relationship between lift vector and AoA

[Floppy_sock]

Hi all,

I'm new here so if there's a better place to post this question please let me know.

Alright - this question is somewhat obscure but I'm curious. It evolved from trying to bound forces on the pilot while cornering in WW2 aircraft.

For the sake of discussion I'll use the 190 D-9 as an example. With flaps retracted it had a CLmax of 1.3 at 14 degrees aoa equating to 6.3g at 500km/h.

My question now is - how does that force act on the aircraft. More specifically, how does the lift vector point with respect to the reference axis of the fuselage as aoa increases. I think I've read that the 190 had an angle of incidence of 2 degrees. So at 14 degrees aoa does the reference axis now differ from the vector tangent to our trajectory by 12 degrees?

Does this explain why certain aircraft "felt" like they were pulling more g while pulling the same turn since some aircraft max perform at higher aoa than others?

Edit: I may have answered my own question. The lift vector always points perpendicular to the relative wind. So I think what I wrote above is correct? The AoA should have an effect on the +Gz experienced by the pilot.

 

[pbehn]

Hi welcome to the forum this is an interesting read http://www.wwiiaircraftperformance.org/mustang/Lednicer_Fighter_Aerodynamics.pdf

 

[pbehn]

Does this explain why certain aircraft "felt" like they were pulling more g while pulling the same turn since some aircraft max perform at higher aoa than others?

.

What a pilot "felt" changes from pilot to pilot and also any pilots tolerance of "G" changes for all sorts of reasons, even in the same plane. Seating position had a big effect. Some inexperienced 109 pilots were unnerved when the leading edge slats came out believing they had reached the limit. A pilots willingness to test the limits depends on the situation, test pilots tested turn performance at 20,000ft rather than 200ft for obvious reasons. Some believed that Douglas Bader had an advantage in high G turning because he had his lower legs amputated, no one copied his example as far as I know though. Some planes gave advanced notice of the wing stalling others just let go, so some were easier to take to the edge than others.

 

[drgondog]

Well - maybe to answer your question without diagrams - which you can look up. There are a lot of beginning Aero books - used, cheap.

The Lift and Drag forces are presented about the 25% aerodynamic Chord line, moved from the center of pressure of the wing. The Moment about the aerodynamic chord line is the 'pitching torque created by the resultant forces (top and bottom of the wing) at the two different Centers of Pressure (top and bottom). For convention the Two lift forces are combined into one 'positive vertical as Net Lift' and the Moment. The moment and Lift Vector are then presented at the 1/4 chord point of the wing Mean Aero Chord - and for conventional airfoils is a 'pitch down' moment' tending to force the nose down if not offset by a trim force from the elevator. The same applies to the empennage - with a vertical force (positive or negative) and pitching moment viewed at the aero center of the horizontal stabilizer/elevator combination.

The Gravity force vector is the Weight of the aircraft and is presented at the center of gravity - which is different from the aerodynamic or the center of pressure but close to both. The exact location relative to the aero forces are crucial in calculation of the stab/elevator sizing for pitch authority and stability considerations. For example, when the CG of the aircraft move forward past the actual net center of Pressure, the lift forces of the wing will force the aircraft nose down with far more authority than can be obtained by elevator down force to restore stable horizontal flight. Conversely, if a shock wave moves the center of lift aft - past the CG - same issue (P-38)

The restoring force required to maintain level flight arises from the center of pressure from the horizontal tail/elevator.

In straight and level flight the forces Lift vector of wing, the Lift vector (positive or negative) at the horizontal stab, Weight vector (vertical down) are presented in a side elevation, along with the Moment pitching torque vector about the same location as the Lift and Drag vector. The Drag force is always a vector perpendicular to the Lift and is parallel to the axis of 'travel' but opposite. The Thrust vector is parallel to the axis if 'travel' and forward - opposite of Drag.

Idealized for understanding, the Lift, Thrust and Drag vectors are presented as parallel and vertical to the aircraft centerline of reference, but in reality actual thrust vectors Not quite parallel to the centerline are designed slightly different from centerline - usually to optimize for cruise efficiency.

In banking flight, the Lift and Drag forces are represented for calculation purposes at the C/L of the aircraft when viewed from the front, but perpendicular to the wing. The Weight vector is always pointing down (to center of gravity of the earth). To obtain the Lift forces for level (but banking) flight, you must use trig to calculate the actual Vertical force (vector) applied directly against the force of gravity (weight). The resultant Side Force vector points to the origin of the turn. In basic calcs, neither Thrust nor Drag are necessary as they cancel each other for sustained velocity/altitude banking turn.

