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Sikorsky Ilya Muromets CG

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Old 06-11-2015, 12:30 PM
  #1  
Sal C.
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Default Sikorsky Ilya Muromets CG

Gentlemen.

Igor Sikorsky was an important figure in aeronautics. From the early days of designing fixed wing aircraft in Imperial Russia to his highly successful flying boats and Helicopter designs his namesake company still manufactures, there's no doubt about the mark he has left in Aviation History.
I am working on a very large model of one of his early four engine designs, the Ilya Muromets. This aircraft has made many "Firsts" , such as the first 4 engine transport, first to have heated cabin with electricity and toilet. First to carry 16 passengers aloft. This all took place in 1913!
When WW1 began, he was asked to convert his large aircraft into a Reconnaissance and Bomber aircraft. I am building a 1/6 version of the VEH type Bomber.
I imagine that one of the many attributes that scare off modelers is the very short nose. It just does not look like the aircraft can be balanced correctly.
I was sent this document from someone in St. Petersburg, Russia. It was found in a Russian book that talks about the CG on the Muromets. Evidently, the CG is on the trailing edge of the aircraft's wing.
Can anyone translate or add to this?
Thanks, Sal
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Old 06-11-2015, 08:04 PM
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jeffo
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My guess it would be the YKP point,it never fly at the trailing edge.jeffo
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Old 06-12-2015, 04:02 AM
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Sorry, both text and sketch seem to be clipped what doesn't really help. Nonetheless I dabbled in translating the text and interpreting the sketch:



Relating to the airplane "Ilya Muromets" we obtain the following. If on the wing the increase of lift force DYwing

= 1000, then on the stabilizer we get less increase -- corresponding to the smaller area its equal to 0,23DYwing,

but because the stab has less length (span, or slenderness or aspect ratio?) the increase will be even less, that is 0,17DYwing.



In the drawing, the dimension should be meters.

Quite surely фокус (the middle point) means Neutral Point.

Since цеитр тяжести means center of gravity, the abbreviation ц.т (the right point) should indeed translate to C/G.

Picture 10: Balance plan for the airplane "Ilya Muromets"



If the wing's angle of attack increases...




Here I stopped because it's too tedious for me and the rest doesn't seem to add much to comprehension. I'd say the book explains the simple balance plan sketched here but I can't really interpret the rest of the text so I'm not sure I grasped its meaning, especially since the rest (following the clipping) seems to mention the Neutral Point, variations in angles of attack and resulting lift forces, that is the really interesting things.

Anyway, I'm quite sure that the middle point is the Neutral point and the point to its right is the C/G. The leftmost point simply should be the wing's aerodynamic center (center of lift), which is assumed at quarter chord. Unfortunately, the sketch is rather coarse and suggests the C/G is at the trailing edge, but it's not.

Obviously, they took the leading edge as a reference what at least was quite common practice. The NP is 1.88 meters behind the LE, and the wing's AC is 1.25 m in front of the NP. That means it is 1.88 - 1.25 = 0.63 m behind the LE. Wing chord is 2.5 m and a quarter of that is 0.625, so 1.88 m is just a rounded value.

Now the C/G is 1.6 m behind the wing's AC, giving 0.63 + 1.6 = 2.23 m behind the LE, wheras the TE is 2.5 m behind the LE.

Still that's a pretty rearward C/G, of course, but I think they had the stabilizer produce lift. I feel affirmed by what I believe to understand of the text, even though no incidences are mentioned. The sketch seems to suggest quite some wing incidence, but that doesn't necessarily mean a forward C/G since the thin and highly cambered airfoil has a huge pitching (down) moment. I think that's why they needed a big stab (like on old free flight models).

