Issues still need tobe addressed in BOS

[unreasonable]

The last two points just so that you have more things to think about.

 

Just what we need

 

Seriously though, I agree: when you want to dive fast why would you want lift ? (Except maybe to dive inverted?)

[6./ZG26_5tuka]

Not that Im a serious contributer to the topic but how does keeeping 0G limot lift to 0? Woukd really like to hear an explanation on that.

 

And yes AoA does infact chabge both effective lift and drag of an airfoil. Lift-Drag polars indicate this influrence in form of graphs. Also, each wing has a posotive and negative critical AoA at which lift starts decreasing rapidly with drag increasing rapidly. Usually, when exeeding the crit AoA, aircraft are prone to stalling.

 

WW2 aircraft also possesed wing inclinations of a few desgrees. That meant that even when the aircraff was flying perfectly level the wings kept a small positive AoA against the horizonal airstream.

 

There are even more factors like V-shape, wing size and airfoil confugurations to add to the lift-drag discussion but it would make things way more complicated than they are already.

Edited June 29, 2015 by Stab/JG26_5tuka

[Marauder]

When lift = 0, you've got no lift. So you're free falling. Which means 0g. I've got no idea what's to explain, it's pretty self explanatory?

 

Lift is what creates induced drag, not angle of attack. It's possible to determine a lift - angle of attack relation or a drag - angle of attack relation for any object, but "lift induced" drag still remains lift induced, not AoA induced.

 

It's true that nearly all WW2 aircraft had their wings installed at a positive angle along the aircrafts longitudinal axis, but an aircraft flying level at high speed could well be flying with a negative angle of attack at an even higher negative 'pitch' (for a lack of a better word, the angle of the aircrafts longitudinal axis vs. flight path). For instance, Cl at 0 AoA for the fairly popular ClarkY airfoil (Yak for instance has them slightly modified) is around 0.35. With a Clmax of around 1.4 and a stall speed of 150, you'll be flying negative AoA at speeds of 300 and above.

Edited June 29, 2015 by Marauder

[unreasonable]

Maybe just a definitional issue here: looking at my idiot's guide "Understanding Flight", it makes a distinction between geometric AoA and effective AoA.

Zero effective AoA they define as the AoA which generates no lift. So a zero effective AoA will often (always? unless the wing is upside down?) be a negative geometric AoA.

 

Every book I look at seems to define these terms in a slightly different way   

[Dakpilot]

Don't worry, I think if you search for the names of the guys who did/do the FM's, they are rather well qualified in Aeronautical engineering

 

Cheers Dakpilot

[unreasonable]

Has anyone suggested otherwise?

[6./ZG26_5tuka]

I think we have 2 definitions of lift here. In the previous post you said if one entered a dive with the aircraft being effected by (constant) 0G lift would be 0 as well.

 

Thats why I asked for an explaination, namely to avoid misinterpretations. As I understand it you described "lift" as climb distance, am I correct?

I think we have 2 definitions of lift here. In the previous post you said if one entered a dive with the aircraft kept at 0G lift would be 0 as well.

 

Thats why I asked for an explaination, namely to avoid misinterpretations. As I understand it you described "lift" as the substraction of lift force - downpush = 0 in that particular situation?

 

At unreasonable Zero effective AoA accounts for the airfoil and usually is a very close to 0. As your airfoil is angled on the aircraft however flying level with the airframe being horizontal does mean you wings AoA is higher than 0 because they're angled. In order to fly with an AoA of 0 you'd infact have to push the nose down as many degrees as the wing is technicly angled at so the airfoil is "leveled".

Edited June 29, 2015 by Stab/JG26_5tuka

[unreasonable]

Two ways "lift" is used in the books seem to be:

 

1) The actual total force created by the flying wing

 

2) The vertical component of the force (1).

 

Marauder is talking about 1, I believe. If 1 is zero, you are not actually flying, you are in the same position as a rocket, so g=0 as experienced by the pilot. The effective AoA is zero: ie the wing produces no lift at all.

 

Obviously if you are producing zero lift, the vertical component is also zero, so you will accelerate towards the earth in a ballistic trajectory.

[6./ZG26_5tuka]

That part is sure but in his initial post both were put in wrong order. Anyway, I think we might settle it on what you said to avoid further confusion.

 

(Man this forum doesnt like phone typing )

[Marauder]

Lift is the force perpendicular to the path of the airflow, i.e. perpendicular to the flying direction in this case. In general, g's are also given in that direction, even though acceleration also exists in other directions. So, my statement explained: 0g dive - no acceleration perpendicular to the flight path. No force perpendicular to the flight path. No lift. I hope this clears up this potential misunderstanding.

