In several threads, including A380(NEO) threads, there have been discussions regarding the relation between wing aspect ratio and span loading (for a given weight, span and wing loading these are proportionate) on the one hand and induced drag on the other hand.

Conflicting information can be found on the subject:

https://en.wikipedia.org/wiki/Lift-to-drag_ratio


https://leehamnews.com/2017/07/05/a380p ... -analysis/

The low effective wingspan means the aircraft flies with a high induced drag.

http://www.wainfan.com/wingdes.pdf pages 30, 34 or 36




Would an aero expert be so kind to describe, in blunt terms, the relation (if any) between wing aspect ratio and span loading on the one hand and induced drag on the other hand? I understand it is not very straightforward as it is also related to the lift distribution across the wing.


I reposted this in the technical forums, so we don't need to drag other threads off-topic and this can be discussed in constructive manner.

Please keep the thread civil and on topic please. Thank you.

A drag polar is the sum of profile drag and induced drag. Simply put, induced drag coefficient is the lift coefficient squared divided by the product of pi, aspect ratio and e. E is the span efficiency factor. That may be what you mean by span loading. E is always equal to or less than 1. For a perfectly elliptical lift distribution, e=1.

Lpbri wrote:

A drag polar is the sum of profile drag and induced drag. Simply put, induced drag coefficient is the lift coefficient squared divided by the product of pi, aspect ratio and e. E is the span efficiency factor.

Thank you. I'm asking because we were having a discussion whether or not weight (as in design weight) could have an influence on the induced drag coefficient or not. I would argue, for instance in the case of the A380 and a possible wing redesign, that at a given wing loading and span, a reduction in design weight will result in a higher aspect ratio and thus lower induced drag. Could you confirm whether or not this to be the case?

Lpbri wrote:

That may be what you mean by span loading.

I meant the relation between weight and wingspan, which with a fixed span and wing loading is directly related to aspect ratio.

Lpbri wrote:

E is always equal to or less than 1. For a perfectly elliptical lift distribution, e=1.

Slightly off topic, but is elliptical lift distribution still the optimum? There seems to be a move to non-elliptical lift distributions which more strongly reduce lift towards the wing ends which leads to weaker wing end vortexes. In this approach the AR can be increased because the reduced force of the lift at the wing ends reduces the bending moment on the wing root compensating the extra weight of the aspect ratio extension. For instance: http://www.dept.aoe.vt.edu/~mason/Mason ... SemOct.pdf or: http://hdl.handle.net/2060/20160003578

I have no idea what the view on that is with regards to commercial airliners in aero professional circles.

Taxi645 wrote:

Thank you. I'm asking because we were having a discussion whether or not weight (as in design weight) could have an influence on the induced drag coefficient or not.

Weight will affect induced drag, since weight affects the amount of lift required.

Taxi645 wrote:

I would argue, for instance in the case of the A380 and a possible wing redesign, that at a given wing loading and span, a reduction in design weight will result in a higher aspect ratio and thus lower induced drag. Could you confirm whether or not this to be the case?

A reduction in design weight doesn't automatically mean a higher aspect ratio. Aspect ratio is a property of the wing geometry (span^2 / area).

vikkyvik wrote:

Taxi645 wrote:

Thank you. I'm asking because we were having a discussion whether or not weight (as in design weight) could have an influence on the induced drag coefficient or not.

Weight will affect induced drag, since weight affects the amount of lift required.

Taxi645 wrote:

I would argue, for instance in the case of the A380 and a possible wing redesign, that at a given wing loading and span, a reduction in design weight will result in a higher aspect ratio and thus lower induced drag. Could you confirm whether or not this to be the case?

A reduction in design weight doesn't automatically mean a higher aspect ratio. Aspect ratio is a property of the wing geometry (span^2 / area).

But it does proportionally affect aspect ratio if you keep span and wing loading constant.

vikkyvik wrote:

A reduction in design weight doesn't automatically mean a higher aspect ratio. Aspect ratio is a property of the wing geometry (span^2 / area).

Higher aspect ration ~= more transferred moment in the wing box ~= more stress ~= more material ~= more weight.

there is no linear scaling of wings. The overall optimum design while increasing MTOW will turn to being stubbier.

Threre is this tapering/staging law. improving your building material will allow less taper.

Taxi645 wrote:

vikkyvik wrote:

Taxi645 wrote:

Thank you. I'm asking because we were having a discussion whether or not weight (as in design weight) could have an influence on the induced drag coefficient or not.

Weight will affect induced drag, since weight affects the amount of lift required.