Really annoying is that the C/G is (0.35 m) behind the NP, not ahead of it. In fact, the static margin is -0.35 / 2.5 = -0.14 (-14%) what is perfectly good by absolute value but annoying by being negative. That doesn't have to be wrong, though, if you take the pitching moment and the lifting (and thus pitching) stab into account. The rearward C/G could just mean the airplane was trimmed for very slow speed, even if the low wing loading is taken into account. The wings have a big aspect ratio (10 each) so induced drag should not really be a problem. Could be they knew what they were doing (and they had read their Lilienthal).

I would use thin and highly cambered airfoils for the model as well. They should be just fine at model Re numbers and make the model realistic in flight. You should just look for a big hall to fly it since it won't cope with wind very well. ;-)

Last edited by UStik; 06-14-2015 at 03:10 AM. Reason: corrections
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Old 06-12-2015, 05:13 AM
  #4  
Sal C.
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Thanks gents

Jeffo...CG on the drawing is UT, which is shaded.

UStik. Wow, thanks for the translations. I have built the undercambered wings very similar to the full scale. The model has a huge lifting stabilzer that is also undercambered.
The model is going to be powered by four AXI 4130/20 outrunner motors spinning 16x8 scimitar props.
I love your line "You should just look for a big hall to fly it since it won't cope with wind very well". That's very funny. I know that it can only be flown in very, very light winds. I had to build it, it is an amazing aircraft for its day. Right now I am adjusting the 82 turnbuckles in the tail. Yeesh..... Maybe I should start a thread in a more appropriate forum and show what I have done so far.
Cheers, Sal
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Old 06-12-2015, 06:38 AM
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Yeah, that would be great, I would follow the thread. I didn't even know that Sikorski was that advanced, and a model makes us think about what the designer thought. Maybe there are some aha effects waiting. :-)
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Old 06-12-2015, 06:55 AM
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Sal C.
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Ustik

I would like to start a thread but where should I go:

RC Warbird and Warplanes

Twin and Multi engine RC aircraft

Scratch Building Aircraft design

RC Scale Aircraft

Giant Scale Aircraft

Electric General Discussion

Geez, it's like going into a supermarket here in the good ole US and looking for breakfast cereal. Must be 50 choices.

Sal
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Old 06-12-2015, 07:18 AM
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I know what you mean, the same here, but actually it's all very similar, just different brands. And it's even worse with clothing, many choices but all similar and not any fit.

Seriously, there seems to be no special WWI forum here and there are virtually only WWII models discussed in the Warbirds forum. I did a search for "WWI" and found that most posts are in the RC Scale Aircraft forum. Nothing in the Giant Scale Aircraft forum, even though your model might qualify.

In that light it seems to be the RC Scale Aircraft forum.
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Old 06-12-2015, 08:19 AM
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Sal,this build would be interesting to say the least.If I were you,I would build a smaller prototype with all the same proportions free flight,and experiment with the c.g.before building such a huge model.jeffo
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Old 06-12-2015, 10:09 AM
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Sal C.
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Jeffo

Yep, your the fifth person to tell me the same thing. I would hate to have to build a proof of concept model..I have such little building time as it is.
But I know you're dead on right. Yes, it is huge and yes, I need my head examined.

Sal
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Old 06-12-2015, 06:14 PM
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Sal,it would not have to be a big model,even out of foam.The question is can you make a stab that generates enough lift that it pushes the C.G. back to the trailing edge of the main wing.jeffo
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Old 06-13-2015, 01:46 AM
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Sal, I think I see the problem why you need your head examined. I don't know if I would ever dare to fly such a valuable model and would at least try to minimize all risks and uncertainties. What I regularly do is calculating a model's flight behaviour (and its balance as part of that) in advance (and even try it in a simulator). That works pretty well but has its limitations. The other method is building a chuckie from slab balsa or foam and test-gliding it. My point is that this method has its limitations as well, and both methods especially in this case.

As the sketch from the Russian book suggests, the airfoil pitching moment plays a vital role here. It's not evident in this clipping how the NP has been calculated. But it wouldn't be too hard to do such a calculation for your model with the dimensions from a top view and some airfoil data. (There are even free tools.) Here's the weak spot since there won't be any moment measurements. But you never know and besides the moment coefficient may be guessed what is better than nothing and probably accurate enough.