 

And in order to avoid two more: I meant to say in my post on the top of this page "...no difference between the heavy and the light aircraft in terms of induced drag..." and "...because the penalty of higher induced drag is offset...". Sorry, no edit function.

[6./ZG26_5tuka]

So, my statement explained: 0g dive - no acceleration perpendicular to the flight path. No force perpendicular to the flight path. No lift. I hope this clears up this potential misunderstanding..

 

That's where the problem lies. If you push the plane to a 0G attitude the wings still produce lift. The lift force however is countered by the downpush induced by the pilot, which in your example equals the value of the lift force. If both are interacting parrelel to each other the effective force lifting the aircraft is 0, although lift is not.

 

Sry for the slight distraction.

[Marauder]

Care to share your source and the definition for the "downpush" force? The pilot induces it how?

 

As far as flying as we know it is concerned, the pilot changes the attitude of the aircraft so that the wings do no longer create lift.

[6./ZG26_5tuka]

I finally lost you there... Time to let the horse die I guess.

[unreasonable]

I finally lost you there... Time to let the horse die I guess.

 

Horse revival time....Take a wing, any wing. You know it has some maximum AoA at which it will stall. When stalled it produces no lift.

 

Reducing the AoA from that stall point allows the wing to fly - it produces lift.  Continue to reduce the AoA and you eventually get to a point that it produces no lift. This will be zero effective AoA, some small negative number of geometric AoA for a normal wing. Just like a fin on a rocket has no lift.

 

If you continue to rotate the wing so that the AoA is strongly negative, the wing is flying again, but you had better roll inverted because the direction of lift has reversed, which is why you can fly level upside down.

 

Of course you can dive while still producing lift - the point is that finding the AoA at which lift is zero, by definition eliminates induced drag.

Edited June 29, 2015 by unreasonable

[Marauder]

All true, unreasonable, simply put and well explained, but of course a wing still produces lift when stalled. Just less than at stalling AoA.

[unreasonable]

OK I think I have it - the lift increases with AoA at a given speed, then starts to reduce at the stall point and then with some more AoA the airflow will become completely chaotic and the lift will eventually give out.

 

Anyway, appreciate your patience!

[JG13_opcode]

For clarity, the industry term for the angle at which the wings are installed relative to the horizontal plane of the aircraft is called "Angle of Incidence". So when you fly around in your Cessna in level flight, the wings remain at some small angle of attack like 3 degrees or so. This is the angle of incidence.

[unreasonable]

We are all well OT by now, but if some people who know these things are still reading perhaps they can help me with another puzzle.

 

I have been scribbling the force vector diagrams and there is one that stumps me entirely.

 

Normally you can just draw W straight down, L perpendicular to the flight path and T and D parallel to the flight path, to make a quadrilateral, or W, L and T-D to make triangles.

 

(BTW flight books for dummies really should not start with that diagram) -

 

 

It is far too easy to look at that and fall into the trap of assuming that L=W and T=D in all circumstances, when this is just a special case. This confused me for a while.

 

Once I had worked out that vectors W+L+T+D = 0, and that level flight is a special case, drawing the diagrams is easy and you suddenly see how gliders fly without thrust - at least "downhill", and why it is impossible for them to fly "uphill" at a constant speed.  

 

The one that I do not get is for vertical flight - up or down. In this case, T, D and W are aligned vertically by definition. So the vectors can only sum to zero if L=0: no other vector has a horizontal component. So is there another force that is not captured in the equation? Or is it in fact true that planes in vertical flight paths cannot produce lift?

 

 

 

 

[6./ZG26_5tuka]

As long as the airfoil huts the incoming air at an AoA below the crit AoA it produces lift, or better a force. The vector of this force however changes with the angle of the aircraft.

 

If we take the easiest example (wing with 0° incidence) pulling the aircraft into a 90 ° climb meams the lift force vector has a max. horizontal component, the vertical component is 0.

 

Talking only about vertical forces your effective lift force would equal zero. So it's basicly only Ft - Fd - Fg (T - D - W in your terms).

 

Talking about an moving aircraft with a defined mass you have to add the inertia momentum as another upward facing vector. The simplified vector definition above does realisticly only account when the aircraft has no vertical velocity ( v=0).

Edited June 30, 2015 by Stab/JG26_5tuka

[Dakpilot]

We are all well OT by now, but if some people who know these things are still reading perhaps they can help me with another puzzle.

 

I have been scribbling the force vector diagrams and there is one that stumps me entirely.

