Rocket Science
Steve R. Gunn
Image, Speech and Intelligent Systems Group
Department of Electronics and Computer Science
University of Southampton
August 29, 2001
Version 0.6
Nomenclature
a
A
cm
cp
Cd
d
I
g
l
m
mc
mp
mr
ρ
r
t
ta
tb
tc
td
tf
T
v
y
ya
yb
1
acceleration
cross-sectional area
centre of mass
centre of pressure
drag coefficient
main body diameter
impulse
gravitational acceleration
rocket length
mass
motor casing mass
propellant mass
rocket mass
air density
rocket radius
time
apogee time
burn time
coast time
descent time
flight time
thrust
velocity
altitude
altitude at apogee
altitude at end of burn
m s−2
m2
m
m
m
Ns
m s−2
m
kg
kg
kg
kg
1.2 kg m−3
m
s
s
s
s
s
s
N
m s−1
m
m
m
Introduction
Assuming a vertical trajectory and no external wind forces the dynamical problem can be reduced to one
dimension. Applying Newton’s second law the dynamics of the rocket satisfy
d2 y
dy dy m 2 = −m g + T − k
,
(1)
dt
dt dt 2 Constant Mass and Thrust assumption
2
where
k=
1
ρ Cd A,
2
(2)
ρ is the density of air (1.2 kg m−3 ), Cd is the drag coefficient (about 0.75 for a model rocket), m is the rocket
mass and A is the cross-sectional area of the rocket.
2
Constant Mass and Thrust assumption
The changing mass of the propellant is approximated by its mean value, and m has the form
m = mr + mc + 12 mp .
(3)
The thrust of the motor, T , is assumed to be constant over the burn time, tb .
2.1
Burn Phase
In the burn phase Equation 1 may be rewritten as,
d2 y
m 2 = −m g + T − k
dt
dy
dt
2
,
for 0 ≤ t ≤ tb ,
and when T ≥ m g (i.e. the motor is powerful enough to lift the rocket), has general solution
!
r
k T − mg
m
log cosh
t − C2 .
y = C1 +
k
m
k
(4)
(5)
The boundary conditions of Equation 4 are
y 0 (0) = 0,
y(0) = 0.
(6)
The solution is
y
y0
y 00
!
r
m
T − mg
k
=
log cosh
t
k
m
k
!
r
r
T − mg
k
T − mg
=
tanh
t
k
m
k
!
r
T
k
T − mg
2
=
− g sech
t
m
m
k
(7)
(8)
(9)
or more simply
y
y0
y 00
1
log cosh (θ t)
φ2
θ
= g 2 tanh (θ t)
φ
θ2
= g 2 sech2 (θ t) ,
φ
= g
(10)
(11)
(12)
2 Constant Mass and Thrust assumption
3
where
p
k (T − m g)
θ =
m
r
kg
φ =
.
m
2.2
(13)
(14)
Coast Phase
In the coast phase Equation 1 may be rewritten as,
d2 y
m 2 = −m g − k
dt
dy
dt
2
,
for tb ≤ t ≤ ta ,
(15)
and has general solution
y = C1 +
r
m
k mg
log cos C2 −
t .
k
m
k
(16)
The boundary conditions of Equation 15 are
y 0 (tb ) = yb0 ,
y(tb ) = yb ,
for tb ≤ t ≤ ta .
(17)
The general solution is
r
m
k mg
= C1 +
log cos C2 −
t
k
m
k
r
r
mg
k mg
=
tan C2 −
t
k
m
k
r
k mg
2
t .
