Draining E in turns

[Sable]

Originally posted by Crumpp
Actually,

The Lower wingloaded plane DOES produce more drag.  For induced drag it depends on the Aspect ratio and the wip tip efficiency factor.


Lift-to-Drag = Low Wingloading = MORE REFERENCE AREA = more drag.

Reference area is used in calculating parasite drag, and isn't part of any formula for induced drag.  It isn't even a real value in most cases until the modern era as it is just too hard to determine the actual surface area of a complex object like an airplane when all you have is a tape measure and a slide rule.

Induced drag greatly depends on the ASPECT RATIO and WING TIP "e" factor.  More so on the AR (bigger number) than the "e" factor (usually 1 or less).

Low Wingloading = More lift = More induced drag

[/b]

Well I agree with the first part of this - Aspect Ratio and E factor are important.

But Low Wingloading doesn't mean more lift.  Guess how much lift a 9000 lbs plane has to create to pull 2 Gs?  18,000 lbs!  Doesn't matter whether it has the wings off a 747, or off a dodo.  Either way the same lift has to be produced to pull 2 Gs.

Here's a formula for lift:

CL x 1/2 rho V^2 x S (wing area)

Here's a formula for induced drag:

CL^2 / (Aspect Ratio x E factor) x 1/2 rho V^2 x S

that is

Coefficient of lift squared divided by Aspect Ratio times E factor time 1/2 rho (density of the air) times Velocity squared times wing area

Now look at that.  Supposing our two hypothetical aircraft are producing the same amount of lift from the first formula (both pulling 2 Gs for instance).  Now since the low wingloaded plane has a larger value for S, and 1/2 rho V ^2 is the same for both, then the higher wing loaded plane has a higher CL to reach the same lift.  This is illustrating that the higher wingloaded plane is pulling a higher angle of attack to achieve the same total lift.

Now if we go to calculate the induced drag notice that wing area (S) is still only counted once, but now CL is squared!  That means that a higher CL is going to disporportionally increase the result vs. the trade off for a high S value.  Again this is assuming Aspect Ratio and E factor are the same.

So all else being equal, the lower wingloading is going to produce less induced drag.

Instantaneous turn is NOT sustained turn.  A low wingloaded plane usually has a great sustained turn rate but that has nothing to with cornering speed and instantaneous turn.  In fact most (not all) low-wingloaded planes have low instantaneous turn speeds and are crappy high speed turners.  Just look at the Zeke.

Doubling the Velocity Quadruples the Force = ALL the Force = DRAG and LIFT.

[/B]

Actually the formulas for lift and induced drag don't care about instantaneous vs. sustained.  And the fact that a certain plane with a low wing loading has a poor ability to turn at high airspeed has nothing to do with it's wing loading.  The zeke sucks at high speed turns for two reasons.

1.  At 300mph for instance, the stick forces are so high that you can no longer pull the same number of G's as something like a P-51.  That's because of it's elevator and stick design.  If you gave it hydraulic elevator boost (and wore a G-suit) you could probably rip it's wings off with the amount of G's you could pull.

2. It loses speed like crazy in a hard turn at high speed.  This is because at 300mph on the deck a Zeke is near it's maximum velocity!  All of it's power is being used up overcoming parasite drag, so there is almost none available to overcome the addition of the induced drag from the turn.  The P-51 is making a lot more induced drag, but it has a ton of excess power at 300mph to help balance that.

Now all my discussion earlier was ignoring Aspect Ratio and E factor.  A high aspect ratio is good for reducing induced drag, and a taper wing giving you a high E factor is also good.

Comparing the P-51 and the Fw190 Godo mentioned earlier, they have nearly identical aspect ratios (6 for the 190, vs. 5.9 for the P-51).  Their wings have very similar taper, and fairly similar squared wingtips so their E factor is probably pretty close to even.  The two big differences between them are the P-51's lower wingloading, and much higher top speed (meaning it has much more excess power at higher speeds) as compared to a Fw190A-8.  As a result, the P-51 bleeds less E performing the same high speed, high G turn.

[Sable]

Originally posted by Crumpp
errr -yes.

Induced drag is a function of Aspect ratio and wing tip efficiency.
Lower wingloading produces a higher Cl too.


