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Lateral Position Of A Wing Mounted Engine  
User currently offlineairbuske From United States of America, joined Jun 2007, 466 posts, RR: 0
Posted (4 years 8 months 2 days 3 hours ago) and read 5453 times:

Since the advent of commerical aviation, most commerical airliners have adopted wing mounted engines. What influences the spanwise placement of the engine?

My initial thoughts:

Moving the engines further outboard puts the engines more into the "clean" free stream air and further away from the "dirty" airflow due to fuselage interference giving a potentially more efficient engine which is significantly less susceptible to FOD. The longer moment arm will also provide greater bending relief yielding a lighter wing structure. The tradeoff would be requiring a larger rudder for adequate lateral control during an engine thrust imbalance situation (Vmcg), and this of course translates into a weight penalty.

However, looking at the A330/A340 for example, the rudder on the A340 isn't significantly larger (if at all) than on it's smaller sister. So why are the engines on the A330 not at the #1 and #4 positions? After all, the old generation A340's share the same wing as the A330.

Thanks,

Jinal

[Edited 2010-03-28 10:42:58]

18 replies: All unread, jump to last
 
User currently offlinetdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 1, posted (4 years 8 months 2 days 2 hours ago) and read 5433 times:

Quoting airbuske (Thread starter):
What influences the spanwise placement of the engine?

Tending to drive engines outboard:
-Wing bending relief
-Flutter (mass on the ends lowers the resonant frequencies)
-Ground clearance (assuming some dihedral)
-Noise

Tending to drive engines inboard:
-Adverse yaw in an engine out
-Wing stiffness

Quoting airbuske (Thread starter):
Moving the engines further outboard puts the engines more into the "clean" free stream air and further away from the "dirty" airflow due to fuselage interference giving a potentially more efficient engine which is significantly less susceptible to FOD.

The boundary layer isn't particularly thick abeam the engines...although there's certainly some flow distortion, it's fairly easily dealt with by aiming the nacelles. This is why you often see a little bit of "toe in" on the nacelle. Note that tail-mounted engines are *far* close to the fuselage and don't really have interference issues.

Quoting airbuske (Thread starter):
The longer moment arm will also provide greater bending relief yielding a lighter wing structure.

The is certainly a factor.

Quoting airbuske (Thread starter):
The tradeoff would be requiring a larger rudder for adequate lateral control during an engine thrust imbalance situation (Vmcg), and this of course translates into a weight penalty

It's more of an influence on total fin size than rudder size but this is a significant factor. This is sometimes why you see double-hinged rudders.

Quoting airbuske (Thread starter):
However, looking at the A330/A340 for example, the rudder on the A340 isn't significantly larger (if at all) than on it's smaller sister. So why are the engines on the A330 not at the #1 and #4 positions?

Because the fin on an A330/A340 isn't big enough to counter the much larger adverse yaw of an engine-out A330 if the engines were at the outboard positions.

The amount of installed thrust is always what the aircraft needs with one engine out. As a result, quads like the A340 have 133% of required thrust when all engines are operating (they drop to 100% when an engine goes out) while a twin like the A330 has 200% of required thrust.

In an engine out, the A340 has 2/3 of the thrust on one wing and 1/3 on the other, while the A330 has all of it on one wing and none on the other...thus twins tend to need considerably larger vertical fins.

Tom.


User currently offlineairbuske From United States of America, joined Jun 2007, 466 posts, RR: 0
Reply 2, posted (4 years 8 months 1 day 22 hours ago) and read 5331 times:

Tom, thanks for yet another insightful reply.

Quoting tdscanuck (Reply 1):
It's more of an influence on total fin size than rudder size but this is a significant factor. This is sometimes why you see double-hinged rudders.

Sorry, but I'm having difficulty understanding why the entire fin would be affected. To my understanding, *control* is independent of the stabilizer portion of the lifting surface. Unless the thrust imbalance during an engine out affects the directional stability of the airplane, I don't see how the situation would influence the vertical stabilizer. So could you please explain this further? And what is a "double-hinged rudder"?