 

Normally you can just draw W straight down, L perpendicular to the flight path and T and D parallel to the flight path, to make a quadrilateral, or W, L and T-D to make triangles.

 

(BTW flight books for dummies really should not start with that diagram) -

 

Liftvextors.png

 

It is far too easy to look at that and fall into the trap of assuming that L=W and T=D in all circumstances, when this is just a special case. This confused me for a while.

 

Once I had worked out that vectors W+L+T+D = 0, and that level flight is a special case, drawing the diagrams is easy and you suddenly see how gliders fly without thrust - at least "downhill", and why it is impossible for them to fly "uphill" at a constant speed.  

 

The one that I do not get is for vertical flight - up or down. In this case, T, D and W are aligned vertically by definition. So the vectors can only sum to zero if L=0: no other vector has a horizontal component. So is there another force that is not captured in the equation? Or is it in fact true that planes in vertical flight paths cannot produce lift?

 

 

Some good reading and interesting accessible  resources for further study if people are interested

 

https://www.grc.nasa.gov/www/k-12/airplane/weight1.html

 

https://www.grc.nasa.gov/www/k-12/airplane/fwrat.html

 

http://www.pilotfriend.com/training/flight_training/aero/man_force.htm

 

Cheers Dakpilot

Edited June 30, 2015 by Dakpilot

[ZachariasX]

Maybe just a definitional issue here: looking at my idiot's guide "Understanding Flight", it makes a distinction between geometric AoA and effective AoA.

 

Zero effective AoA they define as the AoA which generates no lift. So a zero effective AoA will often (always? unless the wing is upside down?) be a negative geometric AoA.

 

Every book I look at seems to define these terms in a slightly different way   

 

The AoA at which a wing/profile creates 0 lift depends not always on the angle being 0 degrees as well, but the profile itself. Having the profile chord at AoA 0 degrees might well produce lift. In fact, on some profiles you have to tilt the chord significantly <0 degrees depending on the cambering and  airspeed. The offset of the wing also turns an assessment of "what is 0 degrees AoA" into a discussion...

 

In all, I think the argument in this thread is consistent, as a basic rule it is obvious that added weight requires higher AoA with resulting drag. The devil is however in the detail. Induced drag (the mian culpit that makes you use more throttle with added weight at cruise speed and within weight limits) strongly dependent on the wing shape as much as on the added weight. This is why an old Boeing 767 suddenly has less drag and can fly more efficient with added winglets compared to the original design. Thus, if you add winglets, you suffer less from the same added weight, even though you have the same profile. And the formula just takes the profile into account, and not the entire wing.

 

My main point was to express that the last 5% or so in difference in performance potential might come from a rather unexpected source. This has nothing to do with the the aforementioned relationship of weight and AoA being inherently correct of course.

 

Personally speaking, if a combat sim leaves out all thos subtleties and just goes with the AoA formula, that wouldn't leave me too unhappy, if the freed ressources could be used for other things. There needs to be headroom for a good DM as well...

[unreasonable]

I understand about the AoA issue:  effective AoA not = geometric AoA (or angle of chord profile, or whatever you want to call it). I do not think this has anything to do with AoA as such.

 

The problem is simply that summing T+D+L+W, when T and D are oriented vertically, cannot have L with a horizontal component.

 

Summing these is the condition of zero acceleration, ie constant speed. So imagine an plane with retractable wings diving vertically towards the earth with wings fully retracted. No lift - no wings, sum T+D+W at terminal velocity and you get zero. No problem.

 

Let us assume the AoA of the plane's wings is variable, the wings can be rotated. The wings are opened and aligned at an angle to the wind to generate some lift. The lift is a force perpendicular to the line of flight. So it has an x axis component. Now T+D+W+L is not = 0.

 

The only way I can see to get it there, since W has no x component, is that T and D fall into alignment to complete the triangle. But that means, if T and D are by definition parallel to the glide path, that the plane is no longer in a vertical descent.

 

  I will read all those links, thanks. 

[unreasonable]

Thanks Dakpilot, I had seen a couple of those before, but not seen the key clue.

 

OK, again the problem is the books give oversimplified descriptions that only apply to special cases.

 

In this case that the angle of T and D vectors = climb/glide angle. I thought this was by definition, but it seems to be merely an approximation.

 

So it is possible to fly in a vertical climb/dive angle with lift, so long as your T-D line is offset to compensate.

 

Panic over.

Edited June 30, 2015 by unreasonable

[SR-F_Winger]

So is there a verdict that can be concluded?

 

EDIT: Was that stupid english?

