Landing on a sloping runway with different wind velocities

"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

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

You're right. Do please have a look, since its important
these things be done right. Analysis applies to take-off
distance.

v = velocity
Vw = wind velocity
Vt = take-off speed
g = Acceleration due to gravity (32 ft/sec/sec)
d = distance
D = take-off distance for specific DA from POH on flat runway
S = runway slope
a = acceleration (assumed constant)
A = Acceleration (assumed constant) during TO on flat runway


v(t) = v0 + at (v0 start velocity, v(t) velocity at some arbitrary
t)

then distance travelled over time t will be

d = Integral(0,t){ v(t).dt} = v0*t + a*t*t/2

So, if starting from rest, find d given final velocity and acceleration

d = a*t*t/2
v = a*t

so

d = v**2 / (2*a)

and in particular, if level and no wind

D = Vt**2 / (2*A) or
A = Vt**2 / (2*D)

{Crosscheck my 182: Vt = 100ft/sec, D = 900ft, A=6.1ft/sec/sec, t to
take-off = 17 secs. All seems reasonable}

So lets compare uphill into headwind vs. downhill with tailwind.

Uphill into headwind.

The velocity we need to achieve is less -- only Vt - Vw -- but
acceleration is less. Acceleration is A - sin( S) * (acceleration due
to gravity), subtracting the component of "g" paralell to our take-off
acceleration. Since S is small, sin(S) is approx S and so acceleration
is A - gS.

Downhill with tailwind

The velocity we need to achieve is greater -- Vt + Vw -- but
the acceleration is greater too. a = A + gS as we are accelerating
downhill.

Now, the cross-over point between TO in different directions is
when the distance needed to take off is the same in each case. Greater
headwind or lesser slope makes taking off uphill into a headwind the
optimal & vice versa. This cross-over point is given when

(Vt + Vw)**2 / (A + gS) = (Vt - Vw)**2 / (A - gS)

or

(Vt + Vw)**2 / (Vt**2/(2*D) +gS) = (Vt - Vw)**2 / ( Vt**2/(2*D) - gS)

which reduces to

Vt**3 * Vw / D = (Vt**2 + Vw**2) * g * S

or

Vt * Vw = gDS * ( 1 + (Vw**2/Vt**2))

in limit where Vw Vt

Vw = gDS / Vt

Converting from to ft/sec and "slope" to degrees &
substituting for g

Vw = 32*D*S / (57 * 3 * Vt)

= D * S / ( 5 * Vt)

(Looks like Sparky's formular is right -- I'd done the unit
conversion improperly in the last step in my result above).


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.

I meant, but I wasn't clear, that I'd have expected the break
event point to occur with a higher wind. Before doing this
calculation, I'd have expected, for conditions mentioned, to
have around 10-12 knots before switching runways, not 6.

Part was influenced by Sparky's assertion that one should
always take off downhill unless wind speed is 15 knots or
greater. It would seem that this is dangerous nonsense, unless
the grade is really excessive (6% or 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.

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.

Except that were it negligible, one would never be able
to land!


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:
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.

The majority of the responses that I have seen have concentrated on what
happens AFTER the wheels touch the runway. You also have to consider the
glideslope with a headwind vs a tailwind -- and obstacle clearance. Runway
slope has no effect on the flight path!

The penalty for misjudging a tailwind approach can be horrendous, as much
more runway distance can be consumed in the air than can be regained by an
uphill slope after touchdown. For winds in excess of 10 knots, runway slope
has little bearing (with the exception of some truly horrific slopes on some
pathalogical 1-way mopuntain strips.


On 9 Oct 2006 13:02:24 -0700, "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:
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.