poor lateral control on a slow tow?

At 13:23 04 January 2011, wrote:
OK..........how about this for (simple) explaination?

"Climbing in descending air" (that's what I get from all of the
complicated explainations of down wash, vortex etc.)

I think that if we compared a motor glider climbing at say 50 MPH and
500 FPM to the same glider on tow at the same climb angle and rate,
and if we assume the air behind the tow plane is moving
downward..........

Then the glider on tow would have a larger AoA.....???


Cookie


For the same lift (see previous!) and the same indicated airspeed the AoA
should be the same, because the angle of attack is measured relative to
your motion through the air.

The AoA for a given lift at a given airspeed could only change if
something significant happens to part of the airflow around the glider
wing (for example putting the flaps down, opening the airbrakes, icing up,
hitting the propwash, or flying into a tip vortex*)

The pitch angle relative to the ground on the other hand will be larger
for two reasons - (a) you are climbing and (b) the airmass you are in is
moving downwards.

It's really difficult to explain without something to draw on ... oh for
a whiteboard or at the very least the back of an envelope!

*not necessarily a bad thing: birds use the upwash outboard of a tip
vortex to increase range by flying in a v-formation - so perhaps what we
should be doing is towing two gliders at a time ...

At 13:32 04 January 2011, BruceGreeff wrote:
Thanks Martin

I did use the "vector" word talking about the winch case - because the

cable has mass (our steel cable is ~150Kg so not insignificant) and
there is a pull at a downward angle. At top of launch cable angle
approaches 90 degrees to fuselage - If you want proof look at one of the

videos on you tube.

http://www.youtube.com/watch?v=v2Qh95I_YM0

http://www.youtube.com/watch?v=np8OGPZ2pvE

As Doug Greenwell points out - there is constant acceleration on winch
launch because the flight path is curved, describing a horisontal S. I
don't know how what magnitude the acceleration has, but subjectively it

is only significant in the brief rotation to steep climb, and possibly
on the level out if you are less than smooth...
Generally it is a relatively small change from a little over 1g to a
little under 1g at release.
Anyone have the maths capability to calculate for a known situation?

Cheers
Bruce

On 2011/01/04 1:43 PM, Martin Gregorie wrote:
On Tue, 04 Jan 2011 10:57:01 +0200, BruceGreeff wrote:

Of course - the angle that the flight path can make relative to the
ground is proportional to the excess power available - hence the low
rate of climb behind the cub, versus the extreme angle on a winch.

Aerodynamics guys - Am I confused?

Sounds fair to me except that you omitted two fairly significant
forces:
- the weight of the cable
- the tension in the cable.

Both will add to the load carried by the wing. The tension should add
a
fairly constant load to the wing once the glider has rotated into full
climb since the throttle setting remains fairly constant[*] from
rotation
until the glider is near the top, but the effective cable weight will
increase as more of it is lifted off the ground and then as the whole
cable gets closer to vertical.

[*] this is true on a calm day but is obviously incorrect in the
presense
of turbulence or a significant wind gradient.



--
Bruce Greeff
T59D #1771 & Std Cirrus #57


Hopefully this post will go this time ...

I did try modelling a winch launch some time ago. It's very dependent on
the assumptions you make on piloting technique and winch control (constant
power, tension, cable speed?), but in general once you've transitioned
into full climb the accelerations as felt by the pilot seem to be very
small ( 0.1g).

Hence the danger of overstressing, since you've no physical indication of
the high wing loads due to the cable tension & weight.

On 1/3/2011 11:51 PM, Darryl Ramm wrote:
On Jan 3, 8:54 pm, Eric wrote:

Imagine an extreme tow, a 50 knot airspeed, but climbing at 35 knots (45
degree angle). The tow rope is providing 70% of the force holding the
glider in the air, so the wing needs to supply only 30% of the force.

Or imagine a really extreme, vertical tow: all the force required to
keep the glider moving steadily through the air is provided by the
towrope/towplane, and none by the wing.

I think you are trying to push this argument up an incline with a
rope. :-) But I'll take your points into consideration next time I'm
vertically towing behind a helicopter.

I'm serious! But, let me add this constraint to make the idea easier to
absorb: the glider pilot flies the tow so the rope is always parallel to
the fuselage.

In level flight, the rope pull equals the drag; the lift equals the
glider weight. Rope force vector and weight vector are at right angles.

In a 50 knot airspeed, 35 knot climb (45 degree angle of climb), the
rope vector and the glider weight vector are now at an obtuse angle, so
some of the rope force is supporting the glider.

Stating it another way: we know the rope is pulling a lot harder, but
the glider is not accelerating, so what force is opposing the rope pull?
It can't be additional drag (glider is still going only 50 knots
airspeed); it can't be the lift (regardless of it's value), because
that's acting almost entirely perpendicularly to the rope. So, what
force is opposing all that extra rope pull? I say - it's the weight of
the glider (about 70% of the weight).

