CG--RUDDER EFFECTIVENESS AND Vmc

[wingtip777]

HI ALL!
WHY DOES A REARWARD CG DECREASE RUDDER EFFECTIVENESS AND INCREASE Vmc?

what are the results of both packs inoperative in flight for a b757?

thanks a lot

[Oktas8]

WT777,

The yaw-slip thread on this forum contains the answer you look for in (some would say) painful detail.

Briefly however, a body subjected to a rotational force, such as yawing, will tend to rotate about its own centre of gravity.

An aircraft yawing, due to asymmetric thrust, will tend to yaw about its own normal axis, which passes through the centre of gravity.

To stop this yaw, you will need a moment exerted to oppose the yaw. Moments are equal to the product of a force and its arm (distance between the force and the centre of rotation).

In the case of the aircraft yawing due to asymmetric thrust, the rudder produces a force. The anti-yaw moment is equal to this force multiplied by the distance between the rudder and the c.g.

To prevent yaw, you can either have a rudder displaced a little bit (small force) a very long way from the c.g. (long arm), or you can have a rudder fully displaced (big force) a short distance from the c.g. (short arm).

To oppose yaw, an aircraft with an aft c.g. (small arm) needs the rudder displaced more, than another aircraft in the same situation but with a forward c.g. (larger arm, therefore smaller force required from rudder).

If both aircraft decelerate together, guess which one will run out of rudder authority first? Yup, the one which has the shortest arm (aft c.g.). This is the one which will have the highest Vmc.

No idea about packs on a 757 though... :ok:

cheers,
O8

[7FF]

Wingtip777 Are you referring to Vmcg? If you are then oktas8 detailed response is in my opinion not quite correct as the aircraft while it still on the ground will yaw about the mainwheels not the aircrafts C of G. Once airborn then oktas8 is absolutely correct.:sad:

[wingtip777]

thanks ro both Oktas8 and 7FF!!
THAT IS THE EXACTALLY ANWSER I AM LOOKING FOR , AND I UNDERSTAND IT NOW, THANKS A LOT !

[SR71]

Milt is just about to get a lot busier.

:)

[Milt]

SR71

You must be psychic. Just when I was going to ask you to pull Octas8's castle down.

Octas8

Must butt in here before you pass on too much duff gen. You left the other thread on Yaw and Sideslip too soon.

When considering Vmca and directional stability you cannot use the cg as the axis. Weight and cg don't get a look in until you use a little bank to help with the asymmetric yaw. Even then it is an indirect relationship Try resolving weight through 90 degrees and what do you get? Longitudinal stability is the one that has serious concern with the cg.

The axis for considering forces in the horizontal is the lateral centre of pressure unless manoeuvre is also considered.

The only participation of the cg arises from the fact that total lift acts through a point close to the cg ( vertically displaced).
Consequently when you use any bank to assist in asymmetric control the small resolved portion of lift acts from the centre of total lift most often with its centre above the cg. The effect into the horizontal plane is so insignificant that for practical purposes it can be ignored.

So one can delete all reference to cg entirely.

The text books are seeing the errors of their ways and will be corrected in due course.

Wings 777
You should now be able to work it out for yourself.

If any of the above does not yet make sense to you then you may learn a lot if Octas8 is challenged.

[Oktas8]

Hullo Milt! Welcome to another thread of heresy! I haven't actually left the other thread, but I must confess to being slightly bored with the argument going in circles at the moment.

Firstly - wingtip 777 - my answer is textbook correct in every respect. Milt wishes to challenge the textbooks! :E (Thanks very much for your response by the way - "I understand it now" is flattery for a flight instructor!)

Now back to heresies and heretics, and the subverting of dominant paradigms.

Milt - please note the very careful use of the word tend in my posts. A body does tend to rotate about its own c.g. - I'm sure you've read the same physics texts I have that demonstrate this. An aircraft, being a body with mass, will tend to rotate about its own normal axis, and this simplification is IMO justified for a question such as was posed at the beginning of this thread.

I am aware of the lateral centre of pressure and its effect on an aircraft - an effect which is minimal at small angles of slip, I might add. The initial tendency of an aircraft to yaw, is about the c.g. - the lateral centre of pressure plays an increasingly important role as the yaw / slip is allowed to develop.

Secondly, please do not equate weight with c.g. In my post I did not mention weight, nor did I resolve it in any particular direction. Confusing mass (c.g.) and weight is a beginner's mistake, or an expert's attempt to confuse a beginner - either way, shame on you.