A small test model wouldn't be better either, because you had to replicate the wing and stab airfoils exactly enough so the lift curve slope and moment coefficient (the critical points here) are roughly the same. I doubt that is possible. Bummer.

So if you have a top view and the airfoil shape we could try to calculate (guess) the neutral point and a suitable C/G position. In my experience that is sufficient for the maiden flight. (Even for me considering I'm chicken.) Would be fun to do.


As to the Russian book, after sleeping over it I think it uses the Ilya Muromets as an example to show an unstable condition. The text seems to argue that if and when the airplane is pitched up the lift increase on the wing would be bigger than on the stab so the pitching up is even aggravated. That is instability (we would say lability in German) but that doesn't mean the airplane is not balanced. The moments are always balanced only at a certain pitch angle and corresponding airspeed, but the pilot has to correct immediately if this angle and/or speed is disturbed. Even that isn't too bad, after all we know that from flying helicopters, we are just not used to it in flying airplanes.

Looking at the pictures of the Ilya Muromets I can well imagine it flew that way. The short nose (nearly non-existent) and the big stab let it look like a tandem configuration. Even a tandem would have the C/G ahead of the NP, but in the Web I found a site which says the IM was hard to fly. That's no evidence but at least indication. My point is that means you shouldn't fret about the maiden flight. Even if it's unstable it can be flown and it has been done before. I think even if it's still not balanced it can be flown by an average pilot. What helps is the slow flight speed, the big stab and elevator on a long tail moment arm, and its sluggishness. A small free-flight test model wouldn't help at all because it would simply crash by all means if it's quite balanced yet unstable like (supposedly) the original.

Now that's my new theory. I'm still surprised myself (aha effect). In my Web search I came across a picture of an Ukrainian 1:10 IM model. Maybe you can make out the builder and ask him.

And yet another afterthought: You can buy a perfect pilot for the model, a modern flight stabilizer or at least a gyro for the elevator. Take one that has a "lock mode" like those for helicopter models (heading lock) to have a certain pitch angle locked and controlled by the transmitter stick. Modern fighter airplanes aren't inherently stable as well and have artificial ("electronic") stabilization instead. Maybe the first airplanes weren't stable because they traded stability for more lift (really lifting stab), or they just didn't know better, who knows. Anyway, this should have occured to me earlier...

Like a stabilizer system today turns an actually too small and squirrely beginner model into a stable and calm flying platform, it would make your model fly not only set up unstable but even out-of-balance and even in a bit wind. Duh! :-)

Last edited by UStik; 06-14-2015 at 02:50 AM. Reason: afterthought
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Old 06-13-2015, 08:22 AM
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Sal C.
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UStik

Thank you for your interest and such a comprehensive reply. If you dont mind I would like to use your first name instead of a pseudonym. My name is Salvatore but you can call me Sal.
You have made some valid points and I agree that unless I make a small model with similar attributes, it really will not react the same. The full size aircraft was difficult to fly, however, very few were
lost to flight. The elevators were very heavy and the pilot often needed assistance when landing to pull back on the Deperdussin control column to flair.
I beleive your right on the money with a 3 axis gyro. I am a competent pilot, I fly many different types of models. In fact, I have some excellent German designs. One of my favorite is the ME-163 Komet. I also have a Fokker D-VII and a Gotha G-IV.
I will prepare to make a build thread and post something early next week.
Cheers, Sal
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Old 06-13-2015, 10:27 AM
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Sal, looking forward to your build thread. Hope you didn't take any offense, I was just thinking out loud and had my flying skills in mind. With your experience I don't see any problem for flying the IM, whatever balance and stability. What intrigued me most was the unstable setup which renders a free-flight test model useless, and that led to the idea to let a gyro do the elevator work. I just wouldn't like to be a flight controller, that should be a machine which is better at that than any human being, especially since it's in the airplane. And it will even compensate for a less than perfect balance.
Great 163 by the way!
All the best for your build, Burkhard

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Old 06-14-2015, 07:23 AM
  #14  
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[ATTACH]2102514[/IMG]

This ZIP contains an Excel spreadsheet. It was made by Alasdair Sutherland who is at this forum as well. Just google for his name to find the original file. Now it contains the IM data from Wikipedia. The spreadsheet suggests weight and power for most realistic rendering of an original airplane at model scale (1:6 here).

