Mar 23, 2012, 05:14 PM General Apocalyptics Inc. Joined Jan 2007 3,343 Posts Calculating MAC for twisted swept wings To keep the technical discussion out of Herk's experimental thread, I thought I'd start a new thread. Over the last few weeks I've had a few enlightening moments on the issues that seem to come up with the CG placement on swept flying wings with different lift distributions. The main problem is the Mean Aerodynamic Chord everyone is calculating, using methods developed for wings with small sweep angles and minimal twist, just don't work on a highly swept and twisted flying wing. First, a definition: "The mean aerodynamic chord is the the chord of an imaginary airfoil which throughout the normal flight envleope has the same force vectors as the three dimensional wing." (Perkins and Hage, "Aircraft Performance, Stability, and Control", P89). So the MAC is the chord line we can imagine all the lift, drag, and moment vectors acting through. To see that the lift distribution of a swept, twisted wing is going to effect the position of the MAC, try this little thought experiment: We calculate the MAC position conventionally for a swept wing planform, which assumes a near elliptical lift distribution for the wing. Now, lets change the lift distribution to a Horten Bell Shaped Lift Distribution (BSLD) by changing the twist and keeping the planform the same. The tips are now lifting less and the middle has to lift more to compensate, resulting in the BSLD. Doesn't the point where the lift is acting through move inboard? By definition of MAC, the lift vector must act through the MAC, so the MAC must move inboard as well. If the MAC moves inboard, then it must move forward as well. If you calculate your MAC and CG position conventionally with a Horten lift distribution wing, you will have the CG too far back. On the other hand, if you have a lift distribution that has the tips more highly loaded due to the induced angle of attack because of the sweep, and the centre with less load because of the hole in the lift distribution due to the sweep, won't the MAC move outward? If the MAC moves outward, it will also move back. So a CG at 25% of the conventional MAC may actually be at 21% of the "real" MAC. Unfortunately, calculating the "real" MAC of a swept and twisted flying wing is not as easy as finding the conventional MAC. From Etkin: MAC is defined by: xbar = 2/(C_L*S) * int(C_l_a*c*x*dy, from 0 to b/2) Not elegantly formatted. C_l_a is the section additional lift coefficient (from the spanwise lift distribution of the untwisted wing) x is the chordwise coordinate of the section aerodynamic centre c = chord y = span variable Integrate the location of the section AC weighted by the chord as well as how important that section is to the lift distribution. Do this integration along the span from y = 0 to y = b/2. Perhaps the best that can be done is some rules of thumb for where the MAC is located with a BSLD or a Northrop type lift distribution. I think Herk's testing is showing this effect, by requiring a CG shift for adequate stability when the twist is changed without altering the planform. Kevin
 Mar 23, 2012, 05:43 PM Herk Virginia USA Joined Jun 2007 1,469 Posts Thanks Kevin, This is very interesting. I think it would be especially interesting to see this actually calculated and the longitudinal stability derivative determined for an aircraft for which the data is available and in flight results known. I don't mean big project. A number of very precise Horten designs have been documented and reported on this forum and also on the sailplane forum.
 Mar 23, 2012, 06:02 PM General Apocalyptics Inc. Joined Jan 2007 3,343 Posts I'd have to remember how to integrate! It is so long ago... I think just doing a step-wise integration by even measuring the area under the lift distribution form an XFLR5 plot would work. It is essentially just finding the centroid of the lift distribution. The Horten BSLD MAC location could be estimated by just removing the tips where the local lift approaches zero. The Horten "real" MAC will be well inboard and forward of a MAC calculated conventionally, and the CG must be forward as well. Horten wings essentially perform the same as a wing with less span. If span is a limitation, they are at a disadvantage compared to a wing with tip fins and a more elliptical lift distribution. But the Horten wing will have the same bending moment at the centre as a smaller span wing with an elliptical LD, so in theory they carry the extra span for free. A 3M Horten wing should likely be compared performance wise to a 2.4m or so wing with tip fins. I'll take a shot at calculating a Horten MAC tonight maybe. Kevin
Mar 23, 2012, 09:22 PM
Joined Aug 2006
2,098 Posts
Quote:
 Originally Posted by kcaldwel I'd have to remember how to integrate! It is so long ago...
If you really want to do the integration the whole thing is here scroll down to "Lifting-Line Solution for the Bell-Shaped Lift Distribution" by Reinhold Stadler. It doesn't take sweep effect into account but Reimar Horten couldn't have done that either because nobody could until the mid '50s. His middle effect hypothesis was wrong and ironically proven wrong by his brother in law flying a Horten airplane. The turbulence that Horten predicted at the center section simply doesn't occur. The lift valley is due to decreased circulation which is in turn due to the spanwise migration of the attachment line. He also failed to identify the lift mountains. It's a testament to hard work that he solved the sweep effect problem without a correct theoretical basis.

