Landing on a sloping runway with different wind velocities

"Kyle Boatright" wrote in message
news

"Jerry" wrote in message
. ..
Fly a 1946 BC12-D Taylorcraft. No wind, no slope takeoff and landing is
about 300-400 feet. Stall speed about 35 mph. 10 knot tailwind uphill
estimated landing is 500 ft. Downhill landing 10 knot headwind over
obstacles uses estimated 1000 to 1200 ft of runway. I only takeoff
downhill. Unless you have experience with a slope you will
underestimated the effect of the slope.

On long paved runways, I always land into the wind.

Jerry in NC

OK, your no slope, no wind takeoff and landing are 300-400'.

What is the downhill, no wind takeoff distance?

What is the uphill, no wind landing distance?

KB


I don't takeoff uphill.

Downhill, no wind takeoff is estimated 250 feet.

Remember that I have obstacles at the end of the uphill runway and none at
the end of the downwind runway.

Jerry in NC

"Jerry" wrote in message
. ..

"Kyle Boatright" wrote in message
news

"Jerry" wrote in message
. ..
Fly a 1946 BC12-D Taylorcraft. No wind, no slope takeoff and landing is
about 300-400 feet. Stall speed about 35 mph. 10 knot tailwind uphill
estimated landing is 500 ft. Downhill landing 10 knot headwind over
obstacles uses estimated 1000 to 1200 ft of runway. I only takeoff
downhill. Unless you have experience with a slope you will
underestimated the effect of the slope.

On long paved runways, I always land into the wind.

Jerry in NC

OK, your no slope, no wind takeoff and landing are 300-400'.

What is the downhill, no wind takeoff distance?

What is the uphill, no wind landing distance?

KB


I don't takeoff uphill.

Downhill, no wind takeoff is estimated 250 feet.

Remember that I have obstacles at the end of the uphill runway and none at
the end of the downwind runway.

Jerry in NC

Jerry,

I never asked about uphill takeoff performance.

Again, what is the uphill, no wind landing distance?

Thanks,

KB

"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?

"Kyle Boatright" wrote in message
. ..

"Jerry" wrote in message
. ..

"Kyle Boatright" wrote in message
news

"Jerry" wrote in message
. ..
Fly a 1946 BC12-D Taylorcraft. No wind, no slope takeoff and landing
is about 300-400 feet. Stall speed about 35 mph. 10 knot tailwind
uphill estimated landing is 500 ft. Downhill landing 10 knot headwind
over obstacles uses estimated 1000 to 1200 ft of runway. I only
takeoff downhill. Unless you have experience with a slope you will
underestimated the effect of the slope.

On long paved runways, I always land into the wind.

Jerry in NC

OK, your no slope, no wind takeoff and landing are 300-400'.

What is the downhill, no wind takeoff distance?

What is the uphill, no wind landing distance?

KB


I don't takeoff uphill.

Downhill, no wind takeoff is estimated 250 feet.

Remember that I have obstacles at the end of the uphill runway and none
at the end of the downwind runway.

Jerry in NC

Jerry,

I never asked about uphill takeoff performance.

Again, what is the uphill, no wind landing distance?

Thanks,

KB




Sorry that I misunderstood. Estimated uphill no wing landing is 200-300
feet.

My main reason getting involved with this thread is to point out that there
is no single best answer. Choice depends on aircraft, pilot, surface,
obstacles, slopes. In the past, I thought that you should only land into
the wind.

Jerry in NC

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.

Actually, there was a reasonable mathematical treatment printed in one
of the aviation magazines a couple years back. The upshot was that in
anything less than a gale you should land uphill and take off downhill.
And if it is a gale, maybe you should think about not flying. It takes
an enormous amount of wind to overcome the effect of a sloping runway.

"Jerry" wrote in message
. ..

"Kyle Boatright" wrote in message
. ..

"Jerry" wrote in message
. ..

"Kyle Boatright" wrote in message
news
"Jerry" wrote in message
. ..
Fly a 1946 BC12-D Taylorcraft. No wind, no slope takeoff and landing
is about 300-400 feet. Stall speed about 35 mph. 10 knot tailwind
uphill estimated landing is 500 ft. Downhill landing 10 knot headwind
over obstacles uses estimated 1000 to 1200 ft of runway. I only
takeoff downhill. Unless you have experience with a slope you will
underestimated the effect of the slope.

On long paved runways, I always land into the wind.

Jerry in NC

OK, your no slope, no wind takeoff and landing are 300-400'.

What is the downhill, no wind takeoff distance?

What is the uphill, no wind landing distance?

KB


I don't takeoff uphill.

Downhill, no wind takeoff is estimated 250 feet.

