Author Topic: Inboard vs. outboard spans (Read 1922 times)

Frank Sheridan · « **on:** June 10, 2009, 04:33:39 PM »

Hello all. I have always been interested in the theory behind unequal span wing panels on dedicated stunt planes. I have several questions. First, who was the originator of this technique? I know that this is done to counter the effect of long lines, but how would the flight characteristics differ if you built two identical planes but with different wing layouts pertaining to the spans? How does the designer know how much to differ the inboard span from the outboard? How much is too much? And lastly, are any of the current generation of stunters NOT laid out this way? All the profile planes that I have built (and crashed) had equal span wing panels, but none of them had flaps either.

john e. holliday · « **Reply #1 on:** June 11, 2009, 08:39:51 AM »

I guess the west coast hasn't woke up yet. If you will look at the top view of your profile. A lot of weight is out board of the center line. I have seen and flown planes with full fuselages with no differential in the wing panels to those with over 2 inches of differential like the All American Senior. Each and every designer had his/her reason for the way the plane was built. Todays planes have very little differential in the wing panels.

Some use the flaps for difference in area to keep the wing level while maneuvering. In all there are so many things to be built into a new modern stunt plane to help the trim, I would myself probably not design in different length wing panels. Now we need the experts to straighten us both out on this.

Having fun, DOC Holliday

John Miller · « **Reply #2 on:** June 11, 2009, 11:27:28 AM »

Well, I can't tell you who first began using span differential, only that it was being used back into the late 40's at least.

The thinking, at the time was that the outboard wing was traveling faster than the inboard, and therefore created more lift. The outboard lift was thought to contribute an inboard rolling motion, causeing a lessening of line tension. The early practitioners aparently never really did the math and estimated the amount of assymetry. One rib bay became the standard in the early 50's.

Later, Netzeband, and others did the math and found that substatially less was actually needed. Most designs can use about .5" to .75" assymetry.

Some even went so far as to say that it isn't really needed, and there are designs that use none at all.

The effects of not using assymetry seems to be a need for more tip weight to balance out the plane for circular flight.

Often it's desirable to have more tip weight than is simply needed to balance the plane out. maintaining line tension at certain parts of the hemisphere, is one that comes to mind.

So, what happens when you have more flap area inboard than outboard, and you are using excess tip weight? The wing starts banging the outboard wing in the hard corners.

Should the flaps be equal area? I'd say yes, at least, and a bit more outboard if there is an excess of tip weight. Making the outboqard flap 1/8" wider at the tip is a common way to do this.

Of course, for the common sport flier, it's not as important, so many sport oriented profiles amy indeed be equal span designs.

phil c · « **Reply #3 on:** June 12, 2009, 04:32:03 PM »

Per Wild Bill's calculations, you automatically get about .4in offset on a 48 in. wing and .7 in. or so on a 60 in. wing. That is just due to the outer panel flying faster than the inner panel. So if you build equal span panels and center the wing in the fuselage the MAC is offset outboard.
Gives a very nice force setup, with the thrustline well inboard of the MAC(helps keep the plane out on the lines), but unless you run a sidemounted engine it takes quite a bit of tipweight to trim the plane. Probably why the Ukranainians build their ARF's with side mount engines.

It's also why many profiles sometimes run into trouble. The engine is offset to the right by half the thickness of the fuselage, using up some of the aerodynamic offset you get from a centerline mounted engine. A good reason to make sure that on your profile plane the thrustline goes through the center of the wing at 25% chord. Putting big thick angled wedges under the engine mounts only make matters worse. The thickness of the wedges often completely negates the angle of the engine.

sleepy gomez · « **Reply #4 on:** June 12, 2009, 05:23:37 PM »

Phil, I don't understand how a centered wing can vary the MAC. Thanks

Frank Sheridan · « **Reply #5 on:** June 12, 2009, 07:09:08 PM »

If I am Building a profile plane with equal span panels, then my engine thrustline is already outboard of the longitudinal center of the aircraft by 1/2 of the thickness of the fuselage. So if I am using 1/2" balsa with 1/8" doublers, then the thrustline is offseting the span by 3/8" with the inboard section being the longer of the two. But for this offset to be correct wouldn't the vertical stabilizer have to be mounted on the outboard face of the fuselage slab with a 1/8" shim as well? And if you think about it, doesn't the area of the airfoil that is covered by the fuselage plank get obscured from airflow thereby negating it from the span formulas? I think I'm getting a headache now.

Richard Grogan · « **Reply #6 on:** June 12, 2009, 07:21:12 PM »

Quote from: Frank Sheridan on June 12, 2009, 07:09:08 PM

I think I'm getting a headache now.

I bet!

Frank Sheridan · « **Reply #7 on:** June 13, 2009, 12:26:30 PM »

I knew that if I sat and thought about it I would come up with more questions than answers. On a profile plane the physical location of the side mounted engine will disrupt the stream of air flowing over the airfoil at the wing root, thereby changing the lift coefficient (I always wanted to use that word) of the first couple of inches of the outboard section near the fuselage. Without wind tunnel testing, we will never know what effect this has on the lift of that small section. For the sake of this discussion, I will assume (probably mistakenly so) that this affected section of airfoil can be dropped from the lifting wingspan. So now we have an equal span profile model that has approximately two inches of the outboard span removed at the root. This would effectively yield an unequal span wing, with the longer section being on the inboard side. And now I will go mow the lawn and enjoy some mindless physical activity.

