Paper air plane competition is the most important part in learning the Flight Mechanics. This contest needs us to apply all knowledge about aerodynamics of flying object (Airplane) and also the Mechanic of Flight theory. The idea of flight is kind of a phenomenon. Throughout history, in 1500, Leonardo da Vinci had studied the movement of birds and their flight to design the way enable human to fly. In 19th century, the idea of glider was come from Cayley who attempt by using large glider model of plane. In 1903, the Wright's brothers come out the first actual flight from the ground powered by engine.
Now is the time for us to apply all the knowledge that we had learned from 1st year until now especially in Flight Mechanic course to design a stable paper plane which can fly with the maximum range and the endurance also. In order to design this paper air plane, there are several rule was given to be obey. The rules are the maximum wing span is 1.5 meter and maximum wing loading is 10 kg/m2. Most challenging is the winner is judge from the range and endurance. So, the paper air plane must be stable during flight in order to have a long endurance and range. This is the challenge behind the contest.
Base on theory, the longer wings should allow a longer gliding distance. Longer gliding distances mean the paper airplane will need to fly at a shallower glide angle. A shallower glide angle requires a higher lift to drag ratio. For a steady glide the lift is equal to the weight, so really we are interested in decreasing drag for a longer gliding distance. Drag comes from two basic causes; the first is form drag that is independent of wingspan, and second from induced drag, which is reduced for longer wing spans.
There are several objectives for the paper air plane contest, which are:
To develop creativity through the limited resources.
To design, build and test an airplane using only 80gm A4 paper and adhesive.
To fly airplane in stable and control manner.
A4 80 gm paper
Scissor and cutting tools
4.0 MODEL OF PAPER PLANE
Our design for this contest is based on the UAV aircraft. The model of UAV we had chosen is as below:
This model is RICs UAV. This kind of UAV can fly without need more propulsion system or do not depend on propulsion.
5.0 WORK AND METHOD USED
Firstly we make an internal structure as a support before us covered by skin. After the model has been create, we make s some calculation to predict our first result. The calculation made us to predict the center of gravity point, aerodynamic center, neutral point, stick fixed static margin and other parameters.
Before we finalize the design, we have been designed already finished the first design of our own paper plane. However, the first model unable to fly due to extra weight. It does not glide at all, instead pitching down. The picture below is our first model of paper air plane:
After discussion, we had decided to design the new model of paper plane. The second model as below :
5.1 Wing contribution
Refer to the dimension of the model; our wing is symmetrical and thin airfoil. Hence our calculation on each airfoil is based on Thin Airfoil Theory on Symmetric Airfoil. It states that;
Total circulation around the airfoil, Γ = π α c V (1)
α - angle of attack
c - chord length
V - freestream velocity
Hence, based on Kutta-Juokowski theorem,
Lift per span, L' = ρ V Γ (2)
V - freestream density
Input the equation (2) into (1), yield;
L' = π α c ρ V 2 (3)
All equation above is based on airfoil. If transform over the finite wing, the calculation become;
Total circulation around the airfoil, Γ = Γ (yâ‚€)
yâ‚€ - point along wing span
Note: y - axis is located along wingspan and origin is at root of the wing.
Hence, equation (1) become,
Γ (yâ‚€) = π α c (yâ‚€) V, α and V is constant
c = 0.231y + 297 c = -0.231y + 297
(-420,0) Graph 1 (420,0) y,mm
Equation (2) become ;
L'(yâ‚€) = π α c (yâ‚€) ρ V 2
Total lift on a finite wing;
L = , = 420 mm
= , are constant
= [ ]
C L = , q
Reference area of wing, S = 208740
Hence, C L = 2
- , X ac = distance from aerodynamic centre to the leading edge of aerofoil.
Based on Thin Airfoil Theory on Symmetric Airfoil, it's shown that aerodynamic centre is also point of centre of pressure (cp) of the aerofoil and;
Cmcp = 0
X ac = X cp = c /4
Cmâ‚€ = - , CLâ‚€ = 0
Hence, Cmâ‚€ = 0
5.2 Tail contribution
The concept calculation used same as wing because geometry of the tail is a thin airfoil.
Hence, L =
= , are constant
C L = , q
Reference area of tail, S = 31185
Hence, C L = 2
Total C L = C Lw + nC Lt
5.3.1 Center of gravity
Horizontal stabilizer rod (left)
Horizontal stabilizer rod(right)
Hence, Xcg is equal to 293.6323mm from paper plane nose.
M = 0.184 kg; lt = 332 mm; Sw = 104363.7 mm2; ARw=6.761;
Xac = 274 mm; Xcg = 293 mm;
Assume parameter: η = 1; Cmα = 0;
Neutral point, XNP
; c is mean chord length.
