A balsa airplane - Basic concept and application -

August 03, 2015 (Heisei 27)

Written by Masajiro Sugawara Translated by Shiro Yakumoto

1. Outline 2. Theoretical Study 3. Computational procedure 4. Work of a launch pad 5. Work results 6. Findings

(attention: In this paper, some kinds of reference and data, and the contents are cited and referred. I have responsibility for the wording of an article including a clerical error, misunderstanding, etc. about the contents, and I inform you beforehand that it is not made the author of reference)

1

１. Outline

When I was an elementary school boy, I had a memory that I made some balsa airplanes and flew them with friends in the schoolyard. At that time I didn’t know what kind of shape and body length should be the best. I only imitated a friend's airplane which flew well. The time has passed, and my curiosity has provoked me to examine a balsa airplane theoretically. I thought that the balsa airplane based on the theory would fly well. This paper describes the basic theory of flight and work progress about the balsa airplane subjected on the Hand Launching. I am pleased if the help of work can be given.

2

２. Theoretical Study

At first, I studied the mechanism/ theory of flight, and then studied the stability and a center-of-gravity position, etc. Although I am not an expert on this matter, I managed to examine the basic theory. I would be pleased if readers could give me constructive comments.

3

2.1 Why can Airplane fly?

In this paper, I define the flight as the case where the body carries out advance in the orbits other than a parabola orbit. Based on this definition, the following two points are required for flight.

(1) The power of propelling the body (advance) should exist. (A thrust is made with a propeller, a jet engine, a rocket, the throw raising (Hand Launching) by a hand, an elastic band, etc.)

(2) The lift which raises up the body should exist.(A lift is explained at section 2.1.1)

２.１.１ Outline of Lifting power

If a board is placed into the uniform parallel flow with certain angle shown in Fig. 2.1, fluid will exert power on a board and power will produce perpendicular to the board as the reaction. It is said that the power element parallel to the flow serves as the drag power and the power element perpendicular to the flow serves as the lifting power.

Drag power

Lifting power

Fig. ２.１ Power element in the fluid

4

The thing is not so simple although written in such a way. In the ideal fluid which does not have viscosity nor whirlpool, neither a lifting power nor a drag power is generated. The board only rotates by a couple of force, since the fluid will not produce the whirlpool on the top and the undersurface of the board even if it is placed in a flow with an angle. This phenomenon is well known as D'Alembert's paradox.

If reaction force would be produced even in an ideal fluid to a board, a center-of-gravity position should move in response to a lift. But a board does not become so. Therefore, a simple action-reaction theory cannot explain the lift phenomenon.

5

２.１.１ The outline of a lift (continued)

Then, why does a lift arise? (1) In fact, fluid (air) to which an airplane flies is not an ideal fluid, but a viscous fluid.

In it, the fluid which passed through the up-and-down side of wings produces vortex flow and a circulation flow is also produced along with wings. It is thought that these vortex and a circulation flow produce a lift.

(2) In order to generate a circulation flow efficiently, the wings of an airplane are curving. However, also with flat wings without a curve, it could produce a circulation flow with certain angle to the fluid that arise a lift. Wings of a jet fighter is made thin with few curves. This is because the design of a jet fighter is excelled in the body operability in an air battle.

(3) Even in case of flat wing, a lift arises. An explanation is illustrated in the following 2.1.2 section. Zhukovsky (Joukowsky) transform is required for this explanation. Zhukovsky transform is required for this explanation. If a certain kind of transformation of variables is carried out to a circle, it can change into various type of wings of an airplane.

２.１.２ Zhukovsky transform

The following Zhukovsky transform is one of those change a circle into airplane wings. Before and after the transform of the position z and Z are expressed as following;

)1.1.2(2

z

azZ

If the coordinates of Z and z are set to (X, Y), (x, y) by a complex plane, respectively, it will become below.

)2.1.2(iyxz

iYXZ

Fig. 2.2 Image of Zhukovsky transform Please imagine that the coordinates (x, y) of z before transform

should consider the point on the circumscribed circle of the left figure.

