


Redondo Beach
Joined Feb 2004
18 Posts

Control Line DS potentially more efficient than Soaring DS?
Hi guys i was looking to get some input from some gurus here about an idea that i had on dynamic soaring. I read some old threads about using a control line for DS, and it gave me some ideas.
I understand that the plane must use its own wings to turn itself in order to extract the energy from the air, but it seems to me that this only has to happen when the wind is moving relative to ground. Assumming that the other half of the DS circuit is completed in dead air, it could actually be very beneficial to put the load of the plane on control lines, so as to prevent a very large amound of induced drag from being created. This would reduced the total induced drag seen by the plane by almost 50% over the circuit, thus effectively doubleing the theoretical top speep when neglecting parasitic drag. Of course there will be a *large* amount of parasitc drag due to the added control lines, but this might be more than offset by the huge reduction in induced drag. Check out the thread under dynamic soaring, under the the topic "control line ds" for a picture of a proposed design that could make this all possible: http://www.rcgroups.com/forums/showt...hreadid=158835 Input would be great! 



Punta Gorda, FL
Joined Apr 2002
4,952 Posts

The air speed is proportional to the square root of the coefficient of lift and the induced drag is proportional to the square of the coefficient of lift. Therefore, the induced drag is inversely proportional to the airspeed. As the speed builds up, the induced drag decreases as the airspeed and the line drag increases as the square of the airspeed. The line drag dominates at all but the slowest airspeeds.



Redondo Beach
Joined Feb 2004
18 Posts

Hey Ollie, please check my math:
coefficient of induced drag: CDi = CL^2/(pi*AR*e) and coefficient of lift is: CL = 2*L/(Surface area*rho*V^2) therefore induced drag is: Di = (Surf. Area*0.5*rho*V^2) * (2*L/(Surf. Area*rho*V^2))^2 / (pi*AR*e) which simplifies down to: Di = 4L^2/(Surf. Area * rho * V^2 *pi *AR * e) In the above equation, it is shown that the induced drag is inversly proportional to the square of the airspeed, which would tell you that induced drag is reduced as the airspeed goes up. This is true when lift is staying constant, but lets make one more substitution into the equation: Lift = mass* V^2/R  where R is the radius of the circular path the plane is flying. This is just the equation of centrifugal force which i think we can assume dominates the total amount of lift the wings are generating. Therefore: Di = (4*(mass*V^2/R)^2)/(Surf. Area * rho * V^2 *pi *AR * e) And this then simplifies to: Di = (4*mass^2*V^2)/(R^2*Surf. Area * rho *pi *AR * e) So in the end, a plane that is turning in a constant radius circular path will experience an induced drag that increases with the square of its velocity. Of course the parasitic drag will also increase with the square of the velocity, so now it would be a matter of making the coefficient of parasitic drag of the lines low enough to make it parasitic drag always lower than induced drag. This might be very difficult, however definitely not impossible. 


Joined Nov 2002
86 Posts

Quote:
I would offer two comments. First, it is not safe to assume that the centrifugal force dominates the lift generated by the wing. From the perspective of the plane, centrifugal force is acting sideways, while lift is acting upward. The force the wings exert upward is not related to the magnitude of the centrifugal force. Second, I would contend that it is impossible to reduce the drag of the control lines sufficiently to make a CL plane a viable DSer. banktoturn 



Punta Gorda, FL
Joined Apr 2002
4,952 Posts

Banktoturn has it right. As a first approximation, with control line, the centrifugal force is provided by the tension in the lines, not by banking the wings. Therefore, the lift force is approximately equal and opposite to the vector sum of weight, which does not change with air speed, and the lift, whose vector is perpendicular to the plane of the circle. That means that even if the plane of the circle goes from horizontal to vertical, the lift never exceeds the weight of the model by much for very long. Conclusions in my previous post remain valid.



