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Old Jan 31, 2013, 11:46 AM
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Because of where the yaw string is placed on most gliders, having the yaw string centered in a small-radius turn means that the vertical tail will "see" a sideslip angle where the relative wind is blowing from the ouside of the turn to the inside. This means that the rudder needs to be deflected to the inside of the turn in order keep the yaw string centered (the glider wants to slip due to the difference in sideslip seen by the vertical tail). Just because most gliders will show a "pedals-free" slip doesn't necessarily mean the slip is due to higher drag on the outboard wing.
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Old Jan 31, 2013, 11:51 AM
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Originally Posted by aeronaut999 View Post
PS Shoe re your equations above, the difference in tip speed may get smaller as speed increases, but since lift scales according to airspeed squared, does the difference in lift get smaller as airspeed increases? I have always guessed the answer was "yes" but have never been completely sure...

Steve
I believe the rolling moment due to to the lift difference (if not compensated for by aileron deflection) will remain constant with airpseed, but the rolling moment coefficient will be proportional to:

span*g*sin(bank_angle)/true_air_speed^2

So any need to "high side the stick" will go down like 1/true_air_speed^2
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Old Jan 31, 2013, 11:53 AM
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Originally Posted by aeronaut999 View Post
...
An aircraft that tended to skid rather than slip in a constant-banked turn would experience some very "interesting" effects-- anhedral would create a rolling-out torque, and dihedral would create a rolling-in torque!

Steve
.
Yes it does.
The rudder works backwards on this one! Very difficult to fly, and nearly impossible with the wing upside down AND backwards...
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Old Jan 31, 2013, 01:00 PM
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.
Yes it does.
The rudder works backwards on this one! Very difficult to fly, and nearly impossible with the wing upside down AND backwards...
Ha ha!

Well-- sure the rudder works backwards, but I bet things are still "conventional" in the sense that, if you are banked and aren't making a rudder input, the anhedral tends to make a rolling-in torque. Because the aircraft is tending to slip. Not skid, which would interact with anhedral to make a rolling-out torque.

Umm.. pray tell WHY put the wing on backward as well as upside-down?

Steve
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Old Jan 31, 2013, 01:12 PM
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Why? Because I could!
And that was a conventional first flight test for free-flights.
The wing on backwards and upright flew fine. Upside down was a bit more difficult, and upside down and backwards was a single flight phenomenon!
The K-F wing kinda proved what was pretty much self evident... a plane with that wing section will have more drag than a conventional airfoil... which is prolly why no one uses it.
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Old Jan 31, 2013, 01:13 PM
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span and length and curvature

Quote:
Originally Posted by ShoeDLG View Post
Because of where the yaw string is placed on most gliders, having the yaw string centered in a small-radius turn means that the vertical tail will "see" a sideslip angle where the relative wind is blowing from the ouside of the turn to the inside. This means that the rudder needs to be deflected to the inside of the turn in order keep the yaw string centered (the glider wants to slip due to the difference in sideslip seen by the vertical tail). Just because most gliders will show a "pedals-free" slip doesn't necessarily mean the slip is due to higher drag on the outboard wing.
Yes the vertical tail certainly plays a role. With vertical tail streamlined to the flow, we would still see some slip at the nose (and at the wing) even if there were no drag difference between the wingtips. If the outside wingtip is creating more drag than the inside wing, then the flow will impact the inside or low side of the vertical fin, with the rudder centered, and we will see a larger slip angle at the nose than if both wings were creating the same amount of drag. I suspect this is what we see in reality--and we could check it with a yaw string on the vertical fin-- but I agree that it would be theoretically possible to have the yaw string on the nose showing a slip, with the rudder centered, even if the inside wing were creating more drag than the outside wing, and the airflow were striking the outside or high side of the vertical fin at some small angle.

With my hang glider, the span is large compared to the length, so I'm pretty sure that the slip I see in the yaw string at the nose, can only be caused by the outside wing creating more drag than the inside wing, not by the way the back end of the glider "feels" a different airflow angle than the front end.

Interesting subjects to puzzle over...

Steve
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Old Jan 31, 2013, 01:47 PM
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Min sink airspeed for a glider is slower than the airspeed for L/D_max. This means that if you reduce the glider's airspeed below min sink in level flight, its total drag would go up, not down.

