John,
I'm afraid I don't follow what you've laid out.
Forget the diagram they showed on the television - that part doesn't work.
They insisted on showing the sun at the intersection of the two curves,
directly above the house location - it doesn't work that way but they
thought it would have dramatic appeal.
You are right - as I explained in an earlier posting, you have 2 equations
with 2 unknowns:
sin alt = sin lat sin dec + cos lat cos dec cos (std.time - eot - long +
cent.merid)
This is essentially just the standard equation for the solar altitude, but
expressed in terms of standard time rather than apparent time. There are 7
variables in this equation. Calculate the solar altitude (alt) from the
shadow and known height of the hoop in the photo. The photo gives the
standard time (std.time, expressed in the equation as an angle before or
after standard noon) and the date, which in turn gives the solar
declination
(dec) and equation of time (eot). Knowing that the photo is taken
somewhere
in the general area of Los Angeles tells us the time zone of the location -
therefore giving us the central meridian (cent.merid.) Thus, we know the
values of 5 of these variables. That leaves 2 unknowns in the equation
determined by any one photo. We have 2 different photos, so we have 2
equations in 2 unknowns. Solve these for lat and lon.
Fred Sawyer
----- Original Message -----
From: <[EMAIL PROTECTED]>
To: <[email protected]>
Sent: Tuesday, February 14, 2006 7:37 AM
Subject: Re: Numb3rs
I'm embarrassed to say I still don't get it. When they first showed two
different photos, I thought, "two equations, two unknowns" but then later
there were a bunch of words and the sun going 'round the earth in two
circles almost touching and I got all messed up.
So let me tell you how I would solve it and then you can tell me, "Hey,
that's the same thing only different, yuh retard!" only out of
consideration for the gentility of the list, you can omit the "yuh
retard" bit.
Say you have a shadow angle A at 10 am and a shadow angle B at 1 pm. So
first you go to the standard line of longitude two hours east of your
standard line of longitude and lay the analemma on it and find the
subsolar point for the date. Then around the subsolar point you draw a
circle on the surface of the earth which is the locus of all points at an
angle A from the subsolar point as measured from the center of the earth.
That circle has all the places which have the first (10 am) shadow
length. Then you go to the standard line of longitude one hour west of
your standard line of longitude and do the same thing mutatis mutandis
for the 1 pm shadow angle.
So now you have two circles drawn on the surface of the earth and where
they intersect are the only two possible loci (relative to the standard
line of longitude) of the abode of him who done the dirty deed.
Is there an intuitive way to get from my earthbound way to solve it to
how they solved it on the show?
Thanks,
John Bercovitz
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