Eric has the right idea.

The forces acting on a glider towed upwards in steady climbing flight
are necessarily at equilibrium. The vectors lift + weight + drag +
thrust add up to zero. Otherwise, as some of the dafter ideas proposed
here imply, the glider will accelerate continuously toward outer
space.

Think of pulling a wheeled trolley with a rope up an inclined plane.
The steeper the incline, the greater is the tension in the rope until,
in the limit, with a vertical incline, the rope supports the entire
weight and the wheels none. At practical climb angles (less than 1:5)
the required wing lift should be slightly reduced from level flight by
the cosine of climb angle.

On that basis a glider on tow should stall at, or less than, the same
airspeed as straight and level flight off tow.

I think pilots who feel alarmed about stalling on tow may be misled by
cues from the nose high attitude combined with the need to maintain
position in turbulence behind the tug. Doesn't mean it feels good
though :-)

Ian Grant
(PEng and D2 driver)





On Jan 5, 2:15*am, Eric Greenwell wrote:
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)