

Dec 31, 2007, 06:50 AM  
Joined Dec 2007
11 Posts

Quote:
Steve 

Dec 31, 2007, 02:08 PM  
United States, VA, Arlington
Joined Dec 1996
12,788 Posts

Gotcha  I think. Did you have something to do with a hobby shop east of the River Hull?
Not many managed to get the Ohmlette to fly, period. It was more like 'scrambled' than much else. Sorry, couldn't resist that one Where you at now? D 
Jan 02, 2008, 07:17 PM  
Joined Jun 2005
2,462 Posts

Canard Sailplane Build
The canard sailplane bug has finally weakened my resistance to the point that my sketches are being expanded to full size which is a 72" wing, a 36" canard and a 57" fuselage. These sizes were chosen to fit into my car and to try to keep the weight under two pounds. What you see in the sketches have come from my own experiences and from the suggestions offered by Don Stackhouse, John 235, Dereck, PJ Wright, Capt. Canardly,chuklbac, Geronimo, biber and others all throughout this thread. The airfoils are flat bottom for high lift and easy construction and came from my french curve and just what looked right for me. The thickness to airfoil length is 9. They favor the Bruce Tharpe shapes from my Sig models and have worked well on my other canards. The Medusa 28321500kv 400 size brushless at 2.5 ounces will power it with 200 watts with an 85 or 76 folding prop driven by a 2100 lipo 3 cell. The canard stabilizer will be flat with swept back leading edges. The main wing will have a flat 3 ft. center section with polyhedral sections at 1.5 ft. each. Since I will work hard to avoid work, the simplicity may not be the best design. Since the CoG is only 22" behind the canard LE and the fuselage and rudder is extended 15" behind the main wing, the balancing should not be a problem. It will use two Hitec 55 servos to control the elevator and rudder. Materials are being gathered and construction will begin soon. The fuselage and main wing are not drawn full size yet. I design as I build and feel that my wings will be overkill if the weight comes out over two pounds. I do not have a chuck glider of this and will trust to luck that the model will fly. The main wing will lay flat on the fuselage and the canard wing will be raised to 3.5 degrees positive incidence. The canard area is around 40% of the main wing area which is 10% more than my usual design and I hope the main wing will be loaded enough. The design will allow the TE of the canard to be shimmed UP if necessary. Help along the way would be appreciated. Charles

Jan 02, 2008, 09:42 PM  
United States, VA, Arlington
Joined Dec 1996
12,788 Posts

Not building a 'chucky'  brave you!
I've also learned that having a bolt on tailplane  foreplane in this case  is well worth the two bolts / two nuts / bits of ply weight penalty. Even if it's not drastically wrong, it still allows for shimming the stab, to allow for returning the elevator to 'zero' with respect to the surface's section. Which has to be better than flying around with a little up or down permanently dialed in... Good to see the little grey cells in employment once more! Regards Dereck 
Jan 03, 2008, 03:57 PM  
Joined Jun 2005
2,462 Posts

You got me again, Dereck. I will have to build the chuck glider to see some promise. I was out today flying my Sig Cub rubber powered which has been around over 30 years and it left the park and hit a tree about 10 feet up. Got a stick and retrieved it. These are lots of fun even for grown folks and if it is still available, I recommend it highly. It is simple to build and takes a little work to get the balance and the wing and rudder in position. You don't get this kind of training from ARF RC. Charles

Jan 05, 2008, 06:24 AM  

Charles, Thanks for posting your design. I am just wondering if there is a particular rationale why the elevator isn't used over the full span on the forward wing.
Your fuselage appears to be a bit longer than some of the designs I have seen, but that shouldn't really be a problem. I have tried to make my fuselage a bit more compact, but I don't know if the flight performance is compromised by this choice. If it was my design I'd be inclined to use a bit more tail and rudder area, similar to the canarrow design. If you read the write up on the Canarrow, the designer needed to increase the rudder area on the later prototypes. I also have to wonder if HS55 servos are strong enough to move the larger control surfaces on a glider of this size. 
Jan 05, 2008, 11:37 AM  
Joined Jun 2005
2,462 Posts

