"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