(1) Theory

where, a is a constant relevant to the size of the circumscribed circle in the left figure (a> 0) (x0,y0)

r

r0

( -a,0 ) ( a,0 )

(0,y1)

内接円外接円

y

x

(From section 2.1.2 to 2.1.5, I referred to the book entitled "hydrodynamics known in high school mathematics", Prof. Atsushi Takeuchi of Waseda Univ., Blue back, June, 2014)

6

Circum-scribed circle

Inscribed circle

)(1,

)(1

)(1

)(1

)(

)()(

)()(

22

2

22

2

22

2

22

2

22

22

yx

ayY

yx

axX

yx

aiy

yx

ax

yx

iyxaiyx

iyx

aiyxiYXZ

The following equation (2.1.3) is obtained from a equation (2.1.1) and (2.1.2).

)3.1.2(

Now, in order to actually carry out a Zhukovsky transform, as shown in Fig. 2.2, the inscribed circle which passes along a point (a, 0) is first drawn from the point on a y-axis (0, y1), and the circumscribed circle which passes along a point (a, 0) next is drawn. The point on this circumscribed circle corresponds to (x, y). The center of a circumscribed circle are (x0, y0), and set polar form of this point to r0 *exp(iδ). If the radius of a circumscribed circle is set to R, the relation between point (x, y) and (x0, y0) becomes below.

RrerreorRyyxx ii 0

22

0

2

0 ,,)()( )4.1.2(

Point (x, y) is written by polar form as follows.

)5.1.2(

The following equation is obtained by substituting equation (2.1.5) into (2.1.3).

sinsin,coscos 00 rryrrx

where, r0>0, δis the angle of a point (x0, y0). π/2 ≦δ≦3π/2、 0≦ θ ≦ 2π

7

)6.1.2(

)cos(21)sinsin(

)cos(21)coscos(

0

2

0

2

2

0

0

2

0

2

2

0

rrrr

arrY

rrrr

arrX

(2) Results

(2.1) Flat wing ( r0= 0)

1.0 0.5 0.5 1.0

1.0

0.5

0.5

1.0

2 1 1 2

2

1

1

2

Fig. 2.3 circumscribed circle

Fig. 2.4 Zhukovsky transform

(2.2) Symmetrical wing ( r0> 0 & δ = π)

2.0 1.5 1.0 0.5 0.5 1.0

1.5

1.0

0.5

0.5

1.0

1.5

2 1 1 2

2

1

1

2

It becomes thin flat wings. It becomes symmetrical wings with the upper and lower sides. The thickness of wings depends on the size of r0 in Fig. 2.2.

Analyze are performed using r0 and δin equation (2.1.6) as a parameter. Results are as follows.

x

y Y

Y

X X x

y

8

Fig. 2.5 circumscribed circle

Fig. 2.6 Zhukovsky transform

(2.3) Circle wing (r0> 0, δ = π/2 ) (2.4) General wing (r0> 0 , π/2 < δ < π )

1.0 0.5 0.5 1.0

0.5

0.5

1.0

1.5

1.5 1.0 0.5 0.5 1.0 1.5 2.0

1.5

1.0

0.5

0.5

1.0

1.5

2.0

1.5 1.0 0.5 0.5 1.0

1.0

0.5

0.5

1.0

1.5

2 1 1

2

1

1

It becomes thin convex circle wings. (in case of δ=- π/2, a circle is symmetrical for reverse)

The thickness of wings depends on the size of r0 in Fig. 2.2. Namely, thickness becomes larger according to the size of r0. (in case of -π< δ < -π/2 , it becomes reverse general wings.)

y y

x x

Y Y

X X

9

Fig. 2.7 circumscribed circle

Fig. 2.8 Zhukovsky transform

Fig. 2.9 circumscribed circle

Fig. 2.10 Zhukovsky transform

２.１.３ Circumferential Fluid Flow of Wings Before asking for a lift, I calculated the flow function ψ to see the circumferential fluid flow of wing. The contour line of the flow function ψ serves as a streamline. This relates to a circulation flow required for calculation of a lift. The flow function ψ centering on the circumscribed circle of Fig. 2.2 is described as the following equation.

][)sin(21)sin(

2

rLnRr

RrU )7.1.2(

Where, U：U: fluid speed (body speed when fluid is still) r ：distance from the center of the point besides a circumscribed circle θ ：The angle which the point besides a circumscribed circle makes (radian) R ： radius of a circumscribed circle, α ：angle between fluid and wing η ： angle which shows the curvature of wing ( η= ArcCos[(a-r0cosδ)/R])

２.１.４ Amount of Circulation Γ The amount of circulation Γ is an important element of a lift. It becomes the following.

)sin(4 RU )8.1.2(

As shown in above equation, the amount of circulation Γ is related to the fluid speed U, the radius R of a circle (therefore, size of wings), the angle α of fluid and wing, and the curvature angle η.

