Tony Cox wrote:
Here's a problem which seems to have a non-trivial solution.
At least, I've not been able to find a definitive answer to it, but
what do I know??

Suppose one wishes to land at an airport with a runway
that slopes at X degrees. The wind -- assumed to be directly
aligned with the runway -- is Y knots from the "high" end of
the runway.

Clearly, if Y is positive, one should try to land in the
"up-slope" direction to minimize one's ground roll. One
will be landing "up" and into a headwind. But what if
Y is negative? Clearly, if Y is just a few knots neg, one would
still land "up-slope", because the braking effect of rolling
out up-hill more than compensates for the higher landing
speed due to the tail wind.

If Y is negative and more substantial, which way should
one land? At some point, it makes sense to switch to the
other end of the runway -- landing downhill -- to take advantage
of the (now) headwind. But how does one establish which way
to land, assuming no clues from other traffic in the pattern? The
aim is to select a direction, given X and Y, which would result
in the smaller ground roll.

Rule of thumb responses are interesting, but better would be
a full mathematical treatment. Presumably, a proper treatment
would need to include touch-down speed too, and perhaps
gross weight as well.

Its more than an academic question for me. My home airport
has a 3 degree runway, and some local airports are even
steeper.

You asked for a mathematical solution, so here is my first attempt at
this:

One can write down the equations and solve this mathematically. For the
landing calculations, you will have a differential equation that
describes the forces as:
m dv/dt = -cv^2 + mgsin(t)
where m is the mass of the airplane, c is the aerodynamic drag coeff,
and t is the runway slope.

This can be recast as m dv/dx and solved for x as a function of v.

The touchdown ground speed will be (vs-w) where vs is the touch-down
airspeed, and w is the headwind. You have to assume that the airplane
will decelerate to some final speed vf, at which point braking action
will be used to stop the airplane. The reason you have to assume this
is because aerodynamic drag alone cannot bring an airplane to a
complete stop. It would lead to an infinite solution. The final speed
can be taken as vf = r*vs, where r is a number you can pick. r=0.1
might be a reasonable number; ie the airplane decelerates 90% due to
aerodynamic drag, and then the last 10% is reduced by braking action.

Given an airplane mass and zero-wind landing distance on a flat runway,
you can get the value for c. This can then be used to calculate the
landing distance for any wind or runway slope.

I have the formulas written out on paper, if you want me to post them I
can do that too (it would take some effort to write the equation using
ascii).

For a 2200 lb airplane with a normal landing distance of 1000 ft,
landing speed of 50 knots, a 10 knot wind is equivalent to a downslope
of 0.07-deg. In other words, you can land in 1000 ft if the downslope
is 0.07-deg with a 10-knot headwind. A 20-knot wind is equivalent to
0.12-deg downslope. If you land on a 0.2-deg downslope with no wind,
you will need 2000 ft of runway.

For takeoff performance, the starting equations have to be slightly
different because you have thrust from the engine (which has been
ignored in the landing calculations). I have not done that
calculations, but I am sure it can be done in a similar fashion.