= g sec C2 −
k
m
y
y0
y 00
(18)
(19)
(20)
Solving for the boundary conditions,
yb
yb0
r
m
k mg
= C1 +
log cos C2 −
tb
k
m
k
r
r
mg
k mg
=
tan C2 −
tb ,
k
m
k
(21)
(22)
gives
s
!
k
gk
−1
0
=
tb + tan
yb
mg
m
s
k yb02
m
= yb +
log 1 +
,
k
mg
r
C2
C1
(23)
(24)
and hence the solution is
s
y
y0
y 00
k yb02
gk
(t − tb ) + tan−1
m
s
!!
r
r
k
mg
gk
−1
0
=
tan −
(t − tb ) + tan
yb
mg
k
m
s
!!
r
g
k
k
= − g sec2 −
(t − tb ) + tan−1 yb0
.
m
mg
= yb +
m
log  1 +
cos −
k
mg
s
r
yb0
!!
k

mg
(25)
(26)
(27)
2 Constant Mass and Thrust assumption
4
Simplifying,
y
= g
y0
= g
y 00
1
θ
log
cos
(φ
(t
−
t
))
cosh
(θ
t
)
+
sin
(φ
(t
−
t
))
sinh
(θ
t
)
b
b
b
b
φ2
φ
tanh (θ tb ) − φθ tan (φ (t − tb ))
θ
φ2 1 + θ tan(φ (t − t )) tanh(θ t )
b
b
φ
2
2
φ
2
θ2 θ2 + tanh (θ tb ) sec (φ (t − tb ))
= −g 2 2 .
φ
1 + φθ tan (φ (t − tb )) tanh (θ tb )
(28)
(29)
(30)
At apogee y 0 (ta ) = 0, hence
0
ta
= θ tanh (θ tb ) − φ tan (φ (ta − tb ))
1
θ
= tb + tan−1
tanh (θ tb ) ,
φ
φ
(31)
= ta − tb
1
θ
−1
=
tan
tanh (θ tb ) .
φ
φ
(33)
(32)
and the coast time is given by
tc
(34)
The altitude at apogee is
ya
ya
ya
ya
2.3
s

02
k
y
m
b 
= yb +
log  1 +
k
mg
v
u
u
m
k T − mg
= yb +
log t1 +
tanh2
k
mg
k
(35)
r
!
T − mg
tb 
k
k
m
v
!
r
u
u
m
k
T − mg
m
k T − mg

=
log cosh
t +
log t1 +
tanh2
k
m
k
k
mg
k
1
T
2
= g
log
1
+
sinh
(θ
t
)
b
2 φ2
mg
(36)
k
m
r
!
T − mg
tb  (37)
k
(38)
Descent Phase (Freefall)
In the descent phase Equation 1 may be rewritten as,
d2 y
m 2 = −m g + k
dt
dy
dt
2
,
for ta ≤ t ≤ tf ,
(39)
!
kg
t .
m
(40)
and has general solution
m
y = C1 −
log cosh C2 −
k
r
The boundary conditions of Equation 39 are
y 0 (ta ) = 0,
y(ta ) = ya .
(41)
2 Constant Mass and Thrust assumption
5
The solution is
!
r
m
kg
= ya −
log cosh
(ta − t)
k
m
!
r
r
mg
kg
=
tanh
(ta − t)
k
m
!
r
kg
2
= − g sech
(ta − t) ,
m
y
y0
y 00
(42)
(43)
(44)
or more simply
y
= ya − g
y0
= g
1
log cosh (φ (ta − t))
φ2
1
tanh (φ (ta − t))
φ
= − g sech2 (φ (ta − t)) .
y 00
(45)
(46)
(47)
The flight time, tf , is given by
1
log cosh (φ (ta − tf ))
φ2
1
ya φ2
= ta + cosh−1 exp
,
φ
g
= ya − g
0
tf
(48)
(49)
where we choose the principal range of cosh−1 (x) such that cosh−1 (x) ≥ 0, and the descent time is given by
td
= tf − ta
(50)
=
1
cosh−1 exp
φ
2
ya φ
g
.
(51)
Substituting,
td
=
1
cosh−1
φ
s
1+
Terminal velocity is given by
yt0 = −
T
sinh2 (θ tb )
mg
g
.
φ
(52)
(53)
3 Variable Mass and Thrust
3
6
Variable Mass and Thrust
m
d2 y
= −m g + T − k
dt2
Consider m to have the form
m = mr + mp
tb − t
tb
dy
dt
2
(54)
(55)
General solution for velocity and acceleration can be recovered in the form of Bessel functions, but the altitude
cannot be written down explicitly due to the integral of the velocity being intractable.