２.１.５ Lift (Lift in this case means the power of direction perpendicular to the wing)

The lift L per unit length becomes the following using the fluid density ρ, the fluid speed U, and the amount of circulation Γ.

)sin(4 2 RUUL )9.1.2(10

２.１.６ Streamline of Flat Wing

(2) In case of α= 30 deg. between flat wing and fluid (air)

The left figure shows the streamline in case α is 30 deg. Crowded streamline can be seen in the direction of upper diagonal right place. It shows that the flow of fluid is faster than the other place. Therefore, it turns out that pressure difference arose in respect of the upper and lower sides, power acted on the upper diagonal right, and the lift has arisen.⇒This shows that flat wing, like a balsa airplane, with angle of attack, can arise a lift and the flight of balsa airplane is possible.

(1) In case of a wing and fluid (air) are parallel

The left figure shows the streamline. A center circle should be thought a flat wing. The place where the streamline interval is crowded shows that it is fast-flowing. Figure shows that the interval of a streamline is symmetrical between the upper and lower sides of wing, therefore pressure difference does not arise, and a lift does not arise.

Fig. 2.11 streamline of flat wing (α= 0)

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Fig. 2.12 streamline of flat wing (α= 30 deg)

２.１.７ Lift of General Wings

(2) In case of α= 30 deg. between general wing and fluid (air)

The left figure shows the streamline in case of general wings and fluid (air) are parallel. The place where the streamline interval is crowded shows that it is fast-flowing. Therefore, it turns out that weak pressure difference arose between the wings and the lift has arisen. As shown in Fig. 2.10, the wing has convex curvature and this character can arise the lift even in case there is no angle difference in general wings and fluid. This point is the difference from the flat wings.

Fig. 2.13 Streamline of general wing (with no angle)

(1) In case of general wings and fluid (air) are parallel

The left figure shows the streamline in case α is 30 deg. Crowded streamline can be seen in the direction of upper diagonal right place. It shows that the flow of fluid is faster than the other place. Therefore, it turns out that pressure difference arose in respect of the upper and lower sides, power acted on the upper diagonal right, and the lift has arisen.

Fig. 2.14 streamline of general wing (α= 30 deg)

12

２.１.８ Lift of General Inverse Wings

(2) In case of α= 30 deg. between general inverse wing and fluid (air)

The left figure shows the streamline in case of general wings and fluid (air) are parallel. The place where the streamline interval is crowded under the circle shows that it is fast-flowing. Therefore, it turns out that weak pressure difference arose between the wings and the inverse lift (downward) has arisen. For this reason, if this state will be kept, an airplane crashes into the ground.

Fig. 2.15 Streamline of inverse general wing (with no angle)

(1) In case of general inverse wings and fluid (air) are parallel

The left figure shows the streamline in case α is 30 deg. Crowded streamline can be seen in the direction of upper diagonal right place. It shows that the flow of fluid is faster than the other place. Therefore, it turns out that pressure difference arose in respect of the upper and lower sides, power acted on the upper diagonal right, and the lift has arisen even in case of general inverse wing. This shows that the possibility of inverted flight is suggested.

Fig. 2.16 streamline of general inverse wing (α= 30 deg)

13

２.１.９ Comparison of Lift of Each Wing By the preceding section, the lift power and mechanism were surveyed. In this section, the magnitude of lift was calculated using angle between fluid and wing as a parameter for the flat wing, general wing, and general inverse wing.

Flat wing General wing General inverse wing

wing area m^2 509 512 512

In case of α= 0 deg

lift perpendicular to the wing Kg・m/s^2

0 2.296x106

(234 ton・Force) -2.296x106

(-234 ton・F)

lift perpendicular to the ground Kg・m/s^2

0 2.296x106

(234 ton・F) -2.296x106

(-234 ton・F)

In case of α= 2deg

lift perpendicular to the wing Kg・m/s^2

1.133x106

(116 ton・重) 3.433x106

(350 ton・F) -1.157x106

(-118 ton・F)

lift perpendicular to the ground Kg・m/s^2

1.133x106

(116 ton・重) 3.431x106

(350 ton・F) -1.156x106

(-118 ton・F)

In case of α= 10 deg

lift perpendicular to the wing Kg・m/s^2

5.64x106

(575 ton・重) 7.92x107

(808 ton・F) 3.40x106

(347 ton・F)

lift perpendicular to the ground Kg・m/s^2

5.53x106

(566 ton・重) 7.80x107

(796 ton・F) 3.35x106

(342 ton・F)

Analysis conditions: The numerical value of the following the flight conditions of an actual commercial jet airplane was used. Air density ρ= 0.2552 kg/m^3 (0 degree C, 150 torr), cruising speed U= 250 m/s (900 km/h), wings area ≒510m^2, angle of attack = 2 deg.