Redondo Beach
Joined Feb 2004
18 Posts

I'm sorry, i don't think i made it clear how the control line plane was to be oriented. It is not a normal control line plane. But let me back up a minute to address the issue of the lift of a wing.
Lift is defined as the force of air pressure acting normal to the wing... not to any other direction. It does *not* have to be equal to the weight of the plane, and certainly does not have to mean the force acting to counter the force of gravity. Ollie, surely you must know this. If a radio controlled glider (not a control line plane) is flying in a tight circle and pulling a constant 30 g's, then its wing is constantly producing approximately thirty times as much lift as the plane weighs. Once the wing is producing over 10 times as much lift as the plane weighs (pulling a 10 g turn), i think it is safe to approximate all lift required as that which is needed to balance out the centrifugal force, and negelect the force of gravity. Now let us address a normal radio controlled slope glider flying a D.S. circuit. Let us assume that it is flying a perfectly circular DS pattern... although this isn't usually how the pattern is shaped, it is certainly plausable, and for now lets just assume this to be the way it is for current calculations. In this case, as i showed in my previous post, the glider's induced drag *must* increase with the square of the velocity. This is because the lift needed increases with the square of the velocity. Keep in mind...normal glider... no control lines. Now on the "upper" side of the ds circuit, while it is flying through the air with wind, it is extracting kinetic energy (when referenced relative to the earth) from the wind. When it is in the "lower" side of the ds circuit, flying through dead air, it extracts no kinetic energy from the air (still using earth as reference). It is only "coasting" through this turn, and the object is to lose as little energy as possible through this turn, so that it can punch back through the shear layer again, to extract more kinetic energy(again, relative to earth). If there were no induced drag on the glider on the "lower" side of the circuit, then the glider would be able to maintain a higher amount of energy, and would punch back through shear layer with a higher velocity than it otherwise would. So imagine this: a regular rc glider flying the DS circuit, with a string sticking straight out of the top of the canopy (through the cg), and leading back to the controller. For now just pretend that the string is frictionless. I know that this is far from reality, however lets just pretend. Now lets say that one is able to control the plane so that it uses its own elevator to control its own angle of attack, and therefore its own wings produce its own lift (equal to centrifugal force) on the UPPER side of the DS circuit, and therefore there is no force pulling on said string on the upper side of the ds circuit. Then once the glider punches through the shear layer into the dead air, the controller adjusts elevator to prevent the angle of attack of the wing from going positive, thereby preventing the wing from producing the centrifugal "turning" lift (and the associated induced drag) so that the string is then bearing the full centrifugul force, on the lower side of the circuit only. In this situation the plane will experience almost a 50% reduction in total induced drag around the Full circuit, since the vast majority (95% of it assuming a 20 g turn, 98% assuming a 50 g turn) of induced drag is elliminated on the lower side of the DS circuit. IN THIS CASE THE PLANE WILL REACH A MUCH HIGHER TOP SPEED THAN IF IT WERE REQUIRED TO PRODUCE ITS OWN LIFT OVER THE ENTIRE CIRCUIT!! SOMEBODY OUT THERE TELL ME THAT YOU AGREE WITH THIS PARAGRAPH PLEASE! Now i know that the little pesky control strings will create a very large amount of drag in reality, but the situation described above IS POSSIBLE sans the frictionless control line. Assuming that one could make a control line that is small enough and/or aero dynamically shaped, then the added parasitic drag of the control line/lines could be less then the reduction in induced drag around the circuit, thereby resulting in a faster DS machine! Here is my cheesy proposed design that might need more explaining, but i hope not: 


Punta Gorda, FL
Joined Apr 2002
4,952 Posts

steingold9,
When you didn't explain how your configuration differed from a conventional control line set up, you left your self open to being misunderstood. My previous posts, of course, do not apply to your unconventional control line configuration. However, the drag of the control cables still has to be accounted for quantitatively to determine the practicality of the system. You may have trouble finding a flying site that allows the pilot to stand in a practical position relative to the DS boundary layer. Then there is the problem of launching the plane with enough initial air speed to get through the first circuit. You can take comfort in the fact that a lot of contemporaries thought the great inventors were crackpots until the success of their ideas was demonstrated. The path to success is sometimes strewn with many failures. Edison tried over a hundred filament materials before he found one that was practical for an incandescent lamp. 



Gold, put the 4 control lines inside a single Teflon™tube... Nothing sticks to Teflon™ except frying pans.
Ask Composite Space Technology for their input on the potential. It'll slide right past those pesky air molecules. I've DS's with ukies and the load on the wires can get pretty high, if memory serves. 


Joined Nov 2002
86 Posts

Quote:
Yes, if you could relieve the airplane of the burden of generating the lift to make the turns, you would greatly reduce the induced drag caused by making the turns. So if you could eliminate the drag of the control lines, you could come out ahead, if you could actually get the plane where it needs to be to gain energy from the boundary layer. On the other hand, you are only paying that penalty during the turns, and the wing is optimized to reduce induced drag. Also, whatever weight the control lines add is supported by the aircraft, all the time. This increases induced drag all the time, including when the plane is banked, supporting the weight of the lines with the least efficient lift generating parts of the plane. Now, back to the real world, the control lines are always going to have really high profile drag. A cylinder has the same amount of drag as an airfoil with about 20 times the frontal area. This means that reducing control line diameter is not as helpful as one would like. Airfoilprofiled control lines could help with this problem, if you could keep them facing into the flow, but they would probably be heavier, increasing induced drag. I think it's a clever idea to try to get rid of some of the induced drag caused by the turns, but I don't see how you could ever come out ahead. banktoturn 




That model in the presentation ain't gonna work.
First of all from a practical standpoint you can't have a model that let's the line go slack and then comes up short against the line without catching on things around you or just plain snapping from the impact of reaching the end of it's tether. And then there is the fact that you can't maintain control if the lines aren't tight. The drawing suggests that the model is purely line controlled. I've dynamic soared with control line models in high winds and while it works it's not that fast and they bleed speed off quickly compared to slippery sailplanes. Control line models that break the lines close to the model have been estimated to very quickly reach speeds from 30 to 50% higher than when tethered. Fast combats for example. You can hear the engines wind up horribly when this happens. A sign that the speed has risen well beyond the usual 120 mph or so. Also the induced drag is only high when the G load is high during the bottom turn especially. But line drag is there ALL the time. Even if the induced drag at it's worst point can be shown to be slightly higher than the line drag I'm sure the induced drag would be well, well below the line drag for major portions of the time. 
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