That was a neat answer for an ellipse drawn on paper, as Frank posed the
problem. But, what if the ellipse is marked out on a patch of grass or
a walkway? Here are two websites that together give a graphical
approach to determining the axes of the ellipse:
first find the ellipse center
http://www.had2know.com/makeit/find-ellipse-center.html
then the ellipse axes
http://www.had2know.com/makeit/find-ellipse-axis-focus.html
Don Snyder
P.S. Come to St. Louis for the 2017 NASS Annual Meeting and a peek
at the total solar-eclipse that will occur. C U there (or here since
I'm already there).
On 10/29/2016 7:56 AM, Frank King wrote:
Dear Geoff,
I had a private list of people I
expected to respond and I have been
waiting for your answer!
I think that these instructions might
work if you ventured to the antarctic
circle during the southern winter and
then trecked to a position such that
your latitude is greater than 90-Dec
(so that the midnight sun is visible)
but less than 90. The shadow of the
stake would then be longest and point
to true north at true solar midnight.
Well, that is easily the best answer so
far!
I have just two teensy quibbles:
1. "during the southern winter"
You won't get too much sun then.
I suspect you meant southern
summer!
2. "your latitude is greater than 90-Dec"
This is difficult to express. In the
southern summer the Dec is negative.
Suppose Dec = -20 then you would
require me to be at a latitude
greater than 110 which is a bit of
a challenge!
Taking southern latitudes to be
negative and southern summer
declinations to be negative too,
The condition is:
Dec < -90 -Lat
So if you are at -70 then the
declination would need to be
less than (i.e. more negative
than) -20.
Alternatively you need:
Lat + Dec < -90
or:
Lat < -90-Dec
Almost using your words: you venture to
the antarctic during the southern summer
and choose any day when you have midnight
sun. The shadow of the stake would then
be longest and point to true north at
true solar midnight.
I feel sure that the writer of the
website didn't mean this!!
There is a practical consideration too...
Over a 24-hour period the shadow traces
out an approximate ellipse. [It is not
quite closed because the declination
doesn't stay still but that's not the
problem.] The difficulty is pinpointing
the ends of the major axis. The solar
altitude is changing very slowly at each
end and the end-points are not sharply
presented.
Ancillary Question:
I hand you a perfect ellipse drawn
on a sheet of paper. The ellipse is
unornamented. What is the geometrical
construction required to determine the
axes?
Aren't sundial questions fun?
Frank
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