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poor lateral control on a slow tow?



 
 
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Old January 5th 11, 10:33 AM posted to rec.aviation.soaring
Doug Greenwell
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Posts: 67
Default poor lateral control on a slow tow?

At 09:25 05 January 2011, Derek C wrote:
On Jan 5, 12:00=A0am, "
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

th=
e
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.. =A0Cosine 0 =3D 1 =A0 =A0so lift =3D100%

glid=
er
weight

5 degree climb (reasonable tow climb angle) =A0 Cosine 5 =3D .996 =A0

=A0=
so
lift =3D 99.6% of glider's weight

45 degree climb (unlikely but just for demonstration) =A0 cosine 45

=3D
.
707 =A0so lift would be only 71% of glider's weight

90 degree climb =A0 Cosine 90 =3D o =A0 so lift would be zero.

If we keep the airspeed constant, the drag shoud be constant....so the
only variables are lift and thrust. =A0 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


There are two components to the energy required in this case - (1) the
energy required to overcome friction (which will indeed be slightly less,
because of the reduced reaction force perpendicular to the slope), (2) the
energy required to lift the load up a given height

(NB this assumes that you are pulling the load at a constant speed -
otherwise we would have to take kinetic energy into account as well)

(1) can be reduced to (near) zero by reducing friction - using rollers for
example, or in your alternative example of a bicycle - the equivalent
effect in a glider on tow is reducing drag by careful streamlining or
increased aspect ratio.

(2) is fixed, and independent of speed or slope angle - raising any object
a given height requires a fixed amount of energy (= mass*acceleration due
to gravity*height change).

Both components of the energy input are provided by you pulling the load
up the slope.

A glider on tow is exactly the same. The wing lift corresponds to the
reaction force between the surface and the load. The drag corresponds to
the friction force between the surface and the load. The tug corresponds
to you pulling the load - and is doing all the work against friction and
gravity. The lift/reaction force does no work - all it does is stop the
load sinking into the ground or the glider falling further and further
below the tug.

Imagine a perfect glider with no drag* on tow (= pulling a load up the
slope with no friction, or a perfect bicycle) ... what happens if you
release the rope (or stop pedalling)? If the wing lift were responsible
for the climb rate then you would carry on climbing until you ran out of
atmosphere (or hill)

* fortunately not currently available in the shops, since it would ruin
the sport!



 




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