Slips in turns and landing with winglets

Eric Greenwell wrote on 2/16/2016 7:58 PM:
I would expect the inner winglet to be stalled, and the outer winglet to
be producing outward lift.

Make that: "both winglets to be stalled"


--
Eric Greenwell - Washington State, USA (change ".netto" to ".us" to
email me)
- "A Guide to Self-Launching Sailplane Operation"

https://sites.google.com/site/motorg...ad-the-guide-1
- "Transponders in Sailplanes - Dec 2014a" also ADS-B, PCAS, Flarm

http://soaringsafety.org/prevention/...anes-2014A.pdf

On Tue, 16 Feb 2016 19:58:42 -0800, Eric Greenwell
wrote:



I am curious about why a 20 meter glider would need a lot of yaw to
climb well, when my 18 meter ASH 26E hardly needs any. 20 or 30 degrees
would be a poor choice for the 26E, but you say an Arcus needs that
much?


I guess it's a question of aircraft geometry and cannot be predicted
precisely.

I was surprised how nasty the Arcus was thermalling with the yaw
string centered (some of the PICs I instructed needed full aileron
input and really hard work to keep it in the turn) and how nice it
flew with significant opposite rudder.

I didn't expect that, especially since the Duo Dicus with its very
similar wing geometry doesn't show this behaviour.

Dihedral does not seem to be an important factor: The Arcus with its
huge dihedral behaves similar to the AS 22-2 that has too little
dihedral (and needs nearly full opposite aileron if you turn with the
yaw string centered) and no winglets.


Is that part of the operating manual for the glider? I would
expect the inner winglet to be stalled, and the outer winglet to be
producing outward lift.

I don't think so, but I haven't read the Arcus POH that carefully I
have to admit.

I am pretty sure that the outer winglet is the one that is stalled
(camber is pointing towards the fuselage, therefore reducing stall
AOA).

On Wednesday, February 17, 2016 at 1:05:18 AM UTC+3, Martin Gregorie wrote:
On Tue, 16 Feb 2016 03:26:42 -0800, Bruce Hoult wrote:

On Tuesday, February 16, 2016 at 3:23:18 AM UTC+3, Martin Gregorie
wrote:
1) The energy conservation law would be violated if the energy taken
from the moving aircraft as drag is less than that needed to generate
the side force.

Incorrect. Force is not energy.

If you're extracting work from the system (in this case generating a lift
force that moves the fuselage sideways) you're using energy that wouldn't
be consumed if there was no side force being generated.

Incorrect. Force is not energy.

If you were correct then a 20:1 glider and a 60:1 glider with the same weight and same speed would be burning through energy at the same rate because they're generating the same lift. But that's not the case .. the 60:1 glider is losing energy at 1/3 of the rate of the 20:1 glider.


This is really basic stuff: it should have been covered in School Cert
physics unless you had a really bad science teacher, and is certainly
revisited in the first year of any physics degree cause, mainly to
compensate for really bad science teachers

Thanks for your input. I got the school prizes for top physics and top math student every year, and also scholarships etc. I'm not some dummy.

On Wed, 17 Feb 2016 01:45:59 -0800, Bruce Hoult wrote:

Incorrect. Force is not energy.

You forgot that the side force is affecting the flight path.

Force * distance = work. Doing work uses energy. This is the first law of
thermodynamics: https://en.wikipedia.org/wiki/First_...thermodynamics

If you were correct then a 20:1 glider and a 60:1 glider with the same
weight and same speed would be burning through energy at the same rate
because they're generating the same lift. But that's not the case .. the
60:1 glider is losing energy at 1/3 of the rate of the 20:1 glider.

You're forgetting drag. Two gliders with the same weight will be
generating the same lift if their descent rates are constant. However, if
their descent rates are different but flying speeds are the same, then
the one with the worst glide will have more drag.

Again, speed * drag = work. Doing work requires energy. A glider is a
device for converting potential energy (height) into kinetic energy
(airspeed), so more drag requires more work to be done if the glider is
to move at a given speed. This is why a tail chute drastically increases
the (constant) descent rate while maintaining a constant airspeed. Doing
more work consumes more energy. If it didn't, you'd have invented a
perpetual motion machine.

Thanks for your input. I got the school prizes for top physics and top
math student every year, and also scholarships etc. I'm not some dummy.

You obviously understand maths. Physics, not so much.


--
martin@ | Martin Gregorie
gregorie. | Essex, UK
org |

Le mardi 16 février 2016 23:01:09 UTC+1, Bruce Hoult a écrit*:

If it's a continuous differentiable function then exactly that is a given..


Well, no. The sole fact that the slope at a minimum is zero does not give you much information on the points close to minimum. You need to know a couple of coefficients of the function, and you need to set the "small deviation" in context with "maximum deviation". With a slip angle of 5-10 degree out of a maximum of 90 degree, you can't expand a function into a series based on small values.

But on the physics side:
(1) You fly your glider at minumum sink speed straight ahead with the string centered.
(2) Next, you fly a quarter roll and then fly straight ahead at the same speed as above (that's actually possible).

Now you tell me that there exists a slip angle between (1) and (2) which gives a better minimum sink rate than (1) ? You must be kidding...

The example of Andreas may still be true: If you need full aileron against the turn in order to keep the string straight, inflicting a longer apparent wing chord to to the whole wing by a slip angle, plus the additional drag of the fuselage, might carry less drag penalty than having the inner wing running with a fully deflected aileron over 2/3 or even the whole semi-span.