Another way to imagine the situation, using the helicopter to provide a
50 knot airspeed tow, rope always parallel to the glider fuselage:
* level flight, wing lift = weight of glider
* vertical flight, wing lift = 0 (or the glider won't have right rope
angle)

So, in between level flight and vertical flight, there must be a region
where the wing lift is less than in level flight, right? I'm saying
there is a continuous reduction in the lift the wing must provide as the
climb angle increases.

Only two months till March flying starts...gotta solve this problem
while we still have time!

--
Eric Greenwell - Washington State, USA (change ".netto" to ".us" to
email me)

So, in between level flight and vertical flight, there must be a region
where the wing lift is less than in level flight, right? I'm saying
there is a continuous reduction in the lift the wing must provide as the
climb angle increases.

Only two months till March flying starts...gotta solve this problem
while we still have time!

--
Eric Greenwell - Washington State, USA (change ".netto" to ".us" to
email me)

Yeah....you got it......the lift is the cosine of the climb angle
times the weight.........

level.....0 degrees climb.. Cosine 0 = 1 so lift =100% glider
weight

5 degree climb (reasonable tow climb angle) Cosine 5 = .996 so
lift = 99.6% of glider's weight

45 degree climb (unlikely but just for demonstration) cosine 45 = .
707 so lift would be only 71% of glider's weight

90 degree climb Cosine 90 = o so lift would be zero.

If we keep the airspeed constant, the drag shoud be constant....so the
only variables are lift and thrust. as the thrust vector gets
bigger, the direction of flgith gets steeper climb, and the lift
vector gets smaller.

Cookie

In very slow flight without flaps my 1-35 drops into a stall long
before it gets as bad as a tow does at 60 statute mph. You would
think that I would have stalled out of the tow too. Perhaps the
wallowing around on tow is just the turbulent air on the ailerons and
not an imminent stall at all. (Think rotor in wave or turbulence
behind a hill on a smaller scale)

Are there any reports of incidents where a glider drops into a stall
on a slow tow or are there just complaints of glider pilot annoyance?
(I agree it's not fun)

For example, if the air turbulence was going "down" on the right side
just when you try to bank "left" that would make the controls feel
sluggish. At some angle of bank, assuming that everything else was
symmetrical, the two ailerons would be in different parts of the
turbulence, confusing the situation.

Has anyone tried some flaps in an integrated flap machine (which
reduces stall speed) to see if the wallowing goes away?

Unfortunately, my trailer is in a snowbank.

On Jan 4, 7:13*pm, AGL wrote:

Has anyone tried some flaps in an integrated flap machine (which
reduces stall speed) to see if the wallowing goes away?

With every flapped glider I've flown, negative flap improves aileron
response fairly dramatically. Positive flap does lower the stall
speed a little.

I've flown a 20 meter Nimbus 2C ballasted to 11 lbs/sq ft wing loading
behind a tug pilot accustomed to towing 2-33's. The speed was low
enough to need +1 flap but it didn't wallow. The tug pilot turned off
his radio when he got tired of me yelling for more speed than what he
"knew" was right.

At 22:15 04 January 2011, Eric Greenwell wrote:
On 1/3/2011 11:51 PM, Darryl Ramm wrote:
On Jan 3, 8:54 pm, Eric Greenwell wrote:

Imagine an extreme tow, a 50 knot airspeed, but climbing at 35 knots
(45
degree angle). The tow rope is providing 70% of the force holding the
glider in the air, so the wing needs to supply only 30% of the force.

Or imagine a really extreme, vertical tow: all the force required to
keep the glider moving steadily through the air is provided by the
towrope/towplane, and none by the wing.

I think you are trying to push this argument up an incline with a
rope. :-) But I'll take your points into consideration next time I'm
vertically towing behind a helicopter.

I'm serious! But, let me add this constraint to make the idea easier to

absorb: the glider pilot flies the tow so the rope is always parallel to

the fuselage.

In level flight, the rope pull equals the drag; the lift equals the
glider weight. Rope force vector and weight vector are at right angles.

In a 50 knot airspeed, 35 knot climb (45 degree angle of climb), the
rope vector and the glider weight vector are now at an obtuse angle, so
some of the rope force is supporting the glider.

Stating it another way: we know the rope is pulling a lot harder, but
the glider is not accelerating, so what force is opposing the rope pull?

It can't be additional drag (glider is still going only 50 knots
airspeed); it can't be the lift (regardless of it's value), because
that's acting almost entirely perpendicularly to the rope. So, what
force is opposing all that extra rope pull? I say - it's the weight of
the glider (about 70% of the weight).