Taxi645 wrote:

I would argue, for instance in the case of the A380 and a possible wing redesign, that at a given wing loading and span, a reduction in design weight will result in a higher aspect ratio and thus lower induced drag. Could you confirm whether or not this to be the case?

A reduction in design weight doesn't automatically mean a higher aspect ratio. Aspect ratio is a property of the wing geometry (span^2 / area).

But it does proportionally affect aspect ratio if you keep span and wing loading constant.

I think you may be confusing yourself on this topic and/or not connecting a few of the dots. Let's see if we can help:

For starters, let's show the definitions/formulas of the key variables:

Aspect Ratio (AR) = (Effective Wingspan^2) / Wing Area.

Spanloading = MTOW / Effective wingspan

Wingloading = MTOW / Wing Area

Induced Drag = 0.5*(air density)*(Velocity^2)*(Wing Area)*(Induced Drag Coefficient)

To start answering your questions, in order:

1) For starters, none of the sources you posted are conflicting, though I can see how the might appear to be. For the purpose of explaining how aspect ratio, spanloading, and induced drag are related, let's start on Wainfan slide #34. This will be challenging to explain via written text, without showing the formulas, but here it goes;

Start with the formula for induced drag, which I posted above: Induced Drag = 0.5*(air density)*(Velocity^2)*(Wing Area)*(Induced Drag Coefficient)

At the top of page 34, Wainfan correctly give the formula for the Induced Drag Coefficient = (Lift Coefficient ^ 2) / [(Pi)*(Span Efficiency Factor)*(Aspect Ratio)]. This is where the Leeham analysis stops - from here, you can see that because Aspect Ratio is in the denominator, increasing Aspect Ratio means you have a corresponding decrease in the Induced Drag Coefficient, which will decrease overall Induced Drag. Per the formula for Aspect Ratio I defined above, there are only two ways to increase Aspect Ratio: either make the area of the wing smaller for a given wingspan, or make the span of the wing longer for a given wing area. Leeham is saying that by increasing the span of the wing, you will lower induced drag, because a longer wingspan leads to a higher Aspect Ratio, which in turn decreases the Induced Drag Coefficient.

Wainfan goes one step further, and plugs the formula for the Induced Drag Coefficient back into the Induced Drag Formula. This is on the middle of slide #34. Wainfan works through the algebra to get to a final formula for Induced Drag, which is boxed and bolded at the bottom of the page. From this, you can see that two key terms appear in the numerator and denominator: (Lift^2) / (Wingspan^2). From our aerodynamics basics, we know that in steady, level flight, Lift = Weight, so we can rewrite that as (Weight^2) / (Wingspan^2). Notice anything familiar about this term? If you take the square root of it, you have Weight / Wingspan, which is the exact formula I posted above in the "definitions" for wingloading.

Therefore: to decrease the induced drag of a wing, you must increase the aspect ratio of the wing, or minimize the spanloading of the wing. The way to achieve both a higher aspect ratio and lower spanloading is to increase the wingspan.

2) You are on the right path with your A380 example and the relationship between spanloading, aspect ratio, and wingloading. If you want to hold Wingloading constant, while decreasing the design weight of the aircraft, then you must reduce the area of the wing. If you reduce the area of your wing, while keeping wingspan constant, hen you will have increased your aspect ratio. By keeping the wingspan constant, and reducing the weight of the aircraft, you have decreased your spanloading. Per my last sentance in #1 above, increasing aspect ratio / decreasing spanloading will have the effect of reducing induced drag.

heavymetal wrote:

Taxi645 wrote:

vikkyvik wrote:

Weight will affect induced drag, since weight affects the amount of lift required.



A reduction in design weight doesn't automatically mean a higher aspect ratio. Aspect ratio is a property of the wing geometry (span^2 / area).

But it does proportionally affect aspect ratio if you keep span and wing loading constant.

I think you may be confusing yourself on this topic and/or not connecting a few of the dots. Let's see if we can help:

For starters, let's show the definitions/formulas of the key variables:

Aspect Ratio (AR) = (Effective Wingspan^2) / Wing Area.