Bert
Ventus cm "TW"

On Wednesday, February 17, 2016 at 2:20:30 PM UTC+3, Martin Gregorie wrote:
On Wed, 17 Feb 2016 01:45:59 -0800, Bruce Hoult wrote:
If you were correct then a 20:1 glider and a 60:1 glider with the same
weight and same speed would be burning through energy at the same rate
because they're generating the same lift. But that's not the case .. the
60:1 glider is losing energy at 1/3 of the rate of the 20:1 glider.

You're forgetting drag.

No, I'm not forgetting drag. Drag is *precisely* the difference between the 20:1 and the 60:1 glider. The lift is the same. And both are doing zero work in a direction perpendicular to the flight path. There is a force, but no movement, hence no work and no energy. The energy is only in the drag opposing the direction of movement.


You obviously understand maths. Physics, not so much.

I've got qualifications and experience that say otherwise.

Anyway, I'm out of here. It's obviously long ceased to be productive.

Bruce Hoult wrote on 2/17/2016 1:45 AM:
On Wednesday, February 17, 2016 at 1:05:18 AM UTC+3, Martin Gregorie
wrote:
On Tue, 16 Feb 2016 03:26:42 -0800, Bruce Hoult wrote:

On Tuesday, February 16, 2016 at 3:23:18 AM UTC+3, Martin
Gregorie wrote:
1) The energy conservation law would be violated if the energy
taken from the moving aircraft as drag is less than that needed
to generate the side force.

Incorrect. Force is not energy.

If you're extracting work from the system (in this case generating
a lift force that moves the fuselage sideways) you're using energy
that wouldn't be consumed if there was no side force being
generated.

Incorrect. Force is not energy.

If you were correct then a 20:1 glider and a 60:1 glider with the
same weight and same speed would be burning through energy at the
same rate because they're generating the same lift. But that's not
the case .. the 60:1 glider is losing energy at 1/3 of the rate of
the 20:1 glider.

The gliders are not losing the same amount of energy, even though the
airspeed and weight are identical: the 60:1 glider has 1/3 the drag of
the 20:1 glider, so it is losing 1/3 energy of the 20:1 glider.

Airspeed x drag = energy loss per unit of time

The weight of the gliders is irrelevant.


--
Eric Greenwell - Washington State, USA (change ".netto" to ".us" to
email me)
- "A Guide to Self-Launching Sailplane Operation"

https://sites.google.com/site/motorg...ad-the-guide-1
- "Transponders in Sailplanes - Dec 2014a" also ADS-B, PCAS, Flarm

http://soaringsafety.org/prevention/...anes-2014A.pdf

On Wednesday, February 17, 2016 at 11:00:36 PM UTC+3, Eric Greenwell wrote:
Bruce Hoult wrote on 2/17/2016 1:45 AM:
On Wednesday, February 17, 2016 at 1:05:18 AM UTC+3, Martin Gregorie
wrote:
On Tue, 16 Feb 2016 03:26:42 -0800, Bruce Hoult wrote:

On Tuesday, February 16, 2016 at 3:23:18 AM UTC+3, Martin
Gregorie wrote:
1) The energy conservation law would be violated if the energy
taken from the moving aircraft as drag is less than that needed
to generate the side force.

Incorrect. Force is not energy.

If you're extracting work from the system (in this case generating
a lift force that moves the fuselage sideways) you're using energy
that wouldn't be consumed if there was no side force being
generated.

Incorrect. Force is not energy.

If you were correct then a 20:1 glider and a 60:1 glider with the
same weight and same speed would be burning through energy at the
same rate because they're generating the same lift. But that's not
the case .. the 60:1 glider is losing energy at 1/3 of the rate of
the 20:1 glider.

The gliders are not losing the same amount of energy, even though the
airspeed and weight are identical: the 60:1 glider has 1/3 the drag of
the 20:1 glider, so it is losing 1/3 energy of the 20:1 glider.

Airspeed x drag = energy loss per unit of time

Exactly!

Martin Gregorie claimed that the energy loss was the same if the lift generated was the same.

Some of you boys do too much talking n not enough flying. Theory is all well and good but the proof of the pudding is in the doing. Get in your bird and fly and see the actual differences. But wait, maybe ya'll are just playing on sims lol

On Wed, 17 Feb 2016 14:44:38 -0800, Bruce Hoult wrote:

On Wednesday, February 17, 2016 at 11:00:36 PM UTC+3, Eric Greenwell
wrote:
Bruce Hoult wrote on 2/17/2016 1:45 AM:
On Wednesday, February 17, 2016 at 1:05:18 AM UTC+3, Martin Gregorie
wrote:
On Tue, 16 Feb 2016 03:26:42 -0800, Bruce Hoult wrote:

On Tuesday, February 16, 2016 at 3:23:18 AM UTC+3, Martin Gregorie
wrote:
1) The energy conservation law would be violated if the energy
taken from the moving aircraft as drag is less than that needed to
generate the side force.

Incorrect. Force is not energy.

If you're extracting work from the system (in this case generating a
lift force that moves the fuselage sideways) you're using energy
that wouldn't be consumed if there was no side force being
generated.

Incorrect. Force is not energy.

If you were correct then a 20:1 glider and a 60:1 glider with the
same weight and same speed would be burning through energy at the
same rate because they're generating the same lift. But that's not
the case .. the 60:1 glider is losing energy at 1/3 of the rate of
the 20:1 glider.

The gliders are not losing the same amount of energy, even though the
airspeed and weight are identical: the 60:1 glider has 1/3 the drag of
the 20:1 glider, so it is losing 1/3 energy of the 20:1 glider.

Airspeed x drag = energy loss per unit of time

Exactly!

Martin Gregorie claimed that the energy loss was the same if the lift
generated was the same.

Bull****. Try understanding what you read next time.


--
martin@ | Martin Gregorie
gregorie. | Essex, UK
org |