View Full Version : Help me with a brain teaser - guaranteed to drive aerody gurus insane!?!

24th Aug 2005, 00:55
O.K. - Here's a doozy of a question I was posed recently...

"If two aircraft are at the same altitude, IAS and bank angle, but one is light weight and the other is heavy weight, which statement is true?"

A. The heavy aircraft will have a greater turn radius but a lower turn rate.

B. The heavy aircraft will have a greater radius but the same rate.

C. Both the light and heavy aircraft will have the same rate and radius

D. The light aircraft will have the same radius and but a higher rate.


Any ideas?? My answer was C. I cannot for the life of me justify it aerodynamically, but from experience, I have flown dissimilar formations where the exact question description has occured. By virtue of being "in formation" the same rate & radius has occured. But, as we all know, reality is not always the correct reasoning when it comes to obscure "knowledge based" questions!

Help me sleep at night by justifying it aerodynamically PLEEEEASE:{

24th Aug 2005, 01:38
Refer to this crappy diagram I made (http://www.geocities.com/aircraftfire/temp/turnbank.jpg) :}

r=radius of turn
teta=bank angle
tan(teta)=tangent of angle teta

Centrifugal force (blue arrow) = Weight (red arrow) X tan(teta)

(m V^2) / r = (m g) tan(teta)

r = (m V^2) / ( (m g) tan(teta) )

r = V^2 / ( g tan(teta) )

The radius of turn doesn't depend on the aircraft mass. All other data is the same, so the radius will be the same for the two aircraft.

If the radius is the same, the length of a complete circumference is the same. Since both aircraft have the same speed, they will fly a complete turn in the same time. Their rate of turn is the same.

Using formulae:

omega=turn rate

r = V^2 / ( g tan(teta) )
V = omega r
r = (omega r V) / ( g tan(teta) )
omega = ( g tan(teta) ) / V

Also the turn rate (omega) does not depend on mass.

I hope this helps.


24th Aug 2005, 09:15
Aerolearner - you sir, are a genius...

I salute you!:ok:

24th Aug 2005, 10:15
you sir, are a genius...
Nah, I'm just a rocket scientist ;)

Tonic Please
27th Aug 2005, 12:40
Great response!

Made me laugh. Great explination too. Thanks from me.


L Peacock
27th Aug 2005, 15:06
Only difference is the heavier aircraft will require more AOA.

27th Aug 2005, 18:55
L Peacock,

Only difference is the heavier aircraft will require more AOA.

Not true...

L Peacock
27th Aug 2005, 18:59
Have I got it wrong?
Higher mass, more Alpha for same g?

Squealing Pig
27th Aug 2005, 20:50
L Peacock & swh

Take two identical B737s one at 40T other at 60T both at FL100 250kts level. The 60T aircraft will need a higher nose attitude (higher AOA) leading to greater induced drag as such requiring more thrust.

So the only difference would be attitude (AOA) & thrust.

Flight Safety
27th Aug 2005, 23:19
Squealing Pig is right. I found this other explanation of what aerolearner was communicating.

other explanation (

An aircraft of greater mass needs to generate more lift to maintain a level turn. As shown in the diagram on the right in the above link, lift provides the centripedal acceleration while in a bank, which turns the aircraft's direction of flight. The heavier the aircraft, the greater the lift needed to remain level, thus more centripetal acceleration is available at a given bank angle to turn a heavier aircraft.

As aerolearner said, the mass is cancelled out of the equation, thus the turn radius remains the same.

28th Aug 2005, 01:31
Squealing Pig & Flight Safety,

This used to be a common instructor rating question.

To clearly illustrate my point, compare the Concorde, 767, 777, and 747.

For a given airspeed, the AoA for a turn with a 767, 777, and 747 is fairly similar ( the Cl for the wing section of each aircraft is between 2.5 and 3), however, Concorde being lighter than the 777 & 747, and much the same weight as the 767 would have a much higher AoA (its wing section Cl is less than 1).

So to compensate for having a poor low lift generating wing (i.e. low Cl), to generate lift Concorde needs a much higher AoA.

Even the same model aircraft, a heavier aircraft can have similar or lower AoA though using high lift devices.

Referring back to the original question "If two aircraft are at the same altitude, IAS and bank angle, but one is light weight and the other is heavy weight, which statement is true?"

Therefore the statement

"Only difference is the heavier aircraft will require more AOA."

Is clearly incorrect, in the example I have given above the lightest aircraft may have the highest AoA.


28th Aug 2005, 03:51
What does the SR-71 weigh relative to the Concorde? What was their relative abilty to make a tight turn? ;)

Squealing Pig
28th Aug 2005, 05:34
I do see what your saying swh, however it would be wise on this question to compare eggs with eggs. ie 60T 737 v 40T 737 rather than a light v heavy type.

28th Aug 2005, 06:08
Squealing Pig,

The aircraft type is irrelevant, so is the weight.