Lower wingloading will make up for it in increased Drag.  Not induced drag which is totally seperate.  Drag is a function of AREA.  To be a low wingloading plane usually means more surface AREA.  Induced drag is reduced with speed. HOWEVER, Drag quadruples with velocity.

Crumpp

No a lower wing loading doesn't change the Cl.  Cl is independent of wing loading, and varies based on angle of attack.  Like I just put in my previous post, a lower wingloaded plane has a lower Cl value at a given amount of lift.  

And the mistake you are making is assuming that Parasite drag (form and pressure drag) are impacted by angle of attack.  As the author of that AV8N site says the coefficient of parasite drag is essentially constant in relation to angle of attack.  He even has a chart of it right  here.
The coefficient of parasite drag only just starts to increase before you reach a stall.  Induced drag however increases dramatically with AoA.  That's why the lower wingloaded aircraft bleeds less E - it's pulling less AoA for a given turn!

[dtango]

Mandoble:

Regardless, the Su27 cobra maneuver isn't dependent upon some notion that a larger wing becomes an airbrake absent any other sophisticated engine, aerodynamic, and flight control systems.  

The high AOA and high Cl are functions of aerodynamic improvements and thrust available combined with nifty flight control systems to keep the Su27 from departing flight.  

It is not relevant to lift-dependent profile drag discussions as it relates to WW2 aircraft because the Su-27 super-maneuver would be outside the envelope of capable flight for WW2 fighters.

Your idea that there is some increase in profile drag in a turn is not incorrect.  The problem is that it's contribution is not as substantial compared to the impact of induced drag on the aircraft.  2ndly the change in profile drag with aoa is not dependent on the size of the wings.

There's some oversimplification in my statements but it's not worth the trouble to try and explain the details.

Tango, XO
412th FS Braunco Mustangs

[dtango]

Crumpp:

I'm sorry buddy.  You don't seem to know what you're talking about and are really confusing the issue.

Tango, XO
412th FS Braunco Mustangs

[Crumpp]

Seems to be alot of confusion on exactly what difference between a coefficient and a force and what is effected by what.

http://www.av8n.com/how/htm/4forces.html#fig-coeff-alpha

http://www.av8n.com/how/htm/4forces.html#fig-coeff-ias

http://www.av8n.com/how/htm/4forces.html#fig-power-ias

http://www.av8n.com/how/htm/4forces.html#fig-force-ias

If you conduct a LEVEL turn then you are cashing in your LEVEL Speed to maintain ALTITUDE.  Hence you are cashing in your energy NOT gaining it.

Here is what is going on in a turn:

http://www.av8n.com/how/htm/4forces.html#sec-defer-forces

Which does NOT havemuch in common with High G steep break turns we seen pulled in AH.

Now lets look at a steep turn:

Imagine that you are initially trimmed for straight and level flight at, say, 100 knots. Then you inadvertently enter a steeply banked turn. Figure 6.12 shows the forces acting on the plane in level flight and in the turn. Let’s imagine that the plane weighs exactly one ton. In level flight the downward force of gravity is exactly canceled by the lift produced by the wings, so the wings must be producing one ton of lift. In the turn, though, the wings must produce enough force not only to support the weight of the airplane (vertically), but also to change the airplane’s direction of motion (horizontally). The total force can be quite large: In a 60° turn, two tons of force is required. In a 75° turn, almost four tons of force is required, as shown in figure 6.13.

http://www.av8n.com/how/htm/aoastab.html#sec-spiral-dive

So far it looks pretty close to what your saying.  We can bank sharply and suddenly increase our SPEED!!!  Coooooollll!!

Wait?? Aren't we creating energy??  NO, Here is why!  It ONLY appears we are making a level Turn.  In fact we have lost lots of ALTITUDE.

http://www.av8n.com/how/htm/vdamp.html#sec-vertical-damping

We see that the vertical component of lift is insufficient to balance the weight of the airplane. (Compare this figure with figure 6.12 or figure 6.13.) The unbalanced force will cause the airplane to drop straight down. This is the same as the “albatross effect” discussed in section 5.2. Then, as soon as any appreciable downward velocity develops, the airplane will pitch down and speed up --- because the airplane wants to maintain its trimmed angle of attack11 as discussed in section 6.1.

