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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

aaccelerationm s2

Across-sectional aream2

cmcentre of massm

cpcentre of pressurem

Cddrag coecient

dmain body diameterm

IimpulseN s

ggravitational accelerationm s2

lrocket lengthm

mmasskg

mcmotor casing masskg

mppropellant masskg

mrrocket masskg

air density1.2 kg m3

rrocket radiusm

ttimes

taapogee times

tbburn times

tccoast times

tddescent times

tfflight times

TthrustN

vvelocitym s1

yaltitudem

yaaltitude at apogeem

ybaltitude at end of burnm

Introduction

Assuming a vertical trajectory and no external wind forces the dynamical problem can be reduced to one dimension. Applying Newtons second law the dynamics of the rocket satisfy

d2y= m g + T kdydy,(1)

m dt2dtdt

2 Constant Mass and Thrust assumption2

where1

k = Cd A,(2)

2

is the density of air (1.2 kg m3), Cd is the drag coecient (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 + 1 mp.(3)

2

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,d2y= m g + T kdy2

m, for 0 t tb,

dt2dt

and when T m g (i.e. the motor is powerful enough to lift the rocket), has general solutionmkTm g! .

y = C1 +log coshrt C2

kmk

The boundary conditions of Equation 4 are

y0(0) = 0,y(0) = 0.

The solution is

mkTt!

y=log coshr m g

kmk

kTm g

y0=rT m gtanhrt!

kmk

!

TkTm g

y00= g sech2rt

mmk

or more simply

y=g1log cosh ( t)

2

y0=gtanh ( t)

2

y00=g2sech2 ( t) ,

2

(4)

(5)

(6)

(7)

(8)

(9)

(10)

(11)

(12)


wherepk (T m g)

3

=

m

(13)

r =kmg .

2.2 Coast Phase

(14)

In the coast phase Equation 1 may be rewritten as,

d2ydy2

m= m g k,

dt2dt

and has general solution

m

y = C1 +log cos C2

k


for tb t ta,

krm gt .

mk

(15)

(16)

y0(t) = y0,y(t) = y,fortbtt.

bbbba

The general solution is

mk

C2 rm gt

y =C1 +log cos

kmk

m gkm g

y0=rtanC2rt

kmk

km g

y00=g sec2C2 rt .

mk

Solving for the boundary conditions,

mk

yb=C1 +log cos C2rm gtb

kmk

k

yb0=rm gtan C2 rm gtb ,

kmk

gives

s!

rg kk

C2=tb + tan1yb0

mm g

log s

=yb +m1 +k yb02

C1,

km g

and hence the solution is

s1 +k y02

m rg k1

y=yb +logbcos(t tb) + tan

km gm

s!!

rg kk

m g

y0=rtan(t tb) + tan1yb0

kmm g

s!! .

rg kk

y00= g sec2(t tb) + tan1yb0

mm g

s!!y0k

b m g

(17)

(18)

(19)

(20)

(21)

(22)

(23)

(24)

(25)

(26)

(27)


Simplifying,

y=g1log cos ( (t tb)) cosh ( tb) +sin ( (t tb)) sinh ( tb)

2

y0=gtanh ( tb) tan ( (t tb))

21 + tan( (t tb)) tanh( tb)

2+ tanh2 ( tb) sec2 ( (t tb))

y00= g22

.

21 + tan ( (t tb)) tanh ( tb) 2

At apogee y0(ta) = 0, hence0= tanh ( tb) tan ( (ta tb))

ta=tb +1tan1tanh ( tb) ,

and the coast time is given by

4

(28)

(29)

(30)

(31)

(32)

tc =ta tbtanh ( tb) .(33)

=1tan1(34)

The altitude at apogee is

s

ya

ya

ya

mk yb02

=yb +log1 +(35)

km g

v

tb!

=yb +mlogk Tm gtanh2kTm g(36)

k1 + m gkmk

ur

u

t

v

! +tb!

=mk Tm gmlogk Tm gtanh2k Tm g(37)

k log coshmrktk1 + m gkmk

ur

u

t

1T

ya = glog1 +sinh2 ( tb)(38)

2 2m g

2.3 Descent Phase (Freefall)

In the descent phase Equation 1 may be rewritten as,

d2ydy2

m= m g + k

dt2dt

and has general solution

m

y = C1 log cosh

k


, forta t tf ,(39)

t! .

