Mass of glider
mg := .367kg
23
lwr :=
24
23
lwt :=
24
⋅ 14in = 0.341 m
Length of wing chord at root
⋅ 4 in = 0.097 m
Length of wing chord at tip
lwlr := lwt = 0.097 m
Length of winglet chord at root
lwlt := 1.167in = 0.03 m
Length of winglet chord a tip
bw := 22.016in = 0.559 m
Span of wing from root to tip
bwl :=
23
Span of winglet
4in = 0.097 m
24
−3
δ := 5mm = 5 × 10
thickness of wings
m
2
2
Area of both wings
Aw := 2 ⋅ 191.576in = 0.247 m
2
Cross sectional area of the fuselage from the top
Aftop := 16.357in⋅ 75mm = 0.031 m
Affront := 75mm⋅ 2.629in = 50.082⋅ cm
(
2
Coss sectional are of the fuselage from the front
)
Ac1 := Affront + 2δ⋅ b w + b wl = 115.74⋅ cm
3
Ac2 := Aftop + Aw = 2.784 × 10 ⋅ cm
(
2
)
Awfront := 2 ⋅ δ⋅ b w + b wl = 65.657⋅ cm
ρall := .0167
kg
3
, .0168
m
kg
3
.. 1.225
m
kg
2
Cross sectional area of glider from the front
Cross sectional area of glider from top or bottom
2
Cross sectional area of wings and wingelts from the front
Range variable of air density from 100,000ft to sea level
3
m
2
AR :=
λ :=
bw
Aw
lwt
lwr
= 1.265
Aspect Ratio of the wings
Taper ratio of the wings
= 0.286
2
lμ :=
2 1+λ+λ
⋅
⋅ lwr = 0.242 m
3
1+λ
ρair := 1.225
kg
mean chord length of swept wings
Density of air at sea level
3
m
α0 := 0 ⋅ ° = 0
α5 := 5° = 0.087
Angle of attack = 0
°
Angle of attack = 5
°
°
Range variable of angle of attack in radians (-10 ° to 10 °
)
Acceleration of gravity
α := −.175 , −.174 .. .175
m
g = 9.807
2
s
−5
μair := 1.983⋅ 10
Viscocity of air
Pa s
m
Vd := 20mph = 8.941
s
Desired airspeed of glider determined in simulation
Reynalds Number
ρair⋅ Vd ⋅ lμ
5
Re :=
= 1.334 × 10
μair
Turbulant flow
boundry layer thickness
.382lμ
δb :=
= 8.712⋅ mm
1
Re
5
WingLoading :=
mg
Aw
= 1.485
kg
2
m
Induced Drag
Cl = 2π α
Equation for coefficient of lift of a flat plate
cl0 := 2π α0 = 0
Coefficient of lift at 0
°
cl5 := 2π α5 = 0.548
Coefficient of lift at 5
°
cl( α) := 2π α
Coefficient of lift at a given angle of attack
Cdi =
Cl
2
Equation for coefficent of induced drag
π AR
2
cdi0 :=
cl0
π AR
=0
Coefficent of induced drag at 0
°
= 0.076
Coefficent of induced drag at 5
°
2
cdi5 :=
cl5
π AR
cdi( α) :=
( 2π α)
π AR
2
Coefficent of induced drag at any given angle of attack
Coefficents of Lift and Drag
Predicted Coefficients of Lift and Drag
1
cl( α)
cdi( α)
0.5
0
0
2
4
6
α
deg
Angle of Attack
Does not account for flow seperation at angles of attack above 10*
8
10
Predicted Coefficients of Lift and Induced Drag
1
cl( α)
0.5
0
0
0.1
0.2
0.3
0.4
cdi( α)
Form Drag [1]
cdff := .1
drag coefficient of long, streamlined body at a=0 °
cdfw := .005
drag coefficient of flat plate wings at a=0 °
Friction (skin) Drag
τ = μair⋅
dV
= μair⋅
dy
v
Equation for shear force of air on glider skin
.382lμ
1
Re
cdfr =
τ
5
Equation for coefficent of friction drag
1
2
ρ ⋅V
2 air
2
As := ( 2 ⋅ 22.118 + 2 + 46.926 + 3.058 + 8.3187 + 23.191 + 4.430 + 14.467 + 4 ⋅ 9.879)in + 4 ⋅ Aw
2
As = 1.109 m
Terminal Velocity
Surface area of entire glider
Fg := mg⋅ g = 3.599⋅ N
Force of gravity on glider
1
2
Fd = ρair⋅ v ⋅ cd ⋅ Ac
2
Equation for drag force
Fg = Fd
Equation to find terminal velocity
Given
guess for the program
v := 150mph
coefficnets of drag added together in proportion to the areas they affect
v