Spanloading = MTOW / Effective wingspan

Wingloading = MTOW / Wing Area

Induced Drag = 0.5*(air density)*(Velocity^2)*(Wing Area)*(Induced Drag Coefficient)

To start answering your questions, in order:

1) For starters, none of the sources you posted are conflicting, though I can see how the might appear to be. For the purpose of explaining how aspect ratio, spanloading, and induced drag are related, let's start on Wainfan slide #34. This will be challenging to explain via written text, without showing the formulas, but here it goes;

Start with the formula for induced drag, which I posted above: Induced Drag = 0.5*(air density)*(Velocity^2)*(Wing Area)*(Induced Drag Coefficient)

At the top of page 34, Wainfan correctly give the formula for the Induced Drag Coefficient = (Lift Coefficient ^ 2) / [(Pi)*(Span Efficiency Factor)*(Aspect Ratio)]. This is where the Leeham analysis stops - from here, you can see that because Aspect Ratio is in the denominator, increasing Aspect Ratio means you have a corresponding decrease in the Induced Drag Coefficient, which will decrease overall Induced Drag. Per the formula for Aspect Ratio I defined above, there are only two ways to increase Aspect Ratio: either make the area of the wing smaller for a given wingspan, or make the span of the wing longer for a given wing area. Leeham is saying that by increasing the span of the wing, you will lower induced drag, because a longer wingspan leads to a higher Aspect Ratio, which in turn decreases the Induced Drag Coefficient.

Wainfan goes one step further, and plugs the formula for the Induced Drag Coefficient back into the Induced Drag Formula. This is on the middle of slide #34. Wainfan works through the algebra to get to a final formula for Induced Drag, which is boxed and bolded at the bottom of the page. From this, you can see that two key terms appear in the numerator and denominator: (Lift^2) / (Wingspan^2). From our aerodynamics basics, we know that in steady, level flight, Lift = Weight, so we can rewrite that as (Weight^2) / (Wingspan^2). Notice anything familiar about this term? If you take the square root of it, you have Weight / Wingspan, which is the exact formula I posted above in the "definitions" for wingloading.

Therefore: to decrease the induced drag of a wing, you must increase the aspect ratio of the wing, or minimize the spanloading of the wing. The way to achieve both a higher aspect ratio and lower spanloading is to increase the wingspan.

2) You are on the right path with your A380 example and the relationship between spanloading, aspect ratio, and wingloading. If you want to hold Wingloading constant, while decreasing the design weight of the aircraft, then you must reduce the area of the wing. If you reduce the area of your wing, while keeping wingspan constant, hen you will have increased your aspect ratio. By keeping the wingspan constant, and reducing the weight of the aircraft, you have decreased your spanloading. Per my last sentance in #1 above, increasing aspect ratio / decreasing spanloading will have the effect of reducing induced drag.

Thanks for the very well written post.

That's what I have been trying to say, I see most people here focus mainly on the aspect ratio alone in order to reduce the induced drag. Thing is, aspect ratio gets canceled out further in the equations to get only span loading in its place, just like how weight gets canceled out in free fall equations so that it becomes irrelevant at the end, and gravity becomes the factor that determines how fast objects accelerate while falling rather than weight.

Eyad89 wrote:

heavymetal wrote:

Taxi645 wrote:

But it does proportionally affect aspect ratio if you keep span and wing loading constant.

I think you may be confusing yourself on this topic and/or not connecting a few of the dots. Let's see if we can help:

For starters, let's show the definitions/formulas of the key variables:

Aspect Ratio (AR) = (Effective Wingspan^2) / Wing Area.

Spanloading = MTOW / Effective wingspan

Wingloading = MTOW / Wing Area

Induced Drag = 0.5*(air density)*(Velocity^2)*(Wing Area)*(Induced Drag Coefficient)

To start answering your questions, in order:

1) For starters, none of the sources you posted are conflicting, though I can see how the might appear to be. For the purpose of explaining how aspect ratio, spanloading, and induced drag are related, let's start on Wainfan slide #34. This will be challenging to explain via written text, without showing the formulas, but here it goes;

Start with the formula for induced drag, which I posted above: Induced Drag = 0.5*(air density)*(Velocity^2)*(Wing Area)*(Induced Drag Coefficient)

At the top of page 34, Wainfan correctly give the formula for the Induced Drag Coefficient = (Lift Coefficient ^ 2) / [(Pi)*(Span Efficiency Factor)*(Aspect Ratio)]. This is where the Leeham analysis stops - from here, you can see that because Aspect Ratio is in the denominator, increasing Aspect Ratio means you have a corresponding decrease in the Induced Drag Coefficient, which will decrease overall Induced Drag. Per the formula for Aspect Ratio I defined above, there are only two ways to increase Aspect Ratio: either make the area of the wing smaller for a given wingspan, or make the span of the wing longer for a given wing area. Leeham is saying that by increasing the span of the wing, you will lower induced drag, because a longer wingspan leads to a higher Aspect Ratio, which in turn decreases the Induced Drag Coefficient.