Table 2.1 Comparison of Lift of Each Wing

14

Table 2.1 shows the following findings.

(1) If there are two degree of angle between fluid and wing with commercial cruising speed, a flat wing can arise a remarkable lift (about 1/3 of general wing).

(2) A general wing has convex curvature and this character can arise the lift even in case there is no angle difference in general wings and fluid.

(3) Since a lift is proportional to the second power of speed, it is very sensitive to speed. but When the angle between fluid and wing is small, a lift becomes sensitive to an angle.

(4) In case of α=0 degree for general inverse wing, (fluid and wing are parallel), the inverse power (downward) arises, and the airplane falls down.

(5) In case of α=10 degree for general inverse wing, sufficient Lift power can be produced.

(6) From the consideration mentioned above, it turns out that commercial jet airplane could be able to inverted flight. (It is another question whether it actually flies or not).

15

2.2 Consideration of Flight Stability Etc.

0 5 10 15 200

2

4

6

8

Xcg

Xnp

Sh

Lh

Lv

Sv

Side view

0 5 10 15 200

5

10

15

20

Plan view

MAC

b

Symbols

：center-of-gravity (CG) position

Xcg: distance between CG and tip of

plane

Xnp: distance between neutral point

(NP) and tip of plane

Lh : horizontal stabilizer moment arm

Lv : vertical tail moment arm

Sh : horizontal stabilizer area

Sv : vertical tail area

MAC : Mean Aerodynamic Chord,

(almost equal to mean aerodynamic chord )

b : wingspan

S : Wing area

C : mean aerodynamic chord (= S/b)

2.2.1 Outline of Body

Wing

Horizontal stabilizer

Vertical tail

16

2.2.2 Stability Margin SM

There are many factors related to the stable flight. Of these, the center-of-gravity position Xcg and the neutral position Xnp are the most important factors. The stability margin SM is the index which shows the robustness of the airplane against the perturbation, such as turbulent flow. The SM is defined as following;

)1.2.2(

c

XXSM

cgnp

When the SM is in negative , that is, the position of Xcg is behind the Xnp, the plane will be in an unstable condition and will easily get a damping state or a crash state. On the contrary, the SM is in big positive value (e.g. > 0.3), the plane will also easily get an unstable condition . For this reason, the SM should be set to suitable positive value. The value SM= 0.05 is thought to be the best based on the empirical experiences.

where、c= mean aerodynamic chord (= S/b)

2. 3 Horizontal Stabilizer The purpose of a horizontal stabilizer controls the pitching of an airplane. Since there is no elevator mechanism at balsa airplanes unlike commercial airplanes, it is required to set the optimal value during production.

The following http was cited during the consideration of the SM; http://ocw.mit.edu/courses/aeronautics-and-astronautics/16-01-unified-engineering-i-ii-iii-iv-fall-2005-spring-2006/systems-labs-06/spl8.pdf

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The above-mentioned Xnp is influenced by a horizontal stabilizer and its moment arm Lh (distance between CG and the center of horizontal stabilizer). In order to show the performance of a horizontal stabilizer quantitatively, the Vh is defined as following;

)1.3.2(

cS

LSV hh

h

Based on the empirical experiences, the value Vh between 0.30-0.60 is thought to be good. When the Vh would be set to small value, the control of pitching action would be difficult.

2.3.1 Horizontal Stabilizer Capacity Vh

2.3.2 Aerodynamic chord (AC) of Horizontal Stabilizer Th

)2.3.2()(3

)2)((

10

10101

tt

tttttth

An aerodynamic mean chord th is geometrically carried out as shown in the left figure, or is given by following formula.

where、t0:central AC、t1:side AC (t0>t1)

The equation (2.3.2) is derived using the similarity between the triangle AOC and BO'C in the left figure. Where point "O" is the middle point of central AC and "O'“ is the middle point of side AC.

18

For the Symbols, refer to 2.2.1 please.