Lastly, someone who did not previously understand why c.g. affected Vmc, now does understand, at least to a level that allows them to load the twin correctly before flight. This is also quite important.

cheers,
O8

[Mad (Flt) Scientist]

The axis for considering forces in the horizontal is the lateral centre of pressure unless manoeuvre is also considered.

Technically, it does not matter whether one considers the equations of motion to be solved about any particular reference point; what is important is that all the relevant forces acting upon the aircraft be considered to be acting at their appropriate locations. How you choose to resolve the sum of the forces and moments is irrelevant.

If one considers the simplified Vmca case of wings-level flight, where the yaw control system only generates a couple, then the moment balance about the yaw axis is simple: the engine generates a yawing moment arising from the laterally displaced force axis, and the yaw control must generate a moment equal and opposite to the engine-induced moment. In this case the longitudinal cg position is irrelevant, and the aircraft is found to be stable at a zero sideslip case.

However, there is no yaw control on an aircraft which generates a pure couple; instead, the rudder will generate a sideforce which is removed from the centre of the aircraft; this therefore generates a moment.

If we now seek to trim the aircraft in a wings level condition, we quickly find that a solution at zero sideslip does not exist, as the sideforce exerted by the rudder cannot be balanced. Therefore, in order to trim a Vmca-like condition with zero bank, we must have a (small) angle of sideslip on the aircraft.

In order to determine the trim conditions (steady state solution for the equation of motion) we must obtain solutions to two simultaneous equations, requiring that both sideforce and yawing moment be balanced. (Again, our case is simplified by omitting the roll axis).

CY(total) = CY(beta)*beta + CY(rudder)*rudder = 0

Cn(total) = Cn(beta)*beta + Cn(rudder)*rudder + Cn(OEI) = 0

The aerodynamic derivatives in the above equations are not defined about the c.g., but rather an arbitrary reference point; c.g. has no bearing on these values. The Cn(OEI) engine yawing moment term is not affected by longitudinal c.g. position.

Therefore there is a single solution for beta and rudder which is not dependent upon c.g. position (or mass, or weight).

If now we adopt a slightly more representative Vmca condition and permit a little bank angle to be used, there is now a component of the aircraft weight acting at the cg (of course) in the sideforce axis. This provides a fixed offset to the CY(total) equation, and provides an offset to the Cn(total) equation which is a function of the cg position. (It's the weight component in the sideforce axis multiplied by the distance between the c.g. and the reference point used for Cn, all normalised appropriately).

Note that "lift" doesn't actually enter into this at all - the force and moment balances we're dealing with are in the Y axis direction and about the Y axis.

Adding the roll axis moment balance to the considerations doesn't materially affect the conclusion. The lift, drag and pitching moment don't matter at all, unless we get an indirect effect on derivatives due to changes in e.g. alpha, which again we're neglecting.

So I have to say I'm not convinced that cg is of no effect on Vmca.

Consequently when you use any bank to assist in asymmetric control the small resolved portion of lift acts from the centre of total lift most often with its centre above the cg. The effect into the horizontal plane is so insignificant that for practical purposes it can be ignored.

Bank is a very powerful effect on rudder required to trim, because its affecting the weight component and hence the yawing moment generated by the weight about the reference point. I hope that statement isn't implying that bank isn't important for Vmca.

[Oktas8]

MFT,

I have read your post several times, and you make a lot of sense - there is a lot of imagining of paper aeroplanes with a point mass as c.g. going on in my head.

(Sorry if you've been trying to say the same thing Milt - I can understand MFT's reasoning, but not yours. Perhaps the fault is mine? I understand your desire to say 'I told you so' anyway.)

As I'm not sure I understand the implications of your simultaneous equations, may I ask a question? Cn(total) = Cn(beta)*beta + Cn(rudder)*rudder + Cn(OEI) = 0 It seems to me that Cn(OEI) does not depend on longitudinal c.g. position, but the other two coefficients do. It also seems to me that it's important that Cn(beta) is less than Cn(rudder) for the wings level case - much less, in fact, as Cn(rudder) has to oppose the other two put together in the wings level case.

If Cn(OEI) is independent of longitudinal c.g. (independent of whatever longitudinal reference point you choose to use), and Cn(rudder) is the dominant of the other two expressions, doesn't this mean that the longitudinal reference point is important? Otherwise we could pick the point at the point of origin of force beta, making Cn(beta) zero. Or we could put the reference point far ahead of the aircraft, and Cn(beta) would approximate Cn(rudder). Either way the equation would cease to equal zero, simply by our apparently arbitrary choice of longitudinal reference point. In turn, this implies a fault somewhere in the reasoning.

