Accepting the above if we divide 3,113 miles by 1,982 miles we get a drop/fall/decrease in height in relation to the north pole of 1 mile per 1.57 miles travelled.

I’m going to start with this comment because I think it is the source of your confusion. The “height” of a position is measured against sea level and depends on the terrain. If the earth was a perfectly smooth sphere everything would be at sea level. “Height” doesn’t have anything to do with the curvature of the earth.

Curvature is the rate a line is changing direction at a given point. The “drop” in the curvature of the earth is a change in direction of an arbitrary line on its surface, not a change in height.

Instead of walking from N to E1 imagine walking from N to X. This is half the distance to the equator and represents one eighth (1/8) of the earths circumference ie 3,113 miles.

You have walked 3,113 miles south, not “down”. Eventually you would be going north.

You have two errors. The first is you try to measure curvature using straight line distances. That doesn’t work. Second you are adding the curvature linearly.

Using the equation I gave before: h = r * (1 - cos a). One mile across the surface is .0145 ° change in direction from straight.

The cosine of a .0145 ° angle is .99999997

h=3959*(1-.99999997)

h=3959*.000000032

h=.000127 miles

h=8.04672 in

So in one mile, there is a “drop”, or change in direction of the surface of 8 in. from A to B.

By your logic, then there should be 16 in of drop between A and C, but:

Two miles across the surface is .0288 ° . The cosine of a .00288 ° angle is 0.99999987

h=3959*(1-.99999987)

h=3959*.00000013

h=.000515 miles

h=32.6304 in

To sum up, a sphere or circle isn’t a straight line so the ratio of distance to curvature isn’t 1:1. It’s close to quadratic…the curvature is proportional to the square of the distance.