Landing on a sloping runway with different wind velocities

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

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

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

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

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

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

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

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

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

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

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

Pete

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

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

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

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

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


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

then distance travelled over time t will be

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

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

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

so

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

and in particular, if level and no wind

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

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

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

Uphill into headwind.

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

Downhill with tailwind

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

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

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

or

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

which reduces to

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

or

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

in limit where Vw Vt

Vw = gDS / Vt

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

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

= D * S / ( 5 * Vt)

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


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

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

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

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

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

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

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


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

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

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


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

Pete

"Tony Cox" wrote 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

"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.

"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