Wainfan goes one step further, and plugs the formula for the Induced Drag Coefficient back into the Induced Drag Formula. This is on the middle of slide #34. Wainfan works through the algebra to get to a final formula for Induced Drag, which is boxed and bolded at the bottom of the page. From this, you can see that two key terms appear in the numerator and denominator: (Lift^2) / (Wingspan^2). From our aerodynamics basics, we know that in steady, level flight, Lift = Weight, so we can rewrite that as (Weight^2) / (Wingspan^2). Notice anything familiar about this term? If you take the square root of it, you have Weight / Wingspan, which is the exact formula I posted above in the "definitions" for wingloading.

Therefore: to decrease the induced drag of a wing, you must increase the aspect ratio of the wing, or minimize the spanloading of the wing. The way to achieve both a higher aspect ratio and lower spanloading is to increase the wingspan.

2) You are on the right path with your A380 example and the relationship between spanloading, aspect ratio, and wingloading. If you want to hold Wingloading constant, while decreasing the design weight of the aircraft, then you must reduce the area of the wing. If you reduce the area of your wing, while keeping wingspan constant, hen you will have increased your aspect ratio. By keeping the wingspan constant, and reducing the weight of the aircraft, you have decreased your spanloading. Per my last sentance in #1 above, increasing aspect ratio / decreasing spanloading will have the effect of reducing induced drag.

Thanks for the very well written post.

That's what I have been trying to say, I see most people here focus mainly on the aspect ratio alone in order to reduce the induced drag. Thing is, aspect ratio gets canceled out further in the equations to get only span loading in its place, just like how weight gets canceled out in free fall equations so that it becomes irrelevant at the end, and gravity becomes the factor that determines how fast objects accelerate while falling rather than weight.

Thanks.

One important note/error in my previous post - I can't seem to go back and edit it now - where I say "Notice anything familiar about this term? If you take the square root of it, you have Weight / Wingspan, which is the exact formula I posted above in the "definitions" for wingloading." - that "wingloading" term should be "spanloading" (Weight / Wingspan). That's what I get for not going back and editing before posting. Hopefully not adding to the OP's confusion!

heavymetal wrote:

If you want to hold Wingloading constant, while decreasing the design weight of the aircraft, then you must reduce the area of the wing. If you reduce the area of your wing, while keeping wingspan constant, hen you will have increased your aspect ratio. By keeping the wingspan constant, and reducing the weight of the aircraft, you have decreased your spanloading. Per my last sentance in #1 above, increasing aspect ratio / decreasing spanloading will have the effect of reducing induced drag.

Be interesting to look at the susceptibility of reducing wing area to impact on structure weight.

IMU that is low. you reduce area ( increase wingloading and thus do not decrease the moment transmitted through the wing.
How does load bearing structure mass compare to "area giving" mass ?

Thank you very much heavymetal. Very helpfull.

heavymetal wrote:

Taxi645 wrote:

vikkyvik wrote:

Weight will affect induced drag, since weight affects the amount of lift required.



A reduction in design weight doesn't automatically mean a higher aspect ratio. Aspect ratio is a property of the wing geometry (span^2 / area).

But it does proportionally affect aspect ratio if you keep span and wing loading constant.

I think you may be confusing yourself on this topic and/or not connecting a few of the dots. Let's see if we can help:

For starters, let's show the definitions/formulas of the key variables:

Aspect Ratio (AR) = (Effective Wingspan^2) / Wing Area.

Spanloading = MTOW / Effective wingspan

Wingloading = MTOW / Wing Area

Induced Drag = 0.5*(air density)*(Velocity^2)*(Wing Area)*(Induced Drag Coefficient)

To start answering your questions, in order:

1) For starters, none of the sources you posted are conflicting, though I can see how the might appear to be. For the purpose of explaining how aspect ratio, spanloading, and induced drag are related, let's start on Wainfan slide #34. This will be challenging to explain via written text, without showing the formulas, but here it goes;

Start with the formula for induced drag, which I posted above: Induced Drag = 0.5*(air density)*(Velocity^2)*(Wing Area)*(Induced Drag Coefficient)

At the top of page 34, Wainfan correctly give the formula for the Induced Drag Coefficient = (Lift Coefficient ^ 2) / [(Pi)*(Span Efficiency Factor)*(Aspect Ratio)]. This is where the Leeham analysis stops - from here, you can see that because Aspect Ratio is in the denominator, increasing Aspect Ratio means you have a corresponding decrease in the Induced Drag Coefficient, which will decrease overall Induced Drag. Per the formula for Aspect Ratio I defined above, there are only two ways to increase Aspect Ratio: either make the area of the wing smaller for a given wingspan, or make the span of the wing longer for a given wing area. Leeham is saying that by increasing the span of the wing, you will lower induced drag, because a longer wingspan leads to a higher Aspect Ratio, which in turn decreases the Induced Drag Coefficient.

