aircraft stability and control assignment
TRANSCRIPT
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Longitudinal Static Stability
Set aspect ratio of wing ARw = 12
Given wing span b = 2.8 m
Wing area At sea level take-off:
Sizing of horizontal tail Take off speed V = 1.2 =
From Xfoil plotter, Assume elliptical lift distribution, for finite wing is as follows: To enable the UAV has enough stick fixed degree of longitudinal static stability, static margin is
set to be 0.15.
Assume chord of horizontal stabilizer to be 45% of wing chord,
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From Xfoil plotter,
Length of fuselage Assuming the wing aerodynamic centre is located at quarter of the fuselage measured from nose.
Assuming the fuselage is a symmetrical hemispheric tube. The fuselage contribution to
longitudinal static stability is analysed using Multhopps method as follows:
station x wf x d/d wf2*(d/d)x
1 0.102 0.155 0.357 1.250 0.003
2 0.102 0.155 0.306 1.300 0.003
3 0.102 0.155 0.255 1.400 0.003
4 0.102 0.155 0.102 3.200 0.008
5 0.245 0.155 0.123 0.084 0.000
6 0.245 0.155 0.368 0.251 0.001
7 0.245 0.155 0.613 0.418 0.002
8 0.245 0.155 0.858 0.585 0.003
9 0.245 0.155 1.103 0.752 0.004
sum 0.0298
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Most forward CG limit using servo limit
From Xfoil plotter, ;
station x wf a0w wf2(a0w+it) x
1 0.207 0.155 -12.86 -0.064
2 0.207 0.155 -12.86 -0.064
3 0.207 0.155 -12.86 -0.064
4 0.207 0.155 -12.86 -0.064
5 0.207 0.155 -12.86 -0.064
6 0.207 0.155 -12.86 -0.064
7 0.207 0.155 -12.86 -0.064
8 0.207 0.155 -12.86 -0.064
9 0.207 0.155 -12.86 -0.064
Sum -0.577 From figure 2.12(Nelson), k2-k1=0.944
( )
Plot the above and equation to get
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From graph:
Moment equation: In trimmed flight, :
Assume 1 deg of elevator deflection changes the horizontal tail angle of attack by 0.4 deg, =0.4
From figure 2.21 (Nelson),
For =0.4, ; therefore
y = -0.15x - 0.0486-0.4
-0.2
0
0 0.5 1 1.5 2 2.5
C
m
CL
Cm vs CL
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From the data of the servo used, operating travel is 60deg, therefore maximum available
elevator deflections Most forward cg location ( ) Therefore, permissible cg range: between 0.168c and 0.467c from wing leading edge.
Without servo control With servo control
Most forward CG 0.317c 0.168c
Cruising at 2000ft at 60 km/h:
Sizing of horizontal tail
From Xfoil plotter, Assume elliptical lift distribution, for finite wing is as follows:
To enable the UAV has enough stick fixed degree of longitudinal static stability, static margin is
set to be 0.15.
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Assume chord of horizontal stabilizer to be 37.775% of wing chord,
From Xfoil plotter, Length of fuselage Assuming the wing aerodynamic centre is located at quarter of the fuselage measured from nose.Assuming the fuselage is a symmetrical hemispheric tube. The fuselage contribution to
longitudinal static stability is analysed using Multhopps method as follows:
station x wf x d/d wf2*(d/d)x
1 0.102 0.155 0.357 1.250 0.003
2 0.102 0.155 0.306 1.300 0.003
3 0.102 0.155 0.255 1.400 0.003
4 0.102 0.155 0.102 3.200 0.0085 0.245 0.155 0.123 0.082 0.000
6 0.245 0.155 0.368 0.245 0.001
7 0.245 0.155 0.613 0.408 0.002
8 0.245 0.155 0.858 0.571 0.003
9 0.245 0.155 1.103 0.734 0.004
sum 0.0295
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Most forward CG limit using servo limit
From Xfoil plotter, ;
station x wf a0w wf2(a0w+it) x
1 0.207 0.155 -13.12 -0.065
2 0.207 0.155 -13.12 -0.065
3 0.207 0.155 -13.12 -0.065
4 0.207 0.155 -13.12 -0.065
5 0.207 0.155 -13.12 -0.065
6 0.207 0.155 -13.12 -0.065
7 0.207 0.155 -13.12 -0.065
8 0.207 0.155 -13.12 -0.065
9 0.207 0.155 -13.12 -0.065
Sum -0.588 From figure 2.12(Nelson), k2-k1=0.944
( )
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Plot the above and equation to get
From graph:
Moment equation: In trimmed flight, : Assume 1 deg of elevator deflection changes the horizontal tail angle of attack by 0.4 deg, =0.4
From figure 2.21 (Nelson),
For =0.4, ; therefore
y = -0.15x - 0.0448-0.4
-0.2
0
0 0.5 1 1.5 2 2.5
Cm
CL
Cm vs CL
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From the data of the servo used, operating travel is 60deg, therefore maximum available
elevator deflections Most forward cg location ( ) Therefore, permissible cg range: between 0.170c and 0.456c from wing leading edge.