Quoting tdscanuck (Reply 1):
Because the fin on an A330/A340 isn't big enough to counter the much larger adverse yaw of an engine-out A330 if the engines were at the outboard positions.

The amount of installed thrust is always what the aircraft needs with one engine out. As a result, quads like the A340 have 133% of required thrust when all engines are operating (they drop to 100% when an engine goes out) while a twin like the A330 has 200% of required thrust.

In an engine out, the A340 has 2/3 of the thrust on one wing and 1/3 on the other, while the A330 has all of it on one wing and none on the other...thus twins tend to need considerably larger vertical fins.

Are you implying that if a quad on it's takeoff roll at Vmcg lost both engines on one side of the airplane, the pilot would not have sufficient rudder authority to safely control the aircraft on the ground?

Thanks,

Jinal

(sorry mods if this is off-topic)


User currently offline474218 From United States of America, joined Oct 2005, 6340 posts, RR: 9
Reply 3, posted (4 years 8 months 1 day 21 hours ago) and read 5278 times:

Quoting airbuske (Reply 2):
And what is a "double-hinged rudder"?

The forward and aft portion of the rudder deflect at different angles.


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User currently offlinetdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 4, posted (4 years 8 months 1 day 21 hours ago) and read 5262 times:

Quoting airbuske (Reply 2):
Sorry, but I'm having difficulty understanding why the entire fin would be affected. To my understanding, *control* is independent of the stabilizer portion of the lifting surface.

Control is a function of both. At very low speeds, the rudder alone may not have nearly enough authority on its own to counter the adverse yaw. Put in technical terms, you don't have enough control power to achieve zero sideslip when you're close to Vmca with an engine out and the other at full thrust.

When you deflect the rudder, you're basically altering the chord line of the entire vertical fin. The torque is a function of the size of the rudder, the size of the rudder deflection, and the size of the fin.

When you enter a sideslip, the restoring torque is a function of the entire surface that's at an angle of attack (normally denoted beta for sideslip), not just the rudder.

Quoting airbuske (Reply 2):
Unless the thrust imbalance during an engine out affects the directional stability of the airplane, I don't see how the situation would influence the vertical stabilizer.

It very much affects the directional stability of the airplane. The vertical fin and rudder, in a control sense, are actuators with a finite amount of control power. If the adverse yaw from the engines exceeds the yaw capability of the fin + rudder, you will be unstable in yaw (at least in one direction). Put another way, if you're making the fin and rudder work to counter yaw, that control power is not available to provide directional stability. With enough torque from the engines, you don't have anything left to provide stability.

Quoting airbuske (Reply 2):
Are you implying that if a quad on it's takeoff roll at Vmcg lost both engines on one side of the airplane, the pilot would not have sufficient rudder authority to safely control the aircraft on the ground?

On a quad I'm not quite sure how the math works out, but I suspect that would be the case. On a twin, it's definitely the case. That's why V1 can never be lower than Vmcg.

Tom.


User currently offlineStarlionblue From Greenland, joined Feb 2004, 17110 posts, RR: 66
Reply 5, posted (4 years 8 months 1 day 20 hours ago) and read 5241 times:

I would add that the length of the aircraft also affects the required fin/rudder size. The longer the plane, the smaller the required fin/rudder, all other things being equal, because the moment arm is longer (as Archimedes knew).

That's why the 747SP has a larger fin than the 747-100/200, and a double hinged rudder.



"There are no stupid questions, but there are a lot of inquisitive idiots."
User currently onlinevikkyvik From United States of America, joined Jul 2003, 10239 posts, RR: 26
Reply 6, posted (4 years 8 months 1 day 17 hours ago) and read 5191 times:
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Quoting tdscanuck (Reply 1):
Quoting airbuske (Thread starter):
What influences the spanwise placement of the engine?

Tending to drive engines outboard:
-Wing bending relief
-Flutter (mass on the ends lowers the resonant frequencies)
-Ground clearance (assuming some dihedral)
-Noise

Tending to drive engines inboard:
-Adverse yaw in an engine out
-Wing stiffness
Quoting tdscanuck (Reply 1):
Quoting airbuske (Thread starter):
The longer moment arm will also provide greater bending relief yielding a lighter wing structure.