Slowing down the inboard wing in a level turn forces it to generate the same lift at a lower speed. The inboard wing also has to mirror the rolling moment of the outboard wing (barring a lateral CG shift), and the outboard wing is seeing higher local tip speeds. This means the inboard aileron needs to be deflected downward which creates additional profile and induced drag. I can't see how a steady turn at any speed slower than the speed for L/D_max could result in a decrease in the drag on the inboard wing (or an increase in drag on the outboard wing).

Also I don't think the yawing moment produced by the sideslip "seen" by the tail in a steady turn is related to the ratio of a glider's span to length... it should only be a function of the ratio of the turn radius to the displacement of the vertical tail from the CG.
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Old Jan 31, 2013, 04:05 PM
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Well, sounds like were going to have to film a yaw string mounted on the vertical fin of a glider, to see if it streams inboard or outboard in a no-rudder constant-banked turn. At various airspeeds. I would predict outboard, especially at low airspeeds. Just a prediction.

Roughly speaking, in such a situation if the yaw string on the vertical fin streams outboard, doesn't that mean that the outboard wing is making more drag than the inboard wing? And if the yaw string streams inboard, the reverse? If both wings are making the same amount of drag, won't the vertical fin tend to streamline itself with the local flow (which would cause a yaw string at the nose to stream outboard.)

The way I see it, if the span is long, the drag difference between the wing tips will tend to be larger and also will act at a much greater moment-arm from the CG, so you'll have lots of yaw torque swinging the nose toward the outside of the turn. Meanwhile if the fuselage is short, then the tail is not so far aft that there is lots of curvature in the flight path/ relative wind between the CG and the tail. All of these factors would seem to me to work together to promote an orientation where the airflow is striking the inside / low side of the vertical fin, not the outside/ high side, in a constant-bank no-rudder turn. If I'm wrong about the drag imbalance between the two tips then my picture of things may be way off.

Maybe Blaine Beron-Rawdon's articles on tail size, dihedral, and spiral stability / instability are relevant here..

On my hang glider, a yaw string back at end of the "keel" definitely streams outboard in a constant-banked turn. Even in very low turns (near min sink airspeed). Required roll input is minimal-- typically a slight roll input toward the high wingtip. Emphasis on slight. The roll input involves both a CG shift and an aileron-like deformation of the trailing edges. Constant-bank turns at higher airspeed (lower angle-of-attack) require a roll input toward the low wingtip, presumably due in part to a higher "roll damping effect" due to a higher sink rate through the airmass, as well as a reduced "effective span" due to washout acting to unload the wingtips, but the direction of yaw string deflection both at nose and rear of keel is still the same. Granted the shape of the hang glider is substantially different than a conventional sailplane...

This idea that if you are flying at min sink, slower than best L/D, then the inboard (slower) wing is making MORE drag than the outboard (faster) wing-- I'm not so sure. Certainly that is true when you slow down by increasing angle-of-attack, and speed up by decreasing angle-of-attack. But-- let's take the case of powered level flight-- in the turn, both wings are at the same angle-of-attack, but one is moving faster than the other. Now deflect an aileron downward on the slow wing and deflect an aileron upward on the fast wing, to equalize lift and roll torque. Maybe you end up with much more total drag on the faster wing with the raised aileron, than if you had simply pitched the whole airfoil down to a lower angle-of-attack. And maybe the same also on the slower wing with the lowered aileron-- but maybe you still end up with more drag on the faster-moving wing? I don't know...

All these details aside, I think we agree that in the absence of rudder input, the majority of an aircraft's wing area will feel a slipping flow (yaw strings /telltales deflect toward high wingtip) in a constant-banked no-rudder turn in nearly all cases? Again, if not, we'll no longer get the expected rolling-out torque from any dihedral that may be present... But I have to admit that I don't have feel I have a complete grasp on the cause of this slip in all situations. I'm sure that you can't simply attribute the slip to the fact there is a net horizontal force on the aircraft. Since the aircraft is free to rotate in the yaw axis, a net horizontal force can curve the flight path without making any slip.

Steve
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Old Feb 01, 2013, 05:05 AM
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This idea that if you are flying at min sink, slower than best L/D, then the inboard (slower) wing is making MORE drag than the outboard (faster) wing-- I'm not so sure. Certainly that is true when you slow down by increasing angle-of-attack, and speed up by decreasing angle-of-attack.
Suppose (for the sake of argument) that you have a wing with an untwisted elliptical planform and a lift distribution that is close to elliptical at all angles of attack.