Without servo control With servo control
Most forward CG 0.306c 0.170c
Cruising at 5000ft at 80 km/h:
Sizing of horizontal tail
From Xfoil plotter, Assume elliptical lift distribution, for finite wing is as follows:
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To enable the UAV has enough stick fixed degree of longitudinal static stability, static margin isset to be 0.15.
Assume chord of horizontal stabilizer to be 30.48% of wing chord, From Xfoil plotter, Length of fuselage Assuming the wing aerodynamic centre is located at quarter of the fuselage measured from nose.
Assuming the fuselage is a symmetrical hemispheric tube. The fuselage contribution to
longitudinal static stability is analysed using Multhopps method as follows:
station x wf x d/d wf2*(d/d)x
1 0.102 0.155 0.357 1.250 0.003
2 0.102 0.155 0.306 1.300 0.003
3 0.102 0.155 0.255 1.400 0.0034 0.102 0.155 0.102 3.200 0.008
5 0.245 0.155 0.123 0.080 0.000
6 0.245 0.155 0.368 0.241 0.001
7 0.245 0.155 0.613 0.401 0.002
8 0.245 0.155 0.858 0.561 0.003
9 0.245 0.155 1.103 0.722 0.004
sum 0.0293
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Most forward CG limit using servo limit
From Xfoil plotter,
; station x wf a0w wf2(a0w+it) x
1 0.207 0.155 -13.51 -0.0672 0.207 0.155 -13.51 -0.067
3 0.207 0.155 -13.51 -0.067
4 0.207 0.155 -13.51 -0.067
5 0.207 0.155 -13.51 -0.067
6 0.207 0.155 -13.51 -0.067
7 0.207 0.155 -13.51 -0.067
8 0.207 0.155 -13.51 -0.067
9 0.207 0.155 -13.51 -0.067
Sum -0.606
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From figure 2.12(Nelson), k2-k1=0.944
( )
Plot the above and equation to get
From graph:
Moment equation:
In trimmed flight, :
Assume 1 deg of elevator deflection changes the horizontal tail angle of attack by 0.4 deg, =0.4
y = -0.15x - 0.0437-0.4
-0.3
-0.2
-0.1
0
0 0.5 1 1.5 2 2.5
Cm
CL
Cm vs CL
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From figure 2.21 (Nelson),
For =0.4, ; therefore
From the data of the servo used, operating travel is 60deg, therefore maximum available
elevator deflections Most forward cg location ( ) Therefore, permissible cg range: between 0.189c and 0.452c from wing leading edge.