4 Stability
4
7
Stability
The stability of a rocket is governed by the centre of mass, cm , and the centre of pressure, cp . Defining the
origin as the nose tip of the rocket, the rocket is stable if cp > cm , (i.e. the centre of pressure is behind the
centre of mass). To calculate cm and cp we consider the axially symmetric part of the rocket, Figure 1, and the
fins separately.
Figure 1: Rocket Cross section
Figure 2 illustrates the case for a stable rocket, where the lift force is denoted by a purple arrow and the drag
force by a red arrow. These forces act about the centre of pressure. It can be seen that the three examples are
stable since the lift forces act to restore the rocket to vertical flight, if it is perturbed. Consequently determining
stability requires the calculation of cm and cp .
Figure 2: Rocket Stability
4.1
Centre of Mass
The centre of mass for an axially symmetric rocket is given by
R 2π R l R R(x)
cm
=
=
x %(r, x, θ) r dr dx dθ
0
0 0
R 2π R l R R(x)
%(r, x, θ) r dr dx dθ
0
0 0
R l R R(x)
x %(r, x) r dr dx
0 0
R l R R(x)
%(r, x) r dr dx
0 0
(56)
(57)
where R(x) is the radius of the rocket, and % is the rocket density.
Example 4.1 (Shell).
%(r, x) = %0 δ(r − R(x))
(58)
4 Stability
8
R l R R(x)
cm
=
=
x %0 δ(r − R(x)) r dr dx
0 0
R l R R(x)
%0 δ(r − R(x)) r dr dx
0 0
Rl
x R(x) dx
0
Rl
R(x) dx
0
(59)
(60)
Alternatively, it may be more convenient to calculate the centre of mass of the rocket components separately
to find cm ,
cmn mn + cmb mb + cmf mf + cmm mm + cmi mi
cm =
(61)
mn + mb + mf + mm + mi
where here we choose to separate the rocket into its nose, body, fins, motor and internals (recovery system
and/or payload). However, once a rocket has been built it is trivial to find the cm by finding its balance point.
N.B. typically the cm will decrease slightly during flight due to the combustion of the propellant.
4.2
Centre of Pressure
Following the work of Barrowman (?):
For an axially symmetric body of revolution, the subsonic steady state aerodynamic running normal load is
given by
∂
n(x) = ρ v
[S(x) W (x)]
(62)
∂x
A rigid body has downwash given by
W (x) = v α
(63)
Thus
n(x) = ρ v 2 α
∂S(x)
∂x
(64)
The normal force coefficient, CN is defined by
CN (x)
n(x)
ρ v2 A
2 α ∂S(x)
A ∂x
8 α ∂S(x)
πd2 ∂x
=
1
2
=
=
(65)
(66)
(67)
By the definition of the normal force curve slope
CNα (x)
=
=
∂CN ∂α α=0
8 ∂S(x)
πd2 ∂x
(68)
(69)
In order to obtain the total CNα we integrate over x,
Z
CNα
=
l
CNα (x) dx
(70)
8
(S(l) − S(0))
πd2
(71)
0
=
By definition the pitching moment of the local normal aerodynamic force about the nose tip (x = 0) is
M (x) = xn(x)
∂S(x)
= xρ v 2 α
∂x
(72)
(73)
4 Stability
9
By definition the aerodynamic pitching moment coefficient is
M (x)
ρ v2 A d
2 α x ∂S(x)
A d ∂x
8 α x ∂S(x)
πd3 ∂x
Cm (x) =
1
2
=
=
(74)
(75)
(76)
By the definition of the moment coefficient curve slope
Cmα (x)
=
=
∂Cm ∂α α=0
8 x ∂S(x)
πd3 ∂x
(77)
(78)
In order to obtain the total Cmα we integrate over x,
Z
Cmα
=
l
Cmα (x) dx
(79)
0
=
=
=
=
8
πd3
Z
l
∂S(x)
dx
∂x
0
(
)
Z l
8
l
[x S(x)]0 −
S(x) dx
πd3
0
(
)
Z l
8
l S(l) −
S(x) dx
πd3
0
x
8
{l S(l) − V }
πd3
(80)
(81)
(82)
(83)
The centre of pressure, cp , is defined as,
cp
Cmα
CNα
l S(l) − V
S(l) − S(0)
= d
(84)
=
(85)
5 Nose Shape
5
10
Nose Shape
Following the work of ?? we consider the optimal form for the nose shape. Newton reasoned that the air resistance was caused by particles of air hitting the moving object and used the conservation of momentum. In the
following we consider only convex noses which avoids complications due to multiple impacts from air molecules.