For climb - the aircraft angle of climb between the C/L and the direction of horizontal travel parallel to the earth, is the calculation angle (different from banking angle). Now viewed from the side, the force vectors are the vertical components (to the plane of the earth) of Thrust, drag, lift and weight. The horizontal components of Thrust, drag and lift - no weight vector in horizontal plane.

For real life? The Lift vectors at the center of Pressure are different, high wing - higher, low wing 'lower' tending to skew the CL axis from a circular travel - requiring a trim force obtained from the empennage (stick and rudder) to bring back to a circular path.

For climb, the drag forces from the empennage (negative - pushing elevator down) need to be looked at.

For banking, climbing turns, the necessary variables are far too complex to calculate all the necessary forces, in time sequence, with conventional algebra and trig

I haven't proofed this and not likely to - but hope it helps

 

[drgondog]

Hi all,





For the sake of discussion I'll use the 190 D-9 as an example. With flaps retracted it had a CLmax of 1.3 at 14 degrees aoa equating to 6.3g at 500km/h.

My question now is - how does that force act on the aircraft. More specifically, how does the lift vector point with respect to the reference axis of the fuselage as aoa increases. I think I've read that the 190 had an angle of incidence of 2 degrees. So at 14 degrees aoa does the reference axis now differ from the vector tangent to our trajectory by 12 degrees?

Does this explain why certain aircraft "felt" like they were pulling more g while pulling the same turn since some aircraft max perform at higher aoa than others?

Edit: I may have answered my own question. The lift vector always points perpendicular to the relative wind. So I think what I wrote above is correct? The AoA should have an effect on the +Gz experienced by the pilot.

Different questions. Angle of Attack relative to the aircraft orientation in contrast to the freestream impinging on the leading edge. The AoA is expressed as the angle between the freestream velocity/relative wind and the mean chord line from LE to trailing edge. The lift and drag forces I discussed above are perpendicular (lift) and parallel (drag) when integrated across the wing surface.
The Angle of Incidence is a refinement usually in reference to displacing the LE of that airfoil, down ward, so that the 'adjusted' chord line at a specific location spanwise is presented at slightly less angle of Attack to the freestream. Two degrees angle of incidence refers to a gradual droop of the LE from the Centerline Root Chord as it span increases from CL. Most such changes occurred from C/L all the way out to the tip. For an Fw 190 the increasing 'droop' extends only to the 80% chord line. The reason for such wash out is to delay the stall transition from inboard wing as long as necessary to provide continued roll authority via the un-stalled tip/aileron part of the wing.

The answer to your question is that at 14 degrees the CL 'breaks' in the straight CL vs alpha curve - stall point. With the conditions you described, the inboard lift regions reach stall at the root area first. The Angle of attack (Relative) to root, at the tip) is 12 degrees with a two degree washout. Not yet stalled.



The angle of incidence is more often the relative angle of the horizontal chord line to the centerline of the aircraft. It is subjected to the abrupt down wash over the wing and often lives in the world of 'negative angle of attack to the relative wind.

 

[Floppy_sock]

I appreciate the time you took to respond. I'm not sure if that gets me closer to the answer. I'm not interested in sustained climb/turn diagrams.

Our goal was to try to understand if we could get an upper bound on the G force an aircraft can exert on its pilot. Obviously - the theoretical maximum is the lift generated at CLmax - which is of course an overestimate since it doesn't include the effects of the fuselage etc on the airflow over the wing. If I understand your first response correctly, that answer isn't readily attainable with basic aero relations.

 

[pbehn]

I appreciate the time you took to respond. I'm not sure if that gets me closer to the answer. I'm not interested in sustained climb/turn diagrams.

Our goal was to try to understand if we could get an upper bound on the G force an aircraft can exert on its pilot. Obviously - the theoretical maximum is the lift generated at CLmax - which is of course an overestimate since it doesn't include the effects of the fuselage etc on the airflow over the wing. If I understand your first response correctly, that answer isn't readily attainable with basic aero relations.

That is a completely different question. The maximum G Force a plane can exert in an instantaneous turn in a high speed dive is governed by the structure of the plane and its condition. The maximum sustained turn is much lower and limited by the power and drag. This is why turning fights almost always descend to the ground, you can pull higher G if you trade height for energy.

 

[drgondog]

I appreciate the time you took to respond. I'm not sure if that gets me closer to the answer. I'm not interested in sustained climb/turn diagrams.

Our goal was to try to understand if we could get an upper bound on the G force an aircraft can exert on its pilot. Obviously - the theoretical maximum is the lift generated at CLmax - which is of course an overestimate since it doesn't include the effects of the fuselage etc on the airflow over the wing. If I understand your first response correctly, that answer isn't readily attainable with basic aero relations.

