The yaw/slip thread (merged) aka Aerodynamics 101 [Archive]

Milt

8th Sep 2004, 12:59

John Tullamarine and ROB-x38

That which you are overlooking is the resolved portion of lift from the few degrees of bank you are using to provide the horizontal force necessary to oppose the yawing moment being produced by the offset thrust.

Moment 1 - Thrust X arm to lateral centre of pressure, opposed by Moment 2 - Resolved lift X arm to lateral centre of pressure cancel out to give you straight flight but NOT along the aircraft axis. You can use parallelogram of forces here. Hence a small angle of sideslip which produced an opposite side force to cancel out the resolved lift sideforce. Total Drag balances the Total Thrust. Then a little extra lift please which with its extra induced drag added to the added drag for sideslip probably makes the two methods of slightly banked and wings level a push. This is one for the honours students doing their aero degrees to get their heads around.

Incidently weight and cg doesn't get a show in this area as it is all mostly in the horizontal and you cannot resolve the weight through 90 degrees. The implication of this is that Vmca is not affected by cg position contrary to what you see in many training publications.

Then you have a "form" of balanced flight and the navigator should estimate the sideslip angle to provide for more accurate navigation. Don't some handbooks provide an estimate.? I say "form" to indicate my reluctance to call straight flight while sideslipping balanced but indeed all the forces are in balance.

I guess you might determine the angle of sideslip without a vane on a boom by aligning a DG with an Inertial/GPS accurate heading reference and go asymmetric. You may see the difference between DG and Inertial/GPS when you stabilise.

Tell me where I might have lost my marbles. They are rattling around in my head!

SR71

8th Sep 2004, 14:03

\tan(\alpha)=u/V & \sin(\beta)=v/V

where

V = ui + vj + wk &

V =\sqrt[u^2+v^2+w^2]

The way I see it is once you've got an assymetric condition you've got two options once you've noticed it, whereupon by definition you've got an angle of sideslip, and depending on the stability derivative C_l_\beta (& C_l_r), some roll angle. Which one you use is a matter of preference in most cases but in critical cases where you've let the angle get large and configuration and performance considerations come into play you best choose one not the other:

1) Jump on the appropriate rudder to arrest the yaw. C_n_\beta i.e., static directional stability requirements tend to ensure the contribution of the fin is to oppose any increasing sideslip, and with the effective fin angle of attack increased by the rudder deflection, a flight path with a characteristic \beta will develop where the forces and moments are in balance. Its a dirty flight condition and unless C_l_\beta ~=0, you need an initial amount of aileron to re-gain wings level. The actual value of sideslip that develops is merely a function of how much load you choose to apply to the rudder.

However, if you want to continue in straight flight, the lateral loads on the aircraft must also sum to zero whereupon the value of \beta will probably be fixed and unique.

2) Accomplish the above, but now you've got an aft CG condition and you're at MTOW and you've let \beta get large (on Sector 4, Day 6, 0200Z perhaps?) at 200' when a donk lets go off R/W32 @ AGP, 32C and 975mb. This is approaching a V_mca condition.

JAR 25.149 requires no more than 150lbs/667N of force be applied to the rudder controls in order to restore directional control in this condition and continue in straight flight. Neither may thrust be reduced.

The proviso for no more than 5 degrees angle of bank is presumably because if one supposes that full rudder deflection is required to counteract the asymmetric yawing tendency, you have lost the ability to control the path of the aircraft in azimuth through space via expedient use of the rudder. Hence the allowance for use of bank. Since the lateral component of lift due to bank is proportional to \sin(\gamma) you now have ~=9% of the aircraft lift to aid the process.

The aerodynamic advantages of the use of bank are numerous, and again, because the vertical component of lift whilst banked is proportional to \cos(\gamma) you now still have ~=99.5% of the original lift available for performance at 5 degrees of bank.

My concern is that for the purposes of calculating RTOW's at speeds approaching V_mca (or for that matter any speed), the performance will clearly be affected by whether or not you choose to employ some bank angle. Oddly, there is no performance requirement to be met associated with V_mca.

For presumably 99% of the time, you needn't concern yourself with this matter, but in the interests of discussion, if the only way you can meet a climb gradient requirement is by flying in a particular configuration, it becomes paramount to ensure you do so.

Now if only I could fly within +/- 5 degrees of bank.

:ok:

PS: V_mca certainly is affected by CG location.

From JAR 25.149:

(c) VMC may not exceed 1·13 VSR with –
(1) Maximum available take-off power or
thrust on the engines;
(2) The most unfavourable centre of
gravity;
(3) The aeroplane trimmed for take-off;
(4) The maximum sea-level take-off
weight (or any lesser weight necessary to show
VMC);
(5) The aeroplane in the most critical
take-off configuration existing along the flight
path after the aeroplane becomes airborne, except
with the landing gear retracted;
(6) The aeroplane airborne and the
ground effect negligible; and
(7) If applicable, the propeller of the
inoperative engine –
(i) Windmilling;
(ii) In the most probable position
for the specific design of the propeller
control; or
(iii) Feathered, if the aeroplane has
an automatic feathering device acceptable
for showing compliance with the climb
requirements of JAR 25.121.

It affects the length of the tail moment arm, a crucial component towards directional stability.

:ok:

Tinstaafl

8th Sep 2004, 16:07

Two separate discussion going on now. One about yaw (or not) during a turn, and the other about sideslip (or not not) during asymmetric flight.

My view:

1. Yaw vs turn.

At 0deg AoB the a/c can be turned using 'all rudder' ie a flat turn. In this case the a/c is yawing but not pitching.

At 90deg AoB the a/c can be turned using 'all elevator'. In this case the a/c is pitching but not yawing.

The a/c can't instantly switch from one case to the other. It's a smooth change from one condition to the other, with yaw reducing & pitch increasing as AoB increases.

2. Slip vs asymmetry.

I think we all agree that in a wings level steady heading asymmetric condition there is some amount of slip. The Rel Airflow will be from the 'dead side' as a result of the rudder/fin sideload.

If an a/c is banked it will start to slip in the banked direction ie the relative airflow will have a lateral component (relative to the a/c axes) from the low wing side. This still applies in the asymmetric case. Imagine an asymmetric a/c at a very large AoB. It *will* start to slip towards the low wing. The slip will then result in a yaw towards the low wing. If the low wing is the live engine side then there will be a sufficient AoB that will counter the thrust asymmetry as a result of the relative airflow approaching from the live engine side causing a yawing moment towards the live engine.

It's possible to have sufficient bank that NO rudder at all is needed - ignoring some incredibly high thrust engine that could make the a/c do cartwheels...) albeit at a very high cost in drag. Not much good for performance then. On the hand, Vmc would be nice & low since the rudder hasn't reached its maximum deflection. In fact, it hasn't moved at all... Large angles of band are good for the manufacturer when it comes to certification since many things are related to Vmc so a low Vmc is 'useful'. Not so good for the pilot since it means unusual handling is required - hence the 5 deg AoB certification limitation.

Note the difference in the two cases above. Case 1 (wings level, only rudder used to oppose the asymmetry) the Rel. Airflow has a component from the 'dead side'. Case 2 (AoB, no rudder used to oppose the asymmetry) the Rel. Airflow has a component from the 'live side'.

As AoB increases towards the live side the component from the dead side must reduce until there is no sideslip, then increase again but this time from the live side.

So, it is possible to fly asymmetrically with zero sideslip and this AoB is where best performance will occur. This typically occurs at ~2-3 deg AoB. Any more than this and there will be a small sideslip towards the live side, any less than this and there will be a small sideslip towards the dead side.

Mad (Flt) Scientist

8th Sep 2004, 18:18

In any practical, stable, banked turn, the aircraft will have roll, yaw and pitch rates simultaneously, even though the bank angle and pitch attitude will be fixed and it appears that the only attitude angle change is in heading.

This is because the roll, yaw and pitch rates are defined in an axis system which is fixed in the aircraft, and the bank angle, pitch attitude and heading (yaw attitude) are all defined by the relationship of that aircraft-fixed axis system to the earth-axis system.

The trick is that rate of change of bank angle is NOT mathematically the same as roll rate; nor for the other axes. Those are approximations which hold good for a given level of accuracy, but not to the ultimate.

If you had no angle of attack, such that the aircraft x-axis - about which we measure roll rate - was perfectly aligned with the velocity vector, you'd have no roll rate, but that's all. You'd still have a yaw rate and a pitch rate.

Having spent many happy hours explaining that, no, the simulator is working properly, those rates on the IRU are REAL, down to digging out the relevant equations of motion, I'm pretty sure of that lot.

Milt

9th Sep 2004, 01:32

SR71

Thanks for the hieroglyphics and the erudite discussion.

Unfortunately you didn't explain, in the banked case, how a horizontally derived portion of lift acting laterally on the fuselage can fail to cause sidslip. The contra force resulting from the sideslip is the balancing force is it not?

So I say again with some conviction
ANY imbalance in thrust on a multi, not having centre line engines, which is then flown straight by either rudder or bank or a combination inevitably results in sideslip. It is unavoidable
Then you refute the obvious in believing and stating that there can be an effect of cg position on Vmca. Derive all the weight you can to the horizontal and what do you get !!

In steady straight flight forces in the horizontal act around the lateral centre of pressure - NOT the cg. The cg can be anywhere until manoeuvre comes into the act..

Consider the directional stability of a twin engined dirigible having no apparent weight. Vmca when asymmetric ???

Looks like JAR 25.149 was prepared by someone having the same misconceptions and it may be important that we take steps to correct.

Hope you are enjoying this as much as I am.

We may have to fit a vane on a boom and do it all over again !!

Come in NATPS ETPS USNTPS Edwards NASA

john_tullamarine

9th Sep 2004, 04:14

Isn't this all jolly good fun ?

Milt, was speaking with JL last weekend .. now that I know to whom the username relates .. I shall continue to read your words with the considerable respect they deserve.

Mind you, though, I think we might continue to disagree on the slip question.

bookworm

9th Sep 2004, 06:40

Anybody in Canberra prepared to take Milt for a ride in a twin with a slip string attached?

SR71

9th Sep 2004, 18:11

Well if empirical flight test results demonstrate otherwise, the mathematics requires amendement. One should ensure the facts work with the hypothesis not vice versa!

I make the following comments.

I see this as essentially a lateral/directional stability problem. Normally these calculations are resolved about the CG.

It is the load on the fin that counteracts the component of the lift along the inertial y-axis, thereby allowing straight flight inspite of the bank angle.

Lets assume an a/c with C_l_\beta ~= 0 and where the No 1 eng fails. In this case the roll DOF is not important - we can assume wings level. Assuming the rudder has sufficient authority to counteract the assymmetric yaw, an equilibrium position will develop where the a/c is sideslipped away from the live engine. Unfortunately relative to inertial space, the a/c cannot maintain the runway centreline because resolving the thrust vector and fin load, they both in the same direction.

In order to arrest the azimuthal drift, you need to reverse the sense of the sideslip. This is done via judicious use of the rudder.

However, in reversing the sense of the sideslip, the rudder is now working hard. Not only must it balance the component of thrust resolved in the relevant axis, but its natural propensity to generate a load in the same direction courtesy of the sideslip.

Now assume by this judicious use of rudder, whereupon I'm able to change the flight path vector in azimuth, I've regained the runway centreline.

One way of off-loading the rudder is merely to incline the lift vector in order to generate a component in the direction of the inertial y-axis i.e., bank.

If I do this without reducing the sideslip, i.e., the load on the rudder, the aircraft will now drift in azimuth away from the runway centreline in the direction of the live engine.

Reduce the sideslip and it should be possible to balance the forces again all the way until you have a zero sideslip condition.

You are merely balancing the thrust component, the lift component and the fin sideload along the inertial y-axis via a combination of sideslip and roll angle.

That said, I don't believe you can have a nil sideslip, nil rudder force, angle of bank /= 0 condition and maintain the runway heading because the fin is symmetric. In this case, the only way to counter the tendency of the a/c to drift in the direction of the bank is to counter with rudder. This effectively cambers the rudder whereupon you get a sideload.

As for V_mca....

The way I understand it, you can change the available moment necessary to counteract asymmetric yaw due to an ENG OUT situation via modification of the force or the distance that the opposing aerodynamic surface generates.

Whereupon my conclusion that the CG location is intimately acquainted with the concept....

:ok:

Milt

10th Sep 2004, 04:44

SR71
CC John Tullamarine for your further delight.

Would be most presumtuous for me to claim the maths are wrong. Haven't checked them lately!

And SR71 - I see that you are now starting to see how it all works by saying that "It is the load on the fin that counteracts the component of lift along the y-axis, thereby allowing straight flight inspite of the bank angle". Partially correct. The only way I know of to get a side force from the fin, rudder central, is with sideslip. So - yes - the fin is probably the biggest contributor aided by the rest of the sideslipping fuselage.

My concern is with the few mYths that continue to perpetuate.

There is the one about our present subject and others such as the 95% pilot belief that aircraft rudders opperate in the correct human instinctive sense.
Right NOW we have many heavy pilots rolling out after touch down using rudder in one direction to hold centre line before he changes over to use nose wheel steering, usually in the form of a wheel, and operating the nose wheel steering control in the reverse sense to achieve the same result. He may even be uncoordinated enough to be doing that simultaneously. But we humans are very versatile and it didn't really take more than an hour or two to suppress our prior learning on billy carts, bicycles, cars and motor bikes to operate the damn thing in the aircraft the other way round - did it?

But how we came to accept reverse operation of rudders is a bit like some switches being up for on and across the pond they are down for on. Subject for another thread.

Same with our mental picture of an aircraft's cg position. It seems to take on its own personna when we are considering it in the pitching/longitudinal sense. Long stab is where we first learned about the balance of forces and all that. Then we transferred our attention much later to directional considerations and found it very convenient to bring along that mental picture of the cg. It just doesn't exist in the unaccelerated horizontal. Certainly you can push it sideways a bit by having an unbalanced transverse load. But so what with our present thoughts?