Remember that I have obstacles at the end of the uphill runway and none
at the end of the downwind runway.

Jerry in NC

Jerry,

I never asked about uphill takeoff performance.

Again, what is the uphill, no wind landing distance?

Thanks,

KB




Sorry that I misunderstood. Estimated uphill no wing landing is 200-300
feet.

My main reason getting involved with this thread is to point out that
there is no single best answer. Choice depends on aircraft, pilot,
surface, obstacles, slopes. In the past, I thought that you should only
land into the wind.

Jerry in NC

And thanks for your posts. I am/was curious as to how much of an advantage
a sloping runway could be if you could land uphill and depart downhill.
Sounds like at least 25% reduction in required field length.

That mountainside property is looking better by the day. ;-)

KB

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

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

OK Guys, there is supposed to be parentheses around the last two symbols,
as:

Vbe = (s * d) / (5 * V)

OK, if anyone is still following this thread with interest, I've
just done the calculation myself and come up with

Vbe = (s * d) / (7*V)

which is pretty much the same. For Vbe = V/2 it underestimates
by around 25%, being only truely accurate when Vbe V.

No idea what Sparky's assumptions were, but for mine,
I assumed that the acceleration during take-off is constant,
which seems reasonable with a constant speed prop and
ignoring the deceleration caused by the increase in parasitic
drag with velocity (which is assumed to be much less that the
acceleration the engine is giving).

Note that this isn't really what I was expecting -- I'd have thought
that wind would be more important. For my 182 on a
2degree grade on a hot summer day, I should take off
downhill only if the tailwind is less than 4 knots. Otherwise,
its best to take off uphill and into the wind. I'd really thought
the break-even point ought to be higher!

Now for *landing*, the calculation is likely to be more
involved. For a start, the deceleration profile is more complex.
One has the parasitic drag (proportional to square of airspeed),
and the deceleration due to brakes (which, when maximally
applied, are proportional to the weight of the plane as it is
transferred from the wings to the wheels). The former isn't
by any means negligible. The latter depends highly upon
pilot technique (how fast you can get the nose down) and
runway surface.

When I have a spare moment, I'll crunch the numbers on that too.

"Tony Cox" wrote in message
oups.com...
OK, if anyone is still following this thread with interest, I've
just done the calculation myself and come up with

Vbe = (s * d) / (7*V)

You didn't post the derivation, so it's impossible for anyone to know
whether you did it correctly or not.

which is pretty much the same. For Vbe = V/2 it underestimates
by around 25%, being only truely accurate when Vbe V.

No idea what Sparky's assumptions were, but for mine,
I assumed that the acceleration during take-off is constant,
which seems reasonable with a constant speed prop and
ignoring the deceleration caused by the increase in parasitic
drag with velocity (which is assumed to be much less that the
acceleration the engine is giving).

They seem like reasonable assumptions. Parasitic drag only starts to get
really dramatic at L/D max speed, as induced drag falls off, so drag during
the takeoff run (when both induced and parasitic are minimal) seems
ignorable for the purpose of a rule of thumb.

Note that this isn't really what I was expecting -- I'd have thought
that wind would be more important. For my 182 on a
2degree grade on a hot summer day, I should take off
downhill only if the tailwind is less than 4 knots. Otherwise,
its best to take off uphill and into the wind. I'd really thought
the break-even point ought to be higher!

I don't understand what you mean. The lower the break-even point based on
wind speed, the more important wind is. Expecting the break-even point to
be higher implies that you expected wind to be less important, not more.

Now for *landing*, the calculation is likely to be more
involved. For a start, the deceleration profile is more complex.
One has the parasitic drag (proportional to square of airspeed),
and the deceleration due to brakes (which, when maximally
applied, are proportional to the weight of the plane as it is
transferred from the wings to the wheels). The former isn't
by any means negligible. The latter depends highly upon
pilot technique (how fast you can get the nose down) and
runway surface.

IMHO, the former is just as negligible during landing as it is during
takeoff, assuming you are landing at a typical near-stall airspeed, and for
the same reasons.

I agree that braking depends on pilot technique, but assuming you get the
nosewheel down, the AoA is too low for the wings to be making a lot of lift.
If you don't get the nosewheel down, then you've got induced drag helping to
slow the airplane, offsetting the reduced brake performance. For most of
the rollout, the weight on the ground is less than total weight, I
agree...but again, for the purpose of a rule of thumb I think it's
ignorable.

Pete

Tony Cox wrote:
"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...

Sparky's formula sounds about right. I did it by solving the
differential equations and came up with a 5 knot wind equivalent to a
2-degree runway slope. 1 knot of wind was equivalent to 0.4-deg slope.
So the answer seems to be, land into the wind unless you have a huge
slope.

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.