Stick fixed static margin
SM = -14.4 mm; unstable condition.
5.3.3 Range and endurance
126.96.36.199 Calculation base on Flight Mechanic approach
To find CDo, assume minimum drag condition where,
and , where L = W
(for steady flight)
; assume =,
, assume ,
188.8.131.52 Based on projectile motion
v2 = u2 + 2asy v = 0
0 = u2 + 2(-9.81)(4)
u = 8.8589 ms-1
Sy = ut + ayt
4 = 8.8589t +(-9.81)t2 Figure 5
t = 0.90 s (predicted endurance)
sx = Uxt + axt ax = 0
Sx = Uxt
Sx = 0.9 x 8.8598
Sx = 7.97 m (predicted range)
This calculation is made to predict the range and endurance before the real flight test, but there still some error that occur because of some factor which are the angle of attack not correct, the height of the launcher, disturbance such air gust are not taking into this calculation, by assuming lift and drag are not generated and other factors.
To the controllability of paper plane, we find the moment act on paper plane
where, η = 1,
hence, Cmo = 0, means bad controllability
To find Cmα,
Hence, Cmα give positive value, so the condition is unstable.
To have the negative value of stick fixed static margin, we had increase the weight at the nose of paper plane by trial and error until the paper plane can glide. Hence, the stick fixed static margin will be positive value.
For Cmo = 0, the change the value of Cmo, we had install the elevator to remove the graph line of Cm versus α to the point we need and to make sure the Cmo is a positive value.
By refer to the stable Cm versus α graph, our target is paper plane can be trim (elevator) at low positive angle of attack. Hence, the paper plane can glide.
5.3.4 Effect of lift-to-drag ratio.
Our paper plane does not produce any thrust since there no propulsion system was installing. So, to make paper plane can glide we high endurance and range, lift-to-drag ratio is being considered.
Based on diagram,
So, the ratio is , so Drag is directly proportional to angle of attack. By reducing the angle of attack during launching the paper plane, the drag produce is small. So, paper plane can glide with low drag. In this case, we need the drag as low as possible because we want to have high endurance and range since our paper plane does not have any propulsion system
ENDURANCE x RANGE (m/s)
From the result that we get:
For 1st trial, range is too low because of stall at the low angle of attack, which makes our paper plane suddenly fall. Before this we have try our plane to glide but at the room where no wind and its glider smoothly, but when we try our paper plane during the flight test, there is some disturbance which is wind that made our paper plane fail to glide.
For 2nd trial, we added some weight at our nose of paper plane to change stick fixed static margin by changing the specific gravity position, cg but the same problem still occur. As the result, our paper plane falling down quickly but the range was improved. This is because of the adding weight acting as reducer for the moment stall.
For the last trial, we put more weight at the nose of our paper plane to bring the cg in front of the neutral point to obtain the high value static margin. This time our paper plane does not feel any stall and glide farthest than before with high endurances.
Besides that, our paper plane glides with high range and endurance because of ground effects which produce induced vortex that oppose the vortex that produce by wing. It can be describe as a magnet principles that when the two magnet that have same poles, it will repel each other.
Figure 7 Figure 8
The concept that we use is:
To get a statistically stable aircraft, we need to get < 0 which make the aircraft creates negative pitching moment that make the aircraft nose down. When the aircraft experiencing 'downward gust' disturbance, the aircraft generate positive pitching moment which make the aircraft nose up to return to it equilibrium point. In order to get < 0, we need to put some weight at the front of the paper plane to bring the specific gravity point, cg to the front.
As a whole, the objective of the task which is to develop, design and to fly the paper plane was successfully achieved. Lift and drag is the main things that need to be consider in all kind of transportation especially aircraft. In order to get the maximum range and endurance, lift to drag ratio must be maximum and the structure also need to be rigid.
For future projects, initial analysis and calculations is the most important thing to predict the initial result. Firstly, we need to design the paper plane that have maximum lift to drag ratio, the structure need to be rigid, the weight of the paper plane cannot to be too heavy because it will decrease your aircraft range and endurance, the position of cg need to be less than aerodynamic centre point position in order to get a stable paper plane and others thing. Besides that the neutral point must be negative value in order to achieve a stable flight.
Another important consideration is the use of wing flaps. In order to get the best glide angle from a paper airplane, the back edge of the wing must be bent upward a little (wing flap) to set the optimum gliding speed.
Wingspan is best varied by folding the wings with different body heights; a tall body results in short wings, a short body results in a longer wing span.