μair⋅


.382l


μ

 

1





5

  ρair⋅ v⋅ lμ   

  μair   

1
π
2

  ⋅A 




Fg = ρall⋅ v cdi( α) ⋅  sin − α ⋅ Ac1 + cos( α) Ac2 + cdff ⋅ Affront + cdfw⋅ Awfront +
s
1
2
2
 2




ρair⋅ v
2


(
)
Vt α , ρall := Find( v)
m
Vt 0 , ρair = 98.875
s
(
)
(221.177mph)
Ideal Terminal Velocity
Velocity, [m/s]
Terminal Velocity as a Function of Angle of Attack
80
(
Vt α , ρair
)
60
40
20
0
− 10
−5
0
α
deg
Angle of Attack
5
10
Ideal Terminal Velocity Through the Atmosphere
900
Ideal Terminal Velocity, [m/s]
810
720
630
540
(
)
Vt 0 , ρall 450
360
270
180
90
0
0
0.5
1
1.5
ρall
Density of Atmosphere, [kg/m^3]
Drag force at ideal terminal velocity
1
π
2
Fdt α , ρall := ρall⋅ Vt 0 , ρair cdi( α) ⋅  sin − α ⋅ Ac1 + cos( α) Ac2 + cdff ⋅ Affront + cdfw⋅ Awfront ...
2
 2




Vt 0 , ρair
 μ ⋅

.382lμ
 air 




(
)
( (
))
(





+


(
)
Fdt 5 ⋅ ° , ρair = 134.43 N
(
)
)
1


5
  ρair⋅ Vt( 0 , ρair) ⋅ lμ  
 

μair

  ⋅A
s
1
2
( (
ρ ⋅ V 0 , ρair
2 air t
))
Inital drag force on airframe
turning 5*








adt( α) :=
(
)
Fdt α , ρair − Fg
Initial decceleration on airframe at terminal velocity (in G's)
mg⋅ g
Max. Deceleration of Airframe after Changing Angle of Attack
150
Decceleration, [G's]
125
100
adt( α) 75
50
25
0
0
1
2
3
4
5
6
α
deg
Angle of Attack
7
8
9
10
Initial Force on Airframe at Terminal Velocity After Changing Angle of Attack
600
Force, [N]
400
(
Fdt α , ρair
)
200
0
− 10
−5
0
5
10
α
deg
Angle of Attack
Drag force at any angle of attack, air density, and velocity:
π
1
2
Fd1 α , ρall , v := ρall⋅ ( v) cdi( α) ⋅  sin − α ⋅ Ac1 + cos( α) Ac2 + cdff ⋅ Affront + cdfw⋅ Awfront ...


2
 2




v
 μair⋅ 5.823mm

⋅ As
+

 1 ρair⋅ ( v) 2

2
(
)