Well now we have traded altitude for Airspeed just to bank.  If we maintain altitude then we trade speed for altitude again.  

Energy can niether be created nor destroyed.  You cannot perform a steep bank turn and not trade energy for altitude in one form or another.

In a turn a plane needs a MUCH higher Cl to maintain the turn.  It gets this by angle of attack AND by dropping the nose to gain speed.  It wants to speed up.  However if you drop the nose then you are exchanging altitude for Airspeed.  Hence you are cashing in your Energy NOT gaining it.

If you conduct a LEVEL turn then you are cashing in your LEVEL Speed to maintain ALTITUDE.  Hence you are cashing in your energy NOT gaining it.

It is either maintain your altitude and lose your speed.

Or

It is lose altitude and gain speed.

Can't do both, Physics won't allow it.

Crumpp

[Crumpp]

And the mistake you are making is assuming that Parasite drag (form and pressure drag) are impacted by angle of attack.

Looks like it goes up to me.

http://www.av8n.com/how/htm/4forces.html#fig-coeff-alpha

The problem is that it's contribution is not as substantial compared to the impact of induced drag on the aircraft.

Since an A/C has to gain speed by cashing in level speed or altitude in a steep bank I would say your wrong.

Force of drag goes up.

In a steep bank lower wingloaded A/C do not bleed less Energy.  Lower Wingloaded A/C burn more Energy at speed due to parasitic drag.

http://www.av8n.com/how/htm/4forces.html#fig-force-ias

Crumpp

[Sable]

Originally posted by Crumpp
Looks like it goes up to me.

http://www.av8n.com/how/htm/4forces.html#fig-coeff-alpha

Since an A/C has to gain speed by cashing in level speed or altitude in a steep bank I would say your wrong.

You do realise that the point where it starts to rise is right around the critical angle of attack for most airfoils?  That means that as the wing is stalling the parasite drag starts to increase as a results of angle of attack.  But we aren't talking about stalls, we're talking about energy loss in high speed turns.

And why does the aircraft need to accelerate?  We are talking about deceleration as a result of turning at high speed.

Force of drag goes up.

In a steep bank lower wingloaded A/C do not bleed less Energy.  Lower Wingloaded A/C burn more Energy at speed due to parasitic drag.

Crumpp [/B]

Yes the force of parasite drag goes up with airspeed.  So what?  We all know this.  That's why airplanes have a top speed.  And you are wrong about lower wingloaded A/C burning more energy at speed.  As I will show below parasitic drag doesn't change meaningfully in any turn that isn't right at a stall (and when it does increase, it inevitably affects the higher winloaded plane first as it will stall first at a given speed and load factor).

The original question was: why do some planes bleed energy faster in a high speed turn then others?  Lets go over this scenario again.  

Plane A: FW190A8 - higher wingloading, lower max speed
Plane B: P-51D - lower wingloading, higher max speed

Otherwise their wings and weights are pretty similar.

Both are at 50ft, traveling 350mph.  They both make a 3g turn.
Which one is going to bleed airspeed faster?

First: how much lift are they producing?

Lift = CL x 1/2 rho V^2 x S

Since the weight is similar, and they are pulling the same load factor, they are both making right around the same lift so we can say:  Lift (P-51) = Lift (Fw190)

Now the P-51 has a larger wing area (S) so in order for the values to be equal the Fw-190 needs a higher CL (since they are going the same speed in the same density of air 1/2 rho V^2 is the same for both sides of the equation).  The 190 pilot provides this higher CL by pulling a little more angle of attack too achieve his 3G turn.

Next: How much drag is being created by this extra lift?

Induced drag = CL^2 / (Aspect Ratio x E Factor) x 1/2 rho V^2 x S

Note that CL (the FW's value is higher) is getting weighted heavier then S (the P-51s value is higher) since it is squared in the induced drag formula.  So that means that the Fw190's induced drag is going to be higher assuming Aspect Ratio and E factor are the same (and as I pointed out in my previous post, they are pretty similar with these two aircraft).  So we know that the Fw190 is creating more induced drag in this turn.  Now what about parasite drag?

Parasite drag = CL x 1/2 rho V^2 x A

Now from the link here we can see that CL isn't going to change with angle of attack for either aircraft.  We aren't stalling so we aren't getting close to the 15-17 degree critical angle of attack where CL just starts to increase (and even if we were the increase would still be very small).  Neither aircraft has started decelerating yet so 1/2 rho V^2 is still equal.  And now we come to A, or the reference area.