Sailplane Design Discussion
John, Thank you for giving your thoughts. Elevator size: It just seemed unnecessary to have more than 24" of length to do the job. The elevator size is over 25% of the chord. If it were ailerons, I would have taken them to full length. There is lots of leverage from the CG to the elevators. Fuselage Length: This just came from my sketches on what looked right. Compare it to a conventional sailplane or a Sig Rascal or to my GWS Slow Stick canard which is much more stable than the Cango short coupled. The nose was extended to take the folding prop, the tail was extended to put the rudder farther from the CG and to provide room for the battery as ballast if necessary. Servos: These will certainly be adequate since they are capable of 2 pounds applied to my 0.5" control horns. The slow speed should put very little resistance on them. Rudder size: This is up for discussion since I have little experience with this method. I want to thank you for the Canarrow tip. Look here: http://search.live.com/results.aspx?...rrow+sailplane This canard is really similar to my design in several ways. I want to save weight in every way. The Captain seems to have a smaller rudder and I hope he will offer some advice here. I just wonder if the downwash from the wings can cause the rudder to get turbulance? The rear wing will lay flat on the fuselage. The rudder hinge line is 26.5" behind the CG which is nearly 1/2 of the 54" fuselage length. Any advice here will be appreciated. Charles

Jan 05, 2008, 07:05 PM  
Joined Jun 2005
2,462 Posts

Canard Chuck Glider
Here is the chuck glider. It balances just a little nose heavy about 1/4" forward of the CG. I will test it tomorrow. The fuselage of solid balsa makes it heavy relative to the model to be built. The leading edges were rounded. Charles

Jan 05, 2008, 09:31 PM  
Livermore, CA
Joined Sep 2004
8,580 Posts

Hey Guys,
Our flying group is building different styles of combat planes. Their all based on the WildWing, molded EPP wing core. We decided they all need to be pushers and we're using a 24g outrunner on a 2 cell as the standard motor. So far I've built/building a tractor type(but we decided pusher only) a pusher jet type and a mid prop type. So now I'm thinking a Rutan type canard, using the WildWing semisymmeterical wing core as well. We'd need it short coupled to maneuver fast and light. I'll use EPP for the build. I've read alot of this thread, but it gets intense sometimes. If I could just get the basics, that would help alot. Thanks, Butch Here's some pics of what we're got so far. 
Jan 05, 2008, 10:45 PM  

Quote:


Jan 05, 2008, 10:55 PM  

Quote:


