Ive read a lot of things here on RCU trying to figure this out, but how do you find the CG of a biplane? The plane is somewhat scratch built, although it started life as a BUSA flybaby. After several apparent crashes, things are not as they once were. I have no plans, etc for this model (even if i did, i dont know that they would be much help). Im not an aerodynamic engineer so alot of the explantions are way over my head. I downloaded some software called CMEI that allows me to find the Mean Aerodynamic Chord (whats this?) of a single wing, but what about a double wing? It also seems to need the CG to use? Well, I dont know the CG.. =) Can someone explain in plain english how to go about figuring this out? Another thing I have noticed is that it seems that the actual fuselage has no bearing on determining the CG. Is it based entirely on the wing structure? I really appreciate any input you guys can give me here. I am at a loss!
The calculator I used gave me these results:
Sweep Distance @ MAC (C) = 1
Mean Aerodynamic Chord (MAC) = 8
MAC Distance from Root (d) = 21.5
Balance Point @ Root Chord (CG) = 3

Is the the Balance point the CG? Is that 3 inches from the trailing or leading edge? Also MOST IMPORTANT this is a BIPLANE. So could someone please help me and explain in plain english? So is the CG in the calculation measured on the upper wing or lower wing? Does the stagger effect the cg? I have been to all the calculators and they are no help to me since I dont understand what they mean and none so far have applied to bipes. Links to more calculators honestly, while thoughtful, are pretty useless to me.
Brice D

This may sound WAY to simple, but it works for a good starting point. Place the plane on the ground in a level position. Stand over the plane and make a mental rectangle of the wing planform. Go from the most forward of the wings to the most rearward of the wings. Using this rectangle, figure 25% of the distance from the most forward portion of the rectangle. You can go as far back as 28% for a strating point, but NO MORE.

It's a starting point that has always worked for me idf nothing else was available. I hate doing "plane" geometry You can adjust the C/G to your taste from there after flying it.

A good question, that.
I used to know how to do that, but a recent thread has pointed to problems with the conventional plan-view method.
Looking into it as we speak.
Anyone got a flying Pitts/Eagle?
Where's the c.g. relative to the leading edge of the -lower- wing?

canopy2k,
Hi, I looked into this when researching for my book "Basic Aeronautics for Modellers". I did some working out and found some stuff on the web to back it up. That website has disappeared now (anyone know if 'The designers & builders online Workshop' has relocated somewhere?).
I don't know how to get my diagrams on the Forum but does this text help?

Biplanes
Use the following procedure to find the mean chord of a biplane, or a sesquiplane with unequal wings. For example the Fokker DVII has two unequal wings with normal (positive) stagger, i.e. the top wing is ahead of the bottom wing. I worked out that the top wing supplied 61% of the total wing area, and the bottom wing obviously 39%.
Referring now to Figure 22.7, I joined the mean chord lines of the top and bottom wings. I then divided the gap between the wings in the ratio of the wing areas, 39:61. The mean chord is 39% of the gap from the upper wing (the bigger one). (Some authors bias the mean chord towards the top wing but I have kept it simpler and made generous allowances elsewhere.)

[img][/img]

The CG formulae in Chapter 8 do not apply to biplanes because there is nearly twice as much downwash as the formulae allow for, and twice as much interference to the airflow, i.e. loss of airspeed over the tail. However, I have found that if you reduce both the Tail Volume and the wing Aspect Ratio by dividing them by the number of wings, then putting these reduced numbers into the original formula gives a sensible CG.
In the case of the Fokker, I called the top wing 1 and the lower wing is 0.64 of it, so the number of wings is 1.64. That gave a reduced AR of 3.3 and a reduced V-bar of 0.24 which in my usual Chapter 8 formula gave a CG position at 18% of mean chord, which gave satisfactory handling.

If one wing is swept as on a Pitts, or both are swept like a Tiger Moth, remember to draw on the side view the MEAN Chord of each wing, not its root chord.

Alasdair

Alasdair, after looking into this stuff, I've come back to the graphical method as presented at the Palos r/c site as adequate, and figuring only the wings seen in the plan view as the parameters..
The attached Pitts demonstrates two items... I found a plan for a Pitts in Model Airplane News, March '95 which shows the c.g. at the leading edge of the lower wing.
The Palos method gives this same location for a 30% c.g.
If the "mean wing" is located between the wings per the areas, this ratio is about 97% for a typical Pitts.
Where does the "mean wing" then go?
Another method pointed out by acropilot_ty uses only the stagger between the wings to locate the "mean wing".
Using this method, which ignores area altogether, the "30%" point is always aft of the 30% point found graphically.
As I've mentioned with my biplane Kadet, the lower wing has a "setback" from the upper wing of 3", with a chord of 12" on the lover wing. Upper wing chord is 15"This places the lower wing area completely -under- the upper wing area; cannot be seen in a plan view. The plane can fly sans lower wing and no other changes.
Any method other than the graphical computes a c.g. way too far aft for stability for this biplane/monoplane.
2nd image as a biplane
3rd image as a monoplane.