Wainfan goes one step further, and plugs the formula for the Induced Drag Coefficient back into the Induced Drag Formula. This is on the middle of slide #34. Wainfan works through the algebra to get to a final formula for Induced Drag, which is boxed and bolded at the bottom of the page. From this, you can see that two key terms appear in the numerator and denominator: (Lift^2) / (Wingspan^2). From our aerodynamics basics, we know that in steady, level flight, Lift = Weight, so we can rewrite that as (Weight^2) / (Wingspan^2). Notice anything familiar about this term? If you take the square root of it, you have Weight / Wingspan, which is the exact formula I posted above in the "definitions" for wingloading.

Therefore: to decrease the induced drag of a wing, you must increase the aspect ratio of the wing, or minimize the spanloading of the wing. The way to achieve both a higher aspect ratio and lower spanloading is to increase the wingspan.

2) You are on the right path with your A380 example and the relationship between spanloading, aspect ratio, and wingloading. If you want to hold Wingloading constant, while decreasing the design weight of the aircraft, then you must reduce the area of the wing. If you reduce the area of your wing, while keeping wingspan constant, hen you will have increased your aspect ratio. By keeping the wingspan constant, and reducing the weight of the aircraft, you have decreased your spanloading. Per my last sentance in #1 above, increasing aspect ratio / decreasing spanloading will have the effect of reducing induced drag.

I am curious how wingtip devices effect this. Are the modeled as an increase in effective wingspan, or something else?

Also is induced drag constant with respect to different lift distributions?

heavymetal wrote:

Taxi645 wrote:

vikkyvik wrote:

Weight will affect induced drag, since weight affects the amount of lift required.



A reduction in design weight doesn't automatically mean a higher aspect ratio. Aspect ratio is a property of the wing geometry (span^2 / area).

But it does proportionally affect aspect ratio if you keep span and wing loading constant.

I think you may be confusing yourself on this topic and/or not connecting a few of the dots. Let's see if we can help:

For starters, let's show the definitions/formulas of the key variables:

Aspect Ratio (AR) = (Effective Wingspan^2) / Wing Area.

Spanloading = MTOW / Effective wingspan

Wingloading = MTOW / Wing Area

Induced Drag = 0.5*(air density)*(Velocity^2)*(Wing Area)*(Induced Drag Coefficient)

To start answering your questions, in order:

1) For starters, none of the sources you posted are conflicting, though I can see how the might appear to be. For the purpose of explaining how aspect ratio, spanloading, and induced drag are related, let's start on Wainfan slide #34. This will be challenging to explain via written text, without showing the formulas, but here it goes;

Start with the formula for induced drag, which I posted above: Induced Drag = 0.5*(air density)*(Velocity^2)*(Wing Area)*(Induced Drag Coefficient)

At the top of page 34, Wainfan correctly give the formula for the Induced Drag Coefficient = (Lift Coefficient ^ 2) / [(Pi)*(Span Efficiency Factor)*(Aspect Ratio)]. This is where the Leeham analysis stops - from here, you can see that because Aspect Ratio is in the denominator, increasing Aspect Ratio means you have a corresponding decrease in the Induced Drag Coefficient, which will decrease overall Induced Drag. Per the formula for Aspect Ratio I defined above, there are only two ways to increase Aspect Ratio: either make the area of the wing smaller for a given wingspan, or make the span of the wing longer for a given wing area. Leeham is saying that by increasing the span of the wing, you will lower induced drag, because a longer wingspan leads to a higher Aspect Ratio, which in turn decreases the Induced Drag Coefficient.

Wainfan goes one step further, and plugs the formula for the Induced Drag Coefficient back into the Induced Drag Formula. This is on the middle of slide #34. Wainfan works through the algebra to get to a final formula for Induced Drag, which is boxed and bolded at the bottom of the page. From this, you can see that two key terms appear in the numerator and denominator: (Lift^2) / (Wingspan^2). From our aerodynamics basics, we know that in steady, level flight, Lift = Weight, so we can rewrite that as (Weight^2) / (Wingspan^2). Notice anything familiar about this term? If you take the square root of it, you have Weight / Wingspan, which is the exact formula I posted above in the "definitions" for wingloading.

Therefore: to decrease the induced drag of a wing, you must increase the aspect ratio of the wing, or minimize the spanloading of the wing. The way to achieve both a higher aspect ratio and lower spanloading is to increase the wingspan.