Without servo control With servo control
Most forward CG 0.302c 0.189c
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Summary for size of horizontal tailplane and most forward CG position with servo control
Wing aspect ratio AR 12
Wing area S 0.653 m2
Wing span b 2.8 m
Wing chord 0.2333m
Flight condition Sea level take
off
Cruise at 2000ft
at 60km/h
Cruise at 5000ft
at 80km/h
Horizontal tail volume ratio VH 0.731 0.720 0.708
Aspect ratio of horizontal tail ARt 7.74 10.69 15.98
Area of tailplane St(m2) 0.0854 0.0831 0.0808
Chord of tailplane ct(m) 0.105 0.0881 0.0711
Span of tailplane bt(m) 0.813 0.942 1.136
Area of elevator Se (m2) 0.0171 0.0166 0.0162
Chord of elevator ce (m) 0.0210 0.0176 0.0142
XNP/c 0.467 0.456 0.452
Xcg/c (without servo control) 0.317 0.306 0.302
Xcg/c (with servo control at
maximum deflection of elevator)
0.168 0.170 0.189
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Directional Static Stability
At sea level take-off: Take off speed
= 1.2
=
Use AXI4130/20 motor with 1311 prop and 30RC1700 battery.
The motor has the following characteristics:
Effeciency = 0.86;
Pout=782W
The dynamic thrust supplied is calculated as shown:
Set the spanwise between two engines = 1.4 m
One engine is off, therefore it creates yawing moment.
Yawing moment due to asymmetric thrust:
Assume chord of vertical tail to be 50% of wing chord: From Xfoil plotter,
Assume 1 deg of rudder deflection changes the fin angle of attack by 0.4 deg, =0.4
Assume fuselage is hemispheric tube,
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By referring to figure 2.29 in textbook (Nelson), By referring to figure 2.30 in textbook (Nelson),
(1)
(2) (3) (4)Substitute equation 2,3, and 4 into 1 yielding:
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Cruising at 2000ft at 60km/h:= Use AXI4130/20 motor with 1311 prop and 30RC1700 battery.
The motor has the following characteristics:
Effeciency = 0.86;
Pout=782W
The dynamic thrust supplied is calculated as shown:
Set the spanwise between two engines = 1.4 m
One engine is off, therefore it creates yawing moment.
Yawing moment due to asymmetric thrust:
Assume chord of vertical tail to be 40% of wing chord:
From Xfoil plotter, Assume 1 deg of rudder deflection changes the fin angle of attack by 0.4 deg, =0.4
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Assume fuselage is hemispheric tube,
By referring to figure 2.29 in textbook (Nelson), By referring to figure 2.30 in textbook (Nelson),
(1)
(2) (3) (4)Substitute equation 2,3, and 4 into 1 yielding:
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Cruising at 5000ft at 80km/h:= Use AXI4130/20 motor with 1311 prop and 30RC1700 battery.
The motor has the following characteristics:
Effeciency = 0.86;
Pout=782W
The dynamic thrust supplied is calculated as shown:
Set the spanwise between two engines = 1.4 m
One engine is off, therefore it creates yawing moment.
Yawing moment due to asymmetric thrust:
Assume chord of vertical tail to be 32.35% of wing chord:
From Xfoil plotter, Assume 1 deg of rudder deflection changes the fin angle of attack by 0.4 deg, =0.4
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Assume fuselage is hemispheric tube,
By referring to figure 2.29 in textbook (Nelson), By referring to figure 2.30 in textbook (Nelson),
(1)
(2) (3) (4)Substitute equation 2,3, and 4 into 1 yielding:
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Size of vertical tailplane with one engine out and crosswind of 20km/h
Flight condition Sea level take
off
Cruise at 2000ft
at 60km/h
Cruise at 5000ft
at 80km/h
Vertical tail volume ratio Vvt 0.0572 0.0410 0.0237
Aspect ratio of vertical tail ARvt 5.93 6,53 5.72
Area of vertical tailplane Svt(m2) 0.0807 0.0569 0.0326
Chord of vertical vertical tailplane
cvt(m)
0.117 0.0933 0.0754
Span of vertical tailplane bvt(m) 0.692 0.610 0.432
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Longitudinal Dynamics
Sea level take off
Assuming the flight is at low speed (M
and the flight is in zero angle of attack (straight
and level).