Understood - that said you need the basis for V-n diagrams to get there from 'here' even in approximate format. If you are an aero or very familiar with the math behind the V-n diagrams I apologize in advance,

The instantaneous G force is readily attainable by posing bank angles. Instantaneous and Sustained G forces require that the aircraft doesn't break up and that the G force sustained doesn't exceed the structural limits - as shown in a V-n diagram for your specific aircraft.

There is a reasonably accurate equation for circular flight that does enable you to calculate and plot such V-n diagram up through the range of structural failure for a solid, impervious, non- deforming object that doesn't yield, and doesn't fail. The equations are developed assuming perfect circular flight near stall, with accurate CLmax in that region for that aircraft in perfect wind tunnel condition. Is that what you want? That is actually what airframe manufacturers calculate for you to present in Pilots Handbook.

The boring part
n=L/W : for steady level flight n=W/W=1; for accelerating flight n=L/W as in climb or turn where the Lift is greater than the Weight. In curvilinear flight
F=Sqrt(L^2 - W^2) directed to center of curve for the perfectly flown aircraft. but now with n=L/W, F=W*Sqrt(n^2-1)
now from Newton F=m*V^2/R = (W/g)*V^2/R

R=V^2/(g*sqrt(n^2-1)

To simplify the resultant equations below for high G capable aircraft, the equations of motion require that you assume that n+1=n and n-1=n to render the equations to straightforward solution without progressive iterations; "n" in this case is the upper positive and negative load factor. E.G. 7.3 G + 1 ~ 7.3G and -3.0 G-1~ 2.3G. (Limit G for P-51H at 9600 GW)

The equations for turn radius R= V^2/g(n-1) and turn rate w= g(n+1)/V for circular flight path. Assuming the simplification above R=V^2/g*n and w=g*n/V where g=force of gravity, n= load factor, and V= flight velocity.

Going further to move from a ball rotating around a string to a theoretical airplane, the equations of Lift relating to velocity V ;
L=1/2*rho*V^2 *S*CLmax and then V^2=2*L/rho*S*CLmax

Further minimum turn R is now = 2 * L/(rho*g*CLmax*S) ------> R=(L/S)*2/(rho*g*CLmax)

Finally nmax=L/W=(1/2*rho*V^2*S*CLmax)/W = (1/2*rho*V^2*CLmax/(W/S))

and,

Vmax = Sqrt((2*nmax)*(W/S)/(rho*CLmax)) -------------> substitute n in progressive G load factors for each velocity until the Yield Stress Limit Load is reached and you have your V-n plot. On the left side is 'stall', on the right side is 'no problem'.

Reality - upper wing usually stalls first, the relative 'real freestream' when chasing something in front - is turbulent with fluctuating 'relative air impinging on your wing'. Your propwash on the inner third of your wingspan adds to the fluctuating relative air on the wing and in 'reverse' (depending on the prop rotation) from one wing to the other. Stall in a high g manuever is of course never close to symmetrical so calculating CLmax for turn is Not accurate when using published stall speeds for 'clean' aircraft.

Net, I'm not sure that even the most sophisticated, currently available, aero/aerodynamic models using CFD with Navier Stokes subsets to predict pressure distributions and boundary layer separation would accurately reflect 'real life'. You are looking at Chaos Theory to determine probability distributions of flow separation/re-attach behavior - which is way out of my pay grade.

 

[Peter Gunn]

Bill: "There is a reasonably accurate equation for circular flight that does enable you to calculate and plot such V-n diagram up through the range of structural failure for a solid, impervious, non- deforming object that doesn't yield, and doesn't fail. The equations are developed assuming perfect circular flight near stall, with accurate CLmax in that region for that aircraft in perfect wind tunnel condition. Is that what you want? That is actually what airframe manufacturers calculate for you to present in Pilots Handbook.

The boring part
n=L/W : for steady level flight n=W/W=1; for accelerating flight n=L/W as in climb or turn where the Lift is greater than the Weight. In curvilinear flight
F=Sqrt(L^2 - W^2) directed to center of curve for the perfectly flown aircraft. but now with n=L/W, F=W*Sqrt(n^2-1)
now from Newton F=m*V^2/R = (W/g)*V^2/R

R=V^2/(g*sqrt(n^2-1)

etc. etc. etc..."

Me:

 

[BiffF15]

Bill: "There is a reasonably accurate equation for circular flight that does enable you to calculate and plot such V-n diagram up through the range of structural failure for a solid, impervious, non- deforming object that doesn't yield, and doesn't fail. The equations are developed assuming perfect circular flight near stall, with accurate CLmax in that region for that aircraft in perfect wind tunnel condition. Is that what you want? That is actually what airframe manufacturers calculate for you to present in Pilots Handbook.