2) You are on the right path with your A380 example and the relationship between spanloading, aspect ratio, and wingloading. If you want to hold Wingloading constant, while decreasing the design weight of the aircraft, then you must reduce the area of the wing. If you reduce the area of your wing, while keeping wingspan constant, hen you will have increased your aspect ratio. By keeping the wingspan constant, and reducing the weight of the aircraft, you have decreased your spanloading. Per my last sentance in #1 above, increasing aspect ratio / decreasing spanloading will have the effect of reducing induced drag.

I don't disagree with anything in this post but, knowing the OP's background thought Here, I have to add one critical, oft-ignored, aspect of induced drag:

INDUCED DRAG INCREASES AS AIR DENSITY DECREASES.

Why is this so important to point out?
Because absent this factor, one can double-count the effect of weight reduction by both (1) decreasing weight and (2) increasing L/D. This is the critical error that I'm fairly certain Taxi645 has been making in our discussions of A380 revisions.

To see why weight reduction doesn't change OPTIMAL L/D, one has to recognize the following:

• 1. All modern airliners fly optimally at about same wing lift coefficient: ~.5.
• 2. Reducing weight of a plane with given wing dimensions means the plane flies higher in lower-density air (unless engines are climb-limited or ceiling is reached).
• 3. Flying higher means pressure drag decreases linearly with weight.
• 4. An x% decrease in weight means X^2 decrease in induced drag AT THE SAME ALTITUDE. By climbing to maintain Cl, however, induced drag increases by the linear inverse of air density. Thus we have a quadratic weight-related denominator and a weight-caused numerator that is functionally linear due to maintaining constant wing Cl. THESE CANCEL OUT TO A FUNCTIONALLY LINEAR RELATIONSHIP BETWEEN INDICED DRAG AND WEIGHT AT OPTIMAL CRUISE ALTITUDE.

This is why L/D is entirely* based on a plane's 3-d description, independent of its weight.

*there are some minor deviations from this rule, such as the lower effective Reynolds numbers at higher cruise altitude.

Reducing the A380's weight would help It, but would not impact its L/D. Mission fuel burn delta would be basically linear with the delta to (OEW+payload).

Note: the foregoing doesn't apply to takeoff performance, as runways don't change altitude according to plane weight or wing loading. Thus high-AR and/or low spanloading shows the most dramatic benefit for takeoff, especially for V2-limited twins (this the 777x's lower thrust despite a longer fuselage, less rotation angle).

matt641 wrote:

3. Flying higher means pressure drag decreases linearly with weight.

I sort of skipped a couple steps here that are obvious to me but probably not to a newcomer or just someone learning.

Less weight and constant Cl means either you keep climbing from level cruise (lift is now more than weight) or you generate less lift per area of wing somehow. How? Use less dense air. I.e. climb a bit higher until level cruise occurs once more at Cl = ~.5.

The negative delta to lift is equal to the negative delta to weight, all these are proportionally equal to the negative delta in air density required to achieve the negative lift delta.

But guess what else has a negative delta proportional to air density? Your main drag! I.e. pressure drag.


To be clear, my first dozen or so posts on this site had all this stuff wrong in almost the same way as Taxi645, then I did what he is doing: asked around and read up.

Thank you for joining in Matt. I was hoping you would and part of the reason to open the thread was to have a place to discuss this.

Matt6461 wrote:

[*]2. Reducing weight of a plane with given wing dimensions means the plane flies higher in lower-density air (unless engines are climb-limited or ceiling is reached).

What do you mean by : "given wing dimensions"? I'm asking because what I'm proposing is a reduction in weight with an equal reduction in wing surface and at equal span an equal increase in aspect ratio.

kitplane01 wrote:

I am curious how wingtip devices effect this. Are the modeled as an increase in effective wingspan, or something else?

I'm obviously a non expert on this, but I have read the wing tips height is counted for as extension of effective wingspan with a factor of 0.45. Wingtip height below wing (scimiter), probably less. Although this is of course a very crude way to look at aerodynamics.

kitplane01 wrote:

Also is induced drag constant with respect to different lift distributions?

I would expect not. In case of a span limited plane like the A380, I think I've read there is more lift generated at the root in order to make the pressure gradient less steep towards the wing ends to reduce wing end vortices. Again I'm no expert so probably best to wait to get a reply from someone more knowledgeable.

Matt6461 wrote:

This is why L/D is entirely* based on a plane's 3-d description, independent of its weight.