EddyR · « **Reply #8 on:** June 15, 2009, 09:45:51 PM »

I believe Walter Musciano was the first to publish a model with a asymmetrical wing. In his article on his Stunt Streak he said it was built that way so it didn't need tip weight.
Ed

Mark Scarborough · « **Reply #9 on:** June 15, 2009, 10:51:21 PM »

Frank,
go a step beyond the engine, what about that razor blade spinning in front of the motor, that too disturbs the airflow over the wing,, flaps and tail,, some of that disruption is a positive, and some is a negative....

John Witt · « **Reply #10 on:** July 04, 2009, 09:23:41 PM »

Think about a wing on the end of the wires with no fuselage. Aerodynamic center (center of lift) is offset to the outside because the air is moving faster over the outer part of the wing and producing more lift. When the outer panel is shortened, it is an effort to move the fuselage onto the center of lift--exactly the same as just moving the fuselage outward on the wing.

What is of concern, I think, is to not produce any rolling moment that would decrease line tension. Seems to me that you get that with equal lift force, which is area x velocity, on each wing half. My assumption was that the tip weight was just to counterbalance the weight of the lines and leadouts.

But, if most successful PA airplanes are now using symmetrical layouts, I must be missing something in the above thoughts. I suspect some of the mystery comes from the behavior of the wing at very high angles of attack, approaching or exceeding an accelerated stall, when making square corners. I suspect also that control of the leading edge shape is very important in getting good performance in that regime.

What do you say, guys?

John

Paul Smith · « **Reply #11 on:** July 05, 2009, 10:08:53 AM »

Let us assume that you have a model with a 4-foot wing span flying 60 MPH on 60-foot lines:

Inboard tip is going 58 MPH.
Inboard midpoint = 59 MPH.
Engine = 60 MPH,
Outboard Midpoint = 61 MPH.
Outboard Tip = 62 MPH.

Obviously, the outboard wing is producing substatially more lift and it will roll in at you.

There are two ways to avoid eating the plane:

Add on enough tip weight to nullify the excess lift on the outboard side, or

Locate the fuselage at the true CENTRE OF LIFT, which is about 3/4" to 1.5" inches outboard of the actual centre of the wing.

The latter technique makes the model look lopsided (but everybody's used to it), but saves A LOT of tip weight.

I tried "equal spans" and I didn't like it one bit. I ended up sawing off 1.5 inches of outside wing.

PS: Kit makers often use equal spans to minimize tooling cost.

Bob Reeves · « **Reply #12 on:** July 05, 2009, 05:48:58 PM »

Some time ago I read about Netzeband's figures and have been putting right at 1/2 inch offset in every stunt ship I've built. Thinking this was a good compromise between allot and none it has worked out pretty good for me.

John Miller · « **Reply #13 on:** July 05, 2009, 05:51:10 PM »

I agree, I've been using 3/4 to 1 inch, depending on the size of the plane.

John Witt · « **Reply #14 on:** July 05, 2009, 08:55:44 PM »

My Panther has one inch offset, which is half of one rib bay. I suspect that was for building convenience on the original kit. I generated the rib templates using Solidworks CAD, so I could have gotten any amount. I did the velocity calculation mentioned by Paul and the proportion worked out very close to an inch, so that looked good to me. Seems to work OK, but I'm not a very discriminating pilot, at least right now.

Another thing just occurred to me and that is that the Panther has a swept leading edge and an elliptical shape tip and flaps that are different lengths, areas and shapes. I did not make any attempt to balance the areas, but the plane seems to fly pretty level in turns, in spite of any design/construction errors.

John

Dennis Adamisin · « **Reply #15 on:** July 06, 2009, 07:23:48 AM »

Early CL designs seem to focus on improving line tension by inducing rolling, specifially rolling OUT. The math (per Paul's post) shows why they thought that way too. Bob Palmer even incuded ideas like over/under leadouts and differential flaps. These methods all induced roll but also cause more DRAG on I/B side causing yaw-in. I have a 1959 (?) vintage Sterling Spitfire with about 2" assymetry. It flies well, and delivers good line tension pretty much everywhere, but is somewhat prone to "hinging". That looks a little funny but really does not hurt anything too much.

Some folks have had success by taking the "Classic" era birds and gently enlarging the flap. Thus they try to neutralize the effect of the smaller o/b wing by widening the chord (wider flap) increasing the drag when the controlls are applied (yaw-out) and then make up for the increased lift by simply carrying an extra slug of tip weight.

Back to Paul's math, and using the case of an equal span wing; the differences in speed betwen the i/b and o/b wing mean that there will always be "excess" lift on the o/b wing (easily countered by tip weight) and extra drag inducing a yaw-out (generally seen as a good thing

)

I have gone back and forth betwen equal span and 1" assymetry looking for a "magic bullet". Lately I've been building all equal spans...