But I remain perplexed that in the case being considered of straight balanced unaccelerated flight most of you still want to consider forces acting around a point which is totally irrelevant to the unaccelerated horizontal situation. Vital yes to the vertical situation.

Too few of us have thought much about the lateral centre of pressure of a body moving through the air. Total drag can be said to concentrate from that point just as we represent total lift in the vertical through a point for ease of reference. But it's damned important to keep that point down the back. Otherwise our aircraft will want to turn around and go backwards.

In considering a stable horizontal situation move the cg anywhere you want. Take moments about the tip of the nose if you want. Makes no difference to the balance of the steady horizontal forces. I know - some of you aren't convinced of that yet. So I must insist that you only look at the horizontal only because that is what this discussion is all about.. Strange I didn't have the problem before when comprehending the vertical situation. There was no flow over then from the horizontal to the vertical unless the cg moved sideways. Don't give up - it gets simpler.

Having convinced you that the position of the cg doesn't matter, I will now proceed to examine the horizontal forces for the case of an asymmetric aircraft having some bank into the side having most thrust. Keep it simple - it's a twin.

All of the HORIZONTAL forces involved can be simplified, I hope you will agree, into thrust component, drag, horizontal lift component and the one that most of you want to ignore which is the horizontal force generated by sideslip. In the wings level case that sideforce was obvious because it just must be there. I absolutely know that the rudder is deflected.. But roll over a little and you replace that force largely produced previously by the rudder with total fuselage sideslip so that you can now have a close to central rudder. May look neater to some.

What are the simple moments and how do they get into balance to give us straight "balanced" flight?

1. Thrust component X arm to the line of total drag. OK those two forces, thrust and drag, balance but are endeavouring to yaw the aircraft mightily and spoil your day. Quickly lets have contra yaw to keep us on the straight and narrow. Let's use the next moment.

2. Horizontal lift component X arm to ?? Whoops - where is the opposing force. It has to be around someplace or we have to go sideways and yaw at the same time. So sideways we start to go until the sideforce generated by the sideslip builds up to equal the horizontal lift component. Hey we are back in balance.

Do I see flashes of inspiration towards a better understanding of asymmetric directional stability.

Will you now join with me in my crusade to start a new religion based on the lateral centre of pressure.? OR

Please someone show me where I may be wrong before I take up hydrodynamics instead to see how they do it with boats and dirigibles.

Had to edit to correct spelling of asymmetric which is impossible to spell correctly all of the time - just like parallel - that doesn't look right either.

Smooth landings

bookworm

10th Sep 2004, 06:21

The only way I know of to get a side force from the fin, rudder central, is with sideslip.
No one is suggesting that the rudder is central for zero sideslip. It may be a little less than for the wings level case, but there's still substantial force coming from the vertical tail surface (without sideslip).

All of the HORIZONTAL forces involved can be simplified, I hope you will agree, into thrust component, drag, horizontal lift component and the one that most of you want to ignore which is the horizontal force generated by sideslip.
And you want to ignore the force at the vertical tail...

2. Horizontal lift component X arm to ?? Whoops - where is the opposing force.
At the vertical tail...

Certainly you can push it sideways a bit by having an unbalanced transverse load. But so what with our present thoughts?
I sympathise with your concern that unquestioningly resolving everything about the C of G is unhelpful. However, in the case of the horizontal component of lift, there's a reason for considering the C of G. The total (mainplane + horizontal tail) lift on the aircraft must act at the C of G, otherwise there would be an unbalanced couple with the weight. Thus the horizontal component of lift also acts at the C of G and the C of G position is therefore of significance in assessing asymmetric control.

Milt

10th Sep 2004, 10:01

bookworm

Thanks for highlighting the bits you don't understand.

A sideforce from the fin, rudder central without sideslip ??? You have discovered something unknown in aerodynamics. Patent it quickly.!!

How can you possibly say " there's still substantial force coming from the vertical tail surface (without sideslip)" when there will be absolutely zilch except drag unless there are inequalities in the shape of one side with the other. I'll concede that slightly different airflows along the vertical tail resulting from one engine operating and the other closed down may have some effect. But skip this for now as it only complicates the case on which we seek a conclusion.

For ease of further discussion, which must be very illuminating to the thread visitors, let us just look at the wing down case where we have carefully adjusted the bank to result in a centralised rudder whilst we maintain balanced straight flight.

You say "And you want to ignore the force at the vertical tail..."

Please tell me how I am ignoring it when I have just said "and the one that most of you want to ignore which is the horizontal force generated by sideslip."

Does this not equate to your "the force at the vertical tail.."

This IS the force at the vertical tail which is ONLY achievable from sideslip unless we have a gremlin out there giving it a shove.

Don't forget that there is a reasonable contribution of force from the side of the fuselage to the balancing force at the vertical tail.

If the above does not convince please tell me where I have failed. I have a thick skin.!

We agree on one fact. That is that the total lift acts through the cg (ignoring any vertical component of thrust.)

Pardon me on my insistence on the use of cg intead of CG. It's the same as using MM for mm and maybe I'm a long way behind on an update.

The combination of a relatively small horizontal lift component and the small permissable fore and aft cg range results in there being insignificant effects of cg position on the stabilising couple with the tail force resulting from the sideslip. Because of its insignificance I have taken the liberty of ignoring it for the sake of simplicity.

Perhaps someone will come up with some typical numbers to illustrate that insignificance to satisfy the purists. That will involve complex multiple hieroglyphics.

Wings level, that insignificance disappears.

Incidently have you handled asymmetric situations?

bookworm

10th Sep 2004, 14:36

A sideforce from the fin, rudder central without sideslip ???

One more time: the rudder is not central.

Incidently have you handled asymmetric situations?

Yes, in aircraft in which if you don't eliminate the sideslip you don't get a rate of climb at blue line!

10th Sep 2004, 14:55

The total (mainplane + horizontal tail) lift on the aircraft must act at the C of G, otherwise there would be an unbalanced couple with the weight.

I'd like to modify that to say that "the vertical lift of the mainplane and the vertical lift of the tail must act at the CoG". After all, the vertical aerodynamical force on the tailplane is not directly linked to the force in the aircraft z direction.

I'm almost scared to put myself in the middle of all of this but hey, this place is usually civilized. :p

Regards,
Fred

bookworm

10th Sep 2004, 17:00

For ease of further discussion, which must be very illuminating to the thread visitors, let us just look at the wing down case where we have carefully adjusted the bank to result in a centralised rudder whilst we maintain balanced straight flight.

Ah, perhaps this is the cause of the crossed wires.

If you hold the wings level and use rudder to correct the asymmetry, the aircraft will slip towards the dead engine.

If you use wing down to correct the asymmetry to the extent that you're able to centralise the rudder, the aircraft will slip towards the live engine.

Between the two is a zero-sideslip, wing-down, non-central-rudder scenario.

alf5071h

10th Sep 2004, 18:56

I do not wish to enter such an expert discussion, but if it helps, please see two somewhat more practical articles – with diagrams here: Turboprop PSM+ICR, “asymmetric flight.pdf” and “Asymmetric flight at low airspeed.pdf” (http://uk.geocities.com/[email protected]/alf5071h.htm) The related articles indicate the flight results if you do not get the rudder and bank sorted out.

Milt

11th Sep 2004, 02:29

bookworm , ft, alf5071h
Where are you SR71?
Still with us J T ?

Great to have you joining the fray. You will find yourselves reviewing a lot of aerodynamic basics and may also come to grips with a major myth that has been perpetuated regarding directional stability for far too long.

Most text books continue to make the mistake that the horizontal forces on an aircraft in steady flight have someting to do with weight and the cg. Well in this case only a smidgen.

A few basics first to re-establish the detail of what we are discussing. That is Sideslipping when Asymmetric.

1. The horizontal forces on a twin engined aircraft in steady straight unaccelerated level flight can be accumulated and represented by just two distict forces familiar to us.

1.1 Component of Thrust - Acts straight along an aircraft centre line assuming equal thrust from engines. Vertical positioning of thrust line irrelevant to horizontal. NO relationship to weight or cg

1.2 Total Drag - Acts straight along an aircraft centre line. Equals Thrust. Vertical positioning of drag line irrelevant to horizontal. NO relationship to weight or cg

2. The horizontal forces on a twin engined aircraft having asymmetric thrust flying in steady straight unaccelerated level flight, wings level, can be accumulated and represented by just three distinct forces familiar to us.

2.1 Component of Thrust - Acts along a line parallel to an aircraft centre line. Vertical positioning of thrust line irrelevant. NO relationship to weight or cg.

2.2 Total Drag - Acts parallel to an aircaft centre line. Equals Thrust. Vertical positioning of Total Drag line irrelevant. NO relationship to weight or cg.

2.3 Fuselage Sideforce - Acts from the lateral centre of pressure normal to the aircraft centre line and towards the engine having least thrust. Verical positioning of Fuselage Sideforce line irrelevant. NO relationship to weight or cg.

3. The horizontal forces on a twin engined aircraft having asymmetric thrust flying in steady straight unaccelerated flight, small bank angle, can be accumulated and represented by just four distinct forces familiar to us.

3.1 Component of Thrust - Acts along a line parallel to the aircraft flight path. Vertical positioning of thrust line irrelevant. NO relationship to weight or cg.

3.2 Total Drag - Acts parallel to the aircraft flight path. Equals Thrust. Vertical positioning of Total Drag line irrelevant. NO relationship to weight or cg

3.3 Component of Lift. Acts from an obscure point along the vertical line from the cg and normal to the flight path horizontally towards the side having greatest thrust.Vertical positioning of Component of Lift force line irrelevant. Point of origin related to cg, NO relationship to weight.

3.4 Balancing force to Component of Lift. Acts from the lateral centre of pressure normal to the aircraft flight path and towards the engine having the greater thrust. Verical positioning of Fuselage Sideforce line irrelevant. NO relationship to weight or cg.

Note. 1 bookworm. Deliberate no mention of rudder to keep it simple. Don't care where it is and incidently there is no way for you to accurately trim rudder to nail the min drag you have been chasing in a performance climb. Re-engined Caribou was a classic. Difficult without a boom and vane. Yaw string too gross.

Note 2 The vertical levels/planes of the above forces do not coincide. Resulting vertical moments are well cared for by the aircraft's longitudinal stability. We are strictly horizontal here.

Now - can someone please explain where we can generate that balancing force to the Component of Lift ?? I can assure you that it does not come from the strengths of a family of Gremlins, Fiffinellas and their proginy - Widgets.

Challenge to all who pass this way.

Watch this space for the next exciting discourse on aerodynamics !!

ROB-x38

11th Sep 2004, 07:34

I'll take the plunge. ;)

G'day Milt

In the banked, central rudder, steady, straight, unaccelerated, level flight case there must be a slip to balance the horizontal component of lift. No worries there.

In the banked, deflected rudder, steady, straight, unaccelerated, level flight case can't the balancing force to the component of lift come from the rudder. Why do we have to rely on the keel surfaces and directional stability of the aircraft to balance the forces?

Milt

11th Sep 2004, 10:08

ROB-x38 and all.

Welcome aboard. I think we need a few more than a quorum before we are through with this.

Ok lets go fly your twin.

We have climbed to a safe height, closed down the left engine, upped the power on the right engine and settled into a steady straight cruise with wings level and found that the rudder is trimmed right foot with about one third deflection. Balance ball is in the centre even though we know we are sideslipping by about 3 degrees. Hard to tell though.

The R foot rudder plus a considerable contribution from the fin and rear right fuselage is balancing out the thrust from the R engine which would otherwise be yawing us strongly to the left.

We don't like that amount of rudder applied. It has reduced our options for lower speed flight down around Vmca and as flight testing has shown (and we may instinctively suspect) we are able to carve off some drag by banking a little into the live engine. A bit more aileron than the few degrees we already have applied you say. Well OK we suffer a little extra drag there but hope to lose more than that by a reducing rudder deflection.

So roll over by about 1 degree and retrim all round. What has happened is that we are now back to about 1/4 R foot rudder deflection. The small lift component is now trying to pull us sideways a bit causing additional side force on the fuselage, which has its greatest surface area to the rear, taking over some of the previous task of the rudder. That feels a bit better.

Roll on another 2 degrees. More lift component and greater tendency to slide to the right causing more side force on the rear fuselage.. Compensate for this by taking off some more rudder. Retrim all round for straight level flight.. Now we only have about 1/8 R rudder. But what about that crazy balance ball. Its now hanging off the centre of the instrument by the angle of bank. Still nothing except instinct to indicate we are sideslipping.

Now roll on another 2 degrees. More lift component, more sideways force, more sideslip, more compensating force from the rear fuselage. Retrim all round.
Hey - rudder is just a nat's whisker away from centre. That feels great but that crazy ball is now out further. Want to cover it? It's not much use to you now unless you reset in your mind its current position as being the new datum to use in turns.

Somewhere as we increased the angle of bank we went through the minimum drag point. Maybe we are right on it now in that we have a central rudder, the rear fuselage having taken over from the rudder. I would suspect that a little touch more bank to cause the rudder to be a little to the left would be the minimum but why worry as the shape of the drag curve in this area will be fairly flat.

What happens when we try more bank.?

OK roll on another 10 degrees R to a full 20 degrees. Now it’s a bit of a lead sled. We've had to trim nose up to compensate for the much larger portion of lift we are using out to the side and to keep us going straight you now have almost half right rudder out the other way. Drag has shot up considerably and we have now stabilised just above Vmca for the power we have.. The side slip is gross.

Roll on another 10 degrees. Hey is that full left rudder?? and come on keep your nose up. Don't let it yaw right like that. Hey you're losing it. Look at your speed.
TAKING OVER.