If the wing is operating at a speed below the airspeed for L/D_max, and you slow it down but keep the lift constant by increasing angle-of-attack, then the wing's total drag will go up because the increase in induced drag is greater than the reduction in profile drag (subject to minor caveats).

You could also keep the lift constant (along the semispan) at reduced airspeed by deflecting an aileron downward. Will the addtional drag that results from keeping the lift constant by deflecting an aileron be more or less than the additional drag that results from keeping lift constant by increasing the wing's angle-of-attack?

If the aileron deflection kept the lift distribution elliptical, then the increase in induced drag should be exactly the same in both cases. If the aileron deflection made the lift distribution anything but elliptical, then the induced drag should increase more in the aileron deflection case.

The only way deflecting an aileron to maintain constant lift at reduced speed could reduce the drag acting on the wing semispan is if:

1. Deflecting the aileron improved the lift distribution of the wing (made it more elliptical) to the extent that the improvement in lift distribution overcame the increase in drag due to the increase in the semispan CL. This would only be the case for a wing that was performing very poorly prior to deflecting the aileron.

-or-

2. Deflecting the aileron to achieve the required increase in CL resulted in a reduction of the profile drag compared to increasing angle-of-attack to achieve the same increase in CL (again, unlikely unless the wing was performing poorly prior to deflection), and the associated reduction in profile drag coefficient was enough to overcome the the increase in induced drag (even more unlikely).

It's very unlikely a glider flying below the airspeed for L/D_max would experience a reduction in the drag on the inboard wing and increase in the drag on the outboard wing in a level turn (again, unless its performance was exceptionally poor before entering the turn).

A hang glider is slightly different. If you shift the CG laterally to maintain roll attitude rather than deflecting an aileron down, you are increasing the lift requirement (and induced drag) on the outboard wing, and reducing the lift requirement (and induced drag) on the inboard wing. It seems far more likely that a hang glider would see an overall reduction in drag on the inboard wing and an increase in the drag on the outboard wing in a level turn.

Someone with AVL skills could sort this out quickly and definitively.
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Old Feb 02, 2013, 09:22 AM
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slip in steady turn

Interesting thoughts Shoe...

* For the sake of talking about sideslip, the main point of interest is whether the outboard wing creates more drag than the inside wing. I didn't mean to suggest that the inboard wing necessarily saw a net drag reduction compared to in wings-level flight at the same angle-of-attack.

* Does it change anything if we say we are only interested in the part of the flight envelope that is equal to or faster than the min-sink-airspeed for the aircraft as a whole for the current bank angle, which--due to differences in lift and angle-of-attack across the span-- does not necessarily involve precisely the same angle-of-attack as measured at the wing root that would deliver min-sink in wings-level flight? In other words we likely don't want to slow the inboard wing down too much, even if that's what would be required to put the wing root at precisely the same angle-of-attack as would deliver the min-sink condition in wings-level flight...

* Stepping back to take a more holistic look at things-- consider this aircraft "ATOS VR" or "ATOS VQ"
(photos are now uploaded, see end of post)
http://www.a-i-r.de/med/nws/photos/A...0_20071106.jpg
http://www.a-i-r-usa.com/blob.asp?I=3073
http://www.a-i-r-usa.com/content.asp?ID=4339

The span is very long compared to the nose-to-tail length. There is substantial dihedral. Roll control is via spoilers ("spoilerons"). Side-to-side weight shift is minimal because with this glider, the triangular control frame is not a structural element, and when the pilot pushes sideways on the triangular control frame, the frame moves (activating the spoilers) and the pilot stays mostly fixed near the glider centerline.

Now, I'll bet you anything that the amount of dihedral was selected to give a good amount of rolling-out torque in a constant-bank turn, to balance the rolling-in tendency from the faster-moving outboard wingtip, so that the net roll torque is near zero with neither the left spoiler open nor the right spoiler open. You wouldn't want to be thermalling with a spoiler open most of the time.

I have flown one of the older ATOS gliders (like the shorter-spanned ATOS V in the last uploaded photo) and it seemed that in a constant-banked thermalling turn, no spoiler deflection was needed in either direction.