The is certainly a factor.

In my aircraft design class, I asked a question of one of the professors about engine placement as it related to wing design. The relevant part of my question:

The way it would make sense to me is if, in the spreadsheet at least, engine weight simply allows for a thinner skin thickness (i.e. keeping the same bending deflection with less material).

The relevant part of his answer:

Yes - in a carefully designed wing which is equally stressed along the span, the deflected shape is a function only of the wing depth.

Now airplane wings are probably not equally stressed along the span (or are they?), but since I have no real-world experience in wing design, I was wondering how applicable that is.



How can I be an admiral without my cap??!
User currently offlinetdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 7, posted (4 years 8 months 1 day 16 hours ago) and read 5152 times:

Quoting vikkyvik (Reply 6):
Now airplane wings are probably not equally stressed along the span (or are they?),

They're not. For structural efficiency reasons, you want to be close, but there are other factors driving gauge beyond just loading (impact resistance, manufacturability, etc.) that will cause you to have stress drop off at the ends.

Tom.


User currently offlineairbuske From United States of America, joined Jun 2007, 466 posts, RR: 0
Reply 8, posted (4 years 8 months 1 day 12 hours ago) and read 5097 times:

Quoting 474218 (Reply 3):
The forward and aft portion of the rudder deflect at different angles.

Thanks for the image. I have previously noticed the "double-hinge" rudder on several airplanes (most noticeably on the Concorde, B747 and the A380) but have always seen it referred to as a "split" rudder.

Quoting tdscanuck (Reply 4):
When you deflect the rudder, you're basically altering the chord line of the entire vertical fin. The torque is a function of the size of the rudder, the size of the rudder deflection, and the size of the fin.

When you enter a sideslip, the restoring torque is a function of the entire surface that's at an angle of attack (normally denoted beta for sideslip), not just the rudder.

That makes simple sense now. Deflecting the flapped portion of a lifting surface changes the lift distribution over the entire surface, not just over the flapped surface - so both portions each have their effective control power. The same happens to the pitching moment coefficient of the entire lifting surface (or in the case of the vertical tail, torsional moment). D'oh!   

Quoting tdscanuck (Reply 4):
Put another way, if you're making the fin and rudder work to counter yaw, that control power is not available to provide directional stability. With enough torque from the engines, you don't have anything left to provide stability.

Back to the issue of an adverse yaw situation affecting directional stability : suppose an airplane is in the condition you just described i.e. engine out + full rudder such that all the "control power " is used to counter adverse yaw (zero sideslip) and none of it remains to provide directional stability. Now the same airplane experiences a gust from one side. The fin will see a change in it's angle of attack and the resultant increase in side force will, just as always, turn the airplane turn into the wind until the slip is eliminated. Or....? So how is directional stability affected?

Quoting Starlionblue (Reply 5):
I would add that the length of the aircraft also affects the required fin/rudder size. The longer the plane, the smaller the required fin/rudder, all other things being equal, because the moment arm is longer (as Archimedes knew).

That's why the 747SP has a larger fin than the 747-100/200, and a double hinged rudder.

I'm unsure if the relationship is as plain as you have described it. While for longer fuselages, the moment arm to the cg certainly is longer, the fuselage's contribution to directional instability is also greater. It may very well be that the end effect is what you have described because perhaps the pro usually outweighs the con - I don't know.

On a side note, while it is true that the 747SP has a larger tail, if I'm not mistaken, the double hinged rudder isn't exclusive to the variant. The split rudder is featured on every model in the family including the -8F.

Quoting vikkyvik (Reply 6):
Now airplane wings are probably not equally stressed along the span (or are they?), but since I have no real-world experience in wing design, I was wondering how applicable that is.


As a college student myself, from the very little that I know about structural design, the wing is not stressed equally across the span. This can very easily be illustrated using the age old example of a simple cantilevered beam with a point load acting on the free end. The shear force is constant across the span however the bending moment varies linearly with the maximum value at the fixed end of the beam. As a result, the stresses (more specifically axial stresses) change across the span.