(However, there are non-convex local optima that have a single impact form ()). Using the conservation of
momentum,
mδv
= ma v (1 + cos(2 ϕ))
= ρ A v δt v (1 + cos(2 ϕ))
= 2 ρ A v 2 cos2 (ϕ) δt
(86)
(87)
(88)
Hence,
Fd = m
dv
= 2 ρ A v 2 cos2 (ϕ)
dt
(89)
Now consider a curved surface such that the area dA is varying over the surface,
ZZ
Fd =
2 ρ v 2 cos2 (ϕ) dA
ZZ
= 2 ρ v2
cos2 (ϕ) dA
ZZ
2
= 2ρv
cos3 (ϕ) dS
(90)
(91)
(92)
From Equation 1 and Equation 2 we note that,
Cd
=
=
ZZ
4
cos2 (ϕ) dA
A
ZZ
4
cos3 (ϕ) dS
A
(93)
(94)
Noting that
cos(ϕ) = r
1+
∂z
∂x
1
2
+
∂z
∂y
2
(95)
then
Cd
=
4
A
We define the nose aspect ratio parameter γ =
radius at the base.
ZZ
1+
h
R,
∂z
∂x
1
2
+
∂z
∂y
2 dx dy
(96)
where h is the nose height from tip to base and R is the nose
5 Nose Shape
5.1
11
Radial Nose Cones
Now consider a radially symmetric nose cone of profile z(r), then
Cd
=
=
=
4
A
2π
Z
R
Z
0
0
1+
R
8π
A
Z
8
R2
Z
0
1+
R
0
1+
r
2 dr dθ
(97)
∂z
∂r
r
2 dr
(98)
r
2 dr
(99)
∂z
∂r
∂z
∂r
h
Example 5.1 (Conical Nose). Consider a conical nose z(r) = − R
r, then
Cd
=
4
1 + γ2
(100)
In limh→0 we obtain the solution for a flat nose, Cd = 4 and in limh→∞ we obtain the solution for an infinitely
pointed nose, Cd = 0.
Example 5.2 (Parabolic Nose). Consider a parabolic nose z(r) = − Rh2 r2 , then
log 1 + 4γ 2
Cd =
γ2
2
Example 5.3 (Ogive Nose). Consider an ogive nose (z + h) + r +
Cd
h2
2R
Cd
4
2
=
h2
2R
+
R
2
2
, then
(102)
z+R 2
R
+
r 2
R
= 1, then
(103)
= 2
z+h 2
h
Example 5.5 (Elliptical Nose). Consider an elliptical nose
=
R
2
8 1 + 2 γ2
3 (1 + γ 2 )2
=
Example 5.4 (Hemispherical Nose). Consider an hemispherical nose
Cd
−
(101)
+
r 2
R
= 1 where h > R, then
1 + (2 log γ − 1) γ 2
(104)
2
(1 − γ 2 )
(
Example 5.6 (Conical Frustrum Nose). Consider an conical frustrum nose z(r) =
0
h(r−λR)
(1−λ)R
0 ≤ r ≤ λR
,
λR < r ≤ R
then
Cd
= 4
(λ − 1)2 + λ2 γ 2
(λ − 1)2 + γ 2
(105)
The optimal λ ∈ [0, 1] that minimises Cd is given by,
λopt
Cdopt
p
1 2
= 1+
γ − γ γ2 + 4
2
p
= 2 2 + γ2 − γ γ2 + 4
(106)
(107)
To find the shape with the minimum drag, we can use calculus of variations. The minimiser with respect to
5 Nose Shape
12
y(x) satisfies the Euler Lagrange equation,
2
dz
dr
+
dz 3
dr
2
+ r ddrz2 − 3 r
2 3
1 + dz
dr

d 

dr
d2 z
dr 2
=0