I used to ask if you had a 747 and Seneca at 150 kts were doing a rate one orbit, which would have the larger turn radius ? which would have the higher angle of bank ? which would finish the orbit first ?

They both have the same turn radius, and 22.5 degrees AoB, and finish at the same time.

The aircraft type and weight are irrelevant, the size of the aircraft has no effect on the radius of turn. Two aircraft flying at the same angle of bank and velocity will make the same radius of turn even if one is 1000 times larger than the other.

This is a basic principle that is used for calculation of approach plate splays for aircraft categories and ATC turning requirements.

Falling for this trap is common for people who don’t understand the basic theory, explaining the theory as aerolearner has above reinforces the theory, that’s what teaching is about.


Squealing Pig
28th Aug 2005, 08:52

At no point did I say the radius of turn would be different I was stating that AOA would be different as weight varied.

Refer to the post you made posted 27th August 2005 17:55

L Peacock,

Only difference is the heavier aircraft will require more AOA.

Not true...


L Peacock
28th Aug 2005, 10:41

Think you've gone off at a tangent. It was established early on that turn radius will be unaffected to mass.

My point was that AOA would be affected (increased for increase in mass to maintain same g). Of course it would require a comparison of identical airframe types.

High lift devices would also fudge the issue (again like for like comparison required). But then high lift devices usually have the effect of increasing the wing alpha anyway, so probably reinforce the point.

28th Aug 2005, 11:03
R = TAS^^2 / (g x tan AoB)

And that's it.

The 'g' loading is equal to L/W and thus = sec AoB.

The question was asked when I was going through the CFS course. They used the example of the BBMF Lancaster, with the Spitfire (on the outside) and Hurricane (on the inside) in formation with it and asked which a/c had the greatest AoB.

The answer they wanted was that they would all have the same AoB. But that is incorrect as the radii of turns differ, even though the turn rate must be constant to stay in formation. Quite complicated to work it out - first assume a bank angle and TAS for the Lancaster to obtain the radius of turn and turn rate. Then assume the formation separation distance for the 2 fighters. Work out their radii of turn, substitute into the turn rate and obtain their TAS. Then from TAS and turn radius for each, recalculate the AoB for each......

Woody (RIP) changed the question after that as he didn't really understand algebra - much less trigonometry!

28th Aug 2005, 11:29
SWH, I see your location is 'Some hole' I'm sure thats where your talking from.

28th Aug 2005, 12:00
Squealing Pig, L Peacock, T3HUY,

My humblest apologies, being an engineer it must be my English comprehension skills which has let me down as I didn’t see in the original question posed by Macchi where they asked to compare the angle of attack and mass/weight of on identical airframes in a steady turn.

I will contact my aerodynamics professors that are still alive and advise them they were all wrong, and that for turn performance “heavier aircraft will require more AOA” and “the only diffence [difference] would be attiude [attitude] (AOA) & thrust”.

T3HUY and as you no doubt are aware a swh is short for “some watering hole”, commonly known as a pub, bar, or an establishment for the consumption of fine beverages, and is somewhere one tends to frequent when in need of rehydration. No doubt you have participated in such practices and locations before.


28th Aug 2005, 12:50

i think you are confusing AOB and AOA.

turning or not, if aircraft x has a greater weight than aircraft y, then aircraft x MUST, given the same airspeed, generate a greater lift force to remain in level flight. Other than generating additional lift forces using other means, the AOA (Angle of Attack) must be greater. Straight flight or turning this is still the case.

If I'm wrong, I'd better revoke the 200 or so pilots I gave CPLs to!

28th Aug 2005, 13:29

No confusion on my part at all, an aircraft with an efficient wing like a 738 with winglets would be at a lower angle of attack when straight or on a steady turn than something like a Concorde.

I would also suggest a 738 with winglets would be at a slightly lower angle of attack than a 734 at the same mass as it has a more efficient wing, and could have a slightly higher mass for the same angle of attack as a 734 of a lower mass.

As you would be aware lift is not a sole function of angle of attack or thrust, and my contention is that design engineers will have similar optimized angles of attack in cruise for sub sonic aircraft, they optimize the wing design with the wing section, area, geometry, and high lift devices, and lift dumping devices.

The pilot can modify the airspeed, angle of attack, wing section, area, geometry, high lift devices, and lift dumping devices whilst in flight for the phase of flight they are in.

So I guess I will have to agree to disagree that “Only difference is the heavier aircraft will require more AOA” or ““the only diffence [difference] would be attiude [attitude] (AOA) & thrust” for straight and level or in a steady turn.


28th Aug 2005, 14:04
Don't worry swh....

Some of us understand what you're on about. Because, of course, commercial aircraft wings are all twisted by the same amount aren't they?