Nose down and wings level, power off , I'll get the power coming on gradually as our speed increases. Lets get that other engine going or will we check out your personal minimum control speed. You may even be able to get it back to the stall.

By the way do you know whether you can stall the fin on this type. Fin stall often has nasty consequences.

A bit simplistic to the experienced but we aren't all so endowed.

SR71

11th Sep 2004, 12:01

Milt,

The only way I know of to get a side force from the fin, rudder central, is with sideslip

The rudder doesn't have to be central though...

In fact, throughout this discussion, apart from the flight condition alf5071h's notes refer to, it won't be.

To my mind, it is quite possible to attain banked, zero sideslip flight when asymmetric with rudder deflection.

Regards the effect of CG location on V_mca...

We need force and moment equilibrium for an unaccelerated flight condition.

The point about which you consider these two forces to act makes a crucial difference to whether \sigma M = 0! It won't remain unaccelerated for long.

You can fix your axes system anywhere on the a/c and define things accordingly.

Agreed, for lateral force equilibrium, balance them about whichever origin you like on the a/c x-axis.

Having chosen this point, you then fix the required sideload required by the fin/rudder because presumably your airframe is not aeroelastic and the distance to the fin is fixed.

But, having chosen that point, if you then change it, whilst it does not change the size of the thrust related yawing moment, I cannot see how you won't change the required sideload that needs to be generated by the fin/rudder?

How do you otherwise propose to size your fin/rudder?

I see an analogous relationship between centre of pressure, aerodynamic centre, CG and AOA in the vertical sense and their lateral equivalents, except that, of course, you can't decouple lateral/directional stability.

My $0.02.

john_tullamarine

12th Sep 2004, 01:37

.. still here, Milt, with a smile and a tally sheet .....

13th Sep 2004, 14:36

“2. The horizontal forces on a twin engined aircraft having asymmetric thrust flying in steady straight unaccelerated level flight, wings level, can be accumulated and represented by just three distinct forces familiar to us.

[...]

2.2 Total Drag - Acts parallel to an aircaft centre line. Equals Thrust. Vertical positioning of Total Drag line irrelevant. NO relationship to weight or cg.”

In my world, drag is always parallel to the direction of travel, no matter if it is horizontal or vertical. An academic point to be sure, you’re simply splitting up what I’d call total drag into the components parallel and perpendicular to the fuselage. The summed total will, for the horizontally unaccelerated case, be parallel to the direction of travel.

That said, here’s another version of what I posted above.

To fly straight and level with thrust asymmetry, you need to push rudder into the live engine to counter the moment generated by the thrust asymmetry which will otherwise turn the aircraft towards the dead engine and make it turn flat (at best).

This opposite rudder creates a lateral force towards the dead engine at the rudder. The opposing force to this lateral force will be generated by a side slip into the dead engine and the resulting aero forces on the fuselage. With the right amount of rudder and slip for the airspeed, the moment will be the same as the moment generated by the thrust asymmetry, the forces will cancel each other out and you will be flying straight.

Rather than having this balancing force created rather inefficiently by going sideways, you can bank the aircraft and have it generated by the wings. Generating forces is what airfoils do best as that is what they are built for.

The vertical component of lift from the wings and aero forces on the stabiliser will, together with m*g applied at the CoG (remember, we’re back in the vertical plane for a moment – pun unintentional), have to cancel out summed up as a moment about an arbitrarily chosen point (which never has to be the CoG as the moment is equal about any point on a rigid body... albeit it will begin rotating about the CoG when the summed moments don’t amount to zero). They also have to cancel out with their magnitudes summed up, unless you want to accelerate vertically. This dictates the amount of vertical force on the stabiliser for any angle of bank.

The horizontal force on the stabiliser has to equal the horizontal component of lift generated by the wings. The moment generated by the fixed moment arm between them has to equal the moment generated by the thrust asymmetry. The magnitude of these forces is set by the angle of bank. Alas, the thrust asymmetry dictates the angle of bank. We have very few options. Newton still has the upper hand on us!

We could center the rudder and bank enough to increase the slip into the live engine until the aero force generated on the fuselage equals the horizontal component of the lift and the angle of attack of the fin with no rudder (camber) generates all the torque to counter the thrust asymmetry. However, generating a force through slipping is not likely to be any more efficient now than when we were wings level.

No, keep that foot on the rudder to keep the nose pointed where we’re going. That slip would be creating loads of drag and pointing the remaining of thrust to the side instead of in the direction in which we want to go, just when we lost half the thrust available to begin with and are likely to be desperate for more of it!

Besides, we banked to get away from slipping into the dead engine. It’d be rather daft to start slipping the other way instead. It’s much better to let the airfoil things at the back take care of the force generation. As was pointed out before, that’s what they’re built for. They also have a longer moment arm, meaning they'll do more torque with less drag.

This conclusion valid as long as we’re going for performance. If we are going for maximum control authority, a bit of sideslip is probably preferable. That way, we’ll leave the rudder less deflected and have more travel to use for changing our situation.

If you are hard pressed to find any disagreement with Milt here, you’re right. I’m just looking at the same thing through a different window to broaden the perspective and perhaps get some feedback.

Regards,
Fred

Milt

15th Sep 2004, 02:09

ft and all temporary disbelievers.

Must take more care to avoid misunderstandings ft. My reference to the the vertical through which the total drag acts was to bring to attention that the horizontal forces we have been considering do not act along the same horizontal reference plane. They have vertical separations and most moments resulting are well looked after by longitudinal stability. For now forget I ever introduced that red herring.

All of us involved so far have been in agreement , I am presuming, with the superb logic in your first 6 paras after "That said, here's another version of what I posted above".

BUT then you continue to persist with the all too common proposition that an aircraft can fly straight and balanced with bank applied and not be side slipping. Old beliefs sure do die hard. Let me put it again as simply as I can.

The horizontally resolved portion of the total lift will be acting at right angles to the fore and aft axis emanating from the point through which total lift can be said to act. This will be close enough to the cg for our purposes.

What is the effect of this sideforce? There is ONLY ONE answer. The fuselage is of course forced to move sideways. This inevitably causes sideslip. This sideslip causes weathercocking of the fuselage and if adequate takes over from the rudder and allows us to fly rudder central. The sideslip angle then will be very close to that for wings level straight flight only differing as a result of the "efficiency" of the particular aircraft's vertical tail when sideslipping with and without rudder. The advantage we are endeavouring to achieve is to bring that rudder close to central thus providing us with better options at Vmca.

If you are still somehow skeptical, take the bank angle to an extreme as I did in a simplistic description of flight in an asymmetric twin in a post above.
Eventually with enough bank you reach a point where you have full opposite rudder, to that with which we started, attempting to oppose the now gross tail force due to sideslip. Take it a smidgin further and you have lost control.

Now, having succeeded in convincing you, SR71, John Tullamarine's tally sheet and the all too silent visitors to the thread of the fact that we are sideslipping whenever we fly with bank and fly straight at the same time, perhaps we can spread the gospel, correct the training references and switch to other fascinating aviation myths.

Could I have missed an obscure situation when you can indeed fly straight and level with bank and not be sideslipping?

Our unfinished myth on this thread is that rediculous proposition that steady state directional stability has any relationship to the cg at a particular weight. This one is bound to exercise great minds further. Perhaps we should tidy up and start a specific thread and a new tally sheet for JT.

15th Sep 2004, 06:50

Milt,
if there's a minute horizontal component of (main plane) lift, countered by a force at the (deflected) rudder (equal in magnitude, opposed in direction), why will anything go sideways?

If there is indeed slip, what will then balance the aero force on the slipping fuselage to avoid horizontal acceleration?

Regards,
Fred

Milt

15th Sep 2004, 06:59

ft

You have answered your own question!!

The minute sideways force is balanced by the minute sideslip caused by the rudder.

Think of it as the rudder pushing the tail around to cause a new position of the rear fuselage. It is now sideslipping.

I think you are overlooking the sideforce created by the sideslip.

And try thinking about rudders on boats!!

SR71

15th Sep 2004, 09:26

Let me get this straight...

Lets say I take a plank freely suspended in air in the horizontal plane and push sideways at the CG with a force, F.

Now I push in the opposite direction with a force, F, except that I exert this force at one of the extremeties.

Will the plank translate at all?

I think not.

Milt

15th Sep 2004, 09:45

SR71

Confused a bit by your use of translate - to change to another language or to change to another form.

Presume you mean rotate but if so the question is too simplistic.

What you have described is a perfect moment and if you keep the forces normal to the plank it will spin up like a prop.

If you maintain the direction of the forces then the plank will turn through 90 degrees and stop.

I think I must be wrong in interpreting what you mean.

Help

SR71

15th Sep 2004, 11:52

Milt,

I agree it will rotate but not translate i.e., move in a linear sense.

Assume the length of the plank is 2d, the applied moment is F*d whereupon you will get an angular acceleration.

Now if I apply an opposing moment to prevent any rotation, of magnitude F*d (Note: this doesn't affect the force equilibrium), I now have neither a rotation or a translation.

Agreed?

Isn't this case directly analogous to the one in question?

15th Sep 2004, 12:28

Milt,
now you’re definitely the one confused here. :)

If the force on the rudder is opposed to the horizontal component of lift, any sideslip it creates will be in the opposite direction of the horizontal component of lift, adding to that horizontal force.

Either way, it doesn’t really pertain to the question I raised. My question was: Why would there have to be sideslip when the forces cancel out without it?

Look at the forces and moments involved.

The thrust asymmetry generates a moment. This moment is countered by the moment generated by the horizontal component of lift and the horizontal force at the stabiliser.

The horizontal component of lift is equal and opposed to the horizontal force at the stabiliser.

Thrust equals drag.

M_yaw_total = 0

F_lateral_total = 0

F_longitudinal_total = 0

We have equilibrium, sans any slip. Why do you insist it has to be there?

I’ll see if I can dig up a good WWW reference for all of this.

Regards,
Fred

P.S. In physics, translate is the common term for when you get something moving, as opposed to merely rotating it. Cmpr “translational lift” for helos. ;)

To be dull and hit the dictionary:

Translate
...
4. To transfer from one place or condition to another.
...
7. Physics. To subject (a body) to translation.
...

john_tullamarine

15th Sep 2004, 12:41

Isn't it rather wonderful the way threads wander off-topic into quite interesting areas ? Would anyone believe that the original question was along the lines of

" should a yaw string not be called a slip string since as far as I can tell it shows sideslip not yaw ? "

I think that we are all somewhat in the same ballpark ?

Milt, we probably need you to itemise several things to focus attention on some, if not all, of the main points of apparent difference throughout this thread.

Indeed, Milt's apparent heresy is quite thought provoking ...

(a) what is your definition of balanced flight ? Do you distinguish between the terms slip and sideslip ?

(b) could you characterise the aerodynamic slip vector direction as a function of bank angle ? From this follows a concern to see how the lateral fuse force varies with variation in bank angle. I don't think that anyone will dispute the observations that, if a sideslip is permitted to develop, the lateral drag will soon balance the accelerating force and the thing gets to a steady state or, indeed, that the lateral force component is a real factor generated in the presence of sideslip and has to be part of the equilibrium analysis.

(c) why should we want to operate with sufficient bank to achieve nil rudder deflection ?

(d) if we do see a need to do so, then what are the advantages and disadvantages of so doing compared to, say, wings level or some other angle of bank ?

(e) what sort of typical bank angle would you suggest is appropriate to achieve zero rudder deflection back in the region of published Vmca ? What sort of climb (or descent) performance might you expect for this case ? Some examples might go a long way to convincing us other heretics ... along with some indication of the generality of such examples ? Indeed, how do you see the climb (or descent) performance varying with bank angle .. for steady state cases ?

(f) you observe that the horizontal lift component generates a sideslip ? However, could you address the concern of the heretic members of the opposing camp who might opine that that force doesn't, itself, create any sideslip .. rather the motion resulting from the (unbalanced) set of forces does. If the said force is balanced by another (the lateral fin force, for instance), and presuming that the moment balance is taken care of along the way and that the equilibrium force/moment balance can be achieved at some point where the sideslip angle is zero, where is the lateral force imbalance necessarily needed to create the acceleration and velocity necessary to give rise to the sideslip which one needs to achieve some lateral flow circulation ? .. or is this scenario not achievable ? We should all agree that, in the presence of sideslip, there will be some balance involving whatever forces exist .. including the lateral fuse force. Re-reading the thread just now it may just be possible that you adopt both points of view in different posts ?

Score ? ... I still like catchup's first post the best ..... however, Milt's observation in respect of "yaw slip is showing" rates a very close second .. perhaps a tie ? Did you think I was scoring on technical matters ? .. not a chance .. life in the day job has been a bit torrid with workload this past fortnight .. I crave plain old hedonistic entertainment ...

I must say that this is all rather jolly good fun ... I'm just not entirely sure if Milt is on the level ... or winding us up for his own entertainment by having 2 bob each way on rudder deflection and varying the tale slowly throughout the thread ? ie the other camp's heretics are not overly concerned with the zero rudder deflection case ? Incidentally probably none of us have any difficulty with Milt's scenario in the case of undeflected rudder... it's just that we are probably not overly interested in that particular case ? .. as, in the real world, control in the low speed takeoff is paramount as is climb after the initial dynamic dramas have been sorted out ?

Either way .. all good, clean fun .. and it has challenged everyone to have a bit of a think about what might be going on ...

.. come to think of it ... Milt wins the entertainment stakes ... what say you, Fred ? I have not met Milt, although I know a little of his background .. and I suspect that he would make a very entertaining tutor in the classroom ... he certainly has provided sufficient fuel to the fire to get us all going.

15th Sep 2004, 13:50

JT,
fun indeed. It is interesting how you know that you know things... and then when that knowledge is questioned, it turns out to be belief rather than knowledge until you have thought it over in more detail. It is always useful exercise! Sometimes, and this is my favourite, it turns out you were in fact wrong all along. Gotta love learning! :D

Milt, I still suspect you of pulling my leg. Are you bribed by JT to keep the discussion going for his amusement? ;) Agreed, very high-scoring contribution and definitely a runner-up.

