there is diffraction when light travels inside atmosphere. On Feb 28, 2018 12:16 PM, "Don Kelly" <[email protected]> wrote:
> If the observer is standing at some point on the surface of the earth.and > is facing in a given direction with the straight edge. you are looking as > you describe. If facing another direction, you are , by symmetry, looking > at the same curvature and the same distance to the horizon. You are looking > at a great circle in any direction. Replace the straight edge by a barrel > band of some radius -with you at the center- whichever way you look there > is symmetry which reflects the statements "if the height... radius" In > theory the distance from the eye to the end of a straight ruler is > different from that to the center of the ruler-but does this give a > measurable effect? Not with a 30 cm ruler 2 m above the surface of the > earth. Certainly the horizon is a small circle but looking out in any > direction from a given point close to the surface of the earth is looking > along a line tangent to a great circle. I'm not sure that there is a need > to deal with spherical trig because of the symmetry. However, my last use > of spherical trig was about 1952 so i am rusty (and not standing at the > center of the celestial sphere or even the earthly sphere). > > Don > > > On 2018-02-27 1:04 PM, J. Patrick Harrington wrote: > >> Someone once told me that the only phrase an academic needs in any >> language is "Well, it's not that simple ..." >> >> If the height of the observer is 0, then the horizon is the plane through >> the observer perpendicular to the zenith direction - it's a great circle >> and there is no curvature. But if you're above the surface, the horizon >> is at the distance d where your line-of-sight is tangent to the earth's >> surface. The distance to the horizon is then d=sqrt[2Rh+h^2], from the >> Pythagorean triangle d^2 + R^2 = (R+h)^2 where R=Earth's radius. In J >> >> NB. The angle from the zenith to the horizon will exceed >> NB. 90 degrees by an amount depending upon the height h >> NB. of the observer above the ocean surface. This excess >> NB. is called the "dip" of the horizon. (This formula >> NB. neglects atmospheric refraction which isn't justified!) >> NB. Usage: hdip h --> angle to horizon in excess of 90 deg >> hdip=: monad define >> R_E=. 6378e3 NB. Earth's equatorial radius in meters >> h=. y NB. height of observer in meters >> d=: %: (2*R_E*h)+ h*h NB. distance to horizon >> NB. dip=: asin d% R_E+ h >> dip=: atan d%R_E >> dip%cdtr >> ) >> cdtr=: 1p1%180 NB. convert degrees to radians >> >> E.g., If you're 10m above the surface >> hdip 10 >> 0.10146 >> and >> d >> 11294.3 >> So the horizon is 11.3 km away and appears 0.1 degree below the >> horizontal plane." At this point, we must do some spherical trig. (I've >> taught observational astronomy, so it's no problem for me :-) Imagine >> you are at the center of the celestial sphere. Your staright-edge will >> project on to the sphere as a *great circle*. But your horizon is *not* >> a great circle, it is a "small circle" cut into the celestial sphere by >> a plane that does not pass through the center (you). (A small circle is >> like a circle of constant latitude on the globe away from the equator.) >> So if your straight-edge touches the horizon at two points, it will pull >> away from the horizon it the middle. (Just like your plane doesn't fly >> along a line of constant latitude when it takes a great circle route from >> NY to Spain.) I've tried to impliment the spherical trig here (I'm sure >> it's worked out carefully somewhere.) >> >> NB. arc of horizon = the angular extent of the straight-edge >> NB. Usage: (arc of horizon) curve (dip of horizon) ===> Z >> NB. Z = angle of horizon above straight edge at its midpoint >> NB. all angles in degrees >> curve=: dyad define >> alpha=: cdtr*x >> a=: 0.5p1 - cdtr*y >> s=: acos (*: cos a)+ (*: sin a)*cos alpha >> chs=: cos -:s >> c=: acos (cos a)%chs >> del=: a - c >> del%cdtr >> ) >> >> So for a straight-edge the subtends 60 degrees of the horizon and an >> observer 10 meters above the sea, we find >> >> 60 curve (hdip 10) >> 0.0156959 >> >> The horizon is ~0.016 degree above the mid-point of the ruler, >> about a minute of arc. >> >> For the space station at h= 400 km, we get >> 60 curve (hdip 400e3) >> 2.77154 >> >> an easily noticed ~3 degrees. Or with a wider ruler, spanning 90 deg >> 90 curve (hdip 400e3) >> 7.17807 >> >> I think this makes sense, but I offer no guarantees. >> Patrick >> >> On Tue, 27 Feb 2018, Don Kelly wrote: >> >>> I agree with the series approach but the final result depends on >>> division which has limited accuracy. I would suggest using the following >>> which uses 100*L%R as a rational fraction >>> >>> (10^_2)*(1r2)* 175r637816300 >>> >>> 1.37187e_9 >>> >>> >>> However, the solution is wrong in any case. It is the result of using a >>> circle and a line parallel to the tangent line at any point. In other >>> words, looking at the earth from a great distance. >>> >>> The right answer (for a perfect sphere) is 0 as the curvature of the >>> earth is the same in every direction from the observer. There is no "arch" >>> It seems that the old Greeks had it right- as ships went off into the >>> distance, they didn't just shrink and remain whole but would also disappear >>> gradually from the bottom up. What Patrick Harrington said is at the core >>> of what is actually observed. >>> >>> >>> Don Kelly >>> >>> >>> On 2018-02-26 10:10 PM, J. Patrick Harrington wrote: >>> >>>> The way to avoid loss of accuracy is to expand in a binomial series: >>>> D = R - √(R²-L²) = R[ 1 - (1 - (L/R)^2)^(1/2)] >>>> = R[ 1 - (1 -(1/2)(L/R)^2 + ...)] = (1/2) L*L/R >>>> and thus >>>> ]D=. -:L*L%R >>>> 1.76384e_9 >>>> >>>> On Tue, 27 Feb 2018, Ian Clark 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/forum >>>> s.htm >>>> ---------------------------------------------------------------------- >>>> For information about J forums see http://www.jsoftware.com/forums.htm >>>> >>> >>> ---------------------------------------------------------------------- >>> For information about J forums see http://www.jsoftware.com/forums.htm >>> >>> ---------------------------------------------------------------------- >> For information about J forums see http://www.jsoftware.com/forums.htm >> > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