O

O’

)3.3.2(:)2

()2

(:)2

( 21

020

1 lt

tlLt

t

Where, l2: the length of the segment CO’

)4.3.2(2)(3

)2(

10

10

2

L

tt

ttl

Then、l2 is given as following;

Equation (2.3.2) is obtained by substituting eq. (2.3.4) into eq. (2.3.5). In this connection, in case of parallel shape, th= t0 and in case of triangle shape, th= 2/3*t0.

)5.3.2()(

1

210 tttL

lttt h

、

Also, let suppose Δt is the difference between t/2 and t1/2 at point C, then the following relation is yielded.

About the aerodynamic chord, 「増補改訂 模型飛行機 (理論と実際)」電波実験社1979年刊森照茂氏著was cited.

19

Suppose L is the half length of the horizontal stabilizer. Then, the following relation is now realized.

Then, Δt is given as following;

2.4 Neutral Point Xnp

The neutral point Xnp is also influenced by the quantities of the horizontal stabilizer. The following fitting equation is used.

)1.4.2(2

41

21

21

4

1

hV

AR

hAR

AR

c

npX

Where, AR= wing aspect ratio, ARh = horizontal stabilizer aspect ratio, c= mean AC

Once the neutral point Xnp is determined, then the center-of-gravity Xcg can be uniquely determined using the stability margin SM. For this reason, stable flight can be attained if a wing’s center-of-gravity position would be tuned with the plane’s center-of-gravity position Xcg.

2.5 Vertical Tail Capacity Vv

The main purpose of a vertical tail is to control the yawing of the plane. The vertical tail capacity Vv is an index that shows the performance of a vertical tail. This is defined as follows;

)1.5.2(

bS

LSV vv

v

Please refer to 2.1 about symbols.

From the empirical flight records, numerical value of the Vv between 0.02-0.05 is thought to be suitable for good flight. If this figure is set to be extremely small, the big yawing like a Dutch roll would be occurred.

20

2.6 Wing Design The main purpose of a wing is to generate the lifting power which brings about a rise of the plane. Here, in order to design the multi-dihedral wing, I considered the way how to calculate a center-of-gravity position.

2.6.1 Calculation of Trapezoid Center of Gravity

Center

Line d1

L1 L2

1,,)( 12

1d

LLawhereaxLxy

Suppose that there is a trapezoid as shown in the left figure, and that the wing is symmetrical on both sides of a center line. The Y-axis is set as a space top and the X-axis is set as a horizontal axis. Height y(x) as a function of X is described as follows; where x is the location.

A minute trapezoid area ds and the area S in the minute distance dx in a position x are given below.

2/1)21()(,2/*)}()({)(1

0dLLxdsSdxdxxyxyxds

d

(L2≦L1)

(2.6.1)

(2.6.2)

21

Given thickness of material is set to dp and density ρ, minute mass dw and the moment dm at the position x are given below. (suppose density is uniform.)

dpdsxdwxdm

dpSdwWdpdsdwd

*,1

0

Therefore, whole moment M will be obtained by integrating with x from 0 to d1.

]32

[

]}22{[2

2/*)}()({

3

1

2

111

1

1

0

1

0

1

0

dadLddp

dxadxaxLxdp

dxdxxyxyxdpdmM

d

dd

Then, the center of gravity Cg is given below.

)/(]32

[2/ 21

3

1

2

11 LLdadL

WMCg

(2.6.3)

(2.6.4)

(2.6.5)

22

2.6.2 Calculation of Center of Gravity for multi-dihedral wing

The center of gravity of the whole wing is computed applying the way of the one trapezoid center of gravity as follows.

Body

center

line

2 3

4 5

1

DD4

Cg4

n

i

n

i

Wi

WiDDiCgiCg

1

1}{

n： number of wing junction, Cgi： Center-of-gravity position of i-th wing, DDi：distance from the body center line to the left side of i-th

wing., Wi： weight of i-th wing.

(2.6.6)

23

Where,

３. Computational Procedure

In order to make the good-flight airplane, the computer program based on the theories in the proceeding chapter was created that could easily provide the optimal size of the plane. The computer program can calculate the airplane performance of the original design and that of the optimal design. The users enable to select the option. In the optimal calculation, the sizes of body length, weight, and wing are maintained as the original size. Final optimal values are obtained by iteration using vertical tail length, area, horizontal stabilizer length, and area as parameters. Default values in section 3.1, the flow of the whole calculation in 3.2 and detailed flow of the optimal calculation in 3.3 are explained.

