Do I understand correctly that since 233 is a quarter of 900MHz, it
appears the rf loss will be about 12db less (simple free space) than
if I went to 900, and thus it would take 16 times as much power at
900MHz to achieve the same distance?

Alien-

At these frequencies, propagation is most likely limited to line-of-sight. If
your aircraft flies over the horizon, it won't matter how much power the
transmitter has.

If you do the math, I think you will find line-of-sight distances of many miles
are achievable with relatively low power, assuming the antennas maintain a
favorable orientation. If you couldn't maintain the orientation, or if the
aircraft was tumbling or spinning, you would need many dB of link margin to be
able to maintain the link.

Yes, there is a 20 log F factor that describes the propagation loss. Assuming
the same type of antenna, it is related to wavelength in that the effective
"capture area" of the receive antenna is related to wavelength. The
transmitted signal will have a power density of so many watts per square meter
when it reaches the receive antenna. Power received is the product of power
density and antenna effective capture area, assuming you have a favorable
relative polarization (orientation) of the two antennas.

The 233 frequency seems appropriate for what you want to do as far as
propagation characteristics are concerned. However, here in the US, it is in
the 225 to 395 MHz military communications band that includes government and
civilian aircraft communications among others. It would be a good idea to
coordinate use of the frequency with the closest government or military
frequency manager. If not, you could interfere with aircraft operations you
can't even hear from your location, not to mention possible loss of your
aircraft!

73, Fred, K4DII