Tall Paul,
I didn’t find the Palos site very helpful. They had a good diagram for finding the MAC and aerodynamic centre of a wing, but I think they were a bit vague about getting a CG position. I can’t find my print out. I’ll have another look.

What exactly are we trying to find?
The Aerodynamic Centre of a pair of wings combined as a biplane.
That is defined as the point about which their combined pitching moment does not change with Angle of Attack.

Your method – the Palos website method – many unscientific bodges invented by modellers, may get a usable answer some of the time or possibly even often, but they are not science. They luck assisted by experience.

Similarly your Kadet with the optional extra wing proves nothing.
“The plane can fly sans lower wing and no other changes. “
So What? An anecdote with no experimental aerodynamic data is not science.

"Any method other than the graphical computes a c.g. way too far aft for stability for this biplane/monoplane."
Then the methods you used were unreliable. What were they?

I looked up all the textbooks in my study, and the only one which gives a method for finding the MAC and Aero Centre for a biplane combination is the well known “The Design of the Aeroplane� by Darrol Stinton. His method certainly looks convincing as science, and is remarkable similar to the one on the “Designers and Builders� website (the article is BeDesign Number 17) and the one in my book.

Figure 1

Consider this. I want to make a biplane from two identical wings of chord 4 inches. (draw it out). I place them one above the other, with the leading edges of both on the datum mark, and the AC of the combination is 1� aft of datum. By either method.

I now decide to mount the lower wing staggered aft one chord length, 4�. The mean chord is half way between the upper and lower wings. Its quarter chord point is 3� aft of datum.
By the method you advocate you will draw a chord 8� long and place the AC at 2� aft of the datum. That’s an inch too far back. That’s an error of 25% of the wing’s chord, quite a lot.

Suppose you have this pair of wings flying along with a lift force on each (at 25% chord of each?) and the plane meets an upset which increases the AoA very slightly. At that instant the lift on each wing increases, and the increase is at 25% chord., one 1� aft of datum and the other 5� aft of datum. The sum of these two small lift increases will act where? Where does the resultant of two small and almost equal forces act? That is the definition of the Aerodynamic Centre.

If you answer is not 3� (or even 2.9�) aft of the datum, I give up.

A workable CG is all many people want, but I want to get the reasons right too, so that I can justify using it in my book.
Best regards,
Alasdair

"Consider this. I want to make a biplane from two identical wings of chord 4 inches. (draw it out). I place them one above the other, with the leading edges of both on the datum mark, and the AC of the combination is 1� aft of datum. By either method.

I now decide to mount the lower wing staggered aft one chord length, 4�. The mean chord is half way between the upper and lower wings. Its quarter chord point is 3� aft of datum.
By the method you advocate you will draw a chord 8� long and place the AC at 2� aft of the datum. That’s an inch too far back. That’s an error of 25% of the wing’s chord, quite a lot."
.
You mean like this?

At $79.41 for a copy at Amazon, I won't be getting "The Design of the Aeroplane" anytime soon.
"Designers and Builders" has 1000s of hits... which one is it?

Tall Paul,
I could not find that Designers and Builders workshop again either. Lucky I printed out the pages last time.

Your top diagram is the one I was trying to convey, only you have it labelled wrong. The 8" dimension is NOT the true mean chord.

The 4" line in between the top and bottom wings IS the true mean chord.
Its length does NOT depend on the separation between the two wings. As drawn (correctly) the lines are parallel so the distance between them is constant. Even when the two wings have different chord lengths, the mean chord lies half way, or maybe a third of the way, down (a proportion depending on relative areas) so the gap does not affect the mean chord.

How do you get these diagrams on the Forum page? I don't know how to post them. I have them drawn out in a Word file which I could email to you if you like?

Continue thus:- On the top diagram put a small lift force at c/4 on each wing. They represent the additional lift caused by a small AoA increase. Now find the resultant of these two forces. Draw it on. See where it falls. By definition this resultant acts at the Aerodynamic Centre of the wing pair. The aerodynamic Centre is traditionally at the quarter chord point of the Mean Aerodynamic Chord. That doesn't work by your method. Therefore your method is wrong.

By my method the resultant turns out at the c/4 of the mean chord, and the Moment on the wing equals 1/2rho V^2 times total area times my mean chord times the moment coefficient of the section. That works as it should. Your method gives the wrong moment because you use the wrong mean chord.