(108)

2  = 0
2
dz
(109)

2r
1+
dz
dr
dr
2r
Letting u =
1+
dz 2
dr
dz
dr
2
dz 2
dr
= C1
(110)
dz
dr ,
2ru
2
(1 + u2 )
r
=
=
Using the chain rule
dz
du
= C1
2
1 + u2
1
2 C1
u
1
1
3
u + 2u +
2 C1
u
(111)
(112)
(113)
dr
= u du
dz
du
=
z
=
z
=
z
=
1
d
1
u
u3 + 2u +
2 C1 du
u
Z
1
1
u 3u2 + 2 − 2 du
2 C1
u
Z
1
1
3u3 + 2u − du
2 C1
u
1
3 4
u + u2 − log |u| + C2
2 C1 4
Hence the parametric equations with the gradient as the parameter are,
1
1
3
r =
u + 2u +
2 C1
u
1
3 4
z =
u + u2 − log |u| + C2
2 C1 4
(114)
(115)
(116)
(117)
(118)
(119)
Applying boundary constraints u(RT ) = −1, z(RT ) = 0
2
RT
RT 7
4 4 RT
1
3
−
u + 2u +
4
u
RT 3 4
−
u + u2 − 74 − log |u|
4
4
3
u2 − 1 u2 + 73 − 14 log |u|
− RT 16
C1
= −
(120)
C2
=
(121)
r
=
z
=
=
(122)
(123)
(124)
5 Nose Shape
13
The tip radius, RT , can be found by first solving for the gradient at the base, uR ,
R
−
h
⇒
=
2
1 + u2R
7
3
2
2
4 uR (uR − 1) uR + 3 − uR log |uR |
3 u5R + 4 γ u4R + 4 u3R + 8 γ u2R − 7 uR + 4 γ − 4 uR log |uR | = 0
and then
RT = − 4 R
uR
2
(1 + u2R )
(125)
(126)
(127)
(128)
6 Useful Links
6
Useful Links
• Variable Force and Mass: Rocket Liftoff
• Pete’s Rockets
• Estes Rockets
• NASA Tutorial
• Stability Analysis
• Aerodynamic Links
• Altimeter
14
A Estes Astrocam
Estes Astrocam
Length:
Diameter:
Weight:
Recovery:
Fins:
Maximum Altitude:
Recommended Engines:
18.38” (46.7 cm)
1.34” (34.0 mm)
2.7 oz (76 g)
12” (30 cm) parachute
Plastic molded
500 ft (152 m)
C6-7
Astrocam Altitude
Astrocam Velocity
250
100
B4 Motor
B6 Motor
C6 Motor
B4 Motor
B6 Motor
C6 Motor
200
50
ms−1
150
m
100
0
50
0
0
2
4
6
8
s
10
12
14
−50
16
0
2
4
6
(a) Altitude
8
s
(b) Velocity
Astrocam Acceleration
80
B4 Motor
B6 Motor
C6 Motor
60
40
20
ms−2
A
15
0
−20
−40
−60
0
2
4
6
8
s
10
12
(c) Acceleration
Figure 3: Astrocam Kinematics
14
16
10
12
14
16
B Estes Bandit
Estes Bandit
Length:
Diameter:
Weight:
Recovery:
Fins:
Maximum Altitude:
Recommended Engines:
16.6” (42.2 cm)
1.0” (25.4 mm)
1.6oz (45 g)
12” (30 cm) parachute
T3 plastic molded
1000 ft (305 m)
A8-3(First Flight), B4-4, B6-4, B6-6, C6-5, C6-7
Bandit Altitude
Bandit Velocity
400
140
A8 Motor
B4 Motor
B6 Motor
C6 Motor
350
A8 Motor
B4 Motor
B6 Motor
C6 Motor
120
100
300
80
250
ms−1
60
200
40
20
150
0
100
−20
50
−40
0
