Landing on a sloping runway with different wind velocities

[Tony Cox]

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.

[kontiki]

It really depends upon how much of slope. I'm sure there is a given
amount of upslope to a runway that could cancel the effects of a slight
tail wind. Just exactly how much would depend upon a given airplane and
how it is loaded... and how effective the brakes are.

I would prefer to land into a headwind always unless my brakes were
really bad.


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.
>

[Mike Isaksen]

"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"?

[Roy Smith]

In article <%PyWg.2774$YD.1017@trndny09>,
"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"?

You shouldn't need a calculator for this one. 3 degrees is 1:20. Just
like an ILS glide slope.

[Jim Macklin]

A downwind take-off into higher terrain is a bad thing.
There are no hard and fast rules. With a landing on a
one-way runway or with a tailwind, there will be a point at
which a go-around can be made safely and once you are below
that point, a landing is your only option. That's why
mountain and hill flying is a challenge



"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.
|

[Jim Macklin]

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.


"Mike Isaksen" > wrote in message
news:%PyWg.2774$YD.1017@trndny09...
| "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"?
|
|

[Peter Duniho]

"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).

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

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).

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.

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. :)

Pete

[Doug[_1_]]

Uphill or down, land into the wind seems to work best. Same goes with
Seaplanes and rivers, land into the wind. But....sometimes with
obstructions and whatnot it's not safe to land at all. Find another
field.

[Neil Gould]

Recently, Tony Cox > posted:

> 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.
>
Having done that (Jaffrey Airport, NH), all I can suggest is that you go
with the head wind, whichever way that is. My first approach was to land
going "uphill", but on final the wind shifted to a tailwind and it made
things pretty dicey. So, I went around and landed downhill. The roll-out
was a quite bit longer than I'm accustomed to, but other than that it was
uneventful. When I left a couple of days later, the wind favored taking
off "uphill", and that was an experience, as well. It gave the folks at
the ice cream stand about 500' off the departure end something to gawk at.

Neil

[Peter Duniho]

"Peter Duniho" > wrote in message
...
> [...]
> More importantly, the change in acceleration or deceleration is linear,
> while the difference in total speed change is exponential.

I should clarify what I mean, since the way I wrote it seems a little
confusing:

The changes themselves are both linear, but the *consequences* (in terms of
distance added or removed) is exponential when changing the speed.

[Ben Jackson]

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/

[Tony Cox]

"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.

[Tony Cox]

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.

[.Blueskies.]

"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...

[Jose[_1_]]

> 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.

[Jerry]

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.
>

[Kyle Boatright]

"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

[Peter Duniho]

"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

[Edwin Johnson]

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

Where:
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

[Tony Cox]

"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.

>
> Where:
> 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...

[Peter Duniho]

"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

[Tony Cox]

"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.

[Peter Duniho]

"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

[Grumman-581[_3_]]

"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>

[Grumman-581[_3_]]

"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...

[Jerry]

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
>

[Andrew Sarangan[_1_]]

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.

[Kyle Boatright]

"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

[Jay Somerset]

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.

[Edwin Johnson]

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
>
> Where:
> 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

[Jerry]

"Kyle Boatright" > wrote in message
. ..
>
> "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

[Kyle Boatright]

"Jerry" > wrote in message
. ..
>
> "Kyle Boatright" > wrote in message
> . ..
>>
>> "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

[Tony Cox]

"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?

[Jerry]

"Kyle Boatright" > wrote in message
. ..
>
> "Jerry" > wrote in message
> . ..
>>
>> "Kyle Boatright" > wrote in message
>> . ..
>>>
>>> "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

[cjcampbell]

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.

[Kyle Boatright]

"Jerry" > wrote in message
. ..
>
> "Kyle Boatright" > wrote in message
> . ..
>>
>> "Jerry" > wrote in message
>> . ..
>>>
>>> "Kyle Boatright" > wrote in message
>>> . ..
>>>>
>>>> "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

[Tony Cox]

"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
> >
> > Where:
> > 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.

[Peter Duniho]

"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

[Andrew Sarangan[_1_]]

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.
>
> >
> > Where:
> > 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.

[Andrew Sarangan[_1_]]

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.

[Peter Duniho]

"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

[Tony Cox]

"cjcampbell" > wrote in message
oups.com...
>
> 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.