However, the opposition is fierce and albeit close, my final vote goes to Bellerophon, as brief as his participation has been. It is certainly no shame in coming a close second to a statement such as:

“The gliding fraternity use a piece of cotton, stuck to the outside of the canopy to achieve the same result, but on Concorde it tended to come off in the cruise, so a sideslip indicator it had to be!”

I must confess to perhaps being somewhat partial, having previously read very interesting posts by the abovementioned gentleman. Not to mentioned being corrected by him when my view of the world was wro... ahem... departed somewhat from the one considered reality by the majority of the population.

I think we should have established early on what position of the rudder we based our respective views upon though, to avoid a bit of confusion. Centered, towards dead engine or towards live engine.

I do think we'd all have a jolly good time arguing until we ran out of envelopes to scribble on if we could cancel the geographical inconveniences separating us all from a suitable common outlet for malt-based refreshment.

Cheers,
Fred

Milt

16th Sep 2004, 02:31

John Tullamarine and my faithful intriguing respondents.

Should have set up the ground rules up front to avoid the confusion which has been flowing back and forth. Perhaps they should preferrably be called air rules.
Thank you JT for your intervention. Yes I had my time as a QFI and RAAF CFS staffer plus other things.

Firstly what are we ON about. It's Asymmetric Handling and an examination of the forces acting..

The overriding fact to be determined initially is whether an aircraft having any angle of bank and in steady straight balanced flight can achieve that condition without sideslip. This is the area causing the greatest confusion. Please can anyone nominate a circumstance now or forever hold your peace.

That should be a very simple determination, should be instinctive to a pilot and remain an uncomplicated fact provided my definitions agree with yours.

So herewith the definitions I am using and listed in response to JT's (a) to (c)..
These definitions are off the top of the head and almost from the heart. They should be consistent with text books.

(a) Balanced Flight. Any condition of flight when there is a continuing state with all forces acting on the aircraft in equilibrium.
Slip = Sideslip.

(b) My 'aerodynamic slip vector' (don't like the term) is that force normal to the line of flight generated by sideslip acting through the lateral centre of pressure.
Explanations. 1. Force Generated by Sideslip. The difference in the lateral aerodynamic forces resulting from differing airflows on each side of the fuselage.
2. Lateral centre of pressure. Similar to wing lift centre of pressure but instead the point through which horizontal side forces act.
No force without sideslip. Rudder force not considered seperately as it is included in total side force.

(c) Why would I bank to zero rudder? I wouldn't except when IAS is close to Vmca. In some cases there may be a trade off between the total drag wings level and banked.
Explanation. 1. Rudder central may be more efficient with rear fuselage only generating the sideforce.
2. Zero rudder allows for decreased Vmca.

(d) (e) Later coverage. Will only overly complicate at this time and further complicated by published Vmca having included a 2 second delay before taking recovery action.

(f) No definition here and we are splitting hairs which are all having a strong tendency to stand out straight! Simplest explanation I can offer is :-
Roll on bank to produce the horizontal sideforce. Immediately we have departed balanced flight. Initially the unbalanced sideforce is now overcoming aircraft inertia and slowly accelerating the aircraft sideways in the direction of the sideforce until the build up of the force due to sideslip balances that lift sideforce. Now we are back in balanced flight/equilibrium, having established a yawing couple by the two sideforces which balances out the opposing yawing couple by the offset thrust and the total drag. The resolved lift sideforce act through the total lift centre of pressure (almost coincident with the cg) and the sideslip force acts through the lateral centre of pressure which will have moved further aft as the sideslip angle increases. Incidently if the lateral centre of pressure moved forward of the total lift centre of pressure the aircraft would have a strong desire to turn around to go backwards.

Damn it I didn't want it to get that complicated again so soon but I don't know how to avoid it.

Are you having problems with the concept of a lateral centre of pressure? It doesn't generally get too much of a mention. Has more significance with boats than aircraft EXCEPT for our present subject and for those wanting to do 4 point slow rolls at low altitude !

Tear me to bits if you can or have I recruited a few disciples?.

bookworm

16th Sep 2004, 06:57

No force without sideslip. Rudder force not considered seperately as it is included in total side force.

So what you're saying is that if the relative airflow is symmetric (i.e. parallel to the fore-aft axis of the aircraft) but the rudder is providing a lateral force because it is non-central, then you describe that lateral force as a side force and infer by definition that the aircraft is in a sideslip? Is that right?

Milt

16th Sep 2004, 07:49

Bookworm

Sorry no - not saying that.

Ask yourself how you can have symmetric airfow around the fuselage when there is rudder applied.

If the airflow is symmetric then the rudder is central.

Think of the tail as a whole and imagine what happens when you wag the tail with the rudder!! Every wag results in a sideslip. If that doesn't compute nut out what happens when the captain of a ship commands "Right full rudder"

I feel a wag story coming on but will spare you!

Does that help?

16th Sep 2004, 08:31

Ask yourself how you can have symmetric airfow around the fuselage when there is rudder applied.

By having the forces cancel out anyway. Equilibrium is equilibrium regardless of how it is created. A horizontal component of lift to cancel the force contribution of the deflected rudder (which exists - see below). Thrust asymmetry to cancel the moment contribution.

That's the equilibrium situation I described above. Can you tell me why you consider it impossible?

No problems with lateral CoP or knife-edge flight. However, I do have a problem with:

No force without sideslip. Rudder force not considered seperately as it is included in total side force.

That is only valid as long as the rudder is not deflected. Once the rudder is deflected, the crude airfoil made up of the fuselage/fin (as viewed from above) will go from a symmetric one to a cambered one. While the above quote holds true for a symmetric airfoil, it does not for the cambered case when there will be a force generated without any angle of attack (beta).

Heh, a thought goes out to newcrew who’s probably asking himself how he could create this monster through a simple question about a piece of yarn!

Regards,
Fred

Oktas8

16th Sep 2004, 09:31

Hi all, long time listener, first (or seventy-first) caller. I'd like to slip my 2c worth in too if I may, talk about this for a moment, although I don't want to force my viewpoint.

Milt several of your posts are worded carelessly, such as the one below. Since you have thought about this a great deal, is it carelessness or a wind-up? I can't tell...
We have climbed to a safe height, closed down the left engine, upped the power on the right engine and settled into a steady straight cruise with wings level and found that the rudder is trimmed right foot with about one third deflection. Balance ball is in the centre even though we know we are sideslipping by about 3 degrees. Hard to tell though.

The R foot rudder plus a considerable contribution from the fin and rear right fuselage is balancing out the thrust from the R engine which would otherwise be yawing us strongly to the left.
No! R foot rudder gives a force to the LEFT. Due to LEFT slip that this causes, aerodynamic forces on the fin & rear fuselage are generated to the RIGHT, so we need even more rudder than we previously thought to maintain heading (i.e. prevent yaw). Change your plus to a minus above.

How can I stop the aircraft from slipping left? I need a force to the RIGHT. Some horizontal component of lift could provide the necessary.

Why can't I have
-Yawing moments balance by having rudder deflection moment opposing thrust / drag asymmetry moment, and
-Sideways (slip-inducing) forces balance by having horizontal component of lift force opposing rudder deflection force?

cheers,
O8

SR71

16th Sep 2004, 11:25

O8,

I agree.

In fact the paragraph you quote has been amended (in the original post - although slightly earlier than your own post - the plot thickens) by Milt because it represented an impossible flight condition.

It used to read "having shut down the left engine......rudder is trimmed left foot..."! I puzzled about this for a while...only a short while I hasten to add.

Maybe it was just a typo? Milt?

Still, all highly amusing.

:ok:

Milt

17th Sep 2004, 04:03

ft and all

"Ask yourself how you can have symmetric airflow around the fuselage when there is rudder applied."

Thought that to be a very simple proposition that could only have one answer. Tis fascinating how noticeably deep thinking minds introduce new angles on aerodynamic basics. Thanks for the challenges and just when I thought I had it all by the tail anyway when I started it wagging.

Please ignore the odd aircraft that has an offset fin by design. They are asymmetric to start with or perhaps screwed up would be a better term to use. Ohhh sorry I ever mentioned that.

Now I'll ask you to mentally turn our subject typical aircraft on to its side and assume that it is being carried along in balanced flight hanging on a giant sky hook suspended through its cg. Zero all the lift with a bit of negative wing alpha. Now we wag the tail again using the rudder to vary fuselage alpha. Doesn't the fuselage now act like a wing and take on what we can call lift and negative lift when we wag the other way.. What stops it doing the same thing horizontally when we fly it normally? Only difference is we call it all by different names and confuse the multitude, but we are now back with sideslip and the horizontal force resulting from that sideslip.

Maybe I can now stick my neck out with another motherhood statement which I believe is inviolate.

Sideslip of an aircraft having a symmetrical fuselage inevitably produces a sideforce acting through the total fuselage lateral centre of pressure PERIOD

ft I think that answers your second set of points also.

Anyone not enjoying this come join us. I promise to keep this going for a while. The red herrings are a hoot!

newcrew are you still there. Please don't blame yourself for having started all of this. Hang in there.

And I think I hurt the feelings of the ex Concord Captain when I called his sideslip indicator a con. Probably one of Brian Trubshaw's bright ideas.
Are you still there on the side Captain?

Standby Oktas 8 for another startling announcement

17th Sep 2004, 06:56

Now I'll ask you to mentally turn our subject typical aircraft on to its side and assume that it is being carried along in balanced flight hanging on a giant sky hook suspended through its cg. Zero all the lift with a bit of negative wing alpha. Now we wag the tail again using the rudder to vary fuselage alpha. Doesn't the fuselage now act like a wing and take on what we can call lift and negative lift when we wag the other way.. What stops it doing the same thing horizontally when we fly it normally? Only difference is we call it all by different names and confuse the multitude, but we are now back with sideslip and the horizontal force resulting from that sideslip.

Maybe I can now stick my neck out with another motherhood statement which I believe is inviolate.

Sideslip of an aircraft having a symmetrical fuselage inevitably produces a sideforce acting through the total fuselage lateral centre of pressure PERIOD

The fact that a slipping fuselage creates a lateral aerodynamic force is not contested here. The requirement to have that lateral aerodynamic force to maintain straight unaccelerated flight with asymmetric thrust is.

As shown above, equilibrium can be maintained without that lateral force and thus without any slip.

Show me why you think the proposed equations of equilibrium do not hold water, and you’ll convince me. Until then, I think you’ll find it near impossible. :)

Regards,
Fred

Keith.Williams.

17th Sep 2004, 07:24

Let's forget aerodynamics for a moment and get back to basic mechanics.

Consider a pendulum swinging from right to left and back again. At one point in time it is moving to the left. At a later point it is moving to the right. I order to change its condition from moving left to moving right it must (however briefly) have stopped moving. This requirement applies to all physical objects. It makes no difference whether the frame of reference against which we measure the motion is space or the air around us, the fundamental requirement remains the same.

Now let's consider our aircraft. We are initially flying straight and level in still air. We throttle back on the left engine and throttle up on the right. Because the thrust line has moved out onto the right wing, it generates a yawing moment to the left. So the aircraft starts to yaw to the left.

To arrest the left yaw we apply right rudder. This generates a force to the left acting at the tail of the aircraft. This has (at least) two effects. The first is a yawing moment to the right. If we have applied just enough right rudder the right yawing moment generated by the rudder will exactly balance the left yawing moment generated by our right engine, and the yawing will stop. (For purists, yes we will actually need a bit more rudder initially to actually stop the yaw). We now have a situation in which the aircraft is not yawing.

But the second effect of the rudder is a side force to the left. This will cause the aircraft to sideslip to the left.

If we now start to bank the aircraft to the right, we will tilt the lift towards the right. This initially small horizontal component of lift will cancel out part, but not all of the side force from the fin. The overall effect will be to reduce sideslip to the left. If we now gradually increase right bank in small increments the left sideslip will reduce to zero, then be replaced by right sideslip.

Now if we look at this sequence we will see that just like the pendulum, there was a point at which the aircraft was not moving to the right or left. It was not sideslipping at all. If we look a bit closer we will see that this did not happen with the wings level, but with a slight bank to the right. This is the situation in which the aircraft is banked to the right but not sideslipping.

Milt

17th Sep 2004, 07:59

Oktas 8 SR71 ft and the tally sheet.

Have you had another look at your post Oktas8 ? Is that OK True Air Speed 8 or heavily overcast, the reverse of CAVU.

Don't want to embarrass you but this is a very good example of why so many get into trouble when presented with a sudden engine failure on a multi.

Even Milt will admit to some milliseconds of confusion until the direction of swing and the instruments tell him which side it's stopped happening It's not instinctive enough to provide for an automatic reflex. And that is after a lifetime of pushing and pulling all those levers. Do I see a lot of heads nodding?

One night I was in the left seat of a Gooney Bird with another very experienced operator in the right. We were determining the power settings to use on an engine which equated to one with its prop feathered for conditions soon after take off. Just after lift off and according to a pre brief I called 'Now" for a close down of the R engine. A Mixture control was pulled into cut off and the R prop feathering activated. Quite quickly I was engineless. A quick smack of the offending hand removed it from the LEFT mixture control which was rapidly reset to Rich to the accompaniement of the left engine going through some overspeed before the prop governor clawed back the revs. Incidently the power settings turned out to be close to 15 inches and 2,000 RPM.

So much room for error in this area.

That's why the TPs allow a 2 seconds do nothing delay in the determination of Vmcg and Vmca.

Oktas8 - I am back with you.

Come fly with me again in that twin and we'll do it more slowly.