24

３. １ Default Values

No. Material Density Remark

1 Hinoki 0.47 (g/cm^3) for the body, a front tip, and auxiliary material.

2 Balsa 1 0.10 for a wing and a tail( density depends on goods) 3 Balsa 2 0.18

4 Enamel 0.96 Paint

5 Woodwork bond 1.02 Adhesion material

6 Aron Alpha 1.05 quick-drying glue

7 Tape 0.016 For fixation of a wing Unit (g/cm^2)

8 Lead plate 11.34 For weights

9 Air 1.184x10-3 25℃, 760mmHg

10 New M type wing (large-sized) (nwing=1) Wing length = 30 cm

11 P type wing (medium size) (nwing=2) Wing length = 22 cm

12 U type wing (improved medium size) (nwing=3) Wing length = 22 cm

13 VV type wing (minor type) (nwing=4) Wing length = 20 cm

14 V type wing (small size) (nwing=5) Wing length = 17.6cm

25

Yes

No

optimal calculation ?

re-setup of a wing position

set up of stability margin and the calculation of length and area of horizontal stabilizer and a vertical tail using the aspect ratio of them.

Calculation of Xcg, vertical tail and horizontal stabilizer capacity, mass, a

moment arm, etc.

position difference between last time and this time.

No

Yes

Yes

End

Convergence O.K.?

No Start

The input of the amounts of plane parts (body length, a wing, tail length, etc.)

The validity check of input values

OK ?

setup of the kind of default values and wing etc.

Calculation of the area of a multi-dihedral wing with an original size,

volume, mass, etc.

Calculation of the body, the area of a tail, volume, mass, etc.

Calculation and the output of stability with an original size, vertical tail, horizontal stabilizer

capacity, a moment arm, etc.

３. ２ The Whole Calculation Flow

printout and graphics output

26

Start

sh= (widhorc + widhore)/2 * lhor, widhorc=Horizontal stabilizer center chord, widhore=side chord

lhor= arh * tt, arh=Horizontal stabilizer aspect ratio, tt=Aerodynamic mean chord

abarh=|arh2/arh1| arh1:last time's aspect ratio, arh2:This time's value abarh>1.1⇒arh2= arh1*1.1, abarh<0.9⇒arh2= arh1*0.9 arh= arh2

sv= sm*lmain*recv/mar2, sm=Wing area, lmain=Wing length, recv:Desired capacity value of a vertical tail , mar2=Moment arm

lver= sv/widverr,

(A)

３．３ Detailed Flow of Optimal Calculation

1

re-setup of the wing center

position Pmain

Setup the SV of a vertical tail

Setup the height of a vertical tail lver

Setup the aspect ratio arh2 of a horizontal stabilizer

Setup the length of a horizontal stabilizer lhor

Setup the surface area of a horizontal stabilizer Sh

27

Check the height of a vertical tail

Check the height of a horizontal

stabilizer

Calculation of the surface area, volume, and weight

of each part

Moment calculation of

each part

Calculation of the Center-of-gravity position cg

mar1= mar2, mar1: Last time's moment arm, mar2= Leng - widhorc/2 – cg dmar= mar2-mar1 (difference with the last calculation)

cg= tm / tw

tm = m1 + m2 + ……. + m7 , tm=Total moment in a body tip

sa1,sa2,…..sa7, v1, v2,…….v7, w1=v1*ro1, tw = w1+w2+ …..+w7 , tw=total weight

lhor>limhu ⇒ lhor=limhu, sh= sh1, sh1=(widhorc+widhore)/2*limhu lhor<limhl ⇒ lhor=limhl, sh= sh2, sh2==(widhorc+widhore)/2*limhl

lver>limvu ⇒ lver=limvu, sv= widverr*lver lver<limvl ⇒ lver=limvl, sv= widverr*lver

sh1, sh2 are fixed value

２

１

Calculation of the moment arm mar2

28

fl= cl*airdens*ubody^2 *sm* 0.001/2

gb= tw *0.001 *grav

ew= pmain- widmainc/2, xx=cg – ew, rwg= xx/widmainc

vh=sh*mar/sm/tt, vv=sv*mar/sm/lmain

stmt= ¼ + {(1+2/arm)/(1+2/arh)}*[1-4/(arm+2)]*vh - rwg

darh=arh2- arh1

dstmt= stmt – stmt1, admst= |stmfix - stmt|

Calculation of Lifting power fl

Calculation of body gravity gb

Calculation of the center of gravity for the wing front tip ew, the wing xx, and

the wingspan ratio rwg

Calculation of the vertical tail capacity Vv and that of horizontal stabilizer Vh

