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mstram
10th Aug 2001, 16:42
I understand that at a 60deg bank, maintaining altitude the wing pulls around 2g's.

If you relax the back pressure and allow the plane to descend the load factor should decrease ???? correct ??

If that is true at what rate of descent would the load factor be back to 1 ?

What is the formula for this ? .... and a 'rule of thumb' ?

Mike

Tinstaafl
10th Aug 2001, 17:22
Just of the top of my head, I think you'll find that once established in a steady state descent ie not accelerating vertically, that the load factors are unchanged.

Change the condition to one where the a/c is accelerating downwards eg just starting a descent, then the trigonometry for it would 'simply' :p involve subtracting the downwards acceleration from that due to gravity & using the resultant net vector added to the centripetal force vector.

I think (but can't recall exactly) that the formula is:

....tan AoB
LF=---------
....'g'

or some variant of those terms. Too lazy to work out the maths correctly.

I'd say the the 'g' term would be replaced with the net result of gravity's accel - the a/c vertical accel.

ShyTorque
10th Aug 2001, 19:28
I agree that there would be no net effect once a steady descent was achieved. There would be NIL vertical acceleration and therefore NIL load factor effect. The fact that a turn is taking place is of no significance.

A simple analogy : Does my weight appear to change while my car goes down a hill at a steady speed? No, only while it goes over the brow, when I may feel light in the seat. Once on the hill it makes no difference.

Astronauts undergo weightlessness training in specially converted large airliners (cleared out of all furniture and padded walls etc). The load factor is reduced to zero by flying a special profile (an elliptical bunt manoeuvre) which results in a constant vertical acceleration.

This is definitely not flown on commercial passengers flights as it upsets the tea and coffee and makes the cabin crews' dresses float up over their heads.

:D

ShyT

mstram
10th Aug 2001, 20:10
ShyTorque
Ah, the 'vomit comit' !
http://zeta.lerc.nasa.gov/kjenks/kc-135.htm

ShyTorque
11th Aug 2001, 00:04
Yes I would think so! Also the toilets, the toilets!

Captain! Have you seen what the turd and slip indicator is showing!

:eek:

[ 10 August 2001: Message edited by: ShyTorque ]

john_tullamarine
12th Aug 2001, 02:59
I think that a review of what load factor means and a sketch of the vector diagram will show that the first post's position is reasonable ....

mstram
14th Aug 2001, 09:44
John,

Could you expand on that?

I don't understand how at a *given power setting*, that a 60deg *level* bank/turn will have the same load factor when you then relax the back pressure to allow an XXX F.P.M. descent.

The back pressure to maintain the altitude was generating the load factor wasn't it ?

So theefore the load factor should decrease, as you relax the back pressure ?

Mike

richard27
14th Aug 2001, 13:35
at 60 degrees aob, 2g holds level . Less than 2g gives a descent, more than 2g gives a climb. When in a steady descent the vertical component of the lift vector is reduced because the overall lift vector is smaller, therefore the load factor is reduced. To illustrate this, consider a 300000 lb aircraft. In 1 g flight the lift vector value is 300000lbs. At 60 degrees aob the lift vector value is 600000lbs (in level flight) If you descend, the amount of lift generated is obviously reduced, therefore assuming a constant aob, the lift vector is reduced. IE vertical component of lift vector gives load factor.

ft
14th Aug 2001, 14:52
If you are in a 60 degree bank at a fixed RoD with a load factor of less than 2 you will be uncoordinated and slipping. To get back into coordinated flight you'll need to load up again, either by adding back pressure to increase the AoA or by accelerating.

The only way to be coordinated with less than 2g in a 60 degree bank is to be accelerating vertically so you will never have a fixed rate of descent.

The trig behind it: Think of the plane viewed from behind with the lift vector pointing straight out of the top of the plane (as it always is during coordinated flight). The vertical component must equal the weight of the aircraft (unless you are accelerating vertically) and the vertical component of the lift vector is L*cos 60 degrees. L*cos 60=W (W being A/C weight)=&gt; L/2=W =&gt; L/W=2. The load factor is defined as L/W. Q.E.D.

In a STEEP dive or climb part of the aircraft weight will be lifted by drag or thrust (weight vector tilted forward or aft relative to the airframe while the lift vector stays put perpendicular to the airflow) which will change things a bit.

Cheers,
/ft

BEagle
15th Aug 2001, 02:19
Remember that when forces are balanced, no acceleration will occur. But a constant velocity may. So, in wings level flight, if L=W there is [B]nothing[/B to say that a constant rate of climb or descent will not result! In S&L flight, constant tiny corrections of attitude are made to maintain level flight; this changes L so that a correcting force is applied in the appropriate sense; that is then changed when the desired level is regained.

In the extreme case of an aircrfat in the hover (let's say a Harrier for simplicity), all the aircraft's weight is opposed by the engine thrust. To descend, thrust is reduced which causes a downward acceleration; if thrust is then restored to the same value as weight, all that will happen is that the aircraft will continue to descend at the vertical velocity it had at that particular moment!

john_tullamarine
15th Aug 2001, 03:35
mstraam,

Unfortunately I have no idea how to generate pictures in this thing. If you email me, I can draw a suitable picture and send it back to you as an attachment ... might be a few days delay as we are up to our ears in R&R at the moment ...

ft
15th Aug 2001, 11:29
Load factor equals lift divided by weight so the doubling of the lift force to maintain altitude in a 60 degree bank constitutes a doubling of load factor.

This does not however mean that the occupants of the aircraft are under 2g. They will simply be supported by the sides of their seets.

