<rant>
So? What is the significance of Point B? It's a point in space that is level with Point C above a spot on Point C's horizon. There's nothing there but thin air.
So, if the Earth were round, after 100 miles the ground would drop 2000 meters, and in order to be able to see certain low profile figure/point at the horizon you should climb up 2000 meters also.
2000 m looks like a reasonable approximation, and that interpretation is reasonable. Remember, though, that 100 miles is not very far.
Now, if somebody claims something like this...
"As I said before, it's not exponential, it's circular. For short distances (less than about 600 miles or so) the distance squared function that you suggested is an adequate approximation, but its answers degrade rapidly as distance increases from that, and utterly fails when the distance reaches 1/4 of the circumference."
...i just can't refrain myself to answer simply and illustratively (like this):
Bull s h i t t!!!
Maybe you should refrain from glib retorts and actually examine your ideas when a problem is pointed out. If, after examination, you think there is an error with the reply, show specifically what the flaw is. If you can't find an error (or don't understand it), admit it. Simply calling "BS" sounds like you're giving up.
Using the approximation formula on p. 9 of
this exhibit (thanks, Aus!) that
you referenced
here - and used, let's look at a more distant case. Since you deny there's a problem with it, let's calculate its value for the dip one-quarter of the circumference away and compare with more accurate calculations. Fig. 14 from the above is replicated here for convenience.
"In calculating the amount of curvature, or dip below the eye-line of the observer, we have a simple rule, ignoring some small decimal points, namely :— Square the number of miles given as the distance, and multiply the product by eight inches, and divide by twelve, which will give in feet the depth of the dip from the observer’s line of sight. This is true for a globe of 25,000 miles circumference ; thus in six miles there would be a dip of 24 feet, and in twelve miles a dip of 96 feet."
The formula is:
Dip = D
2 * 8 / 12
Where Dip is in feet if D is the distance in miles.
Dip = (6250)
2 * 8 / 12
= 39,062,500 * 8 / 12
= 26,041,666.67 feet
= 4932.1 miles
The author gives two different meanings for "Dip" on the following page, referring to Fig. 14.
"And this is true whether we reckon the dip towards the centre of the globe in the direction of G L, or at right-angles from the line of sight G M."
Examining the second meaning of "Dip" - right angles to the sightline - first, since it's easier:
For a location one-quarter of the way around the circle, the "Dip" would clearly be equal to the radius of the circle.
R = 25,000 miles / (2 pi) = 3978.9 miles
This is an error of more than 953 miles.
The author's first meaning of Dip is even worse - far worse.
Define the point 'O' (capital letter O, not zero) in Fig. 14 as the center of the circle "shewn". It's the intersection of the lines through EK and JL. This defines the right triangle EHO with the right angle at point E and line HO the hypotenuse. Line EO is the radius of our circle. Line HJ is the "Dip" under the second meaning. HJ is the length of the hypotenuse of our right triangle, HO, minus the length of line JO, which is the radius of the circle.
First calculate the length of hypotenuse, HO. This is the adjacent side, EO, divided by the cosine of the angle opposite side EH. Let's call this angle 'a'.
HO = EO / cos(a)
Since EO is the radius R, then
HO = R / cos(a)
Since our point J is one-quarter of the way around the circumference of circle O, a is exactly 90 degrees, and recall that R = 3978.9 miles, so
HO = 3978.9 miles / cos(90)
Uh, oh! cos(90) is zero! Mr. Math may quibble with me here since division by zero is undefined, but this effectively means that line HO is infinitely long. Subtracting our radius R, 3978.9 miles, from infinity is still infinity. Your estimated dip using the approximation formula, 4932.1 miles, is infinitely less than the more accurate calculated value (infinity). This is what's meant by "blowing up".
No bullshit, just facts and a little elementary trig. Instead of calling this "mumbo jumbo", instead ask yourself why you didn't try harder in high-school math. A bit more effort there would have saved you from falling for the quite lame arguments the likes of Rowbotham and "Zetetes" foist off on the ignorant and gullible with some apparent success, even today.
Isn't this question quite legitimate : Should i refrain myself (in the first place) from giving such kind of answers, given that i have to deal with constant lying and mumbo jumbo mathematical calculations which have no value and no common ground with reality whatsoever?
No, that's not a legitimate question since the premise it's based on is wrong. It would seem to be a good idea to refrain yourself for other reasons, though.
[Edit] Adjust width of top figure.