Now the reference area isn't an exact measurement of the surface area of the aircraft ... like I said earlier that would be nearly impossible for anything but a modern supercomputer.  It's a guesstimate that is established at the same time as the CD from wind tunnel testing.  Here's a quote from "The Illustrated Guide to Aerodynamics" by HC Skip Smith:

"The customary procedure for bluff bodies (such as the fueselage, nacelle, or landing gear), is to use the projected frontal area, or maximum cross section.  For thin profile bodies, such as the wing or tail surfaces, the planform (top view) area is normally used."

As you can see the reference area has nothing to do with the orientation of the object, the CD is what would reflect changes to that, and as we can see from the link above it doesn't change meaningfully until after we've stalled the airfoil.

So parasite drag is unchanged at the moment we start the turn.

Finally: Total Drag vs. Excess Power

So now we can see that the Fw190's induced drag increased more then the P-51's and parasite drag was unchanged.  Just by that alone, we can see the Fw190 is going to decelerate faster.  Now consider Excess Power.  At 350mph at 50ft the Fw190 is at Vmax - it can't go any faster under it's own power.  Any manuever that requires more power then it has will result in immediate deceleration.  The P-51 on the other hand is still about 17mph below it's Vmax - it still has excess power to be used to accelerate, or climb, or overcome additional drag from turning.  So not only did the P-51 create less drag turning, it also has more power available to try and counter this additional drag.  Also consider that as load factor increases the higher wing loaded plane will eventually reach it's critical AoA (where the parasite drag that Crumpp mentions actualy starts to have some effect) before the low wing loaded aircraft.  So that is why an Fw190 bleeds more E then a P-51 in a high speed turn.

And before someone asks: Why does the Spitfire bleed so little E even though it's a fairly slow plane and doesn't have the excess power advantage of the P-51?

Because the Spitfire's wing loading is much lower, it's Aspect ratio is higher then either aircraft mentioned, and the E Factor of the elliptical wing is REALLY good.

[bozon]

Originally posted by Crumpp
Since an A/C has to gain speed by cashing in level speed or altitude in a steep bank I would say your wrong.

Force of drag goes up.

In a steep bank lower wingloaded A/C do not bleed less Energy.  Lower Wingloaded A/C burn more Energy at speed due to parasitic drag.

completly wrong from a to z.
I don't even know where to begin, the in-sentence contradiction, the wrong physics or the wrong aerodynamics.

Bozon

edit:
this much is true, but not necessarily:

Lower Wingloaded A/C burn more Energy at speed due to parasitic drag.

[Crumpp]

The original question was: why do some planes bleed energy faster in a high speed turn then others? Lets go over this scenario again.

I understood the original question.  

What I want to know is how can some planes in AH make 90 degree banks in a lag turn, and still have energy to spare??


They cannot.  

completly wrong from a to z.

And why does the aircraft need to accelerate? We are talking about deceleration as a result of turning at high speed.

really?

It's right here.

Imagine that you are initially trimmed for straight and level flight at, say, 100 knots. Then you inadvertently enter a steeply banked turn. Figure 6.12 shows the forces acting on the plane in level flight and in the turn. Let’s imagine that the plane weighs exactly one ton. In level flight the downward force of gravity is exactly canceled by the lift produced by the wings, so the wings must be producing one ton of lift. In the turn, though, the wings must produce enough force not only to support the weight of the airplane (vertically), but also to change the airplane’s direction of motion (horizontally). The total force can be quite large: In a 60° turn, two tons of force is required. In a 75° turn, almost four tons of force is required, as shown in figure 6.13.

We see that the vertical component of lift is insufficient to balance the weight of the airplane. (Compare this figure with figure 6.12 or figure 6.13.) The unbalanced force will cause the airplane to drop straight down. This is the same as the “albatross effect” discussed in section 5.2. Then, as soon as any appreciable downward velocity develops, the airplane will pitch down and speed up --- because the airplane wants to maintain its trimmed angle of attack11 as discussed in section 6.1.

 http://www.av8n.com/how/htm/aoastab.html#sec-spiral-overview

Now check out what speed does to drag:

http://www.av8n.com/how/htm/4forces.html#fig-force-ias

Consquently the amount of energy dissipation goes up:

http://www.av8n.com/how/htm/4forces.html#fig-power-ias

You do realise that the point where it starts to rise is right around the critical angle of attack for most airfoils?