2
4
6
8
10
s
12
14
16
18
20
−60
0
2
4
6
(a) Altitude
8
10
s
(b) Velocity
Bandit Acceleration
200
A8 Motor
B4 Motor
B6 Motor
C6 Motor
150
100
50
−2
0
ms
m
B
16
0
−50
−100
−150
0
2
4
6
8
10
s
12
14
16
(c) Acceleration
Figure 4: Bandit Kinematics
18
20
12
14
16
18
20
C Estes Banshee
Estes Banshee
Length:
Diameter:
Weight:
Recovery:
Fins:
Maximum Altitude:
Recommended Engines:
16.6” (42.2 cm)
1.0” (25.4 mm)
1.6oz (45 g)
12” (30 cm) parachute
T3 plastic molded
1100 ft (335 m)
A8-3(First Flight), B4-4, B6-4, B6-6, C6-5, C6-7
Banshee Altitude
Banshee Velocity
400
140
A8 Motor
B4 Motor
B6 Motor
C6 Motor
350
A8 Motor
B4 Motor
B6 Motor
C6 Motor
120
100
300
80
250
ms−1
60
200
40
20
150
0
100
−20
50
−40
0
2
4
6
8
10
s
12
14
16
18
20
−60
0
2
4
6
(a) Altitude
8
10
s
(b) Velocity
Banshee Acceleration
200
A8 Motor
B4 Motor
B6 Motor
C6 Motor
150
100
50
−2
0
ms
m
C
17
0
−50
−100
−150
0
2
4
6
8
10
s
12
14
16
(c) Acceleration
Figure 5: Banshee Kinematics
18
20
12
14
16
18
20
D Estes Echostar
Estes Echostar
Length:
Diameter:
Weight:
Recovery:
Fins:
Maximum Altitude:
Recommended Engines:
29” (73.7 cm)
1.33” (33.8 mm)
3.6 oz (102.1 g)
18” (46 cm) parachute
Die cut balsa
3074 ft (937 m)
Single Stage Flights: B4-4, B6-4 (First Flight), C6-5
Single Stage Flights with Payload: B4-2 (First Flight), C6-3
Two Stage Flights: C6-0 + B6-6 (First Flight), C6-0 + C6-7
Echostar Altitude
Echostar Velocity
450
120
B4 Motor
B6 Motor
C6 Motor
C6+C6 Motor
400
B4 Motor
B6 Motor
C6 Motor
C6+C6 Motor
100
80
300
60
250
40
ms−1
350
m
200
20
150
0
100
−20
50
−40
0
0
5
10
15
20
−60
25
0
5
10
15
s
s
(a) Altitude
(b) Velocity
Echostar Acceleration
60
B4 Motor
B6 Motor
C6 Motor
C6+C6 Motor
40
20
ms−2
D
18
0
−20
−40
−60
0
5
10
15
20
s
(c) Acceleration
Figure 6: Echostar Kinematics
25
20
25
E Estes Engines
E
19
Estes Engines
Length:
Diameter:
No.
1503
1507
1511
1504
1510
Type
1
2 A3-2T
A3-4T
A10-3T
1
2 A3-4T
A10-0T
Stage
Single
Single
Single
Upper
Booster
I (N s)
1.25
2.50
2.50
1.25
2.50
td (s)
2
4
3
4
0
ml (g)
56.6
56.6
141.5
28.3
141.5
1.73” (4.4 cm)
0.5” (12.7 mm)
Tmax (N)
7.8
7.8
13.3
7.8
13.3
tb (s)
0.36
0.86
0.26
0.36
0.26
me (g)
5.6
7.6
7.9
6.0
6.7
mp (g)
1.75
3.50
3.78
1.75
3.70
Table 2: Estes Blackpowder Mini Motors
Length:
Diameter:
No.