It is true that the Maximum L/D ratio is independent of weight and wingloading, but airplanes don't always fly at the speed that achieves the maximum L/D ratio. When the weight of a plane increases, the speed needed to achieve the maximum L/D for that particular airplane also increases. If planes A and B have the same maximum L/D ratio, the heavier plane would require to cruise at a higher speed in order to achieve the max L/D ratio. Weight does affect the cruise L/D ratio indirectly (not maximum L/D). To illustrate, let's look at the following:

Average L/D ratio at cruise (note I am using average L/D, not maximum L/D):

A340-200: 19.2
A340-300: 19.1
747-200: 15.3
747-400: 15.5

source: https://download.docslide.com.br/getdow ... vV1Q%3D%3D

L/D ratio at FL370:

789: 20.77
78J: 20.55

Source: Ferpe's thread

Well, A342 and A343 share an identical wing in terms of everything. Maximum L/D should be the same for both, but the heavier A343 does have a slightly lower L/D ratio at cruise. The same applies to 742/744 and 789/78J. Since they are flying at almost the same speed, their differences in weight resulted in different L/D ratios.


The best aerodynamic efficiency of an aircraft is when it cruises with a speed such that it delivers the highest value of L\D. The speed at which the L/D ratio has the maximum value is equal to the speed with minimum drag and minimum power. That speed is directly affected by weight and altitude (air density). As I said in previous threads, comparing L/D ratios in planes with different weights and wingloadings can be a pain, it does not tell everything.

Eyad89 wrote:

The best aerodynamic efficiency of an aircraft is when it cruises with a speed such that it delivers the highest value of L\D. The speed at which the L/D ratio has the maximum value is equal to the speed with minimum drag and minimum power. That speed is directly affected by weight and altitude (air density). As I said in previous threads, comparing L/D ratios in planes with different weights and wingloadings can be a pain, it does not tell everything.

Best L/D may be the most aerodynamically efficient, but you don't want to cruise there. Far too slow. Best L/D (Green Dot in an Airbus) is the best speed for a hold (highest endurance) or for a clean noise abatement climb.

Taxi645 wrote:

Thank you for joining in Matt. I was hoping you would and part of the reason to open the thread was to have a place to discuss this.

Matt6461 wrote:

[*]2. Reducing weight of a plane with given wing dimensions means the plane flies higher in lower-density air (unless engines are climb-limited or ceiling is reached).

What do you mean by : "given wing dimensions"? I'm asking because what I'm proposing is a reduction in weight with an equal reduction in wing surface and at equal span an equal increase in aspect ratio.

Well that changes everything lol.
To be fair, you haven't always been clear on that point. We generally refer to an A380NEO as keeping the same wing; I call a rewinged A380 and X or NWO.
In my series of posts on A380X/NWO from a few years ago now, I state that merely reducing wing area and using aggressive tip treatments (plus CFRP) gets very close to a ~90m folding wing desgin.

One problem with a smaller plastic wing is the shorter root chord means a ton of rework to the wing/fuse design and probably relocation of systems stored there. By the time that work is done we're looking at a ~2025 whose cost approaches a clean sheet. And you still have a fuselage that less good than it could have been (though still the best in the world by far).

That and the A380's likely imminent demise are why I've shifted focus to a clean sheet VLA instead of A380X/NWO. But if Airbus could find a way to bridge to, and build, something like we're discussing I'd probably have a heart attack of happiness.

Starlionblue wrote:

Eyad89 wrote:

The best aerodynamic efficiency of an aircraft is when it cruises with a speed such that it delivers the highest value of L\D. The speed at which the L/D ratio has the maximum value is equal to the speed with minimum drag and minimum power. That speed is directly affected by weight and altitude (air density). As I said in previous threads, comparing L/D ratios in planes with different weights and wingloadings can be a pain, it does not tell everything.

Best L/D may be the most aerodynamically efficient, but you don't want to cruise there. Far too slow. Best L/D (Green Dot in an Airbus) is the best speed for a hold (highest endurance) or for a clean noise abatement climb.

Indeed, max endurance is at L/Dmax
Max range is at UL/Dmax (although also referred to as ML/Dmax as its 'M's that really hurt you at speed.
Best economic speed is normally a smidge faster than ML/Dmax as it makes better use of assets if its back on the ground sooner but then cost indexes come in to play and that's a whole other world of complexity.

Fred

flipdewaf wrote:

Starlionblue wrote:

Eyad89 wrote:

The best aerodynamic efficiency of an aircraft is when it cruises with a speed such that it delivers the highest value of L\D. The speed at which the L/D ratio has the maximum value is equal to the speed with minimum drag and minimum power. That speed is directly affected by weight and altitude (air density). As I said in previous threads, comparing L/D ratios in planes with different weights and wingloadings can be a pain, it does not tell everything.