The dihedral is only going to give the expected rolling-out torque if the wing as a whole is flying in a slip (experiencing an airflow from the low wingtip toward the high wingtip). And considering the short distance from nose to tail, if a yaw string on the wing is blowing outboard, it seems likely that a yaw string on the tail also is blowing outboard. There isn't room for the flight path / airflow to curve much between the wing and the tail. Or to put it another way, the yaw rotation of the aircraft during the turn doesn't create a very large difference in yaw string angle between the CG of the aircraft and the tail.

If a yaw string on the tail is blowing outboard, then the tail is not the cause of the slipping flow over the wing. We might see even a larger slipping angle over the wing if we removed the tail.

(Many of these types of aircraft are designed to fly just fine whether the tail is installed or removed. The tail mainly provides resistance to a violent nose-down "tuck" or "tumble" in turbulent air. Certainly that is true of the shorter-span versions such as this ATOS V http://www.a-i-r-usa.com/content.asp?ID=4259 .(Last uploaded photo in this post.) Note also the absence of the vertical tip fins or winglets on the ATOS V-- the vertical tip fins or winglets are not critical to the yaw stability of these aircraft.)

I know some friends flying with these types of aircraft and could probably ask one to fly with a yaw string on nose and tail, and video cameras filming them, to confirm some of the suggestions above.

I guess it is mainly based on exposure to aircraft like this, that gives me a strong intuition than a long-spanned aircraft with a short tail moment arm will likely experiencing a slipping flow over the entire aircraft, including the tail. Due to the fact that the outboard wingtip tends to create more drag than the inboard wingtip, when net roll torque is zero. Again I can't fully explain exactly why this should be so, but I believe that it is.

Now, in the ATOS, because of sweep, drag ends up between the same between both wings in a turn, once the nose has yawed outboard to some given slip angle. If we removed the sweep, the aircraft would be yaw-unstable without a vertical fin. If we kept the fin at the same short moment-arm as the moment arm of the small V-tail in the aircraft in the photos, it would need to be very large to prevent the nose from yawing outboard to some absurd slip angle in circling flight. The slipping flow hitting the inboard side of the fin would be the thing that offsets the increased drag from the outboard wingtip. The bigger the fin, the smaller the slip angle.

If the fin were much further aft on a much longer tail boom things could be quite different-- the curvature of the flight path and airflow between the CG and the fin could possibly, at first glance, seem likely to cause the airflow to strike the outside or high side of the fin. Yet-- even here-- it seems that if both wings are making the same amount of drag, then the center of lateral area (actually laying some ways ahead of the vertical fin) will tend to position itself streamlined (tangent) to the curving flight path so that yaw strings on the aft part of the aircraft are nearly centered. If yaw strings at the rear portion of the aircraft are blowing toward the inside/ low side of the turn, this would seem to suggest that the inboard wing is creating more drag than the outboard wing, so that the fin must contribute a yawing-out torque, and if yaw strings at the rear portion of the aircraft are blowing toward the outside/ high side of the turn, this would seem to suggest that the outboard wing is creating more drag than the inboard wing, so that the fin must contribute a yawing-in torque.

I do recognize that with a long tail boom, it is possible for the rear portion of the aircraft to be nearly streamlined to the flight path / relative wind, or even for a yaw string at the tail to blow slightly inboard, even if there is a substantial slipping angle at the wing (yaw strings at wings blowing outboard). So, a slipping airflow at the rear of the aircraft is not an absolute pre-requisite for a slipping airflow over the wing. So, the fact that we generally see that dihedral creates a rolling-out torque, indicating a slipping airflow over the wing, is not absolute proof in all cases that there is a slipping airflow at the tail as well, or that the outboard wingtip is making more drag than the inboard wingtip. But my intuition is that this is the usually the case.

I intend to film a yaw string on the tail of my 2-meter Spider http://www.soaringusa.com/SPIDER-2M.html to see how it is oriented in a constant-bank turn. Noting the required aileron deflection at the same time. Which to the best of my recollection is often very near zero in a thermalling turn.

A couple more thoughts--

* If only the front portion of the ATOS gliders pictured below experienced a slipping airflow in a constant-banked turn, not the rear portions, then the increased dihedral of the outboard and rearmost wing panels (last 3 rib bays or so) would serve little purpose, at least in terms of the aerodynamics of a constant-banked turn.