We have climbed to a safe height, closed down the left engine and allowed the aircraft to yaw hard left to discover that the angle of Right sideslip is excessive and the rudder has trailed over to left foot. It is holding but we can't leave it like that. There is not enough tail sideforce to the left being produced to get us pointing in the direction close to where we want to go and all of that sideslip is giving us unbalanced lift from the wings so we have to hold level with a fair bit of aileron. Let us unscamble this a bit by helping the tail to do its job by pushing in some R foot rudder. Now we have given the tail some help and we have reduced the angle of RIGHT sideslip down to a lower angle so that we are now flying about 2/3 degrees Right of where the aircraft is pointing.

So you now go and change your plus to minus or explain some more where I might have goofed. It wouldn't be the first time!!

Oh and you'll have to be more specific with your last para. You have forces going every which way and can't get my mind around them.

Carelessness or wind -up? I must admit to a bit of wind up for a while when the directions of some of the functions/forces with the first flight in the twin but no-one picked it so after a couple of days I did an edit. I now see that SR71 was on to it.

Further responses pending to ft and welcome aboard Keith.Williams

Smooth landings.

Cap 56

17th Sep 2004, 08:16

Great discussion.

Now what do you say when the question pops up in an interview?

If you have asymmetric thrust, you will have to counter with rudder to stop the angular motion around the vertical axis going trough the CG.

Then if you want to fly in a straight line you will have to add up (after steady state condition has been achieved) all components of all forces involved in the horizontal plane.

If this sum in not zero, no way you will fly in a straight line.

How do you make this sum Zero?

The only thing you can do is to change the position of the CG and start all over again.

However, you may not be able to change the CG sufficiently enough to achieve the goal of straight flight.

Now is this a nutshell answer or not ?

Since it is not guaranteed that the CG will be in the longitudinal axis I see why Concorde trims the fuel in both directions lat and long.

Since in steady state and straight flight the only force on the slip ball is gravity, the ball can be anywhere depending on the amount of bank you need to set all equations to zero.

With straight line I mean straight direction, the plane could still be climbing or decending in a curved line.

17th Sep 2004, 08:38

Cap,
how would the position of the CoG have any effect on the horizontal forces if you are flying in a straight line? There's no acceleration in any direction except the gravitational acceleration, which by definition is acting perpendicular to the horizontal.

If the moments don't add up, the angular acceleration will be about the CoG. However, the moments can be summed up about any point and will be the same regardless of the position of the CoG.

Often you use the CoG (if known) as the reference point for moment calculations. Since the weight is usually a factor, this means the moment arm from the reference point to the CoG is zero and the weight can be disregarded. The moment is the same regardless of reference point though.

Regards,
Fred

Cap 56

17th Sep 2004, 08:48

how would the position of the CoG have any effect on the horizontal forces

ft

In the case of asymmetric thrust, the position of the CG does affect the required forces does it not ?

If that is correct then while shifting the CoG you will have to change the forces.

Milt

17th Sep 2004, 09:24

Cap 56

Pan Pan Pan

Just when I thought we had all sent cg/CG back to longitudinal now we have Cap 56 trying to slip that interloper back in to steady state directional stab.

Heaven to Betsy

Forgotten now how we managed to get rid of it. I think it was with a few simple inspirational statements such as

"You cannot resolve weight acting through the cg/CG through 90 degrees to the horizontal."

and perhaps

"The only time that mass or weight or cg/CG comes into consideration in the horizontal is when we have acceleration.

But would you mind telling us why you think the rudder yaws an aircraft in steady flight around the CG and perhaps what is the alternative point of rotation?

Oh and by the way did you note the references to the so called slip indicator on Concord on page 1 of the thread?
Any comments particularly if you were/are a Concord pilot.

Just saw the post from ft and you can tell he now has similar pronouncements as above.

17th Sep 2004, 09:35

In the case of asymmetric thrust, the position of the CG does affect the required forces does it not ?

I’m afraid not. All that matters is that the moments (about the vertical) tally up to a grand total of zero.

If that is correct then while shifting the CoG you will have to change the forces.

Milt,
“now”? You’ll be hard pressed to find a post in which I disagreed with the irrelevance of the position of the CoG as far as unaccelerated lateral stability goes. ;)

Any moment applied to a rigid body will, however, strive to rotate it about the CoG. Of course there tends to be numerous other constraints making sure that we get translation as well.

Regards,
Fred

Cap 56

17th Sep 2004, 09:41

Milt

But would you mind telling us why you think the rudder yaws an aircraft in steady flight around the CG and perhaps what is the alternative point of rotation?

I didn't state that. I stated that if you consider the motion around the vertical axis going trough the CoG (that is take as a reference point the CoG ) this is NOT stating the rotation point is the CoG.

I realize that inertial forces and aerodynamic ones are quite different.

But unless you are capable to change the physical dimensions of the plane in mid flight the only thing you have to play with is the CoG and as soon as you bank you do not escape from the fact that gravity will have an influence in balancing all horizontal forces whatever the reference point you will use.

Maybe the way to walk around this problem is to assume the aircraft has no mass at all and after that introduce the concept of CoG. This will put all aero and engine forces on the front first.

All this in steady state of course.

17th Sep 2004, 10:16

But unless you are capable to change the physical dimensions of the plane in mid flight the only thing you have to play with is the CoG and as soon as you bank you do not escape from the fact that gravity will have an influence in balancing all horizontal forces whatever the reference point you will use.

Ah, here I think we have the crux of the matter. Gravity is always acting solely in the vertical. It will never be a factor when considering horizontal forces, whether you are level, in knife-edge flight or anywhere in between. As was stated before, the horizontal plane is per definition perpendicular to the gravitational force so there will never be any component of weight (gravity times mass applied at the CoG) in the horizontal.

If you consider lateral forces in the aircraft reference system, however, you will see a component of weight projected on the aircraft xy plane when you are in a bank or pitched up/down. That is why I have taken great care to be working in the horizontal plane rather than the aircraft lateral plane in this discussion.

Regards,
Fred

Cap 56

17th Sep 2004, 10:45

ft

Thanks, you are right 100 % and I appologize I should have made a drawing first.

So I will rewrite my first post as follows.

Now what do you say when the question pops up in an interview?

If you have asymmetric thrust, you will have to counter live engine(s) with rudder to stop the angular motion around the vertical aircraft axis.

Then if you want to fly in a straight line you will have to add up (after steady state condition has been achieved) all components of all aerodynamic and engine forces involved, in the horizontal plane.

If this sum in not zero, no way you will fly in a straight line.

How do you make this sum Zero?

The only thing you can do is to change the physical dimensions of the aircraft and/or bank the aircraft, so you can keep the moments balanced while at the same time change the forces in order to balance them horizontally.

Changing physical dimensions includes in this context "control deflection"

Since in steady state and straight flight the only force on the slip ball is gravity, the ball can be anywhere depending on the amount of bank you need to set all equations to zero.

With straight line I mean straight direction, the plane could still be climbing or descending in a curved line.

17th Sep 2004, 10:49

Cap 56

:ok:

You could end up in straight flight without rudder I suppose... but it wouldn't be pretty!

Regards,
Fred

Cap 56

17th Sep 2004, 11:03

Time to introduce Vector thrust on commercial planes.;)

Bellerophon

17th Sep 2004, 18:27

Milt

...I think I hurt the feelings of the ex Concord Captain when I called his sideslip indicator a con...are you still there on the side Captain?...

Still here, feelings unhurt, but head now aching! ;)

I'm with Keith Williams on this one.

On several occasions I have demonstrated, entirely to my own satisfaction if not that of my simulator instructor's, that my flying skills - so generously described by him during the subsequent debrief as unbelievable and unique - enabled me to fly a constant-heading engine-out climb, sideslipping in either direction, depending on the amount of bank and rudder applied.

If it is possible to slideslip in either direction in this condition, then it must be possible to find and maintain that position in which there is zero sideslip. Not me personally you understand, but perhaps a luckier pilot. It may be a difficult position to find, but that does not mean it does not exist!

Of greater concern to us was Concorde’s tendency, following an engine failure at high speed, to yaw one way (conventionally) but to roll the other (away from the failed engine); however that is a topic for another day!

Far more knowledgeable posters than I are now involved in this particular debate, and, as I said, my head hurts, so I shall now confine myself quietly to the sidelines.

By the way, do you have a licence to fish in this stream? :D

Regards

Bellerophon.

Cap 56

18th Sep 2004, 12:16

Bellerophon

If it is possible to slideslip in either direction in this condition, then it must be possible to find and maintain that position in which there is zero sideslip. Not me personally you understand, but perhaps a luckier pilot. It may be a difficult position to find, but that does not mean it does not exist!

Zero slip implies that the TAS vector is aligned with the longitudinal axis of the aircraft.

Hence, all aerodynamic forces from all non movable surfaces will cancel each other as far as their contribution to stop the rotation due to the engine failure is concerned.

The only thing remaining to stop the rotation (as far as I know Concorde), are the forces from the elevons, rudder and bank.

In order to fly without slip (and assuming the trust vector is parallel to the TAS vector) the aforementioned control surface forces will have to have to, when added together, create a couple (not a moment) equal to the moment created by the engines.

Assuming the horizontal component of the forces created by the elevons is small. The only option left over is to bank into the live engine and rudder.

The position that gives you a bank angle equal to the deflection of the slip ball should be zero slip straight flight on the condition tha your heading remains steady.

john_tullamarine

18th Sep 2004, 13:03

Cap 56,

I don't know that I'd get at all excited about distinguishing between couples (ie torques) and moments .... but I am intrigued by the suggestion that ..

"The position that gives you a bank angle equal to the deflection of the slip ball should be zero slip straight flight on the condition tha your heading remains steady."

It's a bit late at night here ..

.... but I would have thought that I could cause the bank/ball condition to occur at any bank angle of interest by simply banking .. and keep the heading straight by simply pushing on the appropriate rudder pedal .... and get all sorts of different slip angles depending on what bank I choose to use in the specific exercise ... provided that the end result is a steady state flight condition?

.. or is it just the case that it is a bit late at night here ?

I suspect that we are looking for a situation where -

(a) forces are in equilibrium (ie balance out) so that the aircraft stays at the same speed while not going sideways .. although a little bit of going up is desirable ...

(b) moments are in equilibrium (ie balance out) so that the aircraft doesn't spin slowly like a drunkard trying to find his way back from the pub to the marina late at night ...

And if, by some serendipitous sprinkling of pixie dust, that situation can end up with an aircraft under some modest measure of control with a tad spare performance after the initial surprise associated with the failure .. (excluding sim exercises where we are only surprised if the organist in the back forgets to fail the engine) .. then we ought to have a moderate content, assuming the hills and such are not an immediate concern to safety.

I, like B, have, as my first priority, staying blue side up ... and then worrying about whatever level of finesse I might be capable of on any given occasion to get some modest level of performance out of the wee beastie ..

18th Sep 2004, 14:16

When in doubt, google. ;)

Here are the pictures to go with the words (http://www.av8n.com/how/htm/multi.html). Hm, gotta put that as a bookmark methinks, it's a good site. And, I repeat, it has pictures. Little airplanes all over the place. You can never have too many little airplanes!

<stares off into the distance, dreaming of a world full of little airplanes, all with force and moment arrows attached>

Cap 56

18th Sep 2004, 14:16

john_tullamarine

Only a couple will create the condition where there is on force perpendicular to the speed vector, hence straight flight.

john_tullamarine

18th Sep 2004, 21:44

Cap 56,

Unable to link to Fred's reference at the moment ... perhaps that will lift the veil of confusion from my eyes in due course. And, of course, lots of little piccies are always a GOOD THING when trying to work out what's what .. unless, of course, you are Stephen Hawking and can do it all in your head .. what an incredible mind that man must have ..

Presuming your "on" is intended to be "no", that's fine for a simple torque ... no contrary view from me .... but an aeroplane in flight has a myriad of forces so the simple couple thing is a not-very-useful artefact of imagination, I suggest ? .. and, in any case, how does that relate to the suggestion in your previous post ?

Oktas8

19th Sep 2004, 00:08

Thanks for the clarification Milt. Please excuse my quoting you in big blocks - normally I hate that, but in this case the page is so long that scrolling up & down is tedious , so using quotes is easier.

We have climbed to a safe height, closed down the left engine, upped the power on the right engine and settled into a steady straight cruise with wings level and found that the rudder is trimmed right foot with about one third deflection. Balance ball is in the centre even though we know we are sideslipping by about 3 degrees. Hard to tell though.

In steady straight wings level flight with the LEFT engine shut down, you will be slipping LEFT with rudder force to the LEFT. Directional stability forces - those tending to make the aircraft yaw into a sideslip - will be to the RIGHT.

We have climbed to a safe height, closed down the left engine and allowed the aircraft to yaw hard left to discover that the angle of Right sideslip is excessive and the rudder has trailed over to left foot. ... There is not enough tail sideforce to the left being produced to get us pointing in the direction close to where we want to go and all of that sideslip is giving us unbalanced lift from the wings so we have to hold level with a fair bit of aileron. Let us ... push in some R foot rudder. Now we have given the tail some help and we have reduced the angle of RIGHT sideslip down to a lower angle so that we are now flying about 2/3 degrees Right of where the aircraft is pointing.

You describe a different situation here: it will not be straight wings level flight! If the wings are level, you'll be turning into the dead engine. If the flightpath is straight, you will be banking into the live engine. Using some R foot rudder to give the tail some help will do as you say, but the wings still won't quite be level, or the aircraft will not quite be keeping straight. Also of course the balance ball won't be centred in this case.

That's the last time I'll quote you - promise! - as I can see the discussion is moving on to other things. I won't clarify what I meant earlier, simply because that marvellous site bookmarked above says it all much better than I can, with pictures too!

cheers,
overcast8

Cap 56

19th Sep 2004, 10:52

john_tullamarine

Bellerophon has given the discussion a bit of a puch, he wants to know if assymetric thrust straight flying is possible with zero slip.

It an ideal condition since it would give you min drag.