Best L/D may be the most aerodynamically efficient, but you don't want to cruise there. Far too slow. Best L/D (Green Dot in an Airbus) is the best speed for a hold (highest endurance) or for a clean noise abatement climb.

Indeed, max endurance is at L/Dmax
Max range is at UL/Dmax (although also referred to as ML/Dmax as its 'M's that really hurt you at speed.
Best economic speed is normally a smidge faster than ML/Dmax as it makes better use of assets if its back on the ground sooner but then cost indexes come in to play and that's a whole other world of complexity.

Fred

Guys I said "optimal" L/D instead of max L/D for a reason.

Everything I said about Di and air density applies whether you're calibrating max endurance, range, or cost-indexed LRC.

Eyad89 wrote:

A342 and A343 share an identical wing in terms of everything. Maximum L/D should be the same for both, but the heavier A343 does have a slightly lower L/D ratio at cruise. The same applies to 742/744 and 789/78J. Since they are flying at almost the same speed, their differences in weight resulted in different L/D ratios

No. L/D should not be determined by wing alone and nothing in my post suggests that.

A simple stretch always has lower L/D than the base model. You've added wetted area without increasing span; that alone reliably predicts lower L/D. This is true whether were talking optimal L/D for max range cruise, max L/D, or even takeoff L/D.

Eyad89 wrote:

If planes A and B have the same maximum L/D ratio, the heavier plane would require to cruise at a higher speed in order to achieve the max L/D ratio

This is basically* wrong. I'm trying to guess what's behind your intuition but having trouble at it.
Two B789's of different gross weight will fly at almost the exact same speed and identical L/D. The heavier will not fly faster to generate more lift; it will fly lower in denser air.

*the plane flying higher (lighter plane) will, in fact, often fly slightly slower than the heavier plane. But this due to slower speed of sound in colder climes. That diffrlerence disappears, however, once both are in the isotermic stratosphere (above FL 37 or so).

Matt6461 wrote:

Well that changes everything lol.

Good that we've got that straightened out!

Matt6461 wrote:

Eyad89 wrote:

If planes A and B have the same maximum L/D ratio, the heavier plane would require to cruise at a higher speed in order to achieve the max L/D ratio

This is basically* wrong. I'm trying to guess what's behind your intuition but having trouble at it.
Two B789's of different gross weight will fly at almost the exact same speed and identical L/D. The heavier will not fly faster to generate more lift; it will fly lower in denser air.

*the plane flying higher (lighter plane) will, in fact, often fly slightly slower than the heavier plane. But this due to slower speed of sound in colder climes. That diffrlerence disappears, however, once both are in the isotermic stratosphere (above FL 37 or so).

I did not come up with anything from my tuition, I took everything I wrote directly from a couple of aerodynamics textbooks lol.

If you plot drag vs speed, you will get a U-shaped graph facing upwards. The point at which drag is minimum in that U-shaped graph (minimum drag speed) is also equal to the speed with maximum L/D. Now, if weight is increased, the whole graph is displaced upward and to the right. That increases the speed with minumum drag and maximum L/D.

For small changes of weight (up to 20%), the speed at which drag is at its minimum and L/D at its maximum will change by half the percentage change in weight, and so does the whole graph. Minimum drag speed is propotional to the square root of the aircraft weight.

Basically, as in your example, two 789 with slightly different weights flying at the same speed and alttitude would have slightly different L/D at that time( but their maximum L/D would still be the same).

Eyad89 wrote:

Basically, as in your example, two 789 with slightly different weights flying at the same speed and alttitude would have slightly different L/D at that time( but their maximum L/D would still be the same)

I have bolded where you're either missing my point or getting something wrong.

I don't know how I can be any clearer about this: THE LIGHTER 789 ALWAYS CRUISES HIGHER THAN THE HEAVIER 789.

...subject to restrictions on airspace or other miscellanea.

Everything you've said about the drag/speed curve is only true AT CONSTANT ALTITUDE.

And keep in mind I'm talking about optimal L/D for long range cruise, not max L/D - as I said previously (your remarks suggest you still think we're talking about max L/D).

Taxi645 wrote:

Matt6461 wrote:

Well that changes everything lol.

Good that we've got that straightened out!

Yep. Just be sure to specify that your conception of a plane labeled "A380NEO" includes a completely new wing. Most would agree that's so substantially beyond a NEO that it should have a different name. Semantics, I know, but it helps avoid confusion.