Would you clarify this problem?

Tks,
O8

[SR71]

IMHO there is a term missing from the moment equilibrium if one asserts that the rudder is not a pure moment producing control surface...

[Mad (Flt) Scientist]

As I'm not sure I understand the implications of your simultaneous equations, may I ask a question?
Cn(total) = Cn(beta)*beta + Cn(rudder)*rudder + Cn(OEI) = 0

It seems to me that Cn(OEI) does not depend on longitudinal c.g. position, but the other two coefficients do. It also seems to me that it's important that Cn(beta) is less than Cn(rudder) for the wings level case - much less, in fact, as Cn(rudder) has to oppose the other two put together in the wings level case.

If Cn(OEI) is independent of longitudinal c.g. (independent of whatever longitudinal reference point you choose to use), and Cn(rudder) is the dominant of the other two expressions, doesn't this mean that the longitudinal reference point is important? Otherwise we could pick the point at the point of origin of force beta, making Cn(beta) zero. Or we could put the reference point far ahead of the aircraft, and Cn(beta) would approximate Cn(rudder). Either way the equation would cease to equal zero, simply by our apparently arbitrary choice of longitudinal reference point. In turn, this implies a fault somewhere in the reasoning.

None of the terms in that equation are dependent upon the c.g. position. That obviously seems unintuitive, because it's causing problems in another thread, so let's look at it in more detail.

When I measure the forces and moments on an aircraft ina wind tunnel, I use an arbitrary reference point for all forces and moemnts. I don't use the c.g. position, because I don't KNOW the cg.

When a simulation model of an aircraft is produced, specifically the "flight model" which is the six degree of freedom model of the aerodynamic and other forces on the aircraft, all the forces and moments are STILL referenced to the (same) arbitary reference point.

The cg (inertial) effects are added separately.

To consider moving the reference point. Suppose I use the nominal' reference point for forces and moments - 25% mean aerodynamic chord is typical. I then construct my yawing moment balance about that reference point.

Cn_25=Cn_25(beta)*beta + Cn_25(rudder)*rudder + Cn_25(OEI);
where "_25" indicates we're using the 25% reference point.

If I now wish to calculate the yawing moment about the aircraft c.g., I have to translate the point of action of the forces and moments from my 25% point to the cg; let's assume it's at 15%.

Cn_15 = Cn_25 + CY_25 * (25%-15%)*chord

I can then dimensionalise the Cn_15 term into an actual moment, N_15, and then determine the acceleration of the aircraft about the yaw axis:

yaw accel=N_15/Iyy

Now, cg mattered in that lot only because of the transfer of the sideforce term. But we assumed in that particular case that the aircraft wasn't banked, so the residual sideforce on the aircraft is zero. Therefore CY_25 was zero. Therefore the Cn_15=0 condition for trim is identical to the Cn_25=0 condition. Unless there is a lateral force component (which occurs in trimmed flight only with a steady bank angle) then the cg has no effect in determining the trimmed condition in the yawing axis.

Of course, if the yaw accel is non zero - i.e. the aircraft is still moving about the yawing axis - the c.g. matters to determine the motion. But if the aircraft is not accelerating, then the cg has no effect for the special zero bank case.

-------
It's extremely important when discussing the equations of motion and using these to determine the behaviour of the aircraft to be as precise as possible about axes etc.; the impossible can be made to appear intuitive if the definitions are carelessly used.

SR_71:
The only thing missing is the rolling moment due to rudder, Cl(rudder) and that\'s because I\'ve chosen to ignore the rolling axis for simplicity. Imagine the rudder point of action to lie on the aircraft stability x-axis if you wish a practical case.

Cn(rudder)*rudder is the yawing moment produced by the rudder, and is the only rudder term needed in the yawing moment equation.

CY(rudder) is the rudder sideforce derivative; when that term is non-zero then the rudder produces sideforce and yawing moment (i.e. is not producing a pure moment) and in that case we find that to "solve" the sideforce equation for the equilibirum case another source of sideforce is required. Sideslip, through the CY(beta)*beta terms, is the example used.

[Milt]

Mad (Flt) Scientist

Most of us haven' t a clue on the actual relative effect on Vmca banked by moving the cg from forward to aft limits.

My experience in this area has been in the determination of fore and aft limits by handling throughout relevant flight envelopes..