28th Aug 2005, 20:15
Providing that aircraft x and aircraft y are the same type, then for the same airspeed and same c of g the heavier aircraft will require a higher AOA. If your talking different types with different wings and specifications then you cant relate the AOA in that way, it is too broad a statement.

28th Aug 2005, 22:01
I do not even know what algebra is but I figured it out the other way.
The size of an airplane notwithstanding, the same laws of gravity apply to everyone. And all planes fly in the same air.
Soi, the same bank angle and the same speed should produce similar results.
But I reckon that at first sight I figured the bigger AC had a bigger turn radius.

29th Aug 2005, 07:23

with regard to the AOA, of course we are talking about identical aircraft (as ifleeaircraft says) and not only that, but referring to that same particular aircraft, as even the same models have manufacturing differences, which would lead to slight variations.


Piltdown Man
29th Aug 2005, 09:18
The maths is damn good stuff, but if C was not the correct answer, how would air-to-air refuelling or aerotowing be possible?

30th Aug 2005, 04:04
Piltdown man,

that's exactly the point of my opening thread! I know that "C" is the answer (been there, done that) but I wanted it demonstrated aerodynamically and/or mathematically by those who can. I don't think anyone has actually suggested it was anything but C, but there have been a few differing opinions on the physical aspects of why this is so.

Ultimately, when asked the question, I can't really say the answer is "C - because, well, because it just is alright!!!". Hence my request for the maths behind it.

Thanks for all the feedback guys, very interesting input - and sorry about opening a mini can of worms there. I knew it would cause some angst amongst the aerody gurus. :ok:

Righto, next question: "How does an autofocus camera work?":8

30th Aug 2005, 09:00
humblest apologies, being an engineer it must be my English comprehension skills which has let me down

I've never read a thread with so much argument, where everyone was, in fact, right!!

SWH, the point others where making was that
all else being equal i.e same aircraft type, same configuration, that an aircraft being operated at a heavier weight requires a larger AOA maintain altitude.

30th Aug 2005, 17:58
all else being equal i.e same aircraft type, same configuration, that an aircraft being operated at a heavier weight requires a larger AOA maintain altitude.


I would agree with you if you included the CG position being the same.

Putting fuel in an aft trim tank can reduce the angle of attack, do you remember the effect of CG position on stall speed ?


1st Sep 2005, 15:54

Had to think about this for a couple of days, is this how it goes?

A conventionaly configured (e.g. forward wing, rear stabiliser) aircraft requires a down force from it's tail in order to fly level, and this force must be counteracted by lift from the wing.

A rearward c.g, therefore, reduces this trim force, and therefore will increase the total lift force for a given A of A. Thus an aircraft with a rearward c of g will produce less induced drag and stall at a lower speed.

I would imagine the opposite would be true for a canard or three-lifting-surface configuration.

HOWEVER a particuar wing stalls at a particual A of A regardless of weight or c of g.

Am I close?

2nd Sep 2005, 12:11
Okay Maccherscmitt, you figgered the right question thru the turning nomogram/ line astern refuel caper.
The correct answer to your other question is the same: point and shoot!

2nd Sep 2005, 13:18
HOWEVER a particuar [particular] wing stalls at a particual [particular] A of A regardless of weight or c of g.

Yes, a stall will occur when the critical angle of attack is reached in any phase of flight, That being said, an aircraft with a forward CofG will do so at a higher speed S&L everything else being the same.

{will just talk conventional geometry for simplicity all else being equal i.e same aircraft type, same configuration, same mass}

Aircraft A

With a forward CG you will reach that AoA at a higher speed due to the higher down force (lift component perpendicular to the relative air flow in the same direction as gravity) needed to be generated by the tail plane to counter the lift - weight (mass x g) couple .

This is an aerodynamic moment about the CG which effectively increases the weight of the aircraft by the amount of down force generated.

Aircraft B

Another method to generate the same couple with an aircraft of the same mass is to redistribute the mass so part of the fuel load fuel is carried in the tailplane. The additional weight (mass of fuel x g) in the tailplane can be sufficient to counter part or all of the lift weight couple (for this example I will take the extreme, and have zero down force by the tail).

In straight and level flight, the lift needed to be generated by
Aircraft A = mass x g + amount of down force by the tail
Aircraft B = mass x g

So aircraft B will have a lower AoA, as the total lift needed to be generated in just mass x g, this will apply in all phases of flight.

Another way to look at it is that an aircraft with an unfavorable CofG location (read forward limit) can have a higher AoA than that of a aircraft with a higher mass with a favorable CofG position, in any phase of flight.

Going back to what you said before “all else being equal i.e same aircraft type, same configuration, that an aircraft being operated at a heavier weight requires a larger AOA maintain altitude”, so another consideration is CofG location.

Long range aircraft about these days use the optimisation of CofG in flight to reduce the induced drag, this compounds as less fuel is needed not only for the reduction of induced drag, also the reduced fuel load.