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[
]
det(sI-A)=0
The solutions yield the eigenvalues:
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Phugoid mode:
Short period
Cruise at 2000ft at 60km/h:
det(sI-A)=0
The solutions yield the eigenvalues:
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Oscillation characteristics:
Cruising at 5000ft at 80km/h:
det(sI-A)=0
The solutions yield the eigenvalues:
Oscillation characteristics:
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By approximation
Parameter Obtained:
m q U Thrust S Cd Cl Cdo Clo
sea
level
10 108.8
7
13.33
2
58.655
87
1.225 0.6532
4
0.8247
87
1.3794
3
0.8796
43
0.6
200
0 ft
10 160.6
2
16.67 46.910
62
1.007 0.6532
4
0.5132
49
1.0733
11
0.5464
59
0.6
500
0 ft
10 260.0
7
22.22 35.193
52
0.736
4
0.6532
4
0.2963
59
0.8260
84
0.3160
32
0.6
Cla Cma c Lf Iy
4.319 -0.616 0.233 1.87 2.91
4.5189 -0.64627 0.233 1.87 2.66
4.5972 -0.65838 0.233 1.87 2.34
Sea level:
Phugoid mode
Xu = -(CD + 2CDO)/Um
= -(0.824 + 2(0.879))/(13.32*10)=-1.3784
Zu = -(Cl + 2Clo)/Um
= -(1.379 + 2(0.6))/(13.32*10)
= -1.3759
Wph = = 1.006
Sph = -Xu/(2Wph)
=-(-1.3784)/(2(1.006))
= 0.685
Short mode
Zw=-(Cla+Cdo)(q*S)/(U*m)
=-(4.319 +0.879)(108.87*0.653)/(13.32*10)
=--2.7732
Mw=Cma(q*S*c)/(U*Iy)
=(-0.616)( 108.87*0.653*0.233)/(13.32*2.91)
=-0-0.26
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Mq=-lf*Cla(lf/U)*q*S/Iy
=-1.87*(4.319)(1.87/13.32)(108.87*0.653)/( 2.91)
=-27.7
Wnsp= = =8.966
Ssp=-(Mq+U*Mw+Zw)/(2Wnsp)
=-(-27.7+13.32*-0.263+-2.773)/(2*8.966)
=1.896
2000ft
Xu = -(CD + 2CDO)/Um
= -(0.513 + 2(0.879))/(16.67*10)
=-1.01
Zu = -(Cl + 2Clo)/Um
= -(1.073 + 2(0.6))/(16.67*10)
= -1.431
Wph =
= 1.0917Sph = -Xu/(2Wph)
=-(-1.01)/(2(1.0917))
= 0.551
Short mode
Zw=-(Cla+Cdo)(q*S)/(U*m)
=-(4.518 +0.5464)(169.62*0.653)/(16.67*10)
=-3.188
Mw=Cma(q*S*c)/(U*Iy)
=(-0.64)( 160.62*0.653*0.233)/(16.67*2.66)
=-0.502
Mq=-lf*Cla(lf/U)*q*S/Iy
=-1.87*(4.518)(1.87/16.67)(160.62*0.653)/( 2.66)
=-37.3
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Wnsp= = =11.17
Ssp=-(Mq+U*Mw+Zw)/(2Wnsp)
=-(-37.3+16.67*-0.356+-3.1882)/(2*11.17)=2.0776
5000ft
Xu = -(CD + 2CDO)/Um
= -(0.296 + 2(0.3160))/(22.2*10)
=-0.70985
Zu = -(Cl + 2Clo)/Um
= -(0.826 + 2(0.6))/(22.2*10)
= -1.545
Wph = = 0.826
Sph = -Xu/(2Wph)
=-(-0.7098)/(2(0.826))
= 0.4291
Lateral-Directional Dynamics
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Since the UAV is symmetrical, therefore
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det(sI-A)=0
The solutions yield the eigenvalues:
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Cruising at 2000ft at 60km/h
det(sI-A)=0
The solutions yield the eigenvalues:
,
,
,
Cruising at 5000ft at 80km/h
det(sI-A)=0
The solutions yield the eigenvalues:
,
, ,
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Summary
Sea level take-off Cruise at 2000ft at
60km/h
Cruise at 5000ft at
80km/h
Longitudinal
motion
0.293 4.02s
eriod 10.73s
0.375 Lateral-directional
motion
eriod
, , , , , ,