C2 rk g(40)

m

y0(ta) = 0, y(ta) = ya.(41)

2 Constant Mass and Thrust assumption5

The solution is

t)!

y=ya mlog coshrk g(ta(42)

km

k g

m g

y0=rtanhr(ta t)!(43)

km

(ta t)! ,

y00= g sech2rk g(44)

m

or more simply

=1log cosh ( (ta t))(45)

yya g

2

y0=1tanh ( (ta t))(46)

g

y00= g sech2 ( (ta t)) .(47)

The flight time, tf , is given by

0=ya g1log cosh ( (ta tf ))(48)

2

1y 2

tf=ta +cosh1expa,(49)

g

where we choose the principal range of cosh1(x) such that cosh1(x) 0, and the descent time is given by

td=tf taexp.

1y 2

=cosh1a

g

Substituting,

1s

td =cosh11 +Tsinh2 ( tb)

m g

Terminal velocity is given byg

y0=.

t

(50)

(51)

(52)

(53)

3 Variable Mass and Thrust6

3 Variable Mass and Thrust

d2y kdy2

m= m g + T(54)

dt2dt

Consider m to have the formtb t

m = mr + mp(55)

tb

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 Stability7

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

cm =2 l R(x) x %(r, x, ) r dr dx d

000

2lR(x)%(r, x, ) r dr dx d

R 0R0R0

Rl RR(x)Rx %(r, x) r dr dx

=R0 lR0 R(x) %(r, x) r dr dx

R0R0

(56)

(57)

where R(x) is the radius of the rocket, and % is the rocket density.

Example 4.1 (Shell).

%(r, x) = %0 (r R(x))(58)

4 Stability8

l R(x)% (rR(x)) r dr dx

cm =R0lR00R(x)x%00(rR(x)) r dr dx(59)

Rl0xRR(x) dx

=R0l R(x) dx(60)

R0

Alternatively, it may be more convenient to calculate the centre of mass of the rocket components separately to find cm,cm =cmn mn + cmb mb + cmfmf + cmm mm + cmi mi(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 byn(x) = v[S(x) W (x)](62)

x

A rigid body has downwash given by

W (x) = v Thusn(x) = v2 S(x) xThe normal force coecient, CN is defined by

CN (x) =n(x)

1 v2A

2

=2 S(x)

Ax

=8 S(x)

d2x

By the definition of the normal force curve slope

(63)

(64)

(65)

(66)

(67)

CN CN (x) = =0

8 S(x) = d2 xIn order to obtain the total CN we integrate over x,Z l

CN =CN (x) dx

0

=8(S(l) S(0))

d2

By definition the pitching moment of the local normal aerodynamic force about the nose tip (x = 0) is

(68)

(69)

(70)

(71)

M(x) =xn(x)(72)

=x v2 S(x)(73)

x

4 Stability

By definition the aerodynamic pitching moment coecient is

Cm(x) =M(x)

1 v2A d

2

=2 x S(x)

A dx

=8 x S(x)

d3x

By the definition of the moment coecient curve slope

Cm Cm (x) = =0

8 x S(x) = d3 xIn order to obtain the total Cm we integrate over x,

Cm =Z0l Cm (x) dx

8lS(x)

=Z0xdx

d3x

8l

=([x S(x)]0l Z0S(x) dx)

d3

8l

=(l S(l) Z0S(x) dx)

d3

=8{l S(l) V }

d3

The centre of pressure, cp, is defined as,

cp=dCm

C

N

=l S(l) V

S(l) S(0)

9

(74)

(75)

(76)

(77)

(78)

(79)

(80)

(81)

(82)

(83)

(84)

(85)

5 Nose Shape10

Nose Shape

Following the work of ?? we consider the optimal form for the nose shape. Newton reasoned that the air resis-tance 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,

mv=ma v (1 + cos(2 ))(86)

= A v t v (1 + cos(2 ))(87)

=2 A v2 cos2() t(88)

Hence,

Fd = mdv= 2 A v2 cos2()(89)

dt

Now consider a curved surface such that the area dA is varying over the surface,

Fd=ZZ2 v2 cos2() dA(90)

=2 v2ZZ cos2() dA(91)

=2 v2ZZ cos3() dS(92)

From Equation 1 and Equation 2 we note that,

Cd=4ZZ cos2() dA(93)

A

=4ZZ cos3() dS(94)

A

Noting that1

cos() =(95)

r1 + xz2yz2

then+

Cd =4ZZ1dx dy(96)

A1 +z2+z2

xy

We define the nose aspect ratio parameter = Rh , where h is the nose height from tip to base and R is the nose radius at the base.