With a wingspan of nearly 200 inches (5 meters) and only 22.7 lbs (8.7 kg) the 1:6 model would be like a huge bag of air. Probably it will come out a bit heavier but will still be sort of a powered glider, even if with a bad glide angle. Scale power would be only 0.47 hp but more wouldn't hurt. The model would have parkflyer wing loading and fly at parkflyer speed (18 mph). So much about handling.

Anyway, if we take the balance plan for granted the distribution of load (weight) on wing and stab is clear:

Take the quarter chord point (leftmost point) for the pivot of a lever. The airplane's weight pushes down on this lever in the C/G (rightmost point) at a lever arm of 1.6 m, the stab lifts at a lever arm of 12.5 m. The lever arm ratio 1.6 / 12.5 = 0.128 is the same for original and model. It means the stab lifts 12.8% of the weight, so the wing has to lift the rest - 87.2%.

If I got the book correctly, the stab's area is 0.23 (23%) of the wing area. Due to the stab's low aspect ratio it's less effective than the wing (only 17% effect instead of 23%) so it needs more angle-of-attack than the wing to produce a certain lift. Since 18.7% of the total area (23% of the wing area) have to produce only 12.8% of the whole lift there should be still less AoA needed on the stab than on the wing. Correspondingly (I think), the sketch in the book shows less incidence of the stab than of the wing.

Just a nice consideration while having my sunday afternoon coffee... :-)



The evening coffee was less successful:

The neutral point (middle point in the sketch) is 1.25 m aft of the wing's AC, that's exactly 10% of the distance between wing and stab (ACs) - curiously. Only guessing, but this could be even plausible. The NP is approximately the balance point between the wing area and the effective stab area. Assumed the reduced effect of 17% compared to 23% is due to small aspect ratio, a factor of 0.59 - which is a fair estimate of the effect of wing downwash - would make that 10%. They might even have a fuselage effect factored in, which usually shifts the NP forward, even if I'm not sure if that applies to the IM. If in this case the NP is shifted rearward, the 0.54 estimate from my textbook could apply.
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Old 06-14-2015, 08:33 PM
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All good discussion. UStik, I think you have inferred so, but I just wanted to specify that when we talk about CG of an airplane, we also should be talking about the neutral point and static margin of the entire airplane. Not just the wing. So the size of the horizontal stab and the distance from wing to tail have a great influence on the position of the Neutral Point and on the acceptable CG position. In the model world, the Goldberg Sailplane has a large stab and long tail. The Sailplane flies very well with the CG at the wing trailing edge.

In the pictures shown on the original post, the horizontal stab looks huge. I was a little surprised that the numerical data shows the stab area to be only 23% of the wing area. UStik correctly points out that the lower aspect ratio of the horizontal stab diminishes it's effectiveness somewhat. But we also have to remember that the biplane configuration diminishes the effective wing area and aspect ratio somewhat as well.

I've not found a handy online neutral point calculator for biplanes. Does anybody know of one ? That would make life simpler for us.

There is a photo on Wikipedia that is telling. Notice the angle of attack and the downward curvature of the stabilizer/elevator. That says the stab is pretty heavily loaded. And look at the two guys standing on top of the fuselage ! Either the pilot was about to lose control or the airplane tolerates a fairly aft CG. There's no record of a crash immediately after the picture was taken, so I guess the airplane remained controllable in spite of the two "pedestrians" on board.