Quote:
 It is essentially just finding the centroid of the lift distribution.
That's been done for an elliptical lift distribution and 3 bells. Of course there is one unique centroid for any given shape but an infinite number of polygons that can be drawn around a given center point. So just having the CG in a specific spot doesn't insure that the lift distribution is a bell

--Norm
Last edited by nmasters; Mar 25, 2012 at 11:39 AM.
Mar 23, 2012, 11:47 PM
General Apocalyptics Inc.
Joined Jan 2007
3,343 Posts
Quote:
 Originally Posted by nmasters If you really want to do the integration the whole thing is here scroll down to "Lifting-Line Solution for the Bell-Shaped Lift Distribution" by Reinhold Stadler. It doesn't take sweep effect into account but Reimar Horten couldn't have done that either because nobody could until the mid '50s. His middle effect hypothesis was wrong and ironically proven wrong by his brother in law flying a Horten airplane. The turbulence that Horten predicted at the center section simply doesn't occur. The lift valley is due to decreased circulation which is in turn due to the spanwise migration of the attachment line. He also failed to identify the lift mountains. It's a testament to hard work that he solved the sweep effect problem without a correct theoretical basis. That's been done for an elliptical lift distribution and 3 bells. Of course there is one unique centroid for any given shape but an infinite number of polygons that can be drawn around a given center point. So just having the CG in a specific spot doesn't insure that the lift distribution is a bell --Norm
Thanks Norm!

I have seen your post before, but I never really clued into what it was saying until the last few days.

I knew some smart people must have been down this road before me and left breadcrumbs, but it has been fun actually understanding it. And now I don't have to disappear into the Great Integral Swamp, where I'm sure I'd be lost in seconds these days.

I wonder where the "real" MAC ends up with the Northrop lift distribution. It is quite far from elliptical or BSLD by the looks of it. XFLR5 shows that the N9M is pitch stable with the CG at 25% MAC, especially at low alpha. I'll CNC a wing soon, just to try it.

Does anyone know if the N9M just had linear twist from root to tip?

