Landing on a sloping runway with different wind velocities

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

"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

"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

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...

"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

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.

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