* In general, don't we find that increasing the size of an aircraft's vertical fin tends to decrease sideslip while turning, which can lead to an increase in spiral instability (because dihedral now is less effective)? To me this suggests in a no-rudder turn, the airflow usually strikes the inside/ low side of the vertical fin, not the outside/ high side. Which suggests that the outboard wingtip is creating more drag than the inboard wingtip. Granted, if the bank angle and turn rate are increasing not constant, we have a demand for an increasing rate of yaw rotation, which means that yaw rotational inertia tends to promote sideslip even if the wingtips are making the same amount of drag. Also, in a case of spiral instability, I guess we aren't constraining the left and right wings to create the same amount of lift. More appropriate would be to ask, "In an aircraft with dihedral, does increasing the size of the vertical fin tend to mean that we tend to need to give more rolling-out input to keep the bank angle constant while circling at low airspeed with rudder centered?" If so, this suggests that the fin is acting to limit sideslip, suggesting that the airflow is striking the inside or low side for the vertical fin (yaw string on fin streams to outside), suggesting that the outboard wingtip is creating more drag than the inboard wingtip.

Interesting stuff...

Steve
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Old Feb 02, 2013, 11:21 AM
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another thought

Another thought-- in general, doesn't the wingtip tend to operate at a lower l/d ratio than the wing root? And doesn't this intrisically tend to mean that the outboard wingtip tends to make more drag than the inboard wingtip in a constant-banked circle? To take an extreme case, imagine that the wing spar were extended to protude a half-span beyond each wingtip. All drag, no lift. The faster-moving extension on the outboard tip will make more drag than the slower-moving extension on the inboard tip. Isn't a real-world wing just a lesser example of this extreme case? Particularly if we have some washout. It seems that it would take a pretty idealized wing to not have more drag on the outboard wingtip than the inboard wingtip, in a constant-banked turn. Just some intuitive thoughts not a full technical explanation... steve
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Old Feb 02, 2013, 03:35 PM
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more drag & more lift - same thing - all depends on how you use it - -it is moving faster -
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Old Feb 03, 2013, 09:34 AM
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Causes of sideslip

Sorry, this has turned into a very long post, but there's something on my mind I'm going to try to put into words--

We all know that dihedral interacts with sideslip to make a roll torque toward wings-level, and this tends to have a stabilizing effect on an aircraft. Really what I'm after, is "what is the cause of the sideslip?".

As noted in earlier posts, I think the causes can be broken down into what we might call "static" causes that operate any time the aircraft is turning, even at constant bank angle, slip angle, turn rate, airspeed, and sink/ climb rate; versus "dynamic" causes that depend on some sort of temporary imbalance in forces or change in bank angle and turn rate.

My recent long posts were looking at "static" causes. My visualization is that the increased drag of the faster-moving outboard wingtip tends to promote a slip whenever an aircraft is banked and turning, even while the aircraft is slowly returning toward wings-level after a disturbance. Also, even if both wings are making the same amount of drag, the curving flow (due to the curving flight path of the turn) can interact with the vertical fin to create some sideslip as measured at the wing.

These "static" causes can operate even as an aircraft is slowly rolling to wings-level after a disturbance. They don't depend upon the fact that the turn rate is increasing, or the fact that the vertical component of the lift vector is smaller than it ought to be, or anything like that. All that is required is that the aircraft be banked and turning.

As for "dynamic" causes, I think a key point is whether the aircraft is placed in a situation where there is a DEMAND FOR AN INCREASE IN YAW ROTATION RATE. If so, yaw rotational inertia tends to promote a temporary sideslip. If not, we can have all kinds of imbalances in forces without creating any sideslip.

For example even a normal, constant-banked, constant-rate turn involves an imbalance in forces. The vector sum of lift and gravity is a net horizontal force acting perpendicular to the flight path. This is the centripetal force that drives the turn. But since the required yaw rotation rate is constant, there is no demand for an increase in yaw rotation rate, and yaw rotational inertia has no tendency to drive a sideslip.

I think a lot of aerodynamics texts or stability and control texts come up short on this score. We are often told than since the weight vector includes a spanwise component when the aircraft is banked, this tends to cause a sideslip. In some instances we are told that a key point is that in an uncommanded bank, there is (initially) no increase in the total lift vector (G-load initially stays at 1 G), so we have an "uncompensated" portion of the weight vector, and it is only the spanwise component of this "uncompensated" portion of the weight vector that drives the slip.

