Landing on a sloping runway with different wind velocities

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.

"Andrew Sarangan" wrote in message
oups.com...

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.

That doesn't seem right to me. How much headwind do you
need to land in the normal distance if the slope is 2 degrees when
you need 20 knots for 0.12 degrees?

Tony Cox wrote:
"Andrew Sarangan" wrote in message
oups.com...

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.

That doesn't seem right to me. How much headwind do you
need to land in the normal distance if the slope is 2 degrees when
you need 20 knots for 0.12 degrees?

You are correct, I had a typo when I computed the equations in Excel.
The correct answer is a 10-knot wind is equivalent to 4-deg slope.
20-knot wind is equivalent to 8-deg slope. So it seems each knot is
worth 0.4-deg of runway slope. This is rather surprising to me because
it seems that runway slope is almost irrelevant to landing distance
compared to the effects of wind.

"Andrew Sarangan" wrote in message
ups.com...
[...] This is rather surprising to me because
it seems that runway slope is almost irrelevant to landing distance
compared to the effects of wind.

As I pointed out earlier, this makes perfect sense. Landing distance is a
function of deceleration and of initial kinetic energy. Slope affects
deceleration in a linear way, while wind affects the kinetic energy in an
exponential way. Furthermore, the net effect of the wind is doubled when
comparing headwind to tailwind operations.

In the example I used, for a typical light airplane, a 10 knot wind produces
a landing distance that is different by a factor of two, comparing headwind
to tailwind. That is, it takes twice as much distance to come to a stop
landing with a tailwind than landing with a headwind. It would take a
pretty significant slope indeed to increase deceleration to the point that
the landing distance was cut in half.

Pete

Andrew Sarangan wrote:
Tony Cox wrote:
"Andrew Sarangan" wrote in message
oups.com...

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.

That doesn't seem right to me. How much headwind do you
need to land in the normal distance if the slope is 2 degrees when
you need 20 knots for 0.12 degrees?

You are correct, I had a typo when I computed the equations in Excel.
The correct answer is a 10-knot wind is equivalent to 4-deg slope.
20-knot wind is equivalent to 8-deg slope. So it seems each knot is
worth 0.4-deg of runway slope.

Which, I should say, is pretty much in agreement
with my analysis. A 2-degree slope is equivalent
to 6 knots.

This is rather surprising to me because
it seems that runway slope is almost irrelevant to landing distance
compared to the effects of wind.

To me as well. This has been an interesting thread.