Straight flight is only possible if the sum of the projection of all forces on the horizontal plane has no componrent perpendiculal to the TAS vector. Hense in the case of zero slip, perpendicular to the longitudinal axis.

Somehow you will need to balance the moment of the live engine.

I think what Bellerophon is looking for is possible but only at one single point (or confined area around this point) and speed and it may well be somewhere verh high above sea-level.

A point where rudder and bank forces are able to cope with the assymetric thrust.

Cap 56

19th Sep 2004, 14:47

without puching the TAS vector in another direction. For that you need a couple or something very close to it if you want to remain pragmatic.

Cap 56

19th Sep 2004, 16:34

And believe me, I am not worried about the real exercise, just a touch of bank into the live is fine by me.

19th Sep 2004, 17:27

he wants to know if assymetric thrust straight flying is possible with zero slip

Cap56,
that's exactly the question that most of the thread so far has been concerned with and hardly a new angle. ;)

(Just trying to avoid having the thread go around the same circle once more)

Regards,
Fred

alf5071h

19th Sep 2004, 19:36

Forgive me if I have lost the plot; I expect that many others are in a similar position.
However, to answer the question: “if asymmetric thrust straight flying is possible with zero slip – in that this condition would give you min drag”.

Surely, min drag will depend on the control surface positions required to give zero slip. Control surface deflections may give large increases in drag. In some aircraft, the resultant drag leads to a compromise of allowing some slip to minimise the value of total drag, seeking a minimum. A practical example (although not directly slip / drag related) was for the BAe 146 where asymmetric climbs are flown wings level (for ease of flight handling) and the small increase in roll spoiler / aileron drag is accepted; the large fin and effective rudder provides good lateral / directional control, but with slip.

Thus the solution sought depends entirely on the aircraft type; – the application of the preceding theory.

john_tullamarine

19th Sep 2004, 21:09

Alf ... if, by going around in circles, the message eventually becomes clear ... then that is probably a reasonable thing .. ? .. and I concur with your thoughts entirely ..

(a) one does the sums with one's aerodynamicist hat on in the back room

(b) one confirms a range of things in the tunnel

(c) one goes out and flies the climbs to find just where the optimum configuration etc lies

(d) one considers what is a reasonable compromise considering the numbers, workload, etc., etc.

Milt

20th Sep 2004, 01:07

A review.

This voice in the wilderness is returning and reiterating.

For too long now there has been almost aviation wide belief that we pilots can eliminate sideslip and continue to fly an unbalanced aircraft straight down its fore and aft axis.
All I am saying is THAT IS ABSOLUTELY IMPOSSIBLE.

So I am seeking a concensus amongst a few thinkers and movers to change the belief. The books are mostly wrong so there is no point in referring to a bunch of little diagrams to prove your points. You have to go beyond those books.

Here are a few motherhood statements I have used so far.

The overriding fact to be determined initially is whether an aircraft having any angle of bank and in steady straight balanced flight can achieve that condition without sideslip. This is the area causing the greatest confusion. Please can anyone nominate a circumstance now or forever hold your peace.

ANY imbalance in thrust on a multi, not having centre line engines, which is then flown straight by either rudder or bank or a combination inevitably results in sideslip. It is unavoidable

What is the effect of any resultant sideforce on a fusealge? There is ONLY ONE answer. The fuselage is of course forced to move sideways until it returns to balance.. This inevitably causes sideslip. This sideslip causes weathercocking of the fuselage and if adequate takes over from the rudder and allows us to fly rudder central. The sideslip angle then will be very close to that for wings level straight flight only differing as a result of the "efficiency" of the particular aircraft's vertical tail when sideslipping with and without rudder. The advantage we are endeavouring to achieve is to bring that rudder close to central thus providing us with better options at Vmca.

Two new pronouncements.

Any dissimilarity of one side of an aircraft with the other with an aircraft in steady straight flight will result in that aircraft's direction of flight NOT being coincident with the fore and aft axis. OR

Any aircraft having total drag offset from the fore and aft axis will be unable to have a direction of flight along its fore and aft axis.

How can it be otherwise?

heedm

20th Sep 2004, 03:22

As is common with these types of threads, definitions are vital. If we're using the same terminology to imply slightly different meanings, then the arguments are completely lost.

I take sideslip to mean an angle in the horizontal plane between the longitudinal axis and the flight path relative to the air.

If I'm flying a twin with the left engine out then in the normal rudder position, there will be dynamic pressure on the right side of the fuselage. If I have enough right rudder to stabilize at a point where there is even a small amount of dynamic pressure on the left side of the fuselage, then the obvious continuity of this process would imply that there is a rudder position in between where there is no dynamic pressure on either side of the fuselage. That point meets my definition of zero sideslip.

That point would probably be a minimum for profile drag, but if it takes a significant amount of rudder, then I'd be surprised if it were also a minimum for total drag.

Matthew.

john_tullamarine

20th Sep 2004, 04:34

Zero slip (presuming that the heretics of the first part are able to convince the heretics of the second part that such might exist) may constitute minimum drag but there are other contributors to drag in addition to the slip angle and related airflow .. a case of establishing the story for the aircraft.

Definitions in this thread are certainly important and one of the values of the thread is that people are challenged to think about what is being said and what it all might mean ..

.. chaps who, like me, are bears of very little brain .. with the shape to match .. will probably just get confused ... but, hopefully, some folks out there will benefit.

Keith.Williams.

20th Sep 2004, 07:06

Hello again Milt,

You have previously accepted that it is possible to have left sideslip or right sideslip with the left engine out, depending upon what combination of rudder and bank angle you are using. As stated in my pevious post, this means that there must be a combination of rudder and bank angle at which the aircraft is not slipping in either direction.

It is all very well for you to repeatedly pronounce that bank without slip is impossible. But how do you account for the situation above?

Can you describe some process whereby the aircraft changes from left sideslip to right sideslip without actually passing through a zero sideslip condition?

If on the other hand you accept that the zero sideslip condition exists, do you believe it be with wings level or with some degree of bank angle towards or away from the live engine?

bookworm

20th Sep 2004, 07:22

For too long now there has been almost aviation wide belief that we pilots can eliminate sideslip and continue to fly an unbalanced aircraft straight down its fore and aft axis.
All I am saying is THAT IS ABSOLUTELY IMPOSSIBLE.

Second call for a twin with slip string and pilot in Canberra please... ;)

Cap 56

20th Sep 2004, 08:03

So in summary;

If the force from the rudder is equal then the one obtained from the bank, while at the same time you balance the moment from the live engine, you have zero slip straight flight.

Whether this is at all possible depends on the type of aircraft and/or you may have to do it with less than max/cont power/thrust.

Hereby you will have to neglect the small force contributions from ailerons and spoilers.

In my opinion the zero slip straight assymetric flight will have a deflection of the slip ball equal to the bank.

20th Sep 2004, 09:06

Alright, summing things up.

Milt:

The overriding fact to be determined initially is whether an aircraft having any angle of bank and in steady straight balanced flight can achieve that condition without sideslip.

...with asymmetric thrust.

ANY imbalance in thrust on a multi, not having centre line engines, which is then flown straight by either rudder or bank or a combination inevitably results in sideslip. It is unavoidable

That’s your hypothesis which is meeting with some opposition.

What is the effect of any resultant sideforce on a fusealge? There is ONLY ONE answer. The fuselage is of course forced to move sideways until it returns to balance..

This is, at least in my opinion, a fact.

Thus far, we agree. If there is such a resultant horizontal side force, a slip will develop until it is cancelled (straight flight being a given condition).

However, I showed above how you can achieve straight, balanced zero-slip flight with asymmetric thrust without such a resultant horizontal side force.

The disagreement boils down to whether this is possible or not. What you need to address is why the condition I gave, which fulfils all the requirements, is impossible. You claim it is impossible. Now tell us why this is so?

How can it be otherwise?

In reply, I quote myself:
The horizontal component of lift is equal and opposed to the horizontal force at the stabiliser.

Thrust equals drag.

M_yaw_total = 0

F_lateral_total = 0

F_longitudinal_total = 0

We have equilibrium, sans any slip.

That’s how it can be otherwise. :)

Questioning common knowledge is fine and something which should always be done. Convention without consideration is a dangerous thing. But if you do think the given model is wrong, you have to point out an error in the reasoning used, or all you will have is an unsubstantiated claim. If you can’t substantiate your hypothesis, a hypothesis is all it will remain. If the calculations given match up, they are either a good representation of reality and thus valid, or there is an error or oversimplification in there somewhere.

As it stands, we have an explanation of why you can have straight, level, asymmetric thrust flight with zero sideslip. We have an alternative hypothesis that this explanation is not correct. We do not have the data to falsify the given explanation. Until we have this data, the given explanation will stand.

I repeat: Where is the error? If it can't be found, I think I'm ready to rest my case. :)

Regards,
Fred

Milt

20th Sep 2004, 09:45

heedm Keith.Williams ft and all.

Glad to have you join in heedm.

My definition would leave out 'in the horizontal plane' and would include 'normal to the vertical axis' and there are probably a few ways of saying the same thing.

Sideslip of an aircraft is the angle between the longitudinal axis and the relative airflow in the plane of reference. That is only me saying the same thing .

I cannot think of any occasion when there may be an angle between relative airflow and flight path. Can you or anyone else?

Now climb into your twin again with its left engine out. Keep wings level for simplicity. To stabilize in straight balanced flight, as you say, there is dynamic pressure on the right side of the fuselage to balance the tendency to yaw left. That pressure represents a finite force which MUST NOT be diminished, not even by a milligram, if you are to continue to maintain straight balanced flight. Don't you now have a hell of a problem in using rudder any which way to get anywhere near zero sideslip. There is no way to get there. Immediately you change the rudder angle by a nat's whisker you have changed that finite side force and you are no longer in straight balanced flight. Give the aircraft a very long tail and you could reduce that side force requirement considerably and hence the sideslip but you can't just wish it to go away.

I see confusion in appreciating a requirement for a balance of forces. An overly concentration on balances of moments and couples/torques.

Please anyone prove to me where I am wrong so that I can crave forgiveness for being so persistent.

Keith.Williams.

Cannot imagine accepting the possibility of having sideslip either way so I guess I have to apologise for not having refuted your contention. I do so now.

So back to square 1.

Any aircraft in straight and level balanced flight having total drag offset from the fore and aft axis will be unable to have a direction of flight along its fore and aft axis. This applys equally to wings level and banked flight.

Fred
Just saw yours and the response is so simple that I am surprised you missed it.

Your quote - don't know how to put it between lines yet like you pros.

"The horizontal component of lift is equal and opposed to the horizontal force at the stabiliser"
Where are you getting that "horizontal force at the stabiliser" if it is not being generated by the very sideslip you think you have made go away. Fred you can't make it go away else you will have a strong tendency to become a frisby.
Convinced yet?

20th Sep 2004, 10:21

"The horizontal component of lift is equal and opposed to the horizontal force at the stabiliser"
Where are you getting that "horizontal force at the stabiliser" if it is not being generated by the very sideslip you think you have made go away. Fred you can't make it go away else you will have a strong tendency to become a frisby.
Convinced yet?

Milt,
as you deflect the rudder, you turn the fin into a cambered airfoil. Thus it generates a force without any sideslip.

Coincidentally, that behaviour is also required if you are ever to get out of zero-slip flight using the rudder. I don't think we need to debate whether that is possible? ;)

I shall continue waiting.

Regards,
Fred

Keith.Williams.

20th Sep 2004, 11:09

Hello again Milt,

It is of course quite possible that I have got it all wrong and I am quite happy to be swayed by a suitably logical argument.

You said in your most recent post "Cannot imagine accepting the possibility of having sideslip either way...." I, and I suspect many other readers, believe that the following situations are (at least theoretically) possible.

Situation 1.
Let's suppose our left engine fails and we put in right rudder to arrest the left yaw. We then increase rudder so that we yaw the nose to the right of our original flight path. If we get the rudder just right it appears to me that we will now be sideslipping down our original track. That is sideslipping toward our dead engine. Our lateral stability would tend to roll us away from this sideslip, but we should be able to hold the wings level by applying a bit of aileron towards the dead left engine

Situation 2.
Let's suppose that instead of doing the above we bank the aircraft hard towards the live engine. The tilted lift force will cause us to sideslip towards the live right engine.

Which of these situations do you consider to be theoretically impossible and why?

john_tullamarine

20th Sep 2004, 11:29

Milt,

"My definition would leave out 'in the horizontal plane' and would include 'normal to the vertical axis' and there are probably a few ways of saying the same thing."

While one can resolve the vector to the aircraft plane, why bother for this qualitative discussion ? Depending on whether the bird is going up or down at the time, the vector elevation will follow the performance .. with a definite effect on forces developed by the fuselage. ... and, in any case, how does this alter the situation ?

"there is dynamic pressure on the right side of the fuselage to balance the tendency to yaw left."

Are we considering only the skidding zero rudder deflection situation here ? Why would not the pilot input rudder to reduce the skid (slip) ?

"Immediately you change the rudder angle by a nat's whisker you have changed that finite side force and you are no longer in straight balanced flight"

If we ignore minor effects (such as rolling moments) due to rudder deflection, surely the rudder input simply alters the lateral force system along with the yawing moment balance ? Why is this fundamentally different to the free/fixed rudder case ? Why can we not input an additional force in the opposite sense by playing with bank angle ?

"I see confusion in appreciating a requirement for a balance of forces. An overly concentration on balances of moments and couples/torques."

The two considerations (forces and moments) need to be balanced for steady flight (which would then be straight). It is implicit that varying the rudder input requires an opposing lateral force adjustment via a bank variation. Perhaps you can offer an explanation of how the observed airflow direction as rudder and bank is varied fits into your thesis ?

"Any aircraft in straight and level balanced flight having total drag offset from the fore and aft axis will be unable to have a direction of flight along its fore and aft axis. This applys equally to wings level and banked flight."