This is a different conclusion to the emerging consensus
here -- that slope really isn't that important and you should
almost always take off into the wind.

Do you have a reference?

[Andrew Sarangan[_1_]]

Tony Cox wrote:
> "cjcampbell" > wrote in message
> oups.com...
> >
> > 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.
>
> This is a different conclusion to the emerging consensus
> here -- that slope really isn't that important and you should
> almost always take off into the wind.
>
> Do you have a reference?


As Peter Duniho pointed out it makes intuitive sense that slope really
isn't that important compared to wind. Slope affects the deceleration,
while wind affects the touchdown speed. The rolling distance increases
as the square of the speed, but only linearly with deceleration. One
can do a more accurate modeling (as I did) but I think this explanation
is enough to make a qualitative argument. Now I am curious about the
mathematical analysis that is being referred to.

[Tony Cox]

"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
>
>

[Peter Duniho]

"Tony Cox" > wrote in message
oups.com...
> You're right. Do please have a look, since its important
> these things be done right. Analysis applies to take-off
> distance.

Note: frankly, I don't consider myself entirely qualified to check your
work. I simply meant that someone should. :) A thorough review would
require some period of time without distractions, and a clear head, both of
which are in short supply for me right now (I've been fighting off a
somewhat tenacious cold, and even on a good day my environment is filled
with outside interruptions). Indeed, if I had the ability to do a truly
fair review, I think I'd be able to do the derivation myself. :)

Anyway, that said I have some questions (which may indeed expose my lack of
qualification :) )...

> then distance travelled over time t will be
>
> d = Integral(0,t){ v(t).dt} = v0*t + a*t*t/2

This part I agree with.

> So, if starting from rest, find d given final velocity and acceleration
>
> d = a*t*t/2
> v = a*t

As long as you restrict "d" and "v" to mean "d at time t" and "v at time t",
I'll agree with this.

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

This part, I don't get. Using the two equations you provide previously, I
get "d = v*t/2", which is consistent with my distant recollection of basic
high school physics. That is, the final distance traveled accelerating from
zero is half the distance one would travel had they been at the final speed
the whole time.

I don't see where you managed to make distance proportional to the square of
the final speed, nor do I understand why it is that your final calculation
for "d" no longer takes into account the time during the takeoff. I get
some feeling that maybe this has to do with factoring "t" out of the
equation, but I'm not sure about that, nor am I clear on why you'd want to
(since it seems to me that later derivations would be easier with the
simpler formula of "v*t/2"...if it's valid to remove "t" at this point, then
it will just cancel out later in the other derivations anyway).

> 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}

How are you doing the cross-check? Are you calculating A from the second
equation, or from the POH numbers? If from the second equation, I don't see
how that's useful for checking the equation, and if from the POH numbers, I
don't find that the equations check out.

If I assume Vt as 100fps (or 101-1/3 fps, assuming 60 knot takeoff speed),
and use 6.1 fps^2 as given for the accleration, I get 820 feet, rather than
900 feet for the takeoff distance. That's a pretty significant error, IMHO.

I can do a variety of other calculations, making different assumptions
regarding what numbers you've derived and what numbers you've pulled from
the POH, and in each case I don't find the numbers consistent.

So, perhaps you could be more explicit about what numbers you're taking from
the POH, and how you apply those to do your cross-check. (Of course, as I
mentioned, I'm not clear on how you got these equations in the first
place...if they are incorrectly derived, then the cross-check here is
obviously moot :) ).

> [...]
> Vw = gDS / Vt
>
> Converting from to ft/sec and "slope" to degrees &
> substituting for g
>
> Vw = 32*D*S / (57 * 3 * Vt)

I've glossed over the actual derivation. I figure if I'm mistaken about the
simpler reduction of the integral, I can't really expect to correctly check
the break-even derivation. (And of course, if I'm not, the break-even
derivation is moot, even if done correctly).

I believe that the unit conversions are done correctly. You glossed over
some steps, but in the end I come up with a similar number as you...in fact,
the final denominator constant comes out closer to 5 using more precise
conversions.

So at this point, my intuition tells me that you've done the derivation
correctly, but that you made it harder on yourself than was necessary. :)

I also looked back on a previous comment I made about the intuitive nature
of having the takeoff speed in the denominator, and I still am having
trouble with that. This is relevant somewhat to this:

>> > 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.

