I had to laugh when I watched the show last night, too. To a 100th of a
degree? As if the basketball hoop's pole is not tilted at all or bent
(not to mention the litany of other sources of observational error)?!
Of course these and many other expected sources of observational error were
pointed out ad nauseam to the show's writers - but there is something called
suspension of disbelief that needs to be used in such instances. If you
really want to get picky, it should be pointed out that even with a
precision of a 100th of a degree, it would not be possible to pick out a
single house.
... The declination with the calculated solar altitude, will give you the
information needed to determine latitude.
How? A single application of the standard formula relating altitude,
declination, latitude and hour would also require that you know the local
solar time of the event in order to find latitude. And that of course would
require that you know the longitude adjustment.
We can also find the local solar time of the event.
An explanation of how would be helpful at this point.
I provided the basic approach in an article in The Compendium.
sin alt = sin lat sin dec + cos lat cos dec cos (std.time - eot - long +
cent.merid)
Charlie calculates 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.)
We have an equation with two unknowns: latitude (lat) and longitude (long.).
Graphically, we can represent this equation by using the magic of television
to draw a curve in the sky above all the places whose latitude and longitude
satisfy the equation.
If we do the same thing for the second photo we obtain a slightly different
curve. The two curves will intersect at a single point - directly above the
location at which the photos were taken.
Fred Sawyer
----- Original Message -----
From: "Sara Schechner" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>; <[email protected]>
Sent: Saturday, February 11, 2006 2:15 PM
Subject: Re: Numb3rs
Hi John,
You wrote: "OK, then what do we do to compute latitude and longitude to a
hundredth of a degree?"
I had to laugh when I watched the show last night, too. To a 100th of a
degree? As if the basketball hoop's pole is not tilted at all or bent
(not to mention the litany of other sources of observational error)?!
Still the method is the same as if one is computing the length and
direction of a shadow for a given latitude and longitude. Usually we
start with lat. and long. and the shadow's length and position are
unknown. In this case, we work backwards. The lengths of the shadow and
pole give you the altitude of the sun by simple trig. On a given date,
the sun will have this altitude twice (morning or afternoon). We know the
date and approx. time from the camera. The date allows us to look up the
sun's declination. The declination with the calculated solar altitude,
will give you the information needed to determine latitude. We can also
find the local solar time of the event. Corrections for the equation of
time will offer us mean clock time at that location. If the camera's
time stamp were precise for that time zone (a big if), we could determine
how far east or west we were of the center of the zone and so compute
longitude.
All the best,
Sara
Sara Schechner, Ph.D.
David P. Wheatland Curator
Collection of Historical Scientific Instruments
Department of the History of Science
Harvard University, Science Center 251c
Cambridge, MA 02138
Tel: 617-496-9542
Fax: 617-496-5932
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