Now ofcourse the true load distribution on a wing is much more complex than that - for one, the lift is distributed across the entire span and secondly, the shape of the distribution is determined primarily by the wing's planform. And then you have the prescence of heavy objects such as fuel tanks and main landing gears which have their own intertial loading.

Theoretically, only in the case of an elliptical wing planform will the wing loading be uniform across the span. And even in this case, the shear force and bending moment vary parabolically (quadratic and cubic functions respectively). For all other planforms, the lift distribution will not be linear with the root working much harder than the tip and this just aggravates the change in stresses across the span.

Onto your question/comment about wing skin thickness and engine placement, when the wing skin is designed to serve as a load bearing member a.ka. "stressed skin", it is unlikely that placing dead weight such as engines in the wing will have any appreciable effect on the thickness of the wing cover skin panels. This is because the skin acts as the loadpath for wing torsional stresses which are relatively unaffected by inertial loading since typically, the shear center of the wing section sits at the wing's center of gravity. *waits for a B.S. flag*

Placing the dead weight in the wing will however alleviate some of the shear and axial stresses due to vertical shear load and bending moment on the wing. Lower axial stresses means smaller and lighter spar caps and spanwise flanges. Lower shear stresses means a thinner and lighter web. But to maximize on these reliefs/benefits, the dead weight items must be placed as far outboard as possible; which is part of the reason why I asked my original question about spanwise engine placement.

Jinal

[Edited 2010-03-29 00:51:10]

User currently offlineStarlionblue From Greenland, joined Feb 2004, 17110 posts, RR: 66
Reply 9, posted (4 years 8 months 1 day 11 hours ago) and read 5048 times:

Quoting airbuske (Reply 8):
Thanks for the image. I have previously noticed the "double-hinge" rudder on several airplanes (most noticeably on the Concorde, B747 and the A380) but have always seen it referred to as a "split" rudder.

There's a difference. A split rudder is simply a rudder in two parts for redundancy. There is a horizontal split. A double hinged rudder has a vertical split to enable further deflection. You could have a split and double-hinged rudder with four surfaces.

So while the 747SP has a double-hinged rudder the rest of the 747s have a split rudder.



"There are no stupid questions, but there are a lot of inquisitive idiots."
User currently offlinemusang From United Kingdom, joined Apr 2001, 872 posts, RR: 7
Reply 10, posted (4 years 8 months 1 day 9 hours ago) and read 5002 times:

Quoting airbuske (Reply 8):
(most noticeably on the Concorde,

I don't think the Concorde rudders are double hinged. There are two, so they're horizontally split.

Regards - musang


User currently offlineairbuske From United States of America, joined Jun 2007, 466 posts, RR: 0
Reply 11, posted (4 years 8 months 1 day ago) and read 4812 times:

Quoting Starlionblue (Reply 9):
So while the 747SP has a double-hinged rudder the rest of the 747s have a split rudder.
Quoting Starlionblue (Reply 9):
I don't think the Concorde rudders are double hinged. There are two, so they're horizontally split.

Now I see the difference. Thanks for clearing it up. A double hinged rudder is basically a double slotted flap?

Jinal


User currently offlineStarlionblue From Greenland, joined Feb 2004, 17110 posts, RR: 66
Reply 12, posted (4 years 8 months 21 hours ago) and read 4774 times:

No. A double-slotted flap is a flap with two slots (gaps) in it. A double-hinged rudder consists of one rudder mounted on the fin and a second rudder section mounted on the first.

Anyone have a good picture?



"There are no stupid questions, but there are a lot of inquisitive idiots."
User currently offline474218 From United States of America, joined Oct 2005, 6340 posts, RR: 9
Reply 13, posted (4 years 8 months 21 hours ago) and read 4757 times:

Quoting airbuske (Reply 11):
A double hinged rudder is basically a double slotted flap?

Except there is no "slot".

Quoting Starlionblue (Reply 12):
Anyone have a good picture?

Reply 3?


User currently offlineStarlionblue From Greenland, joined Feb 2004, 17110 posts, RR: 66
Reply 14, posted (4 years 8 months 20 hours ago) and read 4749 times:

Quoting 474218 (Reply 13):
Quoting Starlionblue (Reply 12):
Anyone have a good picture?