IMHO, your original terminology was indeed unclear. :) Expecting a higher
break-even point is IMHO suggesting that wind is LESS important, not more.
The more important something is, the less of it you need in order to make a
difference. :)

Which is why I still don't see how increasing the takeoff speed should cause
the break-even point for the wind to go down as it does in this formula.
Intuitively, it seems to me that as takeoff speed goes up, the influence of
the wind on the takeoff performance goes down, and so the wind should be
LESS important, meaning you need MORE of it before you need to takeoff into
the wind rather than with it, not less of it.

I'm wondering if this intuition is incorrect based on other aspects of the
formula. That is, if derived correctly, the formula accounts not only for
wind but also for acceleration performance. And perhaps the higher the
takeoff speed, the more significant the acceleration performance is, and
thus the slope factors in more heavily. That is, even as wind becomes less
influential, for some reason the acceleration becomes even more so.

This is all hand-waving, and I haven't convinced myself that it makes any
sense. Still, it's the best I can come up with. :)

> [...]
>> > 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 disagree. It seems like you are asserting that if parasitic drag is
negligible to the calculation, one couldn't slow down to land. But that
ignores induced drag, which dominates at landing speeds (at the same time
that parasitic drag does indeed become a negligible factor).

I don't see any reason to think that for the purpose of an approximation,
one cannot simply ignore parasitic drag. The formula you've derived ignores
all sorts of similarly measurable-but-insignificant things, and I don't see
the idea of ignoring parasitic drag any different.

Pete

[Tony Cox]

"Peter Duniho" > wrote in message
...
> "Tony Cox" > wrote in message
> oups.com...
> > You're right. Do please have a look, since its important
> > these things be done right. Analysis applies to take-off
> > distance.
>
> Note: frankly, I don't consider myself entirely qualified to check your
> work. I simply meant that someone should. :) A thorough review would
> require some period of time without distractions, and a clear head, both
of
> which are in short supply for me right now (I've been fighting off a
> somewhat tenacious cold, and even on a good day my environment is filled
> with outside interruptions). Indeed, if I had the ability to do a truly
> fair review, I think I'd be able to do the derivation myself. :)

No matter. I'm not submitting a "homework assignment" !

Its just that since this challenges "conventional" wisdom (Sparky's &
at least one poster here) someone ought to look at it! It seems
on the face of it that Sparky Imeson is dispensing dangerous
snake oil by claiming that runway slope is by far the most
significant -- its not, at least according to this calculation.

>
> Anyway, that said I have some questions (which may indeed expose my lack
of
> qualification :) )...
>
> > then distance travelled over time t will be
> >
> > d = Integral(0,t){ v(t).dt} = v0*t + a*t*t/2
>
> This part I agree with.
>
> > So, if starting from rest, find d given final velocity and acceleration
> >
> > d = a*t*t/2
> > v = a*t
>
> As long as you restrict "d" and "v" to mean "d at time t" and "v at time
t",
> I'll agree with this.

I think that was clear in the key, but you're right in your
analysis.

>
> > so
> >
> > d = v**2 / (2*a)
>
> This part, I don't get. Using the two equations you provide previously, I
> get "d = v*t/2", which is consistent with my distant recollection of basic
> high school physics.

Look at it this way...

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

> That is, the final distance traveled accelerating from
> zero is half the distance one would travel had they been at the final
speed
> the whole time.
>
> I don't see where you managed to make distance proportional to the square
of
> the final speed, nor do I understand why it is that your final calculation
> for "d" no longer takes into account the time during the takeoff. I get
> some feeling that maybe this has to do with factoring "t" out of the
> equation, but I'm not sure about that, nor am I clear on why you'd want to
> (since it seems to me that later derivations would be easier with the
> simpler formula of "v*t/2"...if it's valid to remove "t" at this point,
then
> it will just cancel out later in the other derivations anyway).

I've factored it out. The POH doesn't give you "time for takeoff"
but it *does* give you Vt and D.

>
> > 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}
>
> How are you doing the cross-check? Are you calculating A from the second
> equation, or from the POH numbers? If from the second equation, I don't
see
> how that's useful for checking the equation, and if from the POH numbers,
I
> don't find that the equations check out.