Calculation the stability margin stmt

Difference calculation of a horizontal stabilizer aspect ratio darh

Calculation of the stability margin difference dstmt, and the desired value

difference admst

３

２

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Pmain<0. Printout the error message, and stop the calculation Yes

No

stmt2=stmt, arh1=arh2, arh2=arh2-dstmt*0.5

|darh|<judarh

Yes

No stmt1=stmt, judarh= Convergence judgement value

for horizontal stabilizer

admst<judstm

Yes

No dpmain= (stmfix-stmt)/stmfix * 0.9 pmain= pmain + dpmain Judstm= Convergence judgement

value for stability margin

Graphical output End

Iteration over ?

Failure of optimal calculation

Yes

No

Go to (A)

３

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Preservation of calculation results

Calculation completed with no problems

Printout of calculation results

３．４ Example of Optimal Calculation

0 5 10 15 200

5

10

15

20

0 5 10 15 200

5

10

15

20

0 5 10 15 200

1

2

3

4

5

6

7

0 5 10 15 200

1

2

3

4

5

6

7

Body weight ：6.05 g, stability margin SM：0.05 Cg：8.74 cm, Np：8.94 cm Moment arm ：13.01 cm Vh：1.59 cm, Vv= 0.055 Lh6.36cm, Vh= 0.9

Body weight ：6.15 g, stability margin SM：-0.45 Cg ：6.46cm, Np：4.64 cm Moment arm ：15.29 cm Vh：2.00cm, Vv= 0.081 Lh：7.5cm, Vh= 1.25

Body length=23 cm, Weight =1.0 g Wing span =20 cm Central AC = 4 cm, Side AC = 2 cm Area =65.9 cm^2 thickness =0.2 cm Vertica tail center=2.0 cm Ditto side =1.5 cm Horizontal stab center=2.5 cm Ditto side =2.0 cm

The optimal design The original design

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4. Work of Launch Pad Although a launch pad is unnecessary for hand launching, but it is very convenient for people who aims to film the flight from the launching to the landing by oneself. I made the launch pad which was cheap, light and easy making and operation.

4.1 Material

No. Material 数量 Price (Yen)

備考

1 Board (90cmx3cmx0.5cm) Three sheets

300 Japan cedar material was used. (It will OK, if it is a light thing)

2 Hex nut W 1/4 one 100 1/4inch , (It's equivalent to the screw of a camera tripod.)

3 Brass 100 For rubber band stops (four pieces are required)

4 Chain (1m) 150 For airplane stops

5 Wood screw (3mmx12mm) 100 For launch pad assemblies

6 Rubber band 100

7 Quick-drying glue 100 For washer and hex nut adhesion

8 Round shape washer 100 (It is used between board and hex nut. One piece is OK)

9 Double clip One For airplane stops

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4. 2 Work Procedure

(1) (1) First, combine three sheets of a board in shape U and stop with screws. In that, drill the holes with a gimlet beforehand.

(2) 発射台の中心を測り、底板に丸形ワッシャを接着します。

(3) 丸形ワッシャの部分に、六角ナットを接着します。

(4) 真ちゅうヒートンを左右先端、後端に1個づつ、計4個取り付けます。

(5) 輪ゴムを編み(今回、3本1組で5段編成)、先端の真ちゅうヒートンに掛けます。

(6) くさりを50センチ位に切り、後端のヒートンに掛けます。(後端のもう一方のヒートンはくさりの長さ調整に使用します。)

(7) くさりの先端にダブルクリップを接続します。これで完成です。

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4. 3 発射台制作写真

台を上から見た所

台を横から見た所 (カメラの三脚

を接続した所)

後端を横から見た所 (くさりが約50セン

チ有り、真ちゅうヒートンで長さを未調整の場合)

バルサ飛行機を装荷した所 (真ちゅうヒートンで 長さを調整後)

長さ調整部分の写真 (2つ有るヒートンの一方

にくさりを掛けて、長さを調整します。)

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4. 4 発射台の設定について

本項については、雲台が古いタイプで２次元的にしか動かない三脚をお持ちの方への参考資料です。(雲台が３次元的に可動な三脚をお持ちの方はスキップしてください。)