To come up with some examples of this line of argument for this post, I googled "dihedral sideslip weight". Here are some of the hits:

http://www.flightglobal.com/pdfarchi...0-%202149.html

http://people.clarkson.edu/~pmarzocc/AE430/AE-430-7.pdf (scroll down to the frame entitled "dihedral effect" on page 10)

http://books.google.com/books?id=vRO...weight&f=false

(See figure 3.92 on page 294--
This is from the textbook "Performance, Stability, Dynamics, and Control of Airplanes" by Bandu N. Pamadi.)

I think we can find something similar in Martin Simon's "Model Aircraft Aerodynamics". And in many other stability-and-control chapters in aerodynamics textbooks.

Here's one that gives a slightly different twist-- here the claim is that the sideslip is driven by a horizontal "uncompensated component of lift", rather than by a component of the weight vector:

http://en.wikipedia.org/wiki/Dihedra...he_spiral_mode (See Fig. 1)

I think there is an unintentional smoke-and-mirrors trick going on when we break down the vectors in this way and claim to find a cause for sideslip. If the spanwise component of the "uncompensated portion of the weight vector" contributes to sideslip when we accidentally enter a bank, wouldn't it be equally valid to argue there is also going to be a sideslip whenever we are banked and turning, caused by the spanwise component of the centripetal force that is driving the turn? In other words in turning flight, weight acts downward and lift acts perpendicular to the wingspan, and the vector sum of lift and weight is a horizontal force acting perpendicular to the flight path, and this horizontal force has a spanwise component, so shouldn't this cause a slip?

My answer- not if there is not a demand for an increased yaw rotation rate. And likewise for any spanwise component of any single force vector, whether that be the spanwise component of the lift vector or the spanwise component of the weight vector. That's where these arguments come up short-- they equate a spanwise force vector with a spanwise motion through the air, i.e. a sideslip. That is a fundamental misunderstanding of the nature of curving flight.

These arguments fail to consider the complete and dynamic situation of the aircraft. For example, if we suspended an aircraft in a banked attitude at zero airspeed, and dropped it toward the earth, the only force initially acting on the aircraft is the weight vector. The direction of acceleration is earthwards, and the nose must yaw (and pitch) downward to align the nose with the flight path. This is a demand for yaw rotation. We will see some sideslip until the nose yaws downward.

However if we are flying in a 45-degree-banked turn and then we shove the stick forward to unload the wing to zero G's (zero lift), now will we see a sideslip? Suddenly things aren't so clear. The "spanwise portion of the uncompensated portion of the weight vector" line of argument would say "yes of course". And it is true that the flight path will curve earthward, and the nose must yaw (and pitch) downward to align with the changing direction of the flight path. However, looking down from above, we will see that the by unloading the wing to zero G's, we have dropped the turn rate to zero. There is no more curvature in the flight path, as viewed from above. And before we unloaded the wing to zero G's, we already had a yaw rotation rate going. Now it no longer clear whether the yaw rotation rate actually needs to increase, or decrease, to keep the nose aligned with the flight path as the flight path curves earthwards. A demand for increased yaw rotation would promote a temporary slip, until yaw rotational inertia is overcome by "weathervane" stability (sideways airflow striking fin, etc, acting to increase yaw rotation rate.) . A demand for decreased yaw rotation would promote a temporary skid, until yaw rotational inertia is overcome by "weathervane" stability (sideways airflow striking fin, etc, acting to decrease yaw rotation rate).

What if we do the same experiment at 30 degrees bank? At 60 degrees bank? What if we suddenly pull to double the G-load, rather than suddenly pushing to "unload" the wing?

In actual practice in full-scale aircraft I find that these kinds "push" or "pull" maneuvers create very little tendency for slip or skid.

One sort of dynamic maneuver that does create a slip-- due to a sudden demand for increased yaw rotation rate-- is the kind of "wingover" where you unload the wing to a very reduced G-load (lift force) as you float over the top at steep bank (possibly a vertical bank) with very low airspeed. The airspeed is low, so aerodynamic forces are low, but gravity is pulling (curving) the aircraft's trajectory earthwards in a semi-ballistic arc. The low airspeed means that the arcing trajectory will have a small radius of curvature. At this point in the maneuver the yaw rotation rate must increase if the yaw string is to stay centered. With no rudder input, the yaw string tends to stream toward the high wing.