Why is this offset necessarily so ? Why can not one vary the drag components to alter the drag vector ?

"Where are you getting that "horizontal force at the stabiliser" if it is not being generated by the very sideslip you think you have made go away."

This would be quite OK if we were looking at a rudder free/fixed environment, but why is it that one cannot input rudder to achieve the force desired ? Indeed, the arguments suggests that ailerons and elevators, similarly, can't work ?

Milt

20th Sep 2004, 12:07

ft JT and Keith later after some more use of grey matter and recall from experience. Thanks for the prompts JT. Give me a day to get back.

Fred with your comments we get right back to the rudder wagging the tail.

Consider straight and level balanced non asymmetric flight. Rudder is centre. Now apply a little rudder and the aicraft must yaw into a sideslip as a direct result of the sideforce you have just applied to the tail by the rudder/fin force. Any change to rudder angle under any circumstances changes the side force. Slight pause there to quickly go through all of the circumstances that come to mind.?? . That is the very reason we have a rudder is it not? You should not be considering the rudder/fin as a seperate entity to the tail as a whole.

I am most intrigued that you could consider it to be otherwise.

Can you put that down as a little win for JT's tally sheet or does it still leave you puzzled, confused and mybe unconvinced?

Milt

20th Sep 2004, 14:20

Cap 56 Keith

Sorry I missed you last time round Cap 56.

Some rearguard action I seem to be fighting. !!

I think you are saying

"If the force from the rudder/fin (they work together) is equal to the the force from sideslip caused by the bank, while at the same time you balance the moment from the live engine, you have zero slip straight flight."

I have to presume, as you haven't spelt it out, that you have cancelled out the sideforce due to the bank caused sideslip. Now we have no sideforce to counter the yaw being rapidly developed by the live engine. We are turning into a frisby again. Do you really want to accelerate that frisby by taking the rudder further the WRONG way.

Grab a model and visualise.

Then you say " In my opinion the zero slip straight asymmetric flight will have a deflection of the slip ball equal to the bank."

Absolutely impossible to have zero side slip with bank applied. The very reason we put on bank in the first place was to generate that indispensable sideforce to counter the yaw and allow us to get the rudder close to centre. Apply rudder further the other way and we reduce that indispensable sideforce. How can we get it back. Only means if you insist with the rudder is to bank more and retrieve that same X pnds of sideforce with more sideslip. That force cannot be reduced any which way if we are to stay balanced. Take the rudder misapplication further to a conclusion and you auger in unless your speed is relatively very high, probably beyond performance capability.

Unless you can be convinced of those statements then we have to agree to disagree for now but don't let that stop you from contributing wherever you think you can add to the debate.. So many of our less initiated must be benefitting enormously from the repartee and arguements.

Oh and that slip ball. It always drops by gravity to the lowest point in its curved glass tube which in the case of balanced banked flight will result in the ball being out of centre by the bank angle...

Keith.

Hello again. Keep the propositions coming.

Your situation 1 is now real close to how it happens except for that increased use of rudder. Before you put on that extra rudder we were balanced having arrested the yaw.
So now you go and spoil it all by changing the rudder. The tiniest bit either way will unbalance us again. But if all you want to do is go along your original track/course then by all means use a little rudder to do a slow flat turn to do just that. It is only 2 or 3 degrees anyway. Having turned to the new heading you must return to balance again by returning the rudder to its precise original anti yaw position. Remember if you change CAS or thrust or drag you unbalance and there will be a changed rudder deflection to regain balance.

We have no means of accurately measuring the angle of sideslip so must make a guess for navigation purposes if we need to be so accurate.

Situation 2. Yes that is what we do to get the rudder back close to centre. Usually about 5 degrees thank you which gives us enough horizontal component of lift to cause us to sideslip to the desired extent..

Bank hard, say 30 degrees and it is unlikely that full left rudder will hold the right yaw resulting from a now very high left sideforce due to a gross sideslip angle to the right.. We will rapidly lose it rolling over inverted to the right as the slowing right wing stalls. A very nasty event usually ending in a spin - sometimes inverted.. Don't go there.

Hey - but weren't you trying to find a way to go to zero sideslip. Instead we augered in..

With apologies for having to regurgitate much of what has been stated previously.

I should have quit hours ago and it's now after mid night down here. The subject has become addictive. Just when I wanted to start an argument about rudders operating in the wrong instinctive sense, but that can wait a while.

20th Sep 2004, 14:45

Milt,
I certainly will consider that a small win. I'm not sure if you meant that it should be a small win for me though... ;)

Just recently, you stated:
Where are you getting that "horizontal force at the stabiliser" if it is not being generated by the very sideslip you think you have made go away.

You base your case on there not being any horizontal force at the tail sans sideslip?

Now:
Consider straight and level balanced non asymmetric flight. Rudder is centre. Now apply a little rudder and the aicraft must yaw into a sideslip as a direct result of the sideforce you have just applied to the tail by the rudder/fin force.

And voila! Now you base your argument on a horizontal force sans sideslip!

I hope to be going to Dunnunda over the next few years. Am I getting a cold one for this? :D

I am still puzzled though... as to how you seem to have interpreted something I wrote as building on separating the rudder/fin from the rest of the tail?

I'll check your most recent post later. Yes, this is certainly addictive!

Cheers,
Fred (still waiting)

SR71

20th Sep 2004, 14:51

It is absolutely possible to have zero sideslip and non-zero bank.

Take an axis-symmetric cylinder whose longitudinal axis is aligned with the velocity vector i.e., \alpha & \beta = 0.

Rotate it about this axis and tell me you still believe what you affirm to be true is the case Milt...

People are still confusing the axes systems involved in this problem.

Aerodynamics isn't sensitive to \phi.

The state vector includes six fundamental variables none of which is \phi!

The mathematics never lies.

:ok:

Keith.Williams.

20th Sep 2004, 15:21

OK MIlt, I think we are finally getting somewhere.

To quote from your most recent post.

"Your situation 1 is now real close to how it happens except for that increased use of rudder. Before you put on that extra rudder we were balanced having arrested the yaw.
So now you go and spoil it all by changing the rudder. The tiniest bit either way will unbalance us again. But if all you want to do is go along your original track/course then by all means use a little rudder to do a slow flat turn to do just that. It is only 2 or 3 degrees anyway. Having turned to the new heading you must return to balance again by returning the rudder to its precise original anti yaw position. Remember if you change CAS or thrust or drag you unbalance and there will be a changed rudder deflection to regain balance."

I did not say that this manoeuvre was a good idea nor that it is easy to achieve, but simply that it is possible. Now let's be clear about it, we are talking about a situation in which the aircraft is sideslipping to the left. YES or NO?

And:

"Situation 2. Yes that is what we do to get the rudder back close to centre. Usually about 5 degrees thank you which gives us enough horizontal component of lift to cause us to sideslip to the desired extent.."

And isn't this sideslip to the right? YES or NO?

Unless I am mistaken we now have agreement that it is possible to sideslip either to the left or to the right with a left engine out, depending on the combination of bank angle and rudder.

Now we can get back to my original argument.

If we can sideslip left and we can sideslip right, then presumably we can gradually change from slipping left to slipping right. And at some point during the manoeuvre we will not be slipping at all.

If you look at the situation you will see that this will occur with a slight bank towards the live engine. This means that it is possible to have zero sideslip when banked towards the live engine in asymmetric flight.

20th Sep 2004, 16:43

SR71,
going offtopic for a moment here... I'm curious about the peculiar convention you're using for mathematics here. Where does it stem from, with the \backslashes etc? Never seen anything like it before.

Any pointers to explanations, advantages and historical background? Frankly, I find it rather opaque but I'm always willing to learn something new.

Regards,
Fred

SR71

20th Sep 2004, 20:39

Fred,

No problem. The syntax originates from the days of my doctorate studies.

LaTeX is the de facto standard syntax for the communication and publication of scientific documents. A lot more versatile than anything MS can come up with...

LaTeX is based on the idea that it is better to leave document design to document designers, and to let authors get on with writing documents.

You can check the WWW for numerous references to the project but try here first:

The LaTeX Project (http://www.latex-project.org/)

:ok:

20th Sep 2004, 22:06

Gah! Horrible flashback! Yes, I did write in LaTeX (with the horrible formatting of the logotype) way back when... aaaagh!

Never thought anyone would try writing raw LaTeX in a forum though. You're a deeply disturbed person! It's not exactly intended to be human-readable... :D

Hey, how about converting to PostScript? ;)

Cheers,
Fred

john_tullamarine

20th Sep 2004, 23:07

Milt,

What jolly good fun this all is ...

Cap 56 - "If the force from the rudder/fin (they work together) is equal to the the force from sideslip caused by the bank, while at the same time you balance the moment from the live engine, you have zero slip straight flight."

Milt - " .. you have cancelled out the sideforce due to the bank caused sideslip. Now we have no sideforce to counter the yaw being rapidly developed by the live engine. We are turning into a frisby again. "

Cap 56 addressed the need to keep the moment balance - hence no frisbee excitement. A more appropriate observation would have been that the requirement to maintain zero slip (once obtained) is to maintain a nil residual lateral force and nil moment imbalance.

One should observe that a zero slip condition is likely to be approximate and than there will, in all likelihood, be random minor excursions into a slip regime .. but the general point still remains... or are you simply being pendantic on this point of semantics ?

"Absolutely impossible to have zero side slip with bank applied. The very reason we put on bank in the first place was to generate that indispensable sideforce to counter the yaw and allow us to get the rudder close to centre."

Generally accepted for symmetric thrust .. but the crux of the discussion for a thrust asymmetry. Again, why is there a need to seek the Holy Grail of near zero rudder deflection ?

"Having turned to the new heading you must return to balance again by returning the rudder to its precise original anti yaw position. Remember if you change CAS or thrust or drag you unbalance and there will be a changed rudder deflection to regain balance."

Can not the pilot play with bank-related forces in addition to, or in lieu of, rudder-generated forces ?

Fred,

First shout is on me during your forthcoming trip ...

SR71,

Why don't you come along also and at least we will outnumber Milt.

Milt,

You stir the pot superbly well, good sir.

Milt

21st Sep 2004, 00:32

John Tullamarine

Our worthy moderator has raised 6 points for clarification concerning the complexities of asymmetric handling.

Rather than repeat them item by item which would make this a lengthy rendition I ask participants and other parties of interest to refer back to JT's post hoping that it is not now on a separate page. Damn just saw that it is on the previous page.

Item 1.
In seeking a comprehensive definition of sideslip the term "normal to the vertical axis" was included only for completeness knowing full well that the sharp minds viewing the post would likely be critical of a definition which only related to our strictly horizontal determinations. We well may have to defer to our aerodynamics potentate whoever that may be for the absolute precise definition. Let us agree for now on the simple version applicable to the horizontal.

Item 2.
Don't think that quote was mine but I recall I did respond to it. To have "skidding zero rudder deflection" we must necessarily have bank to cause the skid. The skid is creating the force to stop yaw. Use of rudder to reduce the skid changes the force and immediately puts us back into unbalance. There is NO way to go with rudder and stay in balance unless we readjust the bank angle.

Item 3.
Answered by Item 2 above I believe and I think you have answered your own question correctly. We do play with the bank angle to readjust our rudder position. For any bank angle there will be a precise rudder position for balance. You cannot change one without changing the other as you have said in Item 4..

Item 4.
Airflow directions? That's a curly one and I think too complex for a definitive statement. Instinct from experience tells me that the variations to sideslip as we vary our relationship between bank and rudder will be so infinitesimal that we can confidently say there will be insignificant change in the small range of bank angles of any worthwhile use. Somewhere in there may well be the minimum drag combination we want but then the flatness of the drag curve here may mean that it is of no consequence. We are now into the realms of instrumented flight test or wind tunnels. Would be great to hear from someone who has been there.

Item 5.
The statement is applicable to any condition of balanced flight. Of course we often change the drag vector but usually take care to balance one side of an aircraft with the other. If we install a drag producing device on one side only we have a total drag line offset from the fore and aft axis. We have then created an inability for that aircraft to be able to fly balanced directly in line with its fore and aft axis. If the delta drag is small then the sideslip to compensate will be minuscule and indiscernible to a pilot but it will be there nevertheless. Where it is very discernable though is when we have an engine out causing a large drag line offset.

Item 6.
Why is it that one cannot input rudder to achieve the force desired? John that is precisely what we do. The force desired is that which makes up the total from the tail to offset the yaw. It is a precise amount to achieve balance. Move the rudder that nat's whisker (or should it be gnat's?) either way and you go into unbalance. And yes the same with elevators and ailerons. For any balanced condition there will be a precise position/angle for both. We try to trim to those positions.

What now ? Strong coffee maybe !

Ohhhhhhhhhh now I have a whole new set of items from JT amongst others all waiting with baited breath !

Watch this space.

Keith.Williams

We have been consistently trying to handle an asymmetric twin with its left engine shut down whilst being flow by a succession of terribly mixed up pilots .

In your last post you say " we are talking about a situation in which the aircraft is sideslipping to the left. YES or NO?"

A huge NO NO NO Keith. We are sideslipping to the RIGHT to generate that sideforce to oppose the yaw to the LEFT. They balance out don't you see?

Then "And isn't this sideslip to the right? YES or NO?"

YES by banking into the live right engine.

Now , having recognised your incorrect perceptions above, you will clearly see that there is no way we can get to a condition of zero sideslip.

You must now be my first disciple.! or do I have to prepare more sermons.?

Have you ever flown asymmetric?

SR 71

Cannot see the path of your proposed logic SR 71.

You have blinded me with science to the extent that I am unable to achieve a decode.

Put it in the simple terms directed at the asymmetric twin which is fast running out of fuel!!

Milt

21st Sep 2004, 01:35

Just made 3 separate posts which have somehow become merged.

Perhaps this is what happens with PPRuNe when the same poster post consecutive posts.

