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.