A related maneuver that I like to perform with radio-controlled gliders is to put the glider in steeply banked climb and feed in forward stick. By feeding in forward stick at just the right rate, you can get the glider's airspeed to bleed off to a very low value as it continues on a climbing semi-ballistic path. Then the glider runs nearly out of airspeed and the nose must yaw downward rapidly as flight path curves earthward. Since aerodynamic forces are low at low airspeeds, yaw rotational inertia becomes quite significant. With no rudder input, a yaw string on the nose will stream very strongly toward the high wingtip as the glider goes over the top of the maneuver, and then as the required yawing motion develops, it "overshoots" too far and as a result you can see the nose oscillate left and right several times as the aircraft's trajectory continues earthwards. This oscillation is the dynamic "dance" between yaw rotational inertia and "directional stability" or "weathervane stability". Piloting error-- too little bank, too steep a climb angle-- can lead to a tailslide rather than the desired dramatic yaw rotation, as the aircraft runs out of energy in the steep climb, so I don't do this maneuver in full-scale aircraft!

What I'm trying to say, in a very long-winded way, is beware of arguments that suggest that a spanwise force component will act to cause a sideslip. You have to look at the complete dynamic situation and see whether there is a demand for an increase in yaw rotation rate, or not. Only if there is a demand for an increase in yaw rotation rate will we tend to see a temporary sideslip, due to yaw rotational inertia.

At bank angles up to 45 degrees, an increase in bank angle always demands an increasing yaw rotation rate (except in the zero-G condition), so it would seem that yaw rotational inertia would always tend to cause some sideslip as the bank angle increases. However, if aerodynamic effects (yaw torque due to "twisted lift" due to rolling motion, yaw torque due to aileron deflection) are creating a strong aerodynamic adverse torque, then maybe yaw rotational inertia will actually act to slow things down a bit and reduce the net adverse yaw and sideslip that we see as the bank angle increases!

(One thing that is a little unclear to me-- if an uncommanded increase in bank is caused by a downdraft hitting one wing, does the rolling motion still tend to create adverse yaw due to the "twisting" of the lift vectors? It seems likely not; certainly enveloping one wing in a downdraft changes the direction of the aerodynamic forces!)

In actual practice I suspect that sideslip is usually caused primarily by aerodynamic effects, not inertial effects. And force vector diagrams that show the presence of a spanwise force vector, be it from lift, weight, or something else, tell us nothing about these aerodynamic causes of sideslip.

Steve
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Old Feb 03, 2013, 10:11 AM
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Aero, there's no hand waving going on. Every textbook explains the cause of sideslip accurately and it's very simple: The weight vector is always vertical and lift is always perpendicular to the wing. Therefore any bank results in an uncompensated span wise component of lift. From this you get slip, and from slip you get fuselage weathervaning. The wikipedia link depicts this clearly, and though the other links apply the vector math in a confusing manner to the weight vector, they are nonetheless correct. The other causes you describe (roll rate, yaw rate) exist in reality as well as every good text but are only a miniscule and typically ignored component.
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Old Feb 03, 2013, 10:30 AM
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Originally Posted by vespa View Post
Aero, there's no hand waving going on. Every textbook explains the cause of sideslip accurately and it's very simple: The weight vector is always vertical and lift is always perpendicular to the wing. Therefore any bank results in an uncompensated span wise component of lift. From this you get slip, and from slip you get fuselage weathervaning. The wikipedia link depicts this clearly, and though the other links apply the vector math in a confusing manner to the weight vector, they are nonetheless correct. The other causes you describe (roll rate, yaw rate) exist in reality as well as every good text but are only a miniscule and typically ignored component.
Vespa, I think you and I are just going to have to agree to disagree. Just because there's a net force that acts horizontally in banked, turning flight, doesn't mean that the aircraft tends to be dragged sideways through the air. There's no causal link there at all. The thing is, your line of reasoning ends up predicting that the net yaw torque is not zero in a steady, constant-banked, constant-rate turn. Because you have this constant "weathervane"-based yaw torque going on even in the absence of other aerodynamic effects. It doesn't add up at all. But forget the nuances of aerodynamics and consider again the "unconventional" kinds of turns I outlined in post #35 http://www.rcgroups.com/forums/showp...2&postcount=35 . What you and are disagreeing on here is not aerodynamics but rather the basic physics of moving bodies. Steve
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