Kevin
Mar 24, 2012, 01:03 AM
Joined Aug 2006
2,098 Posts
Quote:
 Originally Posted by kcaldwel I wonder where the "real" MAC ends up with the Northrop lift distribution. It is quite far from elliptical or BSLD by the looks of it. XFLR5 shows that the N9M is pitch stable with the CG at 25% MAC, especially at low alpha. I'll CNC a wing soon, just to try it. Does anyone know if the N9M just had linear twist from root to tip?
Jack Northrop was a fanatic for efficiency. It's very unlikely that he would have intentionaly deviated from an elliptical lift distribution. I used data the N-9M as the test case for a spreadsheet incorporating the Panknin twist formula and altitude adjusting CL. I got the data from "The Flying Wings of Jack Northrop" by G Pape, J Campbell and D Campbel. I don't have the book here so I can't look it up but as I recall there are several dimensioned drawings in there and at least one shows MAC and the normal CG position. I think it was farther aft than 25%. If you can find an accurate drawing with the CG on it just draw the abscissa of the CG out to an intersection with the 1/4 chord line. If the abscissa crosses the 1/4c at 42% of the half span then you know that it was intended to have an elliptical lift distribution. I didn't check it because I'm pretty sure that span efficiency ruled his world. He really had completely different ideas about how this type of plane should work than the Hortens.
 Mar 24, 2012, 01:15 AM internet gadfly Colorado Joined Aug 2006 2,098 Posts Well that was kind of muddled. Northrop didn't understand sweep effect any better than anybody else at that time. Even though R. Horten's middle effect theory was essentially wrong it was based on the observation that swept wings develop a negative pitching moment that's independent of the airfoils used. This lead to the conclusion that there must be some lose of lift in the center section and that lead to a twist and area distribution that would produce bell shaped lift distribution according to the lifting line theory. There, clear as mud Last edited by nmasters; Mar 24, 2012 at 02:12 PM.
Mar 24, 2012, 12:10 PM
less is more
United States, CA, Marina
Joined Sep 2006
2,415 Posts
Quote:
 Originally Posted by kcaldwel A 3M Horten wing should likely be compared performance wise to a 2.4m or so wing with tip fins. Kevin
This may be true although I'd argue it's more like 2.8m. Have you ever consider the benefit of those highly washed out Horten wing tips in terms of reducing wing tip vortices losses?

Just a thought.

Kent
Mar 24, 2012, 12:21 PM
Herk
Virginia USA
Joined Jun 2007
1,469 Posts
Quote:
 Originally Posted by Knoll53 This may be true although I'd argue it's more like 2.8m. Have you ever consider the benefit of those highly washed out Horten wing tips in terms of reducing wing tip vortices losses? Just a thought. Kent
I agree Kent - the Selinger/Horten book (in the design data section near the end) shows induced drag charts. I don't fully understand them (still working on the German) but they seem to indicate a negative induced drag near the tips. I'm still trying to figure out exactly what this means, but from the charts those unloaded tips do seem create a significant reduction in induced drag. Of course a fin could have some similar effect too. So which is better - the profile drag of the Horten tip or the profile drag of a fin.

All of aircraft design is compromise, but I'm nor sure that I want yet to vote on this. I think it depends somewhat on the flight characteristics most desired by the designer.
 Mar 24, 2012, 12:25 PM Herk Virginia USA Joined Jun 2007 1,469 Posts Norm - thanks for mentioning the Northrop book. I have always wanted one. Apparently the publisher reissued the book in the mid 90s. It is in print and available on Amazon.com. http://www.amazon.com/Northrop-Flyin...8&sr=8-3-fkmr2 I think I will be getting one.
Mar 25, 2012, 01:25 AM
General Apocalyptics Inc.
Joined Jan 2007
3,343 Posts
I just remembered a really easy way to integrate the lift distribution to find the MAC, without using math.

It requires a plot of the lift distribution, a pair of scissors, and a pin.

Plot out the local lift on the Y axis, and semi-span on the horizontal. XFLR5 will plot that. I made the background white to save ink, and changed the plot settings to make the plot as big as possible, and with a finer x-axis grid for reference.

Cut out the local lift distribution and the x and y axis. Use your pin along the top edge to find the point where it hangs with the x axis horizontal. That is the span-wise location of the MAC for that lift distribution. You can do it for different AoAs to see how the MAC moves.

My first try at measuring the MAC of an elliptical distribution ended up at 41.9%. Norms drawing shows it at 42.1%. Not bad integration for a pair of scissors and a pin!

http://static.rcgroups.net/forums/at...-184-MAC-2.png

For the N9M with 3 degrees of washout (XFLR5 doesn't show a trim for 4 degrees of washout) at it's trim of 3.5 degrees AoA shows the MAC at 38.9%. Now I'm confused about how it can be stable with the CG at 25% again...

XFLR5 doesn't model the 3D flow so the local lift distribution is only approximate, but it still shows a stable pitch curve and trim point with the CG at 25%.

I'll have to try the Horten distributions tomorrow, and see how close I come to the numbers from the integration.