Another way to imagine the situation, using the helicopter to provide a
50 knot airspeed tow, rope always parallel to the glider fuselage:
* level flight, wing lift = weight of glider
* vertical flight, wing lift = 0 (or the glider won't have right rope

angle)

So, in between level flight and vertical flight, there must be a region
where the wing lift is less than in level flight, right? I'm saying
there is a continuous reduction in the lift the wing must provide as the

climb angle increases.

Only two months till March flying starts...gotta solve this problem
while we still have time!

--
Eric Greenwell - Washington State, USA (change ".netto" to ".us" to
email me)


Right ... it's the weight opposing the rope pull, because the tug is
doing all the work of raising the glider to a higher altitude -
difficult to demonstrate without a diagram!

At 02:13 05 January 2011, AGL wrote:

In very slow flight without flaps my 1-35 drops into a stall long
before it gets as bad as a tow does at 60 statute mph. You would
think that I would have stalled out of the tow too. Perhaps the
wallowing around on tow is just the turbulent air on the ailerons and
not an imminent stall at all. (Think rotor in wave or turbulence
behind a hill on a smaller scale)

Are there any reports of incidents where a glider drops into a stall
on a slow tow or are there just complaints of glider pilot annoyance?
(I agree it's not fun)

For example, if the air turbulence was going "down" on the right side
just when you try to bank "left" that would make the controls feel
sluggish. At some angle of bank, assuming that everything else was
symmetrical, the two ailerons would be in different parts of the
turbulence, confusing the situation.

Has anyone tried some flaps in an integrated flap machine (which
reduces stall speed) to see if the wallowing goes away?

Unfortunately, my trailer is in a snowbank.



That's a useful comment: the aerodynamic modelling I did suggests that
the lateral control problems on tow should be different (worse) than those
in a typical stall, because the wing is stalling at the tips rather than
the root.

No-one has yet admitted to actually stalling or dropping a wing on tow -
so the effect seems to be annoying rather than dangerous.

Flaps should (theoretically) improve matters by (a) reducing stall speed
and (b) shifting the spanwise lift distribution inboard and unloading the
tips. However, if the flaps are integrated with the ailerons then the
associated aileron droop would counteract (b).

At 02:48 05 January 2011, bildan wrote:
On Jan 4, 7:13=A0pm, AGL wrote:

Has anyone tried some flaps in an integrated flap machine (which
reduces stall speed) to see if the wallowing goes away?

With every flapped glider I've flown, negative flap improves aileron
response fairly dramatically. Positive flap does lower the stall
speed a little.

I've flown a 20 meter Nimbus 2C ballasted to 11 lbs/sq ft wing loading
behind a tug pilot accustomed to towing 2-33's. The speed was low
enough to need +1 flap but it didn't wallow. The tug pilot turned off
his radio when he got tired of me yelling for more speed than what he
"knew" was right.



Sorry if this is an obvious question (never flown a flapped glider), but
with an integrated flap system what is the relative movement of the
ailerons and flaps? Presumably the ailerons don't move at all for
negative settings?

Doug

On Jan 5, 12:00*am, "
wrote:
So, in between level flight and vertical flight, there must be a region
where the wing lift is less than in level flight, right? I'm saying
there is a continuous reduction in the lift the wing must provide as the
climb angle increases.

Only two months till March flying starts...gotta solve this problem
while we still have time!

--
Eric Greenwell - Washington State, USA (change ".netto" to ".us" to
email me)

Yeah....you got it......the lift is the cosine of the climb angle
times the weight.........

level.....0 degrees climb.. *Cosine 0 = 1 * *so lift =100% glider
weight

5 degree climb (reasonable tow climb angle) * Cosine 5 = .996 * *so
lift = 99.6% of glider's weight

45 degree climb (unlikely but just for demonstration) * cosine 45 = .
707 *so lift would be only 71% of glider's weight

90 degree climb * Cosine 90 = o * so lift would be zero.

If we keep the airspeed constant, the drag shoud be constant....so the
only variables are lift and thrust. * as the thrust vector gets
bigger, the direction of flgith gets steeper climb, and the lift
vector gets smaller.

Cookie

So according to you, pulling a load up a 10 degree slope should
require less energy than pulling it on the flat! Anybody who has ever
ridden a bicycle can tell you that is not the case!

For a glider on tow, the combined vector of Lift and Thrust (provided
by the tug) has to equal the combined vector of weight plus drag. As
the glider is not rigidly connected to the tug, the extra lift has to
come from its wings (at least at moderate climb angles). For a given
airspeed this can only be done by increasing the angle of attack.
Hence you are closer to the stalling angle.

I am not sure that this is the correct explanation, but it seems to
fit the observed facts.

Derek C