1593
1598
1601
1602
1605
1606
1620
1617
1613
1614
1599
1604
1607
1615
1608
1616
Type
1
2 A6-2
A8-3
B4-2
B4-4
B6-2
B6-4
B8-5
C5-3
C6-3
C6-5
A8-5
B4-6
B6-6
C6-7
B6-0
C6-0
Stage
Single
Single
Single
Single
Single
Single
Single
Single
Single
Single
Upper
Upper
Upper
Upper
Booster
Booster
I (N s)
1.25
2.50
5.00
5.00
5.00
5.00
5.00
10.00
10.00
10.00
2.50
5.00
5.00
10.00
5.00
10.00
td (s)
2
3
2
4
2
4
5
3
3
5
5
6
6
7
0
0
ml (g)
70.8
113.2
113.2
99.1
127.4
113.2
141.5
226.4
113.3
113.2
56.6
42.5
56.6
70.8
113.2
113.3
2.76” (7.0 cm)
0.69” (17.5 mm)
Tmax (N)
12.8
13.3
13.3
13.3
13.3
13.3
22.2
22.2
13.3
13.3
13.3
13.3
13.3
13.3
13.3
13.3
tb (s)
0.20
0.32
1.20
1.20
0.83
0.83
0.60
2.10
1.70
1.70
0.32
1.20
0.83
1.70
0.80
1.68
me (g)
15.0
16.2
19.8
21.0
19.3
20.1
19.3
25.5
24.9
25.8
17.6
22.1
22.1
26.9
16.4
22.7
mp (g)
1.56
3.12
8.33
8.33
6.24
6.24
6.24
12.70
12.48
12.48
3.12
8.33
6.24
12.48
6.24
12.48
Table 3: Estes Blackpowder Motors
Length:
Diameter:
No.
1666
1667
1668
1665
1669
Type
D12-3
D12-5
D12-7
D12-0
D11-P
Stage
Single
Single
Upper
Booster
Plugged
I (N s)
20.00
20.00
20.00
20.00
20.00
td (s)
3
5
7
0
0
ml (g)
396.2
283.0
226.4
396.2
453.1
2.76” (7.0 cm)
0.94” (24.0 mm)
Tmax (N)
28.5
28.5
28.5
28.5
27.6
Table 4: Estes Blackpowder size D Motors
tb (s)
1.70
1.70
1.70
1.70
1.82
me (g)
42.2
43.1
44.0
40.9
44.0
mp (g)
24.93
24.93
24.93
24.93
24.93
F MATLAB Code
F
20
MATLAB Code
function [a, v, y, t] = rocket(m,d,motor,Cd)
%ROCKET
%
% Usage: [a, v, y, t] = rocket(m,d,motor,Cd)
%
% Parameters: m
- Rocket Mass
%
d
- Rocket Diameter
%
motor - Motor Type (e.g. ’C6’)
%
Cd
- Drag Coefficient (default 0.75)
%
% Author: Steve Gunn ([email protected])
if (nargin <3 | nargin>4) % check correct number of arguments
help rocket
else
if (nargin<4) Cd = 0.75;, end
% Motor
mtype =
mTave =
mtb
=
mmc
=
mmp
=
index =
Database
[ ’A’ ’A’ ’A’
’A’ ’B’ ’B’ ’B’ ’C’ ’C’
’D’ ’D’ ’E’];
[
3
6
8
10
4
6
8
5
6
6
12
25];
[0.36 0.20 0.32 0.26 1.20 0.83 0.60 2.10 1.70 3.40 1.70 1.70];
[];
[];
intersect(find(str2num(motor(2:length(motor)))==mTave),find(motor(1)==mtype));
% Constants
rho
= 1.2;
g
= 9.81;
ts
= 0.01;
% Formulas
A
= pi*d*d/4;
k
= 0.5*rho*Cd*A;
T
= mTave(index);
tb
= mtb(index);
theta = sqrt(k*(T-m*g))/m;
phi
= sqrt(k*g/m);
% Air Density
% Gravitational Accleration
% Time Step
%
%
%
%
Rocket Cross-sectional Area
Drag Constant
Average Thrust
Burn Time
% Burn Phase
t1
= 0:ts:tb;
yb
= g*(1/phi^2)*log(cosh(theta*t1));
vb
= g*(theta/phi^2)*tanh(theta*t1);
ab
= g*(theta/phi)^2*sech(theta*t1).^2;
% Coast
tc
=
ta
=
t2
=
yc
=
vc
=
ac
ya
Phase
(1/phi)*atan((theta/phi)*tanh(theta*tb));
tb+tc;
tb:ts:ta;
g*(1/phi^2)*log(cos(phi*(t2-tb))*cosh(theta*tb) + (theta/phi)*sin(phi*(t2-tb))*sinh(theta*tb));
g*(theta/phi^2)*(tanh(theta*tb) - (phi/theta)*tan(phi*(t2-tb)))./ ...