Its just a consistency check to see if things are reasonable. D and Vt
come from the POH, and A = g/5 is about what I feel when I push
the throttle in.

>
> If I assume Vt as 100fps (or 101-1/3 fps, assuming 60 knot takeoff speed),
> and use 6.1 fps^2 as given for the accleration, I get 820 feet, rather
than
> 900 feet for the takeoff distance. That's a pretty significant error,
IMHO.

Well, it was just a rough check. If A had turned out to be 3g or
something
outragous, then we'd know something was up.

>
> I can do a variety of other calculations, making different assumptions
> regarding what numbers you've derived and what numbers you've pulled from
> the POH, and in each case I don't find the numbers consistent.
>
> So, perhaps you could be more explicit about what numbers you're taking
from
> the POH, and how you apply those to do your cross-check. (Of course, as I
> mentioned, I'm not clear on how you got these equations in the first
> place...if they are incorrectly derived, then the cross-check here is
> obviously moot :) ).
>
> > [...]
> > Vw = gDS / Vt
> >
> > Converting from to ft/sec and "slope" to degrees &
> > substituting for g
> >
> > Vw = 32*D*S / (57 * 3 * Vt)
>
> I've glossed over the actual derivation. I figure if I'm mistaken about
the
> simpler reduction of the integral, I can't really expect to correctly
check
> the break-even derivation. (And of course, if I'm not, the break-even
> derivation is moot, even if done correctly).
>
> I believe that the unit conversions are done correctly. You glossed over
> some steps, but in the end I come up with a similar number as you...in
fact,
> the final denominator constant comes out closer to 5 using more precise
> conversions.

Thats good . I think it is right, and that's what Sparky claims
anyway (even if it is only in the early revisions of the MFB)

>
> So at this point, my intuition tells me that you've done the derivation
> correctly, but that you made it harder on yourself than was necessary. :)

Simplifications gladly accepted. I can't see an easier
way.

>
> I also looked back on a previous comment I made about the intuitive nature
> of having the takeoff speed in the denominator, and I still am having
> trouble with that. This is relevant somewhat to this:
>
> >> > 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.
>
> IMHO, your original terminology was indeed unclear. :) Expecting a
higher
> break-even point is IMHO suggesting that wind is LESS important, not more.
> The more important something is, the less of it you need in order to make
a
> difference. :)
>
> Which is why I still don't see how increasing the takeoff speed should
cause
> the break-even point for the wind to go down as it does in this formula.

Remember, its increasing TO speed *with the TO distance
constant*. It means the energy gained/lost by going downhill/uphill
is the same, but the effect of wind is less significant because
you're accelerating more quickly through the range where
windspeed *would* be significant.

[Peter Duniho]

"Tony Cox" > wrote in message
oups.com...
> [...]
>> > d = v**2 / (2*a)
>>
>> This part, I don't get. Using the two equations you provide previously,
>> I
>> get "d = v*t/2", which is consistent with my distant recollection of
>> basic
>> high school physics.
>
> Look at it this way...
>
> d = a*t*t/2 = a*t*a*t/(2*a) = v*v/(2*a)

Okay...not sure why I didn't see that earlier, since I had a suspicion
that's what you had done.

Still, I don't see the value in writing it that way. Probably if I had
bothered to actually do the derivation, I'd understand. If I had to guess,
I'd say it has something to do with keeping acceleration in the equation so
you can include the gravity adjustment for the slope. But with "t" in the
equation, I'd think you could do that as well, and with a slightly simpler
starting equation.

Oh well...doesn't really matter. In algebra, the whole point is that
everything comes out the same, no matter how you "phrase the question". :)
And since I'm too lazy to bother with redoing the derivation starting with
the alternative equation, I don't have any business questioning which
version is easier.

> [...]
> Remember, its increasing TO speed *with the TO distance
> constant*. It means the energy gained/lost by going downhill/uphill
> is the same, but the effect of wind is less significant because
> you're accelerating more quickly through the range where
> windspeed *would* be significant.

Ahh, right. Thanks. I knew I must be missing something. I guess that does
stand to reason, that an airplane with better thrust spends less time
arguing with the wind. :)

Okay...well, looks like Sparky's formula is right, and all his other advice
is off-base. Good to know. :)

Pete

[Tony Cox]

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.