Maybe the steel straight edge was some hundreds of meters long? An interesting point about this is that back when I worked in a machine shop for a summer we would commonly set keyway cuts by eye and one of the master machinists proved to me that it was possible to do so within 0.001" on 1" cold rolled stock with a keyway of .1" or .2" - but that is in the 25,000 nm range. The technique was to get the cutter close to the stock and look, errors were readily apparent when you compared the way the cylinder curved away on the two sides. This has relevance because this was a short straight edge as compared to a larger, nearly tangent circle and differences could be observed that were smaller than expected. Not 2nm, of course. Violet light is 400nm, I think that there is a lower limit of what you can see with light because of the wavelength.
I would be interested in the following - how long would the straightedge have to be before a separation of 1mm could be observed? I have certainly seen 1 meter metal straightedges running around so I would at least use that rather than the 30cm I think you have used. (Unless, of course, that was specified in the book). I was just listening to Expanded Universes by Robert Heinlein and he describes spending 2 days doing a problem of orbital mechanics by hand on paper (before computers and calculators) and having his wife check him - just to be sure that a one line throw away statement in a juvenile book was possible. I guess that is why, despite being very dated (read Starman Jones for a laugh regarding how there have been gross mispredictions about computers) I still find Heinlein's fiction readable - he did what he could to make as much of the science as could be right, right. 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 -- Of course I can ride in the carpool lane, officer. Jesus is my constant companion. ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