Jan 06, 2008, 12:46 AM  

Been meaning to respond to this one about wing spar design for a while now, but I've been busy with other stuff:
Quote:
However, the span/4 number that his method comes up with is conservative (i.e.: predicts a number more severe than reality), so although the resulting spar design will be stronger (and heavier) than necessary, at least it will certainly be strong enough. Quote:
What you have to ask yourself is whether you really want to design for the case of getting into a terminal velocity fullpower dive, then survive yanking full up elevator from that condition? Yes, you can design an airplane to survive that, but you're giving up huge amounts of performance in other portions of the flight envelope to do so. Besides the significant increase in spar weight, that heavier spar also adds a lot of inertia about both the yaw and roll control axes, which significantly degrades the control response and dynamic stability. On airplanes, whenever you add more capability or safety margin in one place, you almost inevitably have to lose some somewhere else. In the case of fullscale aerobatic category aircraft, the FAR's (Federal Aviation Regulations, see FAR Part 23.337) specify a limit load of +6 and 3 G's. The plane should handle repeated applications of that, and be able to survive (with some damage allowed, but it must still be able to land safely) after 1.5 times that much, in other words an "ultimate load" of +9 and 4.5 G's. Some of the unlimited aerobatic aircraft are good for quite a bit more, such as the famous Bucher Jungmeister, which was reportedly good for +/12 G's. BTW, the Jungmeister's four wing panels, covered, painted, and ready to hang on the airplane, weighed only 25 pounds each, yet they supported a 1209 lb gross weight aircraft. At 12 G's that's a total of 14,500 pounds carried on a total of only 100 pounds of wing panels. Not too bad for a little spruce, covered with cotton fabric and oldfashioned butyrate dope. That wire external bracing used by the old biplanes was actually extremely efficient from a structural standpoint. On UAV's there's still no clear set of standards, but 10 G's seems to be the approximate number that folks are settling on for limit loads, which is probably a good number for typical R/C models as well. We don't have a Gmeter on board, nor do we have a real pilot in the cockpit who can read that Gmeter and know to back off on the stick if it starts reading too high (or to back off a little before then because he blacked out from all that G!). On larger winchlaunched R/C sailplanes the strength of the tow line is the limiting factor. On open class models, I typically design for 300 pounds. On 2meter models the induced drag limits the plane's airspeed so that it can't go fast enough on launch to make that much lift. 200 pounds is probably more appropriate there. On a model that will only be Histart launched, the appropriate number is probably somewhere between 50 and 100 pounds, depending on whether you want to include circletowing capability (a technique where you harvest energy from the wind during the climb phase of launch to stretch the rubber back out). Quote:
Quote:
Quote:
Unfortunately, knowing the moment at the root only helps you design the spar at the root. If you carry that same bending moment capability all the way out to the tip, you end up with a much stronger and heavier spar than necessary. If you have access to a CAD system with the ability to measure areas and centroids of shapes, it's pretty quick and easy to figure the lift generated outboard of any point along the wing's span. Draw a halfcircle (representing our assumed elliptical lift distribution), with an area equal to the design load for your wing. For example, if we had a 3 lb. airplane designed for 10 G's, we would design for 30 lbs lift (yes, I'm including the weight of the wing, just for a little extra safety factor). Since area of a circle is Pi*r^2, then r = SQRT (2*area/Pi), so we need the upper half of a circle with radius 4.37", or a diameter of 8.74". Now, if the plane in question is a 2meter, then each wing is 1 meter (39.37") long, so any distances along our halfcircle's bottom edge have to be multiplied by (39.37 / 4.37), or 9.01, to find the corresponding location on the model's wing panel. Next, if we take our halfcircle and section it off with a vertical line through the center into left and right portions, then find the area and centroid of the left portion, we find that the area of that portion is 15 sq. in. (corresponding to 15 pounds, no surprise there), with a centroid 1.855" from the center. This corresponds to (9.01*1.855) or 16.71" out from the root on our model's wing. 15 pounds times 16.71 inches equals 251 in.lbs bending moment at the root. If the root is 10" chord and 9% thick, the upper and lower spar caps are identical, and the distance from the centroid of the upper spar cap to the centroid of the lower spar cap is therefore about 0.8", then each spar cap has to handle a load of (251/.8), or 313 pounds. If your spar caps are 1/4" wide x 1/8" thick, that's a stress of 313 / (.25*.125), or a hair over 10,000 PSI. That's a hair more than spruce can handle in tension, and about double what it can handle in compression. You're going to need more than that, either something like 1/2" wide x 1/8" thick spruce, or something stronger than spruce. Yes, you could get away with a stronger cap on top and a lighter one on the bottom, but the calculations for that get a bit more complicated, since the neutral axis is no longer halfway between the upper and lower spar caps. Now, let's say that we have a polyhedral break 20" out from the root, and we want to know the bending moment at that point in the spar. We divide the 20" by our 9.01 factor, we find that location corresponds to 2.22" out from the center of our halfcircle. We drop in a vertical line at that location, trim it up with our horizontal bottom line and the arc of the circle's perimeter, and then find the area and centroid of that segment of the halfcircle. The area is 5.73 sq.in. (so the portion of the wing outboard of there makes 5.73 pounds of lift if the entire wing is making a total of 30 pounds of lift). The centroid is 3.10" out from the center of the circle, or .88" from the location on the circle corresponding to the poly joint. On our actual wing, that lift is therefore centerd 9.01*.88", or 7.94" outboard of our poly joint. The bending moment at the poly joint is therefore 5.73*7.94, or 45.5 in.lbs.. Using this same method we can find the bending moment (assuming an elliptical lift distribution) at any location on the wing. If we know the allowable stress in our spar cap material and the lever arm between the upper and lower spar caps (determined by the distance between the centroids of the upper and lower spar caps), we can figure the loads and required cross sections for the caps at that location. There is one other load case we should consider. Sometimes, when a model is parked out on the flying field, someone might come along and try to lift the model by one wingtip. This means half the model's weight (including the wing!) gets reacted at one wingtip, and half at the other. Normally this is at about one G, but if they do it abruptly or jiggle the model a bit, that could be as much as two or three G's. The resulting bending moments all along the wing could be higher than it sees in flight under max G's, particularly at the locations in the midspan and outboard regions. For smaller models there is usually a surplus of strength, but for a model with a long, slender wing (like a larger sailplane), it could be a major problem. For example, our 2meter, 3 lb example would see bending moments at the root of 1.5 lbs * 3G's * 39.37", or 177 in.lbs., which is almost as high as at the max G condition in flight. At the poly break in our example it's 1.5 lbs * 3 G's * (39.3720)", or 87 in.lbs., which is almost TWICE the bending moment the polybreak joint sees at max G in flight. Think about that the next time you're about to rearrange a model by lifting a wing tip. 

Jan 06, 2008, 01:33 AM  

Quote:
it might be, that the assumption of the "nearly linear" lift distribution is not exactly like in reality, but it is good enough to come up with easy to use numbers to build a spar, which can handle the (bending) loads, it is designed for. As you said, it is a "conservative approach", which is completely adequate for the usual RCplane. Uli 