John Witt · « **Reply #16 on:** July 07, 2009, 10:01:48 PM »

Assuming equal spans, and the outboard wing with a slightly faster airspeed and consequently more lift, wouldn't the outboard wing also have more induced drag (drag from producing lift) producing a yaw outwards? This would possibly (no calculations by me) counteract any tendency to roll in. Experience may show otherwise, but we don't really have a lot of good data. Wind tunnel anyone? A wind tunnel with variable flow across the test section--that would be cool.

Of course a u-control plane is set up with pretty neutral stability and nearly (in a lot of cases) no rudder area so there may be very little rolling moment induced by the yaw, which leaves the asymmetrical lift dominating. Seems like this might vary with the airplane design, but perhaps there is another handle to tweak.

In an extreme case of asymmetrical lift, I flew sailplanes for a number of years and the combination of tight turns at high bank angles can slow the inboard wing enough to cause a stall. This can cause an over-the-top spin entry, which I have done in Schweitzer 1-23G. However that is over 60 ft of span and with the controls crossed, which is a set up for a spin anyway, but the thing sure did a nice quick roll over. I had a later plane, a PIK-20B, which in the same situation, would just have the nose fall through and start flying again, with no roll or spin. Much better airplane, but 20 years newer design. There is at least one B-52 crash because of this.

John

Bill Little · « **Reply #17 on:** July 09, 2009, 02:46:29 PM »

I want to throw something out there, guys. Well two things:

1. Billy Werwage used a good bit of asymmetry in the '59 Ares, and others, but NO tip weight. (many I-Beamers were the same, no tip weight).

2. The TEOSAWKI has 1/2" LONGER outboard panel. That plane, by all accounts, is a tremendous flying profile.

I ain't no scientist, but I find this interesting.

Thanks!
Big Bear

John Witt · « **Reply #18 on:** July 09, 2009, 04:42:12 PM »

Teosawki--Could be the actual center of mass is still on the MAC if the engine and tank are on the outside of the profile fuselage, which could make the airplane functionally symmetrical. I don't know that model, so is just a thought.

If you calculate the g-force on a model turning a 10 ft radius turn at 50 mph, you have around 20 g. That's a lot of magnification of any tip weight that is greater than the line/leadouts.

John

Paul Smith · « **Reply #19 on:** July 09, 2009, 06:34:24 PM »

Line length comes into the deal, too.

In the past, people had to fly on 50-to-60 lines. Now with the power of 15 cc's, you can go up to the full-legal 70 feet of flying radius. With the longer lines, the percent differance in inboard/outboard wing speed is less, so you won't need quite as much offset to get the CG aligned to the CL.

I still believe that when you load up the G's, as in a sqaure corner, the outboard wing will roll up on you and the induced drag, which pulls the ouboard wing back will help some, but not enough. If you have an efficient design with a L/D ratio of 8, the roll that hurts you will be 8 times greater than the drag that helps.

If you don't believe in shortening the outboard wing, then build equal wings and offset the body 3/4" outboard.

L0U CRANE · « **Reply #20 on:** July 22, 2009, 12:37:03 AM »

Okay, let's look at the basic condition, eh?

Lift (and "induced" drag, which results from producing lift) calculate from velocity. The velocity term is squared. If there is a 10% increase in velocity, both lift and drag increase (1.1 squared = 1.21) over 20% compared to what they were at the original velocity. With larger changes, the effect is even greater, of course: 25% increased V causes over a 50% increase in lift and Induced drag (1.25 squared is 1.5625).

Velocity of a CL model depends on the flight radius and laptime. If the flier's chest is the center of the model's flight circle, we have about 2' from there to the handle, we have the lines length, eye to eye, and the length of the leadouts before we get to the spanwise CG, more or less. Say we have 60' lines I-to-I, and a 2' half span for leadouts. That's arm + lines + leadouts of 64'.

The outboard tip is another ~2' out, at 66' from the center. The inboard tip is, well, duhh, 62' from the center. 66' is 106.5% of 62'. So lift and drag at the outboard tip are 13.5% more than at the inboard tip. But there is also a steady variation of velocity across the entire span - it isn't just at the tips. Somewhere between the tips is a point where lift and drag are 'balanced' each side of the point... ...Where drag doesn't cause yaw, and where lift doesn't cause roll...

A friend who speaks calculus convinced me that the way to find that "dynamic mid-point" is to cube the inboard radius, cube the outboard radius, average them, and find the cube root of the average. Oooh, scary, right? Nah...

62 ^3 = 238328

66 ^3 = 287496

Average = 262912

Cube root of Average = 64.0625

.0625 feet = 0.75"

Since the model in this example has a 4' span, the structural midpoint would have been at 64' radius. It calculates 3/4" further outboard. Now, remember, this is how far the midpoint shifts; that means the inboard panel is ~1.50" longer than the outboard at the point where drag and lift are balanced inboard and outboard...

Other things are involved, of course, but this is the basic idea behind different panel lengths.