Initial Force on Airframe at Desired Velocity After Changing Angle of Attack
5
Force, [N]
4
3
(
)
Fd1 α , ρair , Vd
2
1
0
− 10
−5
0
α
deg
Angle of Attack
5
10
Angle of Attack
Bamboo Properties [2]
N
σbb := 20.27
2
Ultimate bending stress of bamboo
= 20.27⋅ MPa
mm
Radius of bamboo spars
rb := .125in = 3.175 mm
Carbon fiber Properties [3]
σbc := 89000psi = 613.633 ⋅ MPa
Ultimate bending stress of carbon fiber
3
rc :=
in = 2.381 mm
32
Radius of carbon fiber spars
σ=
3
FL
Fb =
3
πr
σ⋅ π⋅ r
Equation for maximum bending force on wings
L
Forces on differnt bamboo spar lengths
Fbb1 :=
σbb⋅ π⋅ rb
3
9.5in
= 8.447 N
Fbb2 :=
σbb⋅ π⋅ rb
3
12in
= 6.687 N
Forces on different carbon fiber spar lengths
Fbc1 :=
σbc⋅ π⋅ rc
3
= 107.874 N
9.5in
Fbc2 :=
σbc⋅ π⋅ rc
12in
3
= 85.4 N
Fbc3 :=
σbc⋅ π⋅ rc
3
24in
= 42.7 N
Fbbmax := Fbb1 + 3Fbb2 = 28.507 N
Force to break bamboo spars (4 spars)
Fbcmax := Fbc1 + Fbc2 + Fbc3 = 235.974 N
Force to break carbon fiber spars (3 spars)
Wbb :=
Wbc :=
Fbbmax
g
Fbcmax
g
= 2.907 kg
Equivalent mass on airframe to break bamboo
= 24.063 kg
Equivalent mass on airframe to break carbon fiber
m
Vcs := 0 , .1 .. Vt 0 , ρair
s
(
σbps =
)
Range variable of velocities from 0 to ideal terminal velocity
3F⋅ L
2⋅ b⋅ d
Equation to find bending stress of polystyrene
2
Experimental results:
bbps := 1cm
Width of polystyrene foam board test articles
Lbps := 5.3cm
Length of polystyrene foam board test articles
Fbps := 4.2N
Force applied when polystryene foam board failed
σbps :=
3Fbps⋅ Lbps
2
3
= 1.336 × 10 ⋅ kPa
Ultimate bending stress of polystyrne foam baord.
2⋅ b bps⋅ δ
l
⌠ wr
2
2σbps⋅ δ

