Landing on a sloping runway with different wind velocities

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

[...]

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.

I'd be suspicious of a formula given without any explanation of its
derivation. This formula in particular seems odd, as the break-even wind
speed as interpreted by you decreases as takeoff speed goes up. This is
opposite what I'd have intuitively thought (that is, an airplane with a
higher takeoff speed is less-affected by wind, requiring higher wind speed
before it matters which way one takes off).

That suggests that maybe the formula as given in the previous post is
correct (that is, you really do multiply the (s * d) / 5 by V) and that the
units are what are missing. Though, why a formula would be given that
requires a unit conversion rather than just including the conversion factor
in the formula, I can't say.

The other thing I'd point out is that even if the formula is correct, it's
obviously an approximation, as the term taking into account runway slope is
stated to be in degrees, but is used in a linear fashion (rather than using
some trigonometric function).

Since you have a copy of Imeson's book, perhaps you'll be able to find the
same formula and see whether the previous post left out some important
information given in the book. Otherwise, I'm not sure I see how to apply
the formula. You make a reasonable attempt to get the result into some
sensible magnitude, but it changes the formula in a way so as to make it
counter-intuitive as to how it applies to different airplanes of different
capabilities.

Pete

"Peter Duniho" wrote in message
...

I'd be suspicious of a formula given without any explanation of its
derivation. This formula in particular seems odd, as the break-even wind
speed as interpreted by you decreases as takeoff speed goes up. This is
opposite what I'd have intuitively thought (that is, an airplane with a
higher takeoff speed is less-affected by wind, requiring higher wind speed
before it matters which way one takes off).

No, I think the inverse relationship is the right one, although
I'm not making any claims that Sparky's formula is correct.
I agree that a higher take-off speed means you're less concerned
about what the wind is doing. In the limit with a truly phenomenal
take off speed, who cares at all about wind or even what grade
you're on? You're cranking out lots of power, ignoring wind
and grade and so the break-even speed would be close to zero.


That suggests that maybe the formula as given in the previous post is
correct (that is, you really do multiply the (s * d) / 5 by V) and that the
units are what are missing. Though, why a formula would be given that
requires a unit conversion rather than just including the conversion factor
in the formula, I can't say.

The other thing I'd point out is that even if the formula is correct, it's
obviously an approximation, as the term taking into account runway slope is
stated to be in degrees, but is used in a linear fashion (rather than using
some trigonometric function).

Since you have a copy of Imeson's book, perhaps you'll be able to find the
same formula and see whether the previous post left out some important
information given in the book. Otherwise, I'm not sure I see how to apply
the formula. You make a reasonable attempt to get the result into some
sensible magnitude, but it changes the formula in a way so as to make it
counter-intuitive as to how it applies to different airplanes of different
capabilities.

I've got the 1998 3rd edition right here, and I still can't find
the formula. In the chapter on "Takeoff", S claims that 1%
downslope is equivalent to 10% more runway, that winds
over 15 knots take off uphill (no grad specified), and that
increased drag on a 1% uphill grade results in 2 to 4 %
increase in takeoff distance and subsequent climb. No idea
what to make of all that. Rather disappointing, I'm afraid.

It makes me discount his supposed formula, as reported,
even if I could find it.

"Tony Cox" wrote in message
ups.com...
No, I think the inverse relationship is the right one, although
I'm not making any claims that Sparky's formula is correct.
I agree that a higher take-off speed means you're less concerned
about what the wind is doing. In the limit with a truly phenomenal
take off speed, who cares at all about wind or even what grade
you're on? You're cranking out lots of power, ignoring wind
and grade and so the break-even speed would be close to zero.

I think you're misinterpreting the break-even point. You are right that,
"In the limit with a truly phenomenal take off speed, who cares at all about
wind", but the formula indicates that as airspeed increases, one must be
concerned with ever-decreasing winds.

That is, the point at which the break-even is near 0 occurs when takeoff
speed is very high. According to the formula, the break-even point is the
wind speed ABOVE which it's important to be taking off into the wind. So
using the inverse relationship, the conclusion is that for airplanes that
can basically ignore wind speed, the wind speed is much more important than
slope even when there's practically no wind.

That doesn't seem right to me.

Still, this is all probably moot since the formula seems to have problems
whether you believe that the V is a denominator or numerator.

I've got the 1998 3rd edition right here, and I still can't find
the formula. In the chapter on "Takeoff", S claims that 1%
downslope is equivalent to 10% more runway, that winds
over 15 knots take off uphill (no grad specified), and that
increased drag on a 1% uphill grade results in 2 to 4 %
increase in takeoff distance and subsequent climb. No idea
what to make of all that. Rather disappointing, I'm afraid.

I agree. Three different rules of thumb, none of which are even close to
being equivalent to each other. They can't all be correct.

It makes me discount his supposed formula, as reported,
even if I could find it.

Yup. Looks like you're back to square one.

Pete

"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.

That doesn't classify as CFIT? grin

"Tony Cox" wrote in message
oups.com...
I just feel more comfortable landing uphill. Perhaps its because
on approach you have a bigger target to hit!

One of the problems with landing downhill is that as you are flaring, the
ground is also dropping out from underneath you and as such, your flare ends
up taking longer... I remember once landing on a downhill runway and it
seemed to take so long to touch down after I flared that I was starting to
wonder if I would *ever* land... Landing at an unfamiliar airport at night
with a downhill runway is also an interesting experience the first time it
happens to you...

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

"Kyle Boatright" wrote in message
...

"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:
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.

"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

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.

On 2006-10-10, Edwin Johnson wrote:
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

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

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

5 = a constant
d = POH distance to liftoff with 0 slope and 0 wind (in feet) under the
existing density altitude condition

After seeing the discussion, it ocurred to me that I should have listed
page, etc. The book is copyrighted in 1998, says:

First Edition
First printing: November 1998

The formula is on page 2-43.

If this formula isn't found in subsequent editions, perhaps Sparky decided
it might not be wise to put it in for various reasons. I'm just passing
along the info in my edition and have seen no other editions.

....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