as straight lines, so you could read the needed antenna azimuth angle trivially.
A quick search turns up this site which will create such a map for you.
I think you mean a central projection (from the centre of the globe onto some plane). Great circles will not be straight on, say, a stereographic projection.
With, of course, the complication that your current position wouldn’t be on the map.
or it would be mapped non-uniquely as a line across the top.
Not in a Mercator projection. Some rectangular projections can include the pole as a line at the top, but in Mercator specifically, the pole is infinitely far away.
You’re right I was mixxing up Mercator with Equirectangual.
Downpayment on a complete analysis:
R=radius, L=latitude, l=longitude
Coordinates: x=Rcos(L)cos(l)
y=Rcos(L)sin(l)
z=Rsin(L)
(picture the globe with three cartesian axes meeting at the center. x and y define the plane of the equator, z goes through the north pole).
A great circle satisfies the equation x^2+y^2+z^2=R^2 and the equation of a plane that goes through the center of the earth. A bit of calculus comes up with the following
x=cos(Lm)Rcos(wtime);
y=Rsin(wtime);
z=Rsin(Lm)cos(wtime);
xdot=-wcosd(Lm)Rsin(wtime);
ydot=wRcos(wtime);
zdot=-wR*sin(Lm)sin(wtime);
where Lm is the maximum (or minimum latitude) and w=V/R (the velocity of the object moving on the great circle divided by the radius of the Earth.