Fbwing := 
dl = 9.69⋅ N
3⋅ b w

⌡l
Bending force to break wing at center of span
wt
The wing technically will break before the bamboo or carbon fiber spars, however since the
bending moment of both wings is centered on midpoint of the spars (inside the bulkhead), the
wings experience much less bending force.
Deceleration Forces vs. Breaking Forces
3
1× 10
(
)
Fd1( 1deg , ρair , V cs)
Fd1( 2deg , ρair , V cs)
Fd1( 4.5deg , ρair , Vcs)
Fd1( 10deg , ρair , Vcs)
Forces, [N]
Fd1 0 , ρair , Vcs
100
10
1
Fbcmax
0.1
Fbbmax
0.01
−3
1× 10
0
20
40
60
80
Vcs
Velocity, [m/s]
σTp := 43.6
N
Tensile strength of craft paper [4]
2
mm
δpaper := .12mm
Thickness of paper
lpaper := 19.349in = 0.491 m
Length of control surface
Acpaper := δpaper⋅ lpaper = 0.59⋅ cm
2
Area of control surface connection
3
Ftpaper := σTp ⋅ Acpaper = 2.571 × 10 ⋅ N
Force required to tear off control surface
Asuming force on control surfaces is akin to a fluid jet striking an angled, flat plate
2
Acs := 19.215in⋅ 1.772in = 0.022 m
Area of 1 control surface
θmax := 60° = 1.047
Max deflection of control surface
ϕ := 90° − 83.2° = 0.119
Sweep angle of control surfaces
100
2
( )
(
)
Fcsmax Vcs := ρair⋅ Acs Vcs ⋅ sin θmax ⋅ sin( ϕ)
Force on 1 control surface at maximum deflection
Fcsmax( 20mph) = 0.221 N
Force on 1 control surface at 20mph
θ := 0 .. 1.047
Range variable of control surface deflection, from 0 ° to 60 °
(
2
)
Force on 1 control surface at any velocity and
delfection
Fcs θ , Vcs := ρair⋅ Acs⋅ Vcs sin( θ) sin( ϕ)
Force on Control Surface at a Given Deflection
0.3
(
)
Fcs θ , V d
0.1
0
0
20
40
60
θ
deg
Deflection
Force on Control Surface at Max Deflection vs Force to Tear it off.
4
1×10
3
1×10
100
10
Force, [N]
Force, [N]
0.2
( )
Fcsmax Vcs
Ftpaper
1
0.1
0.01
−3
1×10
−4
1×10
−5
1×10
0
20
40
60
Vcs
Velocity, [m/s]
80
100
Coefficient of drag of a perpendicular
flat plate
cd := 1.2
Distance from COM
dcom := 7.14in
dcenter := 10.604in +
(
)
75mm
(
M com θ , Vcs := dcom⋅ 2 Fcs θ , Vcs
(
)
)
(
M center θ , Vcs := d center⋅ 2 Fcs θ , Vcs
( )
( )
Distance from center of fuselage
= 0.307 m
2
Pitch moment
)
Roll moment
( )
( )
Max pitch moment
M commax Vcs := d com⋅ 2 Fcsmax Vcs
M cmax Vcs := d center⋅ 2 Fcsmax Vcs
Max roll moment
Max. Pitch and Roll Moments at different Airspeeds
20
Moments, [Nm]
15
( )
10
Mcmax ( Vcs)
Mcommax Vcs
5
0
0
20
40
60
Vcs
Airspeed, [m/s]
80
100
TServoStall1 := 1.8kgf ⋅ cm = 0.177⋅ N⋅ m
Max torque of SG90 Servos [5]
TServoStall2 := 22ozf ⋅ in = 0.155⋅ N⋅ m
Max torque of SM22 Servos [6]
lch := 13.5mm
Length of control horn
lsa := 14mm
Length of Servo arm
dcs := 1.633in = 41.478 mm
Distance between control surface center of
force and control horn
(
)
(
M CSonCH θ , Vcs := d cs⋅ Fcs θ , Vcs
)
Moment of control surface on control horn
lch
M CHonSA θ , Vcs := M CSonCH θ , Vcs ⋅
lsa
(
)
(
)
Moment of control horn on servo arm
Torque Applied to Servos at Different Deflections
0.4
TServoStall1
Torque, [Nm]
TServoStall2
MCHonSA 60° , Vcs 0.3
(
)
MCHonSA ( 45° , Vcs)
MCHonSA ( 30° , Vcs)
0.2
MCHonSA ( 15° , Vcs)
0.1
0
0
10
20
30
40
50
V cs
Velocity, [m/s]
60
70
80
90
100
Given
m
Vcs := 100
s
lch
TServoStall1 = dcs⋅ Fcs 60° , Vcs ⋅
lsa
m
Vservostall60 := Find Vcs = 39.993
s
(
)
( )
(89.461 mph)
Given
m
Vcs := 100
s
lch
TServoStall1 = dcs⋅ Fcs 45° , Vcs ⋅
lsa
m
Vservostall45 := Find Vcs = 44.259
s
(
)
( )
(99.005 mph)
Given
Speeds at which the servos will stall
at a given deflection
m
Vcs := 100
s
lch
TServoStall1 = dcs⋅ Fcs 30° , Vcs ⋅
lsa
(
)
m
Vservostall30 := Find Vcs = 52.633
s
( )
(117.738 mph)
Given
m
Vcs := 100
s
lch
TServoStall1 = dcs⋅ Fcs 15° , Vcs ⋅
lsa
(
)
m
Vservostall15 := Find Vcs = 73.156
s
( )
(163.645 mph)
Given
θ := 5°
lch
TServoStall1 = dcs⋅ Fcs θ , Vt 0 , ρair ⋅
lsa
(
(
))
Max deflection at ideal terminal velocity
θmaxt := Find( θ) = 8.145⋅ deg
(
)
(
M com θ , Vcs := dcom⋅ 2 Fcs θ , Vcs
(
(
M com θmaxt , Vt 0 , ρair
)
)) = 1.601⋅ N⋅ m
m
Vcs := 0 , .1 .. Vt 0 , ρair
s
(
)
reset V cs, Mcom from calculations
Moments of Inertia
Division of wing for moment of inertia calculations. I, II, and III are the main sections while 1, 2, and 3 are
subsections of I
bf := 16.357in = 0.415 m
"Base" of fuselage
hf := 75mm
"Height" of fusealge
bw1 := 12.01in = 0.305 m
Base of each section of wing
bw2 := 1.406in = 0.036 m
bw3 := 2.427in = 0.062 m
hw1 := 22.213in = 0.564 m
Height of each section of wing
hw2 := 22.213in = 0.564 m
hw3 := 22.213in = 0.564 m
Areas of each section of wing
1
2
Aw1 := b w1⋅ h w1 = 0.086 m
2
2
Aw2 := bw2⋅ h w2 = 0.02 m
1
−3 2
Aw3 := b w3⋅ h w3 = 8.695 × 10 m ⋅ 2
2
Wing section 1 cut into more sections around COM
Base of each subsection of wing
b11 := 6.5in = 0.165 m
b12 := 5.51in = 0.14 m
b13 := 5.51in = 0.14 m
Height of each subsection of wing
h11 := 12.021in = 0.305 m
h12 := 12.021in = 0.305 m
h13 := 10.191in = 0.259 m
Areas of each subsection of wing
2
Aw11 := .5⋅ b11⋅ h11 = 0.025 m
2
Aw12 := h 12⋅ b 12 = 0.043 m
2
Aw13 := .5⋅ b13⋅ h13 = 0.018 m
A
⌠ ftop