Edited June 30, 2015 by VSG1_Winger

[6./ZG26_5tuka]

And that exactly is the issue with simplyfied models. They're more for people barely getting in touch with the physics rather than ones that like to dig into it.

 

It's a matter of how you treat lift. In the "simplified" exmaple it's only the vertical lift force component while in reality (as you said) the total lift force is perpendicular to the wing's chord (and the aircraft if incidence is 0°).

Edited June 30, 2015 by Stab/JG26_5tuka

[unreasonable]

So is there a verdict that can be concluded?

 

EDIT: Was that stupid english?

Depends what proposition you thought was on trial

 

Proposition 1: that the FM is guilty of grievous mistakes in its handling of dives and zooms in respect of the Fw190 and Yak - verdict Not Proven

 

Proposition 2: that a heavier aircraft has an advantage in dive and zoom at speeds where thrust is no longer accelerating the aircraft, other things being equal - Affirmed, provided the aircraft has unloaded to eliminate lift.

 

Proposition 3: that the disadvantage of induced drag will outweigh any weight advantage in situations where lift is being produced - Not proven, since we have yet to see any quantification of the trade-off. Solving "the equation" for D when a is fixed might help here: more homework. Clearly there will be some situations were this is true and other where it is not true.

 

Proposition 4: that any of this is tactically relevant for the purposes of a Fw190 or Yak jockey in the context of flying in BoS - Still unresolved. I think it is: many Fw190 pilots are overestimating how much extra distance or speed a dive away will give them, so they are surprised to be caught. Whether the FM is spot on or not, they need to have a clear understanding of what to expect if they try to outdive or out zoom Yaks. I suspect the dives are too short for the benefits of weight to make a big difference.

 

Proposition 5: that the Fw190 is at a huge disadvantage because it is fighting between 2km and 4km altitude in the wrong supercharger gear much of the time...that is a whole different story, as of now merely a hypothesis with which I am entertaining myself. I do not think it would affect Ze_Hairy's observations as he started his tests at 2km IIRC.

[JtD]

...So it is possible to fly in a vertical climb/dive angle with lift, so long as your T-D line is offset to compensate.

Drag by definition is parallel to the direction of the airflow, in the same way as lift is perpendicular. Therefore the only way to produce a steady vertical flight through stationary air would be to have the thrust line offset. This, btw., was in fact the case with many of the WW2 aircraft, they had their engines slightly angled upwards, even providing a little bit of lift even when having the datum line parallel to the flight path. I think that on the Bf109, both the engine and the wings were angled upwards at about 2°.

[unreasonable]

Drag by definition is parallel to the direction of the airflow, in the same way as lift is perpendicular. Therefore the only way to produce a steady vertical flight through stationary air would be to have the thrust line offset. This, btw., was in fact the case with many of the WW2 aircraft, they had their engines slightly angled upwards, even providing a little bit of lift even when having the datum line parallel to the flight path. I think that on the Bf109, both the engine and the wings were angled upwards at about 2°.

I had finally worked out ie that T line and D line are not necessarily the same - initially confused by oversimple approximations in my idiots' guides.

 

But just to be clear, the thrust vector offset has to be on the opposite side of a vertical flight line to counter the lift vector and zero all the vectors. In a vertical steady dive with lift the plane would have to be somewhat tucked.

 

 This is all purely theoretical, I cannot see why anyone would want to achieve this state of flight in a fixed wing aircraft, but it is nice to have the theory right at last.

[RydnDirty]

I remember thinking it was a little counterintuitive the first time I heard that the heavier plane will usually out zoom climb the lighter plane.  Zoom climb being high speed dive followed by a high speed climb that is slowly decelerating.

 

It helps to remind myself that kinetic energy is not linear... Ke quadruples when you double speed.

 

So when we have our yak1 chasing our FW190 at 4000m and they both dive at vertical, the heavier FW accelerates quicker. As they reach the bottom of the dive the FW heavy construction can reasonably do 800+km/h IAS whereas the Yak has only reached 700km/h. He cannot exceed it anyway.   

 

They both pull out and the FW is carrying more Kinetic energy than common sense would suggest.  Ke is not linear. The higher inertia of the  Fw pushes through the drag and resists the deceleration of gravity better than the lighter Yak.  Gravity decelerates them both at 9.8m/s. It doesn't decelerate the heavier plane at 20m/s.

 

The key to a zoom climb is that  it comes after a high speed dive.  The heavier plane is going faster and carrying more Ke into that zoom climb and they both decelerate at 9.8 m/s in the climb plus drag. 

Edited July 20, 2015 by WillyZurmacht