Slips in turns and landing with winglets

On Mon, 15 Feb 2016 14:48:52 -0800, Bruce Hoult wrote:

On Monday, February 15, 2016 at 7:13:43 PM UTC+3, Tango Whisky wrote:
Le lundi 15 février 2016 16:32:27 UTC+1, Bruce Hoult a écritÂ*:

I don't agree.

As I've already said once in this thread, you're going to get a
certain amount of drag from the fuselage anyway. If the presentation
to the airflow for minimum drag generates zero lift then, by the
calculus definition of "minimum" of a continuous function, the first
little bit of lift will not add any drag. The optimum thing to do is
to use it. It might be *very* little, and a very small AoA, but it's
nonzero.

If the presentation to the airflow for minimum drag generates
non-zero lift ... then of course you'll take it!

Yes, the fuse has a low L/D. But that's better than the 0.0 L/D if
you don't take what lift you can from it...

Why do you assume that the fuselage has zero lift in a perfect airflow?

I don't. I present the argument for both cases: zero and non-zero lift
at minimum drag.


Why do you assume that the increase of drag is zero for small slip
angles?

Follows directly from the definition of "minimum" for a continuous
function.

The minimum is, by definition, at the point at which the function (the
drag) has zero change for small changes in the input (the AoA or slip
angle).

But, as soon as the fuselage generates any side-force, its drag will
increase, and by more than the energy needed to generate the side-force.
This is due to two things:

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.

2) The energy consumed will be more than that used to generate the side
force because no process that consumes energy is 100% efficient.


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

On Tuesday, February 16, 2016 at 3:23:18 AM UTC+3, Martin Gregorie wrote:
On Mon, 15 Feb 2016 14:48:52 -0800, Bruce Hoult wrote:

On Monday, February 15, 2016 at 7:13:43 PM UTC+3, Tango Whisky wrote:
Le lundi 15 février 2016 16:32:27 UTC+1, Bruce Hoult a écrit*:

I don't agree.

As I've already said once in this thread, you're going to get a
certain amount of drag from the fuselage anyway. If the presentation
to the airflow for minimum drag generates zero lift then, by the
calculus definition of "minimum" of a continuous function, the first
little bit of lift will not add any drag. The optimum thing to do is
to use it. It might be *very* little, and a very small AoA, but it's
nonzero.

If the presentation to the airflow for minimum drag generates
non-zero lift ... then of course you'll take it!

Yes, the fuse has a low L/D. But that's better than the 0.0 L/D if
you don't take what lift you can from it...

Why do you assume that the fuselage has zero lift in a perfect airflow?

I don't. I present the argument for both cases: zero and non-zero lift
at minimum drag.


Why do you assume that the increase of drag is zero for small slip
angles?

Follows directly from the definition of "minimum" for a continuous
function.

The minimum is, by definition, at the point at which the function (the
drag) has zero change for small changes in the input (the AoA or slip
angle).

But, as soon as the fuselage generates any side-force, its drag will
increase,

Incorrect. Please study your Calculus 101 books a little more closely.

and by more than the energy needed to generate the side-force.
This is due to two things:

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 were correct then wings would not work at all.

For maximum efficiency, every part should (as much as possible) be operated at its angle of attack for best L/D, not at minimum drag.

An L/D of 5 on some part is better than an L/D of 0 on that part. Even if some other part has an L/D of 50. (provided you don't make that other part perform worse, of course)

An L/D of 5 on some part is better than an L/D of 0 on that part. Even if some other part has an L/D of 50. (provided you don't make that other part perform worse, of course)

Total lift is constant. So if you shift a part of the lift to some inefficient means, your overall performance will drop.

On Tuesday, February 16, 2016 at 4:20:33 PM UTC+3, Tango Whisky wrote:
An L/D of 5 on some part is better than an L/D of 0 on that part. Even if some other part has an L/D of 50. (provided you don't make that other part perform worse, of course)

Total lift is constant. So if you shift a part of the lift to some inefficient means, your overall performance will drop.

Incorrect, because total drag is not a constant.

If you can get, say, 1% of your lift from the fuselage without increasing the drag of the fuselage, then decreasing the lift needed from the wings by that 1% will decrease the induced drag from the wings by about the same 1%.

Le mardi 16 février 2016 15:24:00 UTC+1, Bruce Hoult a écrit*:

If you can get, say, 1% of your lift from the fuselage without increasing the drag of the fuselage, then decreasing the lift needed from the wings by that 1% will decrease the induced drag from the wings by about the same 1%.

Absolutely correct, but deconnected from reality. Just because a function runs through a minimum, it's not a given that slight deviations from the minimum on the x-axis yield only slight variations on the y-axis.

But again, do what you like ;-)

On Tuesday, February 16, 2016 at 5:48:00 PM UTC+3, Tango Whisky wrote:
Le mardi 16 février 2016 15:24:00 UTC+1, Bruce Hoult a écrit*:

If you can get, say, 1% of your lift from the fuselage without increasing the drag of the fuselage, then decreasing the lift needed from the wings by that 1% will decrease the induced drag from the wings by about the same 1%.

Absolutely correct, but deconnected from reality. Just because a function runs through a minimum, it's not a given that slight deviations from the minimum on the x-axis yield only slight variations on the y-axis.

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

The entire concept of a polar curve relies on this property.

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.

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

The only thing that aerodynamics adds to basic physics is the knowledge
that any object that generates lift also produces induced drag. There is
no induced drag if Cl = 0, but as soon is Cl is non-zero, induced drag
becomes non-zero as well and is added to the skin friction and shape-
related drag terms.



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

Wasting your time worrying about this. All you really need is the MK IV "high tech" yaw string. I hear it auto coordinates all turns while the low profile adhesive mount reduces drag.

It's the equivalent of giving your glider PED's.

On Tue, 16 Feb 2016 14:37:18 -0800, scohpilot wrote:

Wasting your time worrying about this. All you really need is the MK IV
"high tech" yaw string. I hear it auto coordinates all turns while the
low profile adhesive mount reduces drag.

I know: got one already, plus my bum is finely attuned to understanding
what my Libelle is telling it.



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

Andreas Maurer wrote on 2/15/2016 2:17 PM:
On Mon, 15 Feb 2016 07:16:51 -0800, Eric Greenwell
wrote:


Maybe this is intended as joke, but the last thing you want is a very
low L/D "airfoil" like the fuselage involved in providing any lift.


Depends on the type of glider.

An open class ship needs a lot of sideslip while thermalling.

I learned that a long time ago from Uli Schwenk, who told me that he
learned that from a guy named Klaus Holighaus.

Thermalling with the yawstring 30 degrees outwards makes a huge
difference in climb performance. Huge.

Explanation:
The yaw costs a lot of drag due to the "low L/D fuselage" - but it
saves even more drag because you barely need any opposite aileron
anymore, therefore you get a much better lift distribution (and
therefore much less induced drag) on the wing.


One other glider that needs a lot of yaw in a thermal: Arcus.

I agree there are gliders where that drag would be an acceptable
trade-off, but the pilot I was responding to thought it was a "bonus",
and it's clearly not that.

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? 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.


--
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