Landing on a sloping runway with different wind velocities

On 2006-10-09, Doug wrote:
Uphill or down, land into the wind seems to work best. Same goes with
Seaplanes and rivers,

There's a good trick question. Should you always land uphill at a sloping
seaplane base?

--
Ben Jackson AD7GD

http://www.ben.com/

"Peter Duniho" wrote in message
...
"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).

Me too (on motivation), and I'm supposedly qualified to work
it out for myself, but was wondering if anyone had done it before!

Thinking about it after posting, in practice I think it might get even
more complicated. Ever worked out how accurately you can "spot"
a landing? I think of myself as a pretty average sort of private pilot,
and I'd guess I can pick a touchdown point to around +- 200 feet
or so on a good day. But I know I can do much better (calm wind)
on an upslope runway than a downslope one -- I suppose it is because
with an approximate 6 degree descent path, after factoring in the
runway slope (lets say 2 degrees) any deviation in approach angle
gets magnified by 2 (6+2=8 degree intercept vs. 6-2=4 degrees).
So, the actual "rollout" distance isn't really the thing to look at --
one ought to factor in your own personal "uncertainty" in
touchdown point too and look for the minimum runway length
(rollout + anticipated touchdown point uncertainty) that you'll
use for landing. This would bias in favour of the uphill approach,
regardless of other factors. (Probably. Air speed affects
approach angle too!)

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

I'd like to be definitive, but my airport guide is in the plane,
and I don't seem to be able to find the info through google.
I *thought* it was 3 degrees (and I really ought to know,
having been based there 6 years), but I may be wrong. Anyway,
3% works out as 2 degrees. It's our crosswind runway at
61B (Boulder City, Nevada) if you have a book to hand.
Another local airport is 3.5 units (degrees or grade) at
Temple Bar (U30) which is a wonderful destination if you're
in the area. Wish I had my damn book!! Regardless, these
grades make a substantial difference to one's decision. The
runways, at 3500' or so, leave little room to be sloppy if you're
in a Mooney and landing when the DA is over 5000'.


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).

I've heard that too, and Sedona airport (KSEZ) goes as far
as to mandate which runway to use depending on wind (I think
they have a slope of 1.5 *units*, but their runway is so long it
hardly makes a difference to us little guys).

I think the takeoff calculation ought to be more straightforward.
You can use the whole runway, for a start, and all one needs
to do is to work out how much acceleration you sacrifice by
running up the grade. Clearly, braking effects are irrelevant.
Still, it'd take a bit to figure the math.


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.

I just feel more comfortable landing uphill. Perhaps its because
on approach you have a bigger target to hit!


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.

Me too, with my mother on board as well. Quite a
wake-up call.

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"?

It's our x-wind runway at Boulder City (61b). We may be
3%, or 2 degrees, but I don't have my guide to hand. The
runway is 3700' if I remember.

"Jim Macklin" wrote in message news:OOzWg.1482$XX2.259@dukeread04...
: 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.
:


or called it a mountain...

There's a good trick question. Should you always land uphill at a sloping
seaplane base?

Yes. Unless you're flying a barrel, in which case the tradition is to
land downhill.

Jose
--
"Never trust anything that can think for itself, if you can't see where
it keeps its brain." (chapter 10 of book 3 - Harry Potter).
for Email, make the obvious change in the address.

I have an 1800 ft grass strip that has a 3% grade. There is a 50 to 75 ft
obstruction off the high end of the runway and no obstructions off the end
of the low end of the runway. Slope and obstruction is more important than
wind. I will not land downwing unless the headwind exceeds 15 knots. The
deceleration uphill and acceleration downhill is more significant than the
usual winds. Below 200 ft on approach, you have to be committed to land.
Unless you have a very high power to weight you cannot do a go around.

Jerry in NC

"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.

"Jerry" wrote in message
...
I have an 1800 ft grass strip that has a 3% grade. There is a 50 to 75 ft
obstruction off the high end of the runway and no obstructions off the end
of the low end of the runway. Slope and obstruction is more important than
wind. I will not land downwing unless the headwind exceeds 15 knots. The
deceleration uphill and acceleration downhill is more significant than the
usual winds. Below 200 ft on approach, you have to be committed to land.
Unless you have a very high power to weight you cannot do a go around.

Jerry in NC

You get the big brass ones award. I don't have the stomach for a strip
where I have to commit so early.

In a no wind situation, what is your uphill landing roll on your strip
versus a level strip?

Same question for your downhill takeoff roll.

And, what do you fly?

Thanks for the info..

KB

"Tony Cox" wrote in message
oups.com...
[...]
Thinking about it after posting, in practice I think it might get even
more complicated. Ever worked out how accurately you can "spot"
a landing?

I know what you're saying, but all of the landing calculations are subject
to the variations you're describing. I think the basic calculations ought
to be able to be done independent of those, and you can apply the same
wiggle-room at the end that you have to for the usual landing distance
calculations.

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

I'd like to be definitive, but my airport guide is in the plane,
and I don't seem to be able to find the info through google.
I *thought* it was 3 degrees (and I really ought to know,
having been based there 6 years), but I may be wrong. Anyway,
3% works out as 2 degrees.

Well, 3 degrees is 50% steeper. That's a pretty big difference.