Yes I realize that.  You said this:

As the author of that AV8N site says the coefficient of parasite drag is essentially constant in relation to angle of attack.

It goes up before the stall.  

http://www.av8n.com/how/htm/4forces.html#fig-coeff-alpha


Good explaination Sable.  Now how do you explain the FW-190A had the same turn radius as the P51B?  In the tactical trial the P51B had a "slight" advantage in the turn with the FW-190 test pilot admitting he could no get maximum turn performance.

An FW-190A8 outturned a P51D.

The info is down at the bottom.  

http://www.anycities.com/user/j22/j22/aero.htm

This is who did it.

http://www.anycities.com/user/j22/j22/lednicer.htm

This is where he works:

http://www.am-inc.com/

Guess he has one of those supercomputers.

Crumpp

[F4UDOA]

Crumpp,

His Mustang Cdo is to high.

Also his turn rate data seems strange. How can he show sustained turn rate being equal to instantanious? The best they should be is equal IMHO.

[humble]

So all else being equal, the lower wingloading is going to produce less induced drag. sable


I think this is an accepted truism in aviation engineering/design. tons of actual flight test data exist for all the planes modeled here. The horse died along time ago guys....

[Crumpp]

Also his turn rate data seems strange. How can he show sustained turn rate being equal to instantanious? The best they should be is equal IMHO.

Here is the chart explained:

Instantaneous turn performance is in direct proportion to the wing loading. Lower wing loading equals better instantaneous turn. Sustained turn performance is a combination of Wing loading and power loading. Best sustained turn performance is with a low wing loading and a low power loading. I have not taken stick force gradients and stability into consideration here.

His Mustang Cdo is to high.

I would trust his numbers until someone more qualified comes along.  Given the mans background as an aeronautical engineer and his access to much more powerful fluid dynamics computations than anyone is capable of making here.  This data was written for an article in Sport Aviation Magazine.  Not for some flight sim.  

Cd is a very salamanderly problem to calculate.

The drag coefficient contains not only the complex dependencies of object shape and inclination, but also the effects of air viscosity and compressibility.

http://www.grc.nasa.gov/WWW/K-12/airplane/dragco.html

http://www.grc.nasa.gov/WWW/K-12/airplane/airsim.html

And very importantly you have to know the reference area.
This is also why Lower wingloaded A/C generally speaking produce MORE DRAG (not induced drag which is a different animal).  More Surface = More Drag

I think this is an accepted truism in aviation engineering/design. tons of actual flight test data exist for all the planes modeled here. The horse died along time ago guys....

Doubling the area doubles the drag.

http://www.grc.nasa.gov/WWW/K-12/airplane/sized.html

You need access to engineering fluid dynamics to even come close.  

The Only guys calculations in this whole thing that matter though are HTC's.

The turn sable so nicely calculated is NOT the same dynamics as a steep turn.  No, A lower wing loaded plane could pull gentle 1 G turns all day long and not lose much E.  Crank it over 90 degrees and pull G's in the turn and EVERY A/C is burning Energy.  You will and should wind up with Less Energy than you started the turn with.  

Oversimplifying here so don't jump to your calculators.  Low Wing loaded or High wing loaded, they all pay the piper in one form or another.  Because of the speed increase in a sharp bank, parasitic drag increases and the Low Wing loaded planes pay the price.  Low speed maneuvering High Wing loaded = higher induced drag = crappy sustained turn rate and they pay the price.


Crumpp

[hitech]

Crump, there is no such thing as a 1g turn.
There is no such thing as a 90 deg turn.

The term Turn generaly is taken to generaly meen holding a constant altitude.

[Crumpp]

Crump, there is no such thing as a 1g turn.

There is a gentle low G bank as opposed to a steep high G bank.

Crumpp

[Crumpp]

http://www.av8n.com/how/htm/4forces.html#sec-defer-forces



http://www.av8n.com/how/htm/aoastab.html#fig-spiral-start

Crumpp