Dick


Note, I'm not having any luck adding the photo file. If someone will lead me through the upload process I'll add the photo.
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Old 06-15-2015, 03:03 AM
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You have to "Go Advanced" (button below the edit field) and then "Manage Attachments" (below the edit field in advanced mode) to add pictures, if that is the problem. Manage Attachments is a bit tricky, we had to discuss specific problems and see.

But the IM pictures are not absolutely needed here, we can see them directly on Wikipedia. Another web page describes in the text that the IM even had kind of rails in the fuselage and a trolley so the crew members could reach their station in no time. I think they mean the tail gunner in some versions, who was a noticeable tail load which could be borne by the stab only (I think).

I didn't find a biplane calculator as well but that's not the biggest problem. The only three-view I found is too small and I don't know if it's the version Sal is building, so he had to supply a suitable three-view if we wanted to do our own calculations.

What I did so far is only
1. translate some of the text in the book clipping,
2. interpret the sketch,
3. cross-check the details in text and sketch,
in order to see if I comprehended things.

Last time I translated a Russian technical text was 40 years ago, that made it a bit hard but interesting. The clipping left only a part of the reasoning for us to read so the whole context is lost. Therefore it was not possible for me to really construe the statements in the text.

That's why I simply made the assumption the IM was really set up like described (or what I think is described). Otherwise it could be a what-if analysis in the book as well.
Then I assumed in addition that the few figures given in the clipping are quantitatively correct.
Finally, the balance and the neutral point were calculated/estimated. The NP calculation could not be reproduced completely. (Maybe they factored in the reduced biplane efficiency, we just don't have any information about that in the clipping.)

Result:
These simple calculations at least didn't prove the assumptions outright wrong.
So the IM could have been really set up unstable like shown in the sketch (NP ahead of C/G). We just don't know.
Even that wouldn't be the end of the world, though. Still an average pilot should be able to handle the IM.
That's my assessment, of course, based on the fact that the model is a huge parkflyer (see spreadsheet).
A lock-mode gyro would handle it in any case and would make setup problems (finding the balance point etc.) non-critical.
Period.

That's what I managed to squeeze out of the little information we have.
Anyway, if that's correct, a free-flight test model wouldn't work but using the stabilized real model for tests is easy.
Assuming that, one can ask if calculating the NP (actually a complete stability analysis) is worth the effort.
I'd think not, but I'd still do it as a hobby, provided Sal has a three-view and I find the time.

Burkhard

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Old 06-15-2015, 05:39 AM
  #17  
Sal C.
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Hi Gents

Burkhard, I took a look at the chart for 1/6 scale and there are some figures that do not mesh. For instance, Tail span 22?
The tail on my model is just under 6 feet, just about 70 inches.

Here are some three views that might help.
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Old 06-15-2015, 06:47 AM
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Double post deleted.

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Old 06-15-2015, 07:05 AM
  #19  
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Great Sal, I think I'll use the third one because it's biggest and the second one in addition for the dimensions. Need some time, of course. ;-)

Do you happen to have pictures of the wing and tail airfoils? And do already you have a weight estimate?

Those nonsense values in the spreadsheet are leftover, I just didn't have the values for the IM and left the cells unchanged.

These are vast proportions! I assume you can split the wing in pieces like I saw them in the picture of the museum airplane. And the fuse might be split as well. But a 6 ft tail with bracing? How do you transport and even store it?
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Old 06-15-2015, 08:00 AM
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Sal C.
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Burkhard..

You have been extremely helpful so I will share with you a sneak peak of my progress to date. I plan on splitting the wings into two sections of nearly 8 feet each. The lower and upper wing panels will be complete with engines, struts, support wires. The will be joined to the fuselage at the field. I suspect 10 minute installation for flight. Fuse and tail in one piece. Actually the tail does come off but I am not..repeat..not going to remove the tail for transport as it would take me an hour just to hook up all those damn wires....I transport all my Giant scale models in a large 15 passenger Ford Van. I remove the seats needed to accommodate the planes.
By the way.. the model and all its parts have already been framed up.I have the engine mounts, landing gear to be completed.
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Old 06-15-2015, 09:40 AM
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Thanks, the airfoil picture will do.

