 nonsubscriber, please reply to sender and MMD 
Here's what I can contribute to the discussion. First, I must warn
you I'm an aeronautical engineer, not a musician.
For organ pipes, the wavelength of the sound is proportional to the
length of the pipe. The resonant air columns inside open pipes are
onehalf wavelength long, closed or stopped pipes are onequarter
wavelength long.
Wavelength is just the speed of sound divided by the frequency.
For instance, the wave length of the A4 with a frequency of 440 Hz is
approximately 21/2 feet. The speed of sound at normal temperatures
is about 1100 feet per second, which divided by 440 gives you 21/2.
An open pipe is half the wave length, or 11/4 feet. A stopped pipe
is 1/4wavelength or 71/2 inches long.
When it comes to tuning, the one variable is the speed of sound and
the only thing that affects the speed of sound is temperature; pressure
or density does not affect the speed of sound. The speed of sound is
proportional to the square root of the absolute temperature. To get
absolute temperature you add 273 degrees to temperatures in degrees
Celsius, or add 460 degrees for temperatures given in degrees
Fahrenheit.
This all boils down to a formula to determine the frequency to which
you would tune a pipe so that you get the desired pitch after the
temperature changes. Other responses to this message referred the
reader to various tables. I figured that if you had the basic
equation, some could use it to build their own tables, using either
a calculator or a spreadsheet program.
The formula is: Tuning Frequency = Desired Frequency x Square root
(tuning temperature/playing temperature) [absolute temperatures].
For instance, assume that at the time of tuning it is 60 degrees F.
and the next afternoon, when the instrument will be played, it will be
90 deg. F. And assume you want to tune the A4 so it will play 440 Hz.
You add 460 deg. to both 60 and 90 deg., divide, take the square root,
and multiply by the desired frequency of 440 to get 427 Hz.
427 Hz = 440 Hz x Square Root (460 + 60 / 460 + 90)
This is for degrees Fahrenheit; for degrees Celsius add 273.
Note that the factor is always the same for the pair of temperatures so
you only need a table of factors. Also, since this factor will be the
same for all the notes, it will be the same fraction of a note for all
of the notes.
This is getting to the limits of my knowledge. Hopefully someone can
pick it up from here.
Regards,
Ernie Tangren
Bedford, Texas
[ The simple table given by Audsley seems to be a straightline
[ approximation and it is probably okay for the limited temperature
[ change encountered in a large church. The exact formula above
[ will be better for the large temperature excursions experienced
[ under clear skies at a fairground.  Robbie