Well back to the slipping and skidding and yawing and endeavouring not to lose control.

That right engine continues to run sweetly.

Milt

21st Sep 2004, 02:39

ft Fred

Ref your last post.

Sorry don't know who won what?

You say now "You base your case on there not being any horizonatal force at the tail sans sideslip?"

Sorry again I have lost track of the specific 'case' to which you are referring. Suffice to say - I hope we are remaining strictly in the horizontal plane.

Yes - no horizontal force at the tail without sideslip. Rudder is integral with the tail. Bank or rudder angle or offset drag will each cause sideslip seperately of together in our balanced flight circumstances.

Hey fair go Fred

My quote is quite specific in saying that under normal horizontal circumstances the application of some rudder will cause sideslip which generates our side force.
Where is this horizontal force without sideslip that you are now attributing to me? when you say

"And voila! Now you base your argument on a horizontal force without sideslip!" The only ones I know in this forum are thrust and drag. (Oh --- please ignore the horizontally resolved component of thrust not aligned with the line of drag; let's keep it simple.)

The prospects for a cold one or two Dunnunda are most welcome.

There may be a few wanting to suss out the rebel who undertook to thoroughly question some long established beliefs. I have been doing this ever since my first aerodynamics lecturer tried to convince his class that an aircraft made a balanced turn under the influence of rudder alone. He had nil or little flying experience.

Do we progress?

Milt

21st Sep 2004, 04:01

John Tullamarine

John you presume too much.

At no time have I agreed to the impossible. The impossible is to have zero sideslip under conditions of asymmetric straight balanced flight.

We can't even get close to zero sideslip. Maybe 2/3 degrees for our typical twin. Don't think you will pick it with a yaw string. The SIDE force on the tail is ALL that is stopping us from being that frisby. And at Vmca we are using the total sideforce the tail can muster. There is none left.

Why seek near rudder deflection as in the banked case?

Keeps the flight deck tidier!!! Also we may have run out of rudder trim and that leg is objecting . Real reason is to give us wider options of using over-riding rudder control should we need to use it in a turn or to cope with turbulence or to stay further away from fin stall if this is significant or ---- whatever happens. Penalty may arise from fact that we have had to replace the lift we are using for the sideslip. This and the increased sideslip needed to zero out the rudder we now have adds to total drag.

I really do not know the combination for min drag. It will of course vary with different aircraft and differing directional stabilities.

So someone has come up with the 'rule of thumb' that 5 degrees bank is the best average. But you wouldn't need anywhere near that bank angle for someting like the Concord because the tail's sideforces will vary approximately with the square of the CAS. Slow down to Vmca at low altitude in a Concord with an outer shut down and the others at take off power and you may well benefit from the 5 degrees of bank. My WAG (Wild Arsed Guess) for a Concord's raw Vmca so afflicted would be about 140 Kts in that the engines are fairly close inboard.

I said "Concord's raw Vmca" as I have taken off the 2 seconds do nothing time which allows for the reaction time of the less than average pilot.

Are you there Bellerophon for comments on a Concord's Vmcg and Vmca for a failed outer.?

John - a get together would be a memorable ocassion but I am already considerably outnumbered.

Keith.Williams.

21st Sep 2004, 06:51

Helo again Milt,

You have convinced me (and I suspect a good many other readers) of only one thing:

None there so are blind who those as see will not (anag)

21st Sep 2004, 16:09

To recap:

I asked you to falsify my presented scenario of straight non-slipping flight, as you claim it cannot happen. This scenario builds on the horizontal component of stabiliser force. Your reply was:

Where are you getting that "horizontal force at the stabiliser" if it is not being generated by the very sideslip you think you have made go away.

In other words, you claimed there’s no horizontal force generated at the stab without sideslip. In return, I point out that if you deflect the rudder there will most certainly be a horizontal force even though there’s no sideslip.

Then you counter (?) with:
Consider straight and level balanced non asymmetric flight. Rudder is centre. Now apply a little rudder and the aicraft must yaw into a sideslip as a direct result of the sideforce you have just applied to the tail by the rudder/fin force. Any change to rudder angle under any circumstances changes the side force.

My good man! I do believe you are contradicting yourself slightly!

And this leaves me with my example of straight, non-slipping flight with asymmetrical thrust still unfalsified. And while it is unfalsified, there is non-slipping asymmetrical straight flight.

Regards,
Fred, patiently waiting

(BTW, my definition of slip, for the purpose of this discussion, would be “the angle between the projections of the aircraft x axis and the flight path vector on the horizontal plane”)

Mad (Flt) Scientist

21st Sep 2004, 20:03

Assuming we're talking about civil certified aircraft here, there's something that should be clarified in a previous post...

raw Vmca

There is no such thing.

14 CFR 25.149 (to amdt 25-84 if we're being precise) has no requirements for reaction time in determining Vmc (=Vmca), nor Vmcg. In fact a 2 second reaction time for the latter test would be mightily exciting.

I believe the confusion is arising from the takeoff speeds requirements, 14 CFR 25.107(a)(2) to be precise, which requires that V1 be not less than Vef plus a time interval for pilot reaction, and also 14 CFR 25.107(a)(1) which requires that Vef be not less than Vmcg. Together these paragraphs require that V1 be not less than Vmcg plus a reaction time (2 seconds) BUT that is a reaction to begin aborting the takeoff, not a reaction to keep the aircraft on the runway. That is assumed to be a great deal faster than 2 seconds. (And is shown to be so during normal Vmcg tests)

The only variations which can be applied to the Vmc values are airspeed indication corrections (the technical analysis will be completed with fully corrected speeds, but information for crew will be "indicated" speeds).

john_tullamarine

21st Sep 2004, 22:51

I was wondering how long it would take for a member of the sandpit to take Milt to task on that one ..

A minor point to add to MFS' post.

The 2 second accel-stop pause and have a brief coffee requirement came in at FAR 25 A/L 42. There is no retrospective requirement so aircraft to older frozen design standard amendments remain a bigger problem for line pilots in ASDA-limiting situations.

Milt

22nd Sep 2004, 10:12

Raw Vmca.

Apolgies to all who considered Raw Vmca to be a formalised term.

It is my way of saying that most of the certified speeds for aircaft types have a buffer to give us a reaction time. None of us can react intantaneously and succeed at what I have called my Raw Vmca. Without that buffer you are instantaneously right at the edge of losing it.

For those who consider there is no buffer and then realise that they need around 2 seconds to identify the culprit engine and follow through with appropriate actions (and that will be most of us) then you had better add YOUR 2 seconds to the published IAS/CAS. Your pilot not in command will certainly appreciate you doing just that. So will you if you are down the back.

My mention of the 2 seconds was to cause readers of the thread to ponder their predicament if no buffer exists.

Not having been involved with the determination of Vmcaes/Vmcges for a decade or two I am temporarily outdated on current certification flight tests. The buffer is a judgement made by the TPs who have a responsibility to you for safe operation.

Tinstaafl

23rd Sep 2004, 03:08

From Milt
...the impossible is to have zero sideslip under conditions of asymmetric straight balanced flight... (my emphasis)

I don't think any of us think this is possible. As far as I can see all 'zero sideslip' supporters specify that some angle of bank towards the live engine will be necessary, albeit not much.

Milt

23rd Sep 2004, 03:20

Vmcg and Vmca Speeds and Safety Factors

Some research on this subject since my recent post indicates that the old 2 second delay stipulation has been overtaken by civil type certification guidelines for TPs . The 2 second do nothing is still considered for special circumstances.

The following are my summaries which are presented as a stimulus for your further considerations and mental preparedness should you be suddenly faced with the non blessed event.

In the determination of Vmcg, recovery should be achievable by the average pilot within the confines of a 30 feet diversion from centre line.

In the determination of Vmca, recovery should be achievable by the average pilot within a 20 degree change of heading.

These allowances embrace your reaction buffer which will only be enough if you maintain your skill and intuitive reflexes.

I decry the fact that simulators cannot quite substitute for the real thing in this critical area of flight and blame the bean counters for taking simulation too far.

When was the last time you experienced Captains had one pulled on you on a training flight and how was it?

23rd Sep 2004, 07:36

Tinstaafl,
aaaah, a change of position! Missed that one.

In previous episodes of the 'Yawing, slipping and sliding about' miniseries:
Any multi having engines outside of centre line will be sideslipping during straight flight when there is any imbalance of thrust.

Sorry - any multi, not including those with centre line engines, having ANY measure of asymmetric thrust will be sideslipping in straight flight whether wings level or banked.

Add 'balanced' into the mix and it's a whole new ballgame (pun unintentional). Cunning, Milt, very cunning indeed! :D

Cheers,
Fred

newcrew

27th Sep 2004, 18:54

thanks a ton to all who replied

well belive i am pretty clear wrt the issues re turning and zero sideslip / minimum drag...

the other thing i was interested to know was if any other types had a sideslip indicator fitted ? (concorde style)

anybody know (pictures ?)

thanks

cheers

zzuf

10th Oct 2004, 12:55

Wow!!!
Extract of my Flight Test Report Canberra Asymmetric Handling ETPS 1970.
Results of Tests
Turning During Asymmetric Flight
a. Turns at 1.1 Vmca
Turns at 30 degrees of bank were made in both directions starting from 4000 feet with the left engine set at idle and the right at 7850 RPM and 187k IAS. The rate of climb during these tests was 1800fpm. Turns were easy to make with trimmer settings of; rudder 4 units right and aileron 1 unit right. The sideslip angle was zero and the slip ball position varied between 1/2 and 1 ball width displacement right. The slipball could be centred using full right rudder trim, 3" right rudder pedal travel and a pedal force of 150lbf. This resulted in a sideslip angle of 5 degrees left during left turns and 2degrees left during right turns. These turns were uncomfortable to fly because of relatively high rudder forces and also because the aileron trimmer setting for zero force changed from 3-4 units left during left turns to neutral during right turns. During these tests the yoke position for zero roll rate varied between neutral (right turns) and 10-15 degrees rotation left (left turns).

Discussion of Results and Conclusions
Turns During Asymmetric Flight
Turns at both safety speed (160k IAS) and 1.1 Vmca (187k IAS) with 7850 RPM set were very easy with zero sideslip. When the slip ball was centred (1.1 Vmca) it was necessary to hold high pedal forces, retrim the aileron and accept aileron positions other than neutral for zero roll rate. It is recommended that the aircraft be manoeuvred with zero sideslip using the trim technique described in para 4.7.1 and co-ordinated rudder, if necessary, to keep the slip ball approximately 1/2 ball width to the right

Milt

11th Oct 2004, 01:59

ft and zzuf

OK ft I'm back and inserting balanced into the statement of fact.

Here is the statement again.

Any multi having engines outside of centre line will be sideslipping during straight flight, rudder free, when there is any imbalance of thrust.

or in other words

Any aircraft uniformally shaped around its fuselage centre line which has its line of total thrust offset from that fuselage centre line will be unable to achieve balanced flight, rudder free, in a straight line without sideslip.

Confusion continues to arise because of two differing concepts of balanced flight.

The common concept is that balanced flight is achieved when the balance ball/instrument is centred. All OK when the forces on the aircraft are symmetric and our aim is to fly the aircraft most efficiently in the precise direction of its centre line..

One could say that it is a whole new ball game when the forces are asymmetric.

In the asymmetric case we start with an imbalance of forces and seek ways to bring that imbalance into balance or equilibrium so that we can make progress through the air in a straight line. The only means we have to counter an engine thrust imbalance is to sideslip or to use rudder which creates a counter balancing force at the tail.

zzuf

Wish I'd been your tutor!!

Your review of the above will be fascinating considering your report at ETPS.

zzuf

11th Oct 2004, 04:14

Further from my Canberra Asymmetric Handling Report
Determination of Minimum Control Speed
4.3 Minimim control speed was determined by reducing speed from the minimum trim speed in a straight climb at between 4,000 and 5,000 feet (first cut). The minimum control speed was the lowest speed at which a straight climb could be maintained. These test were made with power settings of 7600 RPM and 7850 RPM on the "live" engine and at bank angles of 0 degrees and 5 degrees towards this engine. Rudder pedal forces measured during these tests were very high and difficult to hold for even a short period.

4.3.3 Minimum Control Speed 7850 RPM Wings Level

Using the technique described in para 4.3 the minimum control speed with the right engine set at 7850 RPM was 175k IAS. The rudder pedal position was full right rudder held by a force of 250lbf. At 175kIAS the slipball was about 1/2 width to the right and the sideslip angle was 3 degrees left. Aileron forces were zero with the yoke position at 10 degrees left and the aileron trimmer position at 2 unit left.

4.3.4 Minimum Control Speed 7850 RPM 5 Degrees Right Bank

With 5 degrees of bank towards the live engine, the speed was reduced to 150k IAS before the aircraft commenced a slow turn to the left. The rudder force was still 250 lbf right rudder and a slight rudder buffet became apparent. The yoke and aileron trimmer positions were 5 degrees and 4 units right respectively.
The sideslip angle was 5 degrees right and the slip ball was at 1/2 the total available travel to the right.

Comment
This is clearly not Vmc - rudder forces way beyond the then BCARD limits.
No CAS quoted - no PEC info with sideslip.
At some bank anle between 0 degrees and 5 degrees right the sideslip direction changed from left to right - at some small angle of bank the sideslip angle was zero.
Bank angle was extremely powerful in reducing the minimum control speed

Of the many jet aircraft types I have flown most have had fairly strong lateral static stability( low speed, low altitude), so during asymmetric flight using a small bank angle towards the operating engine and co-ordinating rudder so the roll rate with neutral aileron is close to zero. Sideslip will be close to zero and the aircraft close to optimum for both handling and performance.
(Not true for most Prop/Turboprop aircraft and not true for all jet aircraft).

Milt, sorry gave up on FTSA years ago.
I will now disappear back into cyberspace.
Cheers