You might also want to take into consideration the height of your eyes above the surrounding terrain.
This might not seem like much, if you are standing on flat ground (for example, pretty much anywhere in Maryland (USA - not the long since vanished country in Africa)). However, if you're standing on a mountain it can be a significant distance. -- Raul On Mon, Feb 26, 2018 at 9:52 PM, Ian Clark <[email protected]> wrote: > In "The Curious Incident of the Dog in the Night-Time", Mark Haddon has his > young autistic hero verify that the earth is not flat by holding up a steel > straight-edge to the horizon. > > In the course of private research into public gullibility, I tried to > replicate Haddon's thought-experiment – and failed! Try as I might, I could > not see the slightest discrepancy between the horizon and my straight-edge. > > I strongly suspect that Haddon hasn't performed the experiment himself, but > simply written down what he thought one "ought" to see. > > So what am I to conclude? > (a) The earth is flat. > (b) The curvature is too small to be seen this way. > (c) Don't believe everything you read in novels. > > Let's go with (b), and try to calculate the height D of the "arch" of the > horizon over the ruler's edge, and show it's too small to observe by the > naked eye. > > There's a diagram at: > > https://whitbywriters.wordpress.com/2018/02/26/the-curious-incident-of-the-dead-flat-horizon > where I discuss the matter for non-mathematical readers. There I baldly > state that D is given by the formula: > > D = R * (1 – cos arcsin L/R) > > where L is the half-length of the straight-edge and R the radius of the > earth. > > Have I got it right? > > I calculated D = 1.4 nm, but omit to say how. (I used J of course). > > R=: 6378163 NB. equatorial radius of earth (m) > L=: 0.15 NB. half-length of metal ruler (m) > cos=: 2&o. > arcsin=: _1&o. > smoutput D=: R * (1 - cos arcsin L%R) > 1.41624e_9 > > Now 1.41624e_9 m is roughly 1.4 nm, less than the width of a DNA molecule. > > But I suspect that my estimate of D is unreliable. Moreover it could be out > by several orders of magnitude. Why? Because I'm calculating the arcsin of > a very small number, and then calculating the cosine of the resulting very > small angle, which may be pushing (o.) beyond its limits. > > Now I could use Pythagoras instead of trigonometry: > > D = R - √(R²-L²) > > which avoids using (o.) and may perform the entire calculation in extended > precision without conversion to (float). I'll also work in nanometres (nm), > not metres (m): > > R=: 6378163000000000x NB. (nm) > L=: 150000000x NB. (nm) > smoutput D=: R - %:(R*R)-(L*L) > 2 > > I thought I was getting 6-figure accuracy, so this result worries me. > 2 nm differs significantly from the result of the trig formula: 1.4 nm. > > Now (R*R)-(L*L) is extended precision, but (%:) is returning (float) not > (extended). Is it also corrupting the result? Is "2" just an artefact of > its algorithm? > > So let's work in picometres, to see if the "2" holds up under greater > precision: > > R=: 6378163000000000000x > L=: 150000000000x > smoutput D=: R - %:(R*R)-(L*L) > 2048 > > It does. This is encouraging. But (%:) still returns (float). > > Let's try different ways of calculating the square root of an extended > precision result: > > sq=: ^&2 > sqr=: %: > smoutput D=: R - sqr (sq R)-(sq L) > 2048 > sqr=: ^&0.5 > smoutput D=: R - sqr (sq R)-(sq L) > 2048 > sqr=: ^&1r2 > smoutput D=: R - sqr (sq R)-(sq L) > 3072 > …which is even more worrying! > > So which is the most dependable approximation to D? > 3 nm > 2 nm > 1.4 nm > none of the above? > > Is it robust enough to warrant my original deduction? > > Have I made a logical error somewhere? > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