設定迎角 θ1

設定ロール角 θ2

仮想延長雲台方向

発射台設定方向

雲台

(1)目標迎角=45度、目標ロール角=0度 設定迎角=45度、設定ロール角=0度

(2)目標迎角=45度、目標ロール角=45度 設定迎角=63.6度、設定ロール角=45度

(3)目標迎角=45度、目標ロール角=75度 設定迎角=87.5度、設定ロール角=59度

(4)目標迎角=0度、目標ロール角=45度 水平発射 設定迎角=45度、設定ロール角=90度

(5)目標迎角=0度、目標ロール角=75度 水平発射 設定迎角=75度、設定ロール角=90度

目標迎角(θp)とは水平面に対する飛ばしたい発射角度で、目標ロール角(θr)とは、機体の中心軸に対する傾きを示します。(後述5.3の表では「左右角度」と表示) 設定迎角(θ1)とは、三脚雲台の水平面に対する角度で、設定ロール角(θ2)は、仮想延長雲台に対する発射台の設定角度です。 これらの関係は以下の式となります。(但し、角度はラジアンです)

)1.4(][

],[2

12

Cos

ArcTan p

p

r

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5 制作機

左の飛行機群は、小生が制作したものです。 大型機、中型機、中小型機、小型機の４種類です。 制作にあたっては、以下の点に留意しました。

(1)着地時に主翼を傷めないように、2mm角の檜材の前縁部を全ての飛行機に取りつけました。

(2)流体抵抗減少及び飛行安定性の為、後縁が直線の多段翼を採用しました。

(3)空中での反転性を良くするため、垂直尾翼に双尾翼を採用しました。 (4)飛ばす場所の大きさに制限が有りますので、直進的に飛ぶことを

目指すのではなく、ブーメランの様に戻ってくるように、垂直尾翼に角度を付けて方向舵の役目を持たせました。

(5)朝露や霜の付着による主翼の濡れを防ぐため、エナメル塗料を使用しました。

(6)主翼位置を動かして最適位置の調整を容易にするため、主翼の底部に主翼の長さより前後で各1 cm程度長い取付部を設け、それを機体に接合し、テープで固定する方法を採用しました。

5.1 制作に当たっての留意点

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M-02機、翼長30cm、機体長30cm 大型機

P-02機、翼長22cm、機体長27cm 中型機

V-03機、翼長20cm、機体長22cm 中小型機

W機、翼長17.6cm、機体長20cm 小型機

5.2 ４種類の飛行機

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5.3 制作機の諸量

小生が制作した飛行機の諸量を以下に示します。 機体全体の所の比は、機体長に対するモーメントアームの比を示します。 水平尾翼の所の尾主翼比は、主翼面積に対する水平尾翼面積の比を示します。 重量の所の錘比は、機体全体の重さに対する錘の比を示します。

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6. 考察

(1) コンピュータプログラミングにより、最適寸法によるバルサ飛行機の制作が可能となり、それを基に実際に種々の飛行機を制作しました。飛行テストを行い、多くの貴重なデータを得ました。その結果、計算上の主翼(前縁)位置、重量等、ほぼ予期したような性能を得ることが出来ました。

(2) 制作が終わっても、直ぐに意図した飛行性能が得られるわけではありませんので、調整方法として、今回、主翼の位置を前後に動かす方式を用いました。具体的には、主翼の底部に主翼の長さより前後で各1 cm程度長い取付部を設け、それを機体に接合し、テー

プで固定する方法です。飛行特性を見て毎回、剥しては接合する遣り方により、最適位置が見つかりました。この方法は主翼を最初から機体に固定して錘の重量で調整する遣り方に比べて簡便だと思います。

(3) 制作上の不確かさ(左右の翼角度(上半角)の制作、接着剤の量の管理)に課題が見つかりました。今後改善が必要です。

―上半角の制作においては、簡単なジグを作り、左右の翼角度(上半角)を同じになるように寸法を管

理したつもりでしたが、実際には微妙な角度のずれが有り、どちらか一方に偏った飛行になるものが有りました。翼角度を目標角度に設定できるしっかりとしたジグの必要性を感じました。

―接着剤の量の管理が不十分で、理論重量をオーバーするものが有り、これについては、接着剤を注入する方法で解決できないかと考えています。(接着剤の注入口を上向きにし、重力落下による意図しない量の注入を防止する。)

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