Determining the forward limit was the easiest because the most critical I found was the flare for landing when it was all too easy to run out of elevator power with control against the rear stops and then needing that extra little bit which was not available. One would back away from that cg position somewhat to provide for better flare when setting the forward limit.. But I digress.

Could you surprise us all with a calculation of the delta IAS on Vmca for a typical asymmetric heavy using 5 degrees of bank covering the full range of cg fore and aft?

If that is too tall an order how about a WAG?

[safetypee]

Sorry Milt I cannot surprise you with data, but to assist the less erudite readers or to add even more confusion to this thread or the one on yaw/slip; see the abbreviated presentation in the ‘Asymmetric Flight’ section here; Vmca.ppt (http://uk.geocities.com/[email protected]/alf5071h.htm) This discusses both longitudinal and lateral cg change. Please acknowledge the author’s contribution who supplied this for the PSM+ICR project, but it was not published.

Thanks to ALF for hosting the files. Please do not shoot the messenger!

[Mad (Flt) Scientist]

OK, WAG time, because we don't actually test minimum control speeds at other than aft c.g., that being the critical case, so I can't just look it up.

Let's assume that the ideal is to have the theoretical Vmca somewhere near stall speed, since that should ensure that the various takeoff speed limits are driven by e.g. 1.2Vs and not by 1.1 Vmca.

Therefore the lift coefficient in "steady flight at the stall" (yes, I know there's no such thing, but this is theoretically) is CLmax. If I bank the aircraft through 5 degrees, then the weight of the aircraft now acts 85degrees in the lift plane and 5 degrees in the sideforce plane. So the amount of sideforce I need to generate is:

Y=mgsin(5deg)

Since the lift and sideforce coefficients are nondimensionalised by the same factor, qS, I can make the links:

mg=qS*CLmax

so

qS*CY=qS*CLmax*sin(5deg)

so to prevent the aircraft accelerating in the sideways direction, we'll need a sideforce coefficient roughly

CY= CLmax * sin(5) ~0.08CLmax.

If we assume a CLmax of about 2, that gives us a CY of about 0.16.

As we move the c.g. back and forward, the couple due to the sideforce will change - the aerodynamic force is defined as acting at 25%cg (standard reference point for most aircraft) while the weight component of course acts at the cg. So the (dimensional) yawing moment is:

weight component * [cg-to-ref-point distance]

We've already defined the weight/sideforce as being about 0.16 CLmax. To get a non dimensional distance for yawing moment we need to refer the distance offset to the wing span.

non-dim offset = [cg-to-ref-point distance]/b
= (cg% - 0.25)*mac/b

b/mac is, of course, the aspect ratio. So the non-dimensional yawing moment due to the cg is:

weight component * [cg-to-ref-point distance] = 0.08CLmax * (cg% - 0.25)/AR

To make my sums easier I'm going to assume an aspect ratio of 8.

Therefore the yawing moment coefficient generated by the weight effect at 5 degrees bank will be

0.08 * 2.0 * (cg%-0.25) /8 = 0.02 * (cg%-0.25)

Therefore the delta yawing moment caused by moving the cg is:

deltaCG * 0.02

If one assumes a cg range of 20%, that means the yawing moment variation between forward and aft c.g would be approximately 0.004. That sounds pretty damned small, but to convert it to something useful, rudder angle, we need to consider how the rudder will generate a corresponding moment:

Cn(due to rudder) = Cn(rudder)*rudder.

Typical values for Cn(rudder) are of the order of 0.001 per degree (I don't have any exact numbers to hand). That would imply that some 4 degrees of rudder would be additionally needed at aft c.g. compared to forward. That's not insignificant.

For an aircraft which was rudder limited with perhaps 30 degrees of maximum rudder, in order to generate the equivalent of 4 degrees more rudder (or about 10% more yawing moment) would require flight about 5% faster - meaning a five knot or so penalty on Vmca, 6 knots or so on V2, etc. No performance engineer will lightly give up that kind of number.

Just to note that, of course, the effect of bank itself, rather than cg is MUCH more powerful than the numbers in my previous post. It's very easy to alter the rudder required to trim at a given speed by banking into and out of the live engine, which is of course why there has to be a bank angle limit in the regs.

PS I plan on checking the maths and assumptions at work tomorrow, just in case I dropped an order of magnitude somewhere by mistake.
OK, the main assumption I was worried about was the rudder derivative, 0.001 per degree; that checks about right on one of our aircraft (it's a function of angle of attack, but that's a good enough number for high alpha)