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I've been thinking a bit more, following the guidance of Herbert and Gary.
Perhaps, having slept on the problem, I'm getting a better idea about how
Wright's method worked, back in 1600, for getting great-circle distance by a
geometrical construction.

Gary's simplified versions of Wright's drawing, that he made back in NavList
3253, labelled "illustration.pdf", have been really helpful.

Why anyone should have wanted to know the great circle distance, London to
Jerusalem, before the days of long-distance air travel, rather defeats me.
Perhaps there were religious arguments involved. Anyway, that was the
example Wright chose, so let's follow it.

Wright starts by picturing the Earth as seen from above the North Pole, with
a scale of longitudes shown around the equator. The Earth is treated as
spherical, and with a radius equal to 1 unit. Radial lines are drawn in,
corresponding to London and Jerusalem, with a longitude difference between
them of 46 degrees, reaching to H and M respectively. He chooses a zero of
longitude that's way out in the Atlantic, but that is irrelevant.

He then draws in points E and N, displaced from H and M by the corresponding
latitudes, 51 deg 32', and 32 deg. These don't represent angles in the plane
of that diagram at all, but actually angles measured upwards from the
equator, as latitudes are. Then, if he drops a perpendicular to each
corresponding radius, meeting it at F and O, that puts F at a radius from
the centre equal to cos 51d 32', and O at a radius of cos 32d. These are
exactly the radii you would see London and Jerusalem placed if you took a
photo of the Earth, from a camera way above the North Pole. So F and O are
projections of the positions of London and Jerusalem, on to the plane of the
Earth's equator.

I was thinking of that Northern hemisphere of the Earth as rather like a
Christmas pudding (to add a seasonal note). Not the Dickensian cannonball
type, put an exact hemisphere, sitting on a flat plate through its equator.
You might add a sprig of holly at the North Pole if you felt so inclined.
You can mark London and Jerusalem in their right places, perhaps with a
couple of sultanas.

Now comes the step I was having such difficulty in visualising. Take a
really sharp knife or a cleaver, and make an exactly vertical cut, passing
through both London and Jerusalem. Remove the smaller piece, and look at
what's left. On that vertical face, you can draw a vertical line, between
the sultana at London and its projection below it on the plane of the
equator, at F. Similarly, another between Jerusalem and point O.

What we are after, is the length of the line between London and Jerusalem,
as it you had bored a dead-straight tunnel through the Earth between those
points. Or more realistically, the immersed depth of a skewer, poked into
the pudding in London, to emerge at Jerusalem. We can get that length by a
bit of Pythagoras, using the horizontal distance between them, which is FO
on Wright's diagram, and the vertical difference, which the difference
between the two sines of the latitudes. We could sketch that scaled triangle
anywhere, to find the length of its diagonal, but it's convenient to do it
using Wright's diagram, even though that was drawn on a different plane.

The two sines are the lengths EF and NO, and we can subtract them by copying
NO on top of EF, starting from E, and the resultant length FQ is their
difference; one side of the right-angled triangle that's needed.
Conveniently, there's already another line drawn in at right-angles to that;
the line between the centre of the circle and point F. So we copy the length
FO on to that line, starting from F, to arrive at a new point P. Now the
right-angled triangle PFQ is exactly the same as we needed to provide the
straight-line distance (through the tunnel) between London and Jerusalem. So
its hypotenuse, PQ, is just that straight-line distance that was needed.

Now, there's one final step. We didn't want the straight-line distance, but
the distance around the curved surface of the Earth, measured in degrees.
This is in yet another, different plane, and we know its chord, which was
that straight-line distance. Simply convert chord to angle, against the
angular scale, by stepping-off a distance, PQ, against that scale, using the
dividers.

If the two points happened to be in different hemispheres, the only
difference would be that the sines of the latitudes would be added rather
than subtracted.

So, after all that, I've convinced myself that Wright's method works, and
that I can follow, more or less, how it works. What impresses me,
particularly, is the precision of the answers that he arrived at, even
though there were so many steps in his procedures, every one of which had to
be got just right.

George.

contact George Huxtable at george@huxtable.u-net.com
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.


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