Kevin

# Images

 Mar 25, 2012, 02:34 AM internet gadfly Colorado Joined Aug 2006 2,098 Posts BTW if you twist a trapezoid and cut it up into strips (the reference lines of the ribs of a tapered wing) and plot the angle of those cut lines the curve through those graphed points is curve that's nearly flat at the fat end of the trapezoid and much steeper at the skinny end. ie even though the leading and trailing edges are strait lines the washout is not linear. I'll post a graph tomorrow if somebody else doesn't beat me to it. --Norm Last edited by nmasters; Mar 25, 2012 at 11:44 AM. Reason: Corected my hillbily spelin'
Mar 25, 2012, 01:45 PM
General Apocalyptics Inc.
Joined Jan 2007
3,343 Posts
I cut-out the BSLD sin^4, and got the MAC at 31.5%. I'm out by 0.9% from the integrated results. I guess that is acceptable error for a pin and scissors.

I'm still scratching my head over how the 25% CG position works on the N9M though. It seems to be behind the lift centre. I must be doing something wrong. The wind tunnel tests and XFLR5 both show the N9M is stable with the CG at the 1/4c of the mean geometrical chord. I wonder if that Canopy kills enough of the centre lift to move the MAC outboard?

Kevin

# Images

 Mar 26, 2012, 11:35 AM General Apocalyptics Inc. Joined Jan 2007 3,343 Posts I have found one flying wing that appears to fly with the CG very near the "geometric aerodynamic centre". I am a bit unsure what terms to use. I'm ending up with a lot of aerodynamic centres! That is the term on Norm's drawing here: http://static.rcgroups.net/forums/at...-184-MAC-2.png Biber's wonderful Mulitbumm (I hope that means something different in German...) looks to me as if it was flying with the CG just about on the 1/4c point of the average geometric chord, well maybe 1.6mm ahead of it: http://home.arcor.de/andre.kubasik/mubu/whatismubu.html I calculate the 1/4c of the average geometric chord to be 191.7mm back from the LE apex, and Biber reported they were flying with the CG at 190mm: http://www.rcgroups.com/forums/showp...6&postcount=92 This is far from a typical flying wing though. It has tip fins, 0.5 degrees of washout in the tip only, negative Cm cambered airfoils, a centre elevator, and basically zero static margin. It was somewhat heartening to find a flying wing that actually worked with the CG near the 1/4c of the average geometric chord. It also seems to confirm that it would be extremely difficult to make a flying wing like the N9M/YB-49 work with it's CG at the 1/4c of the average geometric chord - it has no tip fins, and must have been somewhat pitch stable. They are reported to have 4 degrees of twist with symmetrical airfoils, which would require the CG be ahead of the "real" aerodynamic centre. And from the XFLR5 analysis, it looks like the "real" MAC should be inboard and forward of even the elliptical MAC location. I suspect Herk was right, and that there is no way it can fly with the CG as reported. I'm off to hot wire a 60cm span chuck glider version to see what it looks like though. Kevin Last edited by kcaldwel; Mar 26, 2012 at 11:44 AM.
Mar 26, 2012, 12:50 PM
Herk
Virginia USA
Joined Jun 2007
1,469 Posts
Hey Kevin, My copy of the Northrop book will arrive sometime today. If there is anything useful about the N9M or Bomber in there I'll pass it along.

Here are two models from my shop. Both have fins and both fly with a relatively small SM. The white one is a kit bashed DAW ME-163 made to look sorta-like a DeHaveland Swift. The other is my old HL wing - same one as my avatar.

I remeasured them very carefully today and the DAW model is flying with calculated 6%SM. The handlaunch is actually at 4%SM - which is only about a third of an inch ahead of the calculated neutral point. When I built the HL model the starting SM was about 6% and I tried it on the slope without fins. It was not really controllable. If I knew then what I know now, I would have started with the calculated SM in the 15-20% range. I don't know if that would have worked, but I think the chances would have been better.