5 Nose Shape11

5.1 Radial Nose Cones

Now consider a radially symmetric nose cone of profile z(r), then

42R

Cd =Z0Z0rdr d(97)

A1 +z2

8Rrr

=Z0dr(98)

A1 +z2

8Rr

=Z0rdr(99)

R21 +z2

r

Example 5.1 (Conical Nose). Consider a conical nose z(r) = Rh r, then

Cd =4(100)

1 + 2

In limh0 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) = hr2, then

R2

+ 42

Cd=log 12(101)

2222

Example 5.3(Ogive Nose). Consider an ogive nose (z + h)2+ r +h R2=h+ R2, then

2 R2 R

Cd=81 + 2 2(102)

3 (1 + 2)2

+R2r2

Example 5.4(Hemispherical Nose). Consider an hemispherical nose z R+= 1, then

R

Cd=2(103)

+h2r2

Example 5.5(Elliptical Nose). Consider an elliptical nose z h+= 1 where h > R, then

R

Cd= 41 + (2 log 1) 2

(1 2)2

Example 5.6 (Conical Frustrum Nose). Consider an conical frustrum nose z(r) =0

h(r R)

((1)R

then

Cd= 4( 1)2 + 22

( 1)2 + 2

(104)

0 r R R < r R ,

(105)

The optimal [0, 1] that minimises Cd is given by,

1

opt=1 +2 p2+ 4(106)

2

Cdopt=22 + 2 (107)

p2 + 4

To find the shape with the minimum drag, we can use calculus of variations. The minimiser with respect to

5 Nose Shape

y(x) satisfies the Euler Lagrange equation,

dz+dz3d2zd2z dz 2

drdr+ r dr23 r dr2dr

21 + dzdr 2 3= 0

d2 r dzdr= 0

2

dr1 + dzdr2

dz

2 r= C1

dr

Letting u = dz,1 + dzdr 2 2

dr

2 r u= C1

(1 + u2)2

11 + u22

r=

2 C1u

11

=u3 + 2u +

2 C1u

Using the chain rule dz= u dr

dudu

dz1d1

=uu3 + 2u +

du2 C1duu

z =1Z u 3u2 + 2 1du

2 C1u2

11

z =Z 3u3 + 2u du

2 C1u

z=143 u4 + u2 log |u| + C2

2 C1

Hence the parametric equations with the gradient as the parameter are,

1u31

r =+ 2u +

2 C1u

z=143 u4 + u2 log |u| + C2

2 C1

Applying boundary constraints u(RT ) = 1, z(RT ) = 0

C1=2

RT

C2=RT 7

44u3 + 2u +1

R

r = T

4u

RT34+ u27log u| 1

z = 434 u2427|

= RTu 1 u +3 4 log |u|

16

12

(108)

(109)

(110)

(111)

(112)

(113)

(114)

(115)

(116)

(117)

(118)

(119)

(120)

(121)

(122)

(123)

(124)

5 Nose Shape

The tip radius, RT , can be found by first solving for the gradient at the base, uR,

R=1 + uR22

h32 1)27 uR log |uR|

4 uR (uRuR +3

3 uR5+ 4 uR4 +4 uR3+ 8 uR2 7 uR + 4 4 uR log |uR| = 0

and thenRT = 4 RuR

(1 + u2 )2

R

13

(125)

(126)

(127)

(128)

6 Useful Links14

Useful Links

Variable Force and Mass: Rocket Lifto

Petes Rockets

Estes Rockets

NASA Tutorial

Stability Analysis

Aerodynamic Links

Altimeter

A Estes Astrocam15

Estes Astrocam

Length:18.38 (46.7 cm)

Diameter:1.34 (34.0 mm)

Weight:2.7 oz (76 g)

Recovery:12 (30 cm) parachute

Fins:Plastic molded

Maximum Altitude:500 ft (152 m)

Recommended Engines:C6-7

Astrocam AltitudeAstrocam Velocity

250

100

B4 MotorB4 Motor

B6 MotorB6 Motor

C6 MotorC6 Motor

200

50

150

m1

ms

100

0

50

024681012141650246810121416

00

ss

(a) Altitude(b) Velocity

Astrocam Acceleration

80

B4 Motor

B6 Motor

C6 Motor

60

40

20

2

ms

0

20

40

600246810121416

s

(c) Acceleration

Figure 3: Astrocam Kinematics

B Estes Bandit16

Estes Bandit

Length:