Reply 3?

Good point.  



"There are no stupid questions, but there are a lot of inquisitive idiots."
User currently offlinetdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 15, posted (4 years 8 months 20 hours ago) and read 4739 times:

Quoting airbuske (Reply 8):
Back to the issue of an adverse yaw situation affecting directional stability : suppose an airplane is in the condition you just described i.e. engine out + full rudder such that all the "control power " is used to counter adverse yaw (zero sideslip) and none of it remains to provide directional stability. Now the same airplane experiences a gust from one side. The fin will see a change in it's angle of attack and the resultant increase in side force will, just as always, turn the airplane turn into the wind until the slip is eliminated. Or....? So how is directional stability affected?

If you're at zero sideslip in the static case, you've still got some stability left. The critical case is when you're flying with some permanent beta in order to get enough torque to counter the engine. At sufficiently high torque (or sufficiently low airspeed), the vertical fin will stall before it generates enough torque to counter and you have no more directional stability. This is where Vmca comes from.

Quoting airbuske (Reply 8):
As a college student myself, from the very little that I know about structural design, the wing is not stressed equally across the span. This can very easily be illustrated using the age old example of a simple cantilevered beam with a point load acting on the free end.

I think you're confusing equal loading with equal stress. A wing is most definitely not equally loaded, for all the reasons you suggest. However, for efficient material usage, the gauges of the structural members should be constantly changing to keep the material stresses roughly even throughout the wing. However, you can't do that 100% and there will be some parts of the wing that are operating at higher stress than others.

Tom.


User currently onlinevikkyvik From United States of America, joined Jul 2003, 10239 posts, RR: 26
Reply 16, posted (4 years 8 months 20 hours ago) and read 4723 times:
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Quoting tdscanuck (Reply 7):

They're not. For structural efficiency reasons, you want to be close, but there are other factors driving gauge beyond just loading (impact resistance, manufacturability, etc.) that will cause you to have stress drop off at the ends.

Gotcha. That's sort of what I figured, but didn't know for sure.

Quoting airbuske (Reply 8):
As a college student myself,

I'm no longer a college student, but I haven't gone into the aerospace field (yet?).



How can I be an admiral without my cap??!
User currently offlineairbuske From United States of America, joined Jun 2007, 466 posts, RR: 0
Reply 17, posted (4 years 7 months 3 weeks 6 days 6 hours ago) and read 4482 times:

Quoting tdscanuck (Reply 15):
I think you're confusing equal loading with equal stress. A wing is most definitely not equally loaded, for all the reasons you suggest. However, for efficient material usage, the gauges of the structural members should be constantly changing to keep the material stresses roughly even throughout the wing. However, you can't do that 100% and there will be some parts of the wing that are operating at higher stress than others.

Tom, thanks for correcting my statement. Your post made me realize a simple but fundamental distinction in my train of thought.   

Subconsciously, I was thinking along the same lines. The applied loads vary greatly across the span but the structural designer tries his best to even the stresses out. Typical examples of this are tapering spar caps along the wing span and changing the diameter and spacing of lightening holes in the shear web.


User currently offlineOldAeroGuy From United States of America, joined Dec 2004, 3574 posts, RR: 67
Reply 18, posted (4 years 7 months 3 weeks 5 days 19 hours ago) and read 4385 times:

Quoting tdscanuck (Reply 4):
Quoting airbuske (Reply 2):
Are you implying that if a quad on it's takeoff roll at Vmcg lost both engines on one side of the airplane, the pilot would not have sufficient rudder authority to safely control the aircraft on the ground?

On a quad I'm not quite sure how the math works out, but I suspect that would be the case. On a twin, it's definitely the case. That's why V1 can never be lower than Vmcg.

For either Twin or Quad, Vmcg and the associated V1 is defined considering only one engine inoperative. If a Quad fails two engines on the same wing is probably uncontrollable at V1.



Airplane design is easy, the difficulty is getting them to fly - Barnes Wallis
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