Interestingly, Airnav (which basically just republishes A/FD data) does
publish a runway gradient for Temple Bar, but not Boulder City. Though, it
does specifically warn against takeoffs from runway 33 at Boulder City due
to the gradient. Odd that they wouldn't publish the number.

Anyway, the published number has a very specific FAA definition, but it is
essentially just a percent grade. So if you've got a published number for
Boulder City that's called runway gradient, that's most likely what it is.

However you cut it, a 3% grade is certainly significant. Didn't mean to
imply otherwise. It's just that 3 degrees would have been even more
dramatic.

[...]
I've heard that too, and Sedona airport (KSEZ) goes as far
as to mandate which runway to use depending on wind (I think
they have a slope of 1.5 *units*, but their runway is so long it
hardly makes a difference to us little guys).

Driggs, Idaho is also 1.4%, and at 7300 feet seems quite long. But between
the 6200' airport elevation and the slope, both takeoffs and landings can be
interesting, especially done downwind. Sedona is lower at 4800', but the
runway is only 5100' and the gradient steeper at 1.9%. I'll bet things can
get interesting there too. I think even a little guy would want to take
care there.

I think the takeoff calculation ought to be more straightforward.
You can use the whole runway, for a start, and all one needs
to do is to work out how much acceleration you sacrifice by
running up the grade. Clearly, braking effects are irrelevant.
Still, it'd take a bit to figure the math.

Well, the landing distance calculation would make the same assumptions about
touchdown point and speed that one would make for normal landing distance
calculations as well. I agree that in general, takeoff calculations ought
to be more reliably repeatable than landing calculations, but that's not
unique to the question of sloped runways. So as a starting point, I'd say
one ought to just make those same assumptions and acknowledge that there's
some inherent error in being able to replicate the theoretical numbers.

I just feel more comfortable landing uphill. Perhaps its because
on approach you have a bigger target to hit!

Well, the steeper the hill, the safer you ought to feel. But there's
uphill, and then there's UPHILL. Distance to stop increases with the
square of your airspeed, so even a little extra speed translates into a lot
of extra landing distance. The difference between landing with a headwind
and landing with a tailwind is proportional to *twice* the wind speed (since
it's added as a tailwind and subtracted as a headwind). If your landing
speed is 60 knots and you've got a 10 knot wind, that's almost a DOUBLING of
landing distance comparing headwind to tailwind (70 knots is 40% faster than
50 knots, giving you a 96% increase in stopping distance).

You'd need a pretty good hill to halve that doubled landing distance to get
back to breaking even.

Pete

On 2006-10-09, 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??

Rule of thumb responses are interesting, but better would be
a full mathematical treatment. Presumably, a proper treatment

Tony, in "Mountain Flying Bible" the author Sparky Imeson was once
discussing this and a physics professor handed him a formula which is
published in that book. It is the beakeven wind speed for taking off uphill
into wind and downhill with a tailwind. Hopefully it formats correctly here.
If wind is less, takeoff downhill and if more take off uphill.

Vbe = (s * d) / 5 * V

Whe
Vbe = breakeven speed in knots
s = slope up in degrees
d = POH distance to liftoff with 0 slope and 0 wind in feet
V = volocity of liftoff speed in knots TAS

Sparky says that as a rule of thumb, if wind is less than 15 kts take off
downhill. If more than 15 kts take off uphill, provided obstacle clearance
can be maintained.

As in everything, academics and practical may not be the same, depending on
airplane's performance and pilot proficiency.

I've not tested the above, but Sparky Imeson is a recognized authority in
mountain flying and has done much of it as well as teaching many courses in
such. At least something to consider.

....Edwin
__________________________________________________ __________
"Once you have flown, you will walk the earth with your eyes
turned skyward, for there you have been, there you long to
return."-da Vinci http://bellsouthpwp2.net/e/d/edwinljohnson

"Edwin Johnson" wrote in message
. ..
On 2006-10-09, 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??

Rule of thumb responses are interesting, but better would be
a full mathematical treatment. Presumably, a proper treatment

Tony, in "Mountain Flying Bible" the author Sparky Imeson was once
discussing this and a physics professor handed him a formula which is
published in that book. It is the beakeven wind speed for taking off uphill
into wind and downhill with a tailwind. Hopefully it formats correctly here.
If wind is less, takeoff downhill and if more take off uphill.

Vbe = (s * d) / 5 * V

I looked through my copy of the MFB before posting,
but didn't find anything. Good book, but the information
in there is so disorganized its not surprising I drew a
blank. Sparky could do with a better editor!

When I have a spare moment, I'll see if I can verify
it. Of course, its only for take-off -- landing may be
different.


Whe
Vbe = breakeven speed in knots
s = slope up in degrees
d = POH distance to liftoff with 0 slope and 0 wind in feet
V = volocity of liftoff speed in knots TAS

OK, lets do a reality check. My 182 at full weight
supposedly takes 900' to lift off on a summer's day
at 61B. Runway 15-33 has a 2 degree slope. Lift-off
speed is 60 knots.

Vbe = (2*900)/5 * 60 = 21600

Seems a bit high to me. Perhaps Sparky meant

Vbe = (2*900)/(5*60) = 6 knots.

Well, I suppose its possible, but I'd have thought the
figure a little on the low side.


Sparky says that as a rule of thumb, if wind is less than 15 kts take off
downhill. If more than 15 kts take off uphill, provided obstacle clearance
can be maintained.

In this case, Sparky's ROT wouldn't work out too well...