Hmm, on the contrary I do hope that the acceleration (including load factor) affecting the self-loading cargo and the airframe will be more or less the same at all times. People tend to react poorly to being thrown into walls (or, to be exact, have walls thrown at them). Cases where the load factor and g load on passengers differ significantly (although only very briefly) would be when encountering severe CAT... :D

Isn't it annoying when you're thinking one thing and end up typing something different. :)

Cheers,
/ft

ft
16th Aug 2001, 11:40
Keith,
but those components in the (earth reference) lateral and vertical planes amount to 2 g acceleration perpendicular to the planform of the plane. No drinks will be spilled unless a passenger has had too many drinks before to be able to cope with his/her glass suddenly becoming twice as heavy.

A g meter inside the plane WILL show 2g and the passengers WILL feel 2 g pulling them straight down into their seats rather than being pushed into the armrests.

In reference to another thread, in the cabin's reference system pendulums will hang straight down, tied down helium balloons will rise straight up and any flying birds will fall straight down accelerating at 1g with a very surprised look on their faces unless they figure out to flap twice as hard - especially those carrying coconuts whether they be african or european. :D

If you ARE in fact saying the same thing then I am afraid that I was able to misunderstand your post thoroughly.

Cheers,
/ft

Checkboard
16th Aug 2001, 20:50
The Load Factor is defined as the proportion between Lift and Weight, where the Lift is that part of the force exerted by the wings perpendicular to the flight path.

The load factor in straight and level flight is equal to one, as the Lift must oppose the Weight. Any aircraft in a steady climb or decent required less lift than the straight and level case. (Sounds strange, I know.)

Consider this: as you sit in your chair at your computer, you are experiencing "1 g", all of which is through your seat on the chair. If you now tilt your chair back onto its rear legs, then you are still experiencing "1 g", however, the force is now shared by the seat of the chair, and the back.

It's the same with aircraft, tilt the aircraft back (in a steady climb) and the force is now shared between the thrust and the Lift. Tilt it forward (in a steady descent) and the force is shared by the Lift and the Drag. In either of these cases as the Load Factor only considers the Lift then the Load Factor is reduced.

Same applies in a descending turn or climb.

As an aside, your stall speed is reduced in a climb or descent for the same reason. The stall speed in a verticle climb (no lift required) is zero - something all aerobatic pilots (performing "Stall Turns" or "Hammerheads") know.

Luftwaffle
17th Aug 2001, 07:35
Bank or no bank, if you let the aircraft accelerate long enough it will pass through Vne and the wings will fall off. Don't forget about the square of the airspeed while worrying about the sine of the angle of bank.

mstram
17th Aug 2001, 09:25
Waffle,

This could be the start of a new topic, probably been discussed before though.

My understanding is that the wings will not fall off just by exceeding VNE, it's the manouvering forces, i.e. frantic pullups at that speed which do the damage.
http://www.avweb.com/articles/usual/

ft
17th Aug 2001, 13:35
I can't really see where Vne came into it all, perhaps we're considering different kinds of acceleration? Interesting anyway so I'm in. I seem to spend lots of time waiting these last few days - check my posting history and you'll se. *grin*

Of course you will have some kind of design safety margin above Vne where the wings will not instantly decide that they like the slipstream better than the rest of the airframe - but I for one wouldn't count on it. As they indicated in the AVWeb article, metallurgy is a weird and wonderful world and certainly not one to be trusted. Even at 1g above Vne and all bets are off - you might or you might not start to notice parts falling off.

Frantic pullups above Va, not above Vne,is what you need to be careful about (Va is the speed at which the structural damage load factor intersects the accelerated stall curve). What lacks in the depicted flight envelope diagram (it often does) is the fact that loads generated by the airspeed and loads generated by manouevering add up. I e, the structural damage load factor limit will in fact be lower at Vne than it is at Va - you can be safely within the structural limits pulling n g's at Va (actually by definition you can't go above the load factor limit at Va since you should stall first(*)) while pulling n g's at Vne might leave you sans a wing or two. Graphically, the real maximum load factor line slopes down towards Vne.

Unless I'm completely mistaken this is taken into account when certifying gliders, I wonder why it is frequently left out of these diagrams... probably the depicted load limit is for Vne, adding some extra margin. Does anyone know the details? Is this considered for powered A/C certification?

Cheers,
/ft

(*) I think this is for a smooth pull though, momentarily it probably is possible to exceed the load limit even at Va if you yank back on the yoke/stick.

(Proofreading before posting probably would avoid having to edit every post... :))

[ 17 August 2001: Message edited by: ft ]

LAN
18th Aug 2001, 00:23
As far as I remember (which may not be that far after all), the following applies to FAR23 A/C :

Vd = design divespeed. At this speed, the aircraft must be able to withstand

a) A vertigal gust of 15 ft/sec without exceeding LLF (or does this one apply to Vne
:o ?)

b) Aerodynamic torsional forces and show no undesirable flight characteristics, including flutter.

Vne is thus established as Vd - 10%...

Regarding the descending turn thing : The descent gradient in a 60deg banked, coordinated turn will depend on the L/D ratio of the A/C and the powersetting. Speed changes may cause variation i g-forces, e.g. will a speed reduction result in momentarilu increased G and reduced ROD, but this will only apply to the transition, not to the stabilised case...

(as the humble man ends his sayings)...I believe :D

Oktas8
20th Aug 2001, 02:20
To get a Load Factor of 1.0 in a 60° banked turn, your vertical component of lift is half the total weight.

To get a stable descent when L = 0.5W, your coefficient of drag will have to be substantially larger than coefficient of lift.

There are two ways to get CD higher than CL: fly very fast or very slow!

Very fast - probably above Vne. Oops. :p
Very slow: probably a stalled condition - spinning yes? :(

O8 :)

[ 19 August 2001: Message edited by: Oktas8 ]