Landing on a sloping runway with different wind velocities

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.

It really depends upon how much of slope. I'm sure there is a given
amount of upslope to a runway that could cancel the effects of a slight
tail wind. Just exactly how much would depend upon a given airplane and
how it is loaded... and how effective the brakes are.

I would prefer to land into a headwind always unless my brakes were
really bad.


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.

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

I had to dust off my old HP calc for this and came up with an assumed 3000
ft runway being 157 ft higher at one end than the other?!?. Where is this
"home airport"?

In article %PyWg.2774$YD.1017@trndny09,
"Mike Isaksen" wrote:

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

I had to dust off my old HP calc for this and came up with an assumed 3000
ft runway being 157 ft higher at one end than the other?!?. Where is this
"home airport"?

You shouldn't need a calculator for this one. 3 degrees is 1:20. Just
like an ILS glide slope.

A downwind take-off into higher terrain is a bad thing.
There are no hard and fast rules. With a landing on a
one-way runway or with a tailwind, there will be a point at
which a go-around can be made safely and once you are below
that point, a landing is your only option. That's why
mountain and hill flying is a challenge



"Tony Cox" wrote in message
ups.com...
| 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.
|

He may have meant 3 % grade, that is 90 feet. I once landed
at a strip in Wyoming that was about 300 feet higher on the
south end than the north. It was a one-way runway. Can't
find it on the charts any more, they must have closed it.


"Mike Isaksen" wrote in message
news:%PyWg.2774$YD.1017@trndny09...
| "Tony Cox" wrote in message ...
| Its more than an academic question for me. My home
airport
| has a 3 degree runway, and some local airports are even
| steeper.
|
| I had to dust off my old HP calc for this and came up with
an assumed 3000
| ft runway being 157 ft higher at one end than the
other?!?. Where is this
| "home airport"?
|
|

"Tony Cox" wrote in message
ups.com...
[...]
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.

I have occasionally thought about trying to treat the problem
mathematically, but so far haven't had enough motivation to do so. It's a
very complicated problem, mathematically speaking (assuming you're not
someone who does this sort of math on a daily basis, and I'm not). I
suspect that in addition to looking at touch-down speed and gross-weight,
along with runway slope and wind velocity, you would also have to include
some measure of braking performance (maybe this is somehow derivable from
the POH roll-out distances).

A 3 degree slope sounds pretty steep to me. Are you sure it's not 3%?

As far as general rules of thumb go, the one I've heard is that 1% of slope
is worth about 10 knots of headwind *for a takeoff*. This is not
necessarily applicable to the landing case, which is what you're asking
about, but it's at least related. In this rule of thumb, take the runway
slope in percent, multiply that by 10, and if you've got a headwind less
than that, operating downslope is better for a takeoff (upslope for a
landing, if you apply the same rule of thumb).

Personally, that rule of thumb seems optimistic to me, but I don't have any
good justification for doubting it. Still, it's worth considering the fact
that a headwind or tailwind affects the takeoff or landing differently than
slope. That is, the wind speed affects the total velocity change required,
while the slope affects the acceleration available. Even if you take off
uphill but upwind, while the acceleration will be less, so may the runway
used since you need a lower total velocity change to reach takeoff speed.
Likewise, landing downhill but upwind, yes your deceleration is less but you
also need less reduction in speed to come to a stop.

More importantly, the change in acceleration or deceleration is linear,
while the difference in total speed change is exponential.

To me, that suggests that if you're going to err, it's better to choose the
headwind over slope when in doubt, since a good headwind is beneficial to
the exponentially related parameter, while the slope is only beneficial to
the linearly related parameter.

That said, like I said I haven't taken the time to look at any of this in a
rigorous mathematical way, so I might have made a mistake in some
assumptions. Still, I have to say that the one time I ever took off
downwind but downslope, I sure used a lot more runway than I thought I was
going to.

Pete

Uphill or down, land into the wind seems to work best. Same goes with
Seaplanes and rivers, land into the wind. But....sometimes with
obstructions and whatnot it's not safe to land at all. Find another
field.

Recently, Tony Cox posted:

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.

Having done that (Jaffrey Airport, NH), all I can suggest is that you go
with the head wind, whichever way that is. My first approach was to land
going "uphill", but on final the wind shifted to a tailwind and it made
things pretty dicey. So, I went around and landed downhill. The roll-out
was a quite bit longer than I'm accustomed to, but other than that it was
uneventful. When I left a couple of days later, the wind favored taking
off "uphill", and that was an experience, as well. It gave the folks at
the ice cream stand about 500' off the departure end something to gawk at.

Neil

"Peter Duniho" wrote in message
...
[...]
More importantly, the change in acceleration or deceleration is linear,
while the difference in total speed change is exponential.

I should clarify what I mean, since the way I wrote it seems a little
confusing:

The changes themselves are both linear, but the *consequences* (in terms of
distance added or removed) is exponential when changing the speed.