Don't you see? It doesn't work. It may be useful as a quick dodge to guess a CG, but it is unscientific. You can't use it to calculate anything. If you use it to calculate the Tail Volume Ratio you will get the wrong answer.
Alasdair

Alasdair, I spent a few hours yesterday looking for "biplane design" on the 'net. Lots of hits but none too useful for this subject.
I found one which uses something similar to the "stagger method" combined with area ratioing... it's lurking somewhere on my computer if I can find it.
Found dit...
http://www.weebeastie.com/hatzcb1/BratzCG.html
However... the example drawing.. sans tail, it's either a canard with a very large foreplane, or a tandem.
As a tandem, each wing lifts equally. The front wing also has to counteract the nose-down pitch of both airfoils, if they're cambered, so it is loaded higher. The center of the lifts is nominally ahead of the c.g. to counter the nose-up lift of the aft wing.
.
I see the example as a no-gap tandem/canard, or a 100% stagger biplane.
As a biplane, I would rely on the graphical (unscientific) method to find where the c.g. is.
M.a.c. is nice to know, but it's the c.g that the airplane sees.
As a tandem/canard, I would expect the c.g. to be aft of the trailing edge of the front wing.
.
Image posting..
Chose the "REPLY" box at the lower right corner of the thread listing under the last message.
That brings up a different reply area than the reply area below the messages.
The lower left corner of that area has Upload Images!, which will let you browse your image sources for uploading.

Tall Paul,
Thanks for the tips.
I'll try to insert my diagram when I have more time, though yours was probably better anyway.

I see this wing pair as an extremely staggered biplane with a tail, let's say a tail 15% of the total wing area with its c/4 15" aft of the datum, the LE of the leading wing.

Using your method I believe that the CG will come out a bit far aft. Using mine probably a little forward, but that's the safe side. That is, unless you make another mistake to compensate.

Interesting web page you took me to.
On it, item number 4 locates the AC of the mean chord of the pair of wings just the way I do, I noticed. He has located the LE of the mean chord by proportioning the stagger. He then gets pretty mixed up between Centre of Pressure and Centre of Gravity. Ugly.
(The C of P runs from infinitely far behind a cambered section at zero lift, forward almost to the c/4 (say 27% or 30%) just before the stall, and then usually aft again when the wing stalls)

Dinner time now, must go,
Regards
Alasdair

Alasdair, I've been playing....
Here's Hatz's c.g. drawing, "enhanced".

Model Aviation, August '77 had a construction on the Hatz Biplane..
Anyone got a copy.. to scan...

It's available on the AMA website in the members only section.

If you're not a member, I can email it.

Thanks, I had a "senior moment" there.. I recalled it probably was available.. got it..
The c.g. range on the plan matches that on the full-scale site..
When searching, I was surprised to find how many of them there are!
I'd never heard of it.

Back in 89, when we did some Bucker Jungmans for the TOC, we also wondered about CG -best for aerobatics
The first flights, I watched as Rojecki flew some stuff and we commented on what looked good/bad.
The plane flew fine -tho a bit touchy- and when entering an upright spin it promptly fell through into an inverted slow flat spin.
It did this easily and predictably.
I said "waytail heavy"
Steve said "no"
on landings it floated a lot.
We changed cg to just aft the lower wing LE at the root. about a 2" forward shift ( 11.25" chord).
The plane was now very solid - did all the maneuvers and held spins - any type as input commanded.
I still fly that plane and have changed engines from a 40 cc to an 80 cc- -1-2 lb weight change but the plane still works nicely shifting from 16-18 lbs .
wing loading is in low -mid 20's ( 1790 sq in)
Bipes are strange creatures - The good cg range on this one is very broad.
BTW ,the model is exact scale except for 9% wing clip- a normal proceure on many full scale types.
(some full scale had fully sym airfoils -as did these.)

Tall Paul,
Nice wee aeroplane. A homebuild I assume?

We can find the Aerodynamic Centre of a wing or pair of wings but no way can we discuss CG without knowing details about the tail. I know nothing about this aircraft's tail, so how about adding a tail and discussing the CG of my hypothetical 100% staggered biplane which you also drew and I have tried to upload here.[img][/img]

To sumarise, we have two identical wings, each 4" chord, section does not matter but I drew approx Clark Y, with 4" normal stagger, and a tail whose area is 15% of the total wing area and whose c/4 is 15" aft of the datum (LE of top wing). Let's say for completeness that the wingspan is 26".

Where would the CG go according to the method proposed and used by Tall Paul and others?

Too far aft I would suspect, unless a second error compensates for the first.

Alasdair

Hey, the image upload worked, thanks again!

Tall Paul,
Nice wee aeroplane. A homebuild I assume?