I see why you leave the fuselage and tail together. Even seeing it on the picture is overwhelming. I think you'd have problems with such a huge model and even such a van here. America you've got it better... (Goethe) :-)

And that must be real dedication, or the other way around, the model chose you to be built because it had to.

We'll figure it out, only a matter of time. (Thursday I'm off for a few days.)
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Old 06-15-2015, 10:49 AM
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Dick, my textbook is of limited help. Not very educated guesses:

Lift ratio biplane/monoplane in this case 0.86 (86%). Useful for equivalent wing area.
"Biplane Coefficient" in this case 0.57, hence correction factor for induced drag/AoA 0.79 (coarse) compared to monoplane of same span.

Mean span is (29+21)/2 = 25 m, chord is 2.46 m (ailerons neglected).
Wing area is (29+21)*2.46 = 123 sqm. Ailerons +5.5 sqm makes it 129 sqm.
Effective wing area (0.86) is 105/111 sqm.

Stab area 10*3.6 = 36 sqm.
Ratio is 0.28 (real area) / 0.33 (effective) including ailerons (0.29 / 0.34 w/o).


Wing airfoil has 6.4% camber.
Estimate for zero-lift AoA: -6.4
Estimate for pitching moment coefficient Cmo: -0,16 (Wow!)

Simple equation for equivalent moment coefficient: Cmm = Cmb (Chordb/Chordm) biplane/monoplane

Equivalent monoplane wing:
mean span 25 m ==> 105 sqm / 25 m = 4.2 m chord (that produces the same lift)
aspect ratio should be bigger by correction factor for same induced AoA and downwash:
AR(1) = 25/4.2 = 5.95 | AR(2) = 5.95/0.79 = 7.58
span = sqrt(7.58*105) = 28.2 m | chord = 105/28,2 = 3.73 m
Cmo = -0.16*(2.46/3.73) = -0.106


That'll suffice for NP calculation.
I'll put it into my monoplane calculator - tomorrow.

Last edited by UStik; 06-16-2015 at 02:50 AM.
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Old 06-15-2015, 11:48 AM
  #23  
rgburrill
 
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You could contact the Sikorsky Museum in Bridgeport/Stratford Connecticut. They have a lot of the original info he was able to get out of Russia.
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Old 06-16-2015, 12:40 PM
  #24  
UStik
 
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With all due caution, the book clipping seems to show just a possible balance.
I had to replace the biplane wings by an equivalent monoplane wing so the stability calculations work. They produce quite similar results.

First picture view of the equivalent monoplane model. Assumed weight is 45 lbs.
Second picture C/G (green, 92% chord) aft of NP (red, 78% chord) like in the book. Stab lifts 6.8 lbs.
Third picture with the C/G ahead of the NP (at 64% chord). Important is that the stab is still lifting (3.1 lbs).
I don't think the undercambered stab airfoil would work with negative AoA.
Question is how to get the C/G that far forward.

Pitch damping is huge, so I still think the model would be manageable even with rearward C/G.
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Last edited by UStik; 06-16-2015 at 01:00 PM.
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Old 06-16-2015, 04:05 PM
  #25  
Sal C.
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Good Morning Burkhard.

Nice work.. So, the Russian document could be close to correct...The CG is far back on the wing? Your estimation of 45 pounds is right on. That is what I figure it will be.
Most of the weight on the wings will be forward, close to the leading edge. I have designed the wings to have carbon fiber spar tubes and leading edge. I can install steel music wire
in the leading edge and forward spar which is around 1.5 inches from the leading edge. In addition, the electric motors will be very close to the leading edge.
I don't think that I will have too much trouble with getting the balance close. I can install lead sheet in the cabin as it will have a double wall and I can hide the lead in between.
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