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/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