2
Iwx := 
3 h w1 dAw + 


⌡− A
⌡− A
w
A
⌠ w
(
)
2
 hf 
4
  dAftop = 0.472 m
2
ftop
Second moment of area of wing around x axis (roll)
 ⌠ Aw11

 ⌠ Aw12

 ⌠ Aw13

2
2
2
−3 4






Iwy := 2 ⋅ 
h 11 dAw11 − 2 ⋅ 
h12 dAw12 − 2⋅ 
h 13 dAw13 ... = −5.879 × 10 m
⌡

⌡

⌡

 0

 0

 0

A
A
 ⌠ w2

 ⌠ w3

2
2




+ −2 ⋅ 
b w2 dAw2 − 2 ⋅ 
b w3 dAw3
⌡

⌡

 0

 0

Second moment of area around COM (pitch)
Rgx :=
Rgy :=
Iwx
Radius of gyration about the x axis
= 6.388 m
Ac1
Iwy
Radius of gyration about the y axis
= 0.713 m
Ac1
2
2
Ixx := mg⋅ Rgx = 14.974 m ⋅ kg
2
2
Iyy := mg⋅ R gy = 0.186 m ⋅ kg
Moment of inertia about x axis (roll)
Moment of inertia about y axis (pitch)
Roll rate [7]
p⋅ bw
Clδa = roll authority
= constant
2Vcs
Clp = roll damping
 2Vcs 
p=−
δa

Clp  b 
C lδa
p⋅ bw
2Vcs
b1 :=
b2 :=
bcs :=
=−
C lδa
Clp
75mm
2
75mm
2
75mm
2
δa = θ
c lδa=coefficiemt of lift with aileron deflection
⋅ δa
Inner edge of aileron
+ 1.006in = 0.063 m
Outer edge of aileron
+ 20.213in = 0.551 m
Spanwise location of midpoint of control surface
+ 10.611in = 0.307 m
π⋅ AR⋅ .5[ 1 + cos[ 2 ⋅ ( 0°) ] ]
cla :=
2
1+
1+
AR
4
(
2
⋅ 1 − .0390997
Acs b cs
clδa := cla⋅ λ⋅
⋅
= 0.046
Aw bw
= 1.781
Coeffcient of lift of aileron [8]
[ 2 ⋅ ( 28.4°) ]
) 11 −+ cos
+ 1
cos[ 2 ⋅ ( 28.4°) ]


Coeffient of lift with aileron deflection
Clδa :=
Clp := −
clδa⋅ lwr
Aw⋅ b w
⋅  b2 − b 1

2
2
+

( cla + cdfw)⋅ lwr⋅ bw
24⋅ Aw
−3
3
3
 b2 − b 1  = 1.74 × 10

4( λ − 1 ) 
3⋅ bw
Roll damping
( 1 + 3 ⋅ λ) = −0.107
Redefining θ for program
θ := 0 .. 1.047
(
)
proll θ , Vcs := −
(
Roll authority
Clδa
C lp
)
proll 10° , Vd = 5.221⋅
θ⋅
2 ⋅ Vcs
Roll Rate
bw
deg
Roll rate with 5 ° deflection at 20mph
s
Estimated Roll Rate as a Function of Aileron Delfection
50
45
40
Roll Rate
35
(
) 30
proll θ , V d
 deg 
 
 s 
25
20
15
10
5
0
0
5
10
15
20
25
30
35
40
45
50
55
60
θ
deg
Aileron Deflection
px := −
Clδa 2
1
⋅
= 0.05839
Clp b w
m
Roll rate constant, multiply by control surface deflection
(in degrees) and velocity (in m/s) to get roll rate in deg/s
1] Drag Coefficient. (n.d.). Retrieved from http://www.engineeringtoolbox.com/drag-coefficient-d_627.html
2] Mechanical Properties of Bamboo. (n.d.). Retrieved from http://bamboo.wikispaces.asu.edu/4. Bamboo
Properties
3] More About Hard Fiber, Fiberglass, Garolite, and Carbon Fiber. (2015). Retrieved from
http://www.mcmaster.com/#8549kac/=11z0iim
4] Bm paper white kraft paper 001 kraft paper 300gsm. (n.d.). Retrieved from
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