(1 + (theta/phi)*tan(phi*(t2-tb))*tanh(theta*tb));
= -g*(theta/phi)^2*((phi/theta)^2 + tanh(theta*tb)^2)*sec(phi*(t2-tb)).^2./ ...
(1 + (theta/phi)*tan(phi*(t2-tb))*tanh(theta*tb)).^2;
= m/(2*k)*log(1 + T/(m*g)*sinh(theta*tb).^2);
% Descent Phase
td
= (1/phi)*acosh(exp(k*ya/m)); % Note Matlab returns positive x for x = acosh(y)
tr
= ta+td;
t3
= ta:ts:tr;
yd
= ya - g*(1/phi^2)*log(cosh(phi*(ta-t3)));
vd
= g*(1/phi)*tanh(phi*(ta-t3));
ad
= -g*sech(phi*(ta-t3)).^2;
% Augment Intervals
t
= [t1 t2 t3];
y
= [yb yc yd];
v
= [vb vc vd];
a
= [ab ac ad];
end
Listing 1: Rocket Kinematics MATLAB Code
F MATLAB Code
21
function plotkinematics(name,mass,diameter,motors,Cd)
%PLOTKINEMATICS
%
% Usage: plotkinematics(name,mass,diameter,motors,Cd)
%
% Parameters: name
- Rocket Name
%
mass
- Rocket Mass
%
diameter - Rocket Diameter
%
motors
- List of Motor Types (e.g. {’A8’,’C6’})
%
Cd
- Drag Coefficient (default 0.75)
%
% Author: Steve Gunn ([email protected])
if (nargin <4 | nargin>5) % check correct number of arguments
help plotkinematics
else
if (nargin<5) Cd = 0.75;, end
lw = 2;
c = [’k’ ’r’ ’g’ ’m’ ’b’ ’y’ ’c’];
yf = figure;, hold on
vf = figure;, hold on
af = figure;, hold on
for i=1:length(motors)
[a v y t] = rocket(mass,diameter,char(motors(i)),Cd);
figure(yf), plot(t,y,c(mod(i,length(c))+1),’LineWidth’,lw)
figure(vf), plot(t,v,c(mod(i,length(c))+1),’LineWidth’,lw)
figure(af), plot(t,a,c(mod(i,length(c))+1),’LineWidth’,lw)
end
figure(yf)
eval(legendstring(motors))
title([name ’ Altitude’])
xlabel(’s’), ylabel(’m’)
eval([’print -depsc2 -f ’ name ’y.eps’])
figure(vf)
eval(legendstring(motors))
title([name ’ Velocity’])
xlabel(’s’), ylabel(’ms^{-1}’)
eval([’print -depsc2 -f ’ name ’v.eps’])
figure(af)
eval(legendstring(motors))
title([name ’ Acceleration’])
xlabel(’s’), ylabel(’ms^{-2}’)
eval([’print -depsc2 -f ’ name ’a.eps’])
end
function s = legendstring(motors)
s = ’legend(’;
sep = ’’;
for i=1:length(motors)
switch(char(motors(i)))
case ’D6’
s = [s sep ’’’C6+C6 Motor’’’];
otherwise
s = [s sep ’’’’ char(motors(i)) ’ Motor’’’];
end
sep = ’,’;
end
s = [s ’)’];
end
Listing 2: Plot Kinematics MATLAB Code