We can find the Aerodynamic Centre of a wing or pair of wings but no way can we discuss CG without knowing details about the tail. I know nothing about this aircraft's tail, so how about adding a tail and discussing the CG of my hypothetical 100% staggered biplane which you also drew and I have tried to upload here.[img][/img]

To sumarise, we have two identical wings, each 4" chord, section does not matter but I drew approx Clark Y, with 4" normal stagger, and a tail whose area is 15% of the total wing area and whose c/4 is 15" aft of the datum (LE of top wing). Let's say for completeness that the wingspan is 26".

Where would the CG go according to the method proposed and used by Tall Paul and others?

Too far aft I would suspect, unless a second error compensates for the first.

Alasdair
.
Alasdair
I noted the position of the c.g on the Hatz biplane at the Bratz site where the range is calculated is given first in the usual manner... percentage back from the leading edge of the upper wing.
And the "usual manner" numbers are "..from 18% forward to 35% rearward".
Thru a certain amount of legerdermain, Brooks then locates the virtual mac aft of the leading edge of the upper wing. Then he uses the percentages to find the delta forward and aft of this m.a.c. for the c.gs. relative to the position of the virtual m.a.c. Percentages of a percentage.
The method used at the Palo site (and Jean Claude Etienble, and DJ Aerotech, Martin Simons, Appendix 1) OTOH considers the plan view area, and computes the m.a.c. graphically for a single bay wing, or using moments for multi-panel wings, based on that. The c.g. is then found.
When the percentage numbers of 18 and 35 are related to the plan view area of the wing as given by Brooks and with the Palo r/c method, the positions are identical between the method used by Brooks and the Palo method, which tells me the process "normally" used, based on plan view area, is correct.
Tail areas and moments are NOT involved because "general knowledge" of where the c.g. must be for stability is related to the wing shape only, IOW, for a normal wing-tail configuration.
25% m.a.c. for a stable not-too-manuverable plane, aft of 35% m.a.c. for an "exciting" plane.
In the neutral point and static margin computations (Martin Simons, Appendix 1) the horizontal tail efficiency contribution is a constant variable. "This must be estimated."
Again, IOW, if you have a plane which flies stably, by definition the static margin is "good". And it is aft of the c.g.. How far aft is a subject of what "good" is relative to the degree of stability of the airplane.
And it's a very subjective thing. Pilot skill and experience AND static margin contribute to stable flight.
For other shapes of airplanes, there's other positions for the c.g. based on what experience has found works.
Jiggling the constant variable then affirms the correctness of the position, if the plane flies.
.
The example plane above, a 30% c.g. would be at 2.4" aft of the upper wing leading edge.
.
(Later on, that same day, a plane appeared.... )
Toss testing found the c.g.s as marked by the pins.
Flies quite well with the lower wing back that far!

How about this?
The graphical method of locating the m.a.c. -of a wing- does only that.
There's more to it... wing position... canard, conventional, bipe... etc, but without knowing where the m.a.c. of the wing is to start with, nothing else can be done mathematically.

Tall Paul,
said "without knowing where the m.a.c. of the wing is to start with, nothing else can be done mathematically."

That was MY point, but no matter. I see you turn words into deeds. Well done.
By the way, my proposed CG range is 2.7" to 2.9" from datum. How does it fly at that?

2.7" cooresponds to 17.5% of the true MAC and aims for a Stability Margin of 15% (NP estimated at 32.5%).
My more rearward point at 2.9" aims for a reduced Stability Margin of 10% MAC. I would start at the front point and hope to move it back a little after flight tests.

If your method works for you, carry on. I'll stick to mine.
Where is that Palos site BTW?
Alasdair

The glider as monoplane balances at 2-1/2" aft of the top leading edge.
As the biplane, it balances at 3" aft.
The long tail has lot to do with the aftish normal c.g.
I'm going to do a tail-ectomy to get the c.g to 25% on the monoplane...
..
Moved the horizontal forward one chord... 3".. rebalanced as a monoplane to 25%.. 1"
Added the lower wing @ 4" aft, found the new c.g. without changing weight.. 1-5/8"
Glides well.
.
Palos r/c...
http://www.palosrc.com/instructors/cg.htm
Jean-Claude..
http://perso.wanadoo.fr/scherrer/mat...glish/mce.html
Ask DJ ...
http://www.djaerotech.com/dj_askjd/askdesign.html
Andy Lennon...
Page 8
Martin Simons
page 222..

The wind quieted down, got to toss the plane for some distance.. the flight results are attached...
I've been gathering many versions of how to compute static margin and trying to make sense of the results..
There's incoherence in the information right now between Simons, Lennon, Van Putte...
Have to make a more comprehensive spreadsheet.