The boring part
n=L/W : for steady level flight n=W/W=1; for accelerating flight n=L/W as in climb or turn where the Lift is greater than the Weight. In curvilinear flight
F=Sqrt(L^2 - W^2) directed to center of curve for the perfectly flown aircraft. but now with n=L/W, F=W*Sqrt(n^2-1)
now from Newton F=m*V^2/R = (W/g)*V^2/R

R=V^2/(g*sqrt(n^2-1)

etc. etc. etc..."

Me:
View attachment 556830

This is one of those times where a "double bacon" option needs to be added...

 

[rochie]

Bill: "There is a reasonably accurate equation for circular flight that does enable you to calculate and plot such V-n diagram up through the range of structural failure for a solid, impervious, non- deforming object that doesn't yield, and doesn't fail. The equations are developed assuming perfect circular flight near stall, with accurate CLmax in that region for that aircraft in perfect wind tunnel condition. Is that what you want? That is actually what airframe manufacturers calculate for you to present in Pilots Handbook.

The boring part
n=L/W : for steady level flight n=W/W=1; for accelerating flight n=L/W as in climb or turn where the Lift is greater than the Weight. In curvilinear flight
F=Sqrt(L^2 - W^2) directed to center of curve for the perfectly flown aircraft. but now with n=L/W, F=W*Sqrt(n^2-1)
now from Newton F=m*V^2/R = (W/g)*V^2/R

R=V^2/(g*sqrt(n^2-1)

etc. etc. etc..."

Me:
View attachment 556830

and me

 

[Floppy_sock]

I study compressible flow / dynamical systems as a career so I'm much more familiar with NS and chaos than I am with the basic aero equations (which I know next to nothing about).

Unfortunately that's of no use to me here since I don't have a personal CFD package nor the computational power that I'd need to model any high Re 3-D flow

Now that I read it again, my second response was terribly unclear and, for that matter, rude - my apologies (I really appreciate your in depth responses). What I meant to say was - there's no way of getting the direction of the lift force relative to the reference axis of the aircraft from basic aero.

My thinking was - if the lift vector is not colinear (or parallel) with the Z axis of the aircraft while producing max lift, then the pilot experiences only the projection of the lift force on the Z axis. My intuition tells me it the two should differ by the AoA - but that's only intuition.

If the above line is correct - that would explain why different aircraft exerted more G on their pilots than others. The 109 - for example - produced max lift at a relatively large ~20 deg AoA and had a seat reclined at about ~25 deg w.r.t the reference axis of the aircraft. Compare that to the p51 which had a seat angle of 13 deg and max performed at 15 deg AoA. The projection of the lift force onto the Z axis of the pilot is substantially different between the 2 (about 15% difference).

However - now that I'm writing this - I think I also need to consider the drag + thrust to get the total force on the aircraft - since drag will tend to swing the force vector rearward - i.e. closer to the vertical axis of the pilot.

 

[soulezoo]

I stayed at a Holiday Inn Express last night...

And I'm lost.

 

[Zippythehog]

Weighty

But everything I didn't understand about aerodynamics was (and is) explained by Doc Hurt here
https://www.faa.gov/regulations_policies/handbooks_manuals/aviation/media/00-80t-80.pdf

That said, you'll find one error in the forces of a turn. Hurt states that the result of an unbalanced force is an acceleration. That is true. Unfortunately, and this is true of some FAA pubs, Horizontal Component of Lift (inward turning force) is not countered by centrifugal force. The force we describe as centrifugal force is actually inertia as the matter that is making the arc of the turn wants to depart that arc and travel straight.

Interesting that speed doesn't effect that G loading. The angle of bank does. So a Cessna 150 at 60 degrees of bank pulls two Gs just like the Fw-190D. The Cessna flys slower so has a smaller radius of turn.

I don't know if I've contributed or muddied the waters but the tome in the link would be a great read for you.
Zip

 

[Floppy_sock]

Zippy - awesome link! Thank you!

I'm not quite sure what you mean by error in your second paragraph. Centrifugal/centripetal forces "fictitious" in the sense that they only appear as a consequence of moving to a non-inertial reference frame moving with the rotation but I'm not sure how that plays a part here.

In your second point I think you mean speed does not affect g loading in a sustained turn at a specific angle of bank since we only need a lift force sufficiently large to balance the weight of the aircraft and no more aka at higher air speeds your aoa can decrease to sustain the turn. This is speed independent. However, maximum lift is certainly a function of velocity - a second order relationship at that.

 

[Zippythehog]

Glad you enjoy the link.
And
Yup

 

[Peter Gunn]

This is one of those times where a "double bacon" option needs to be added...

Buddy you're too kind, just happy to bring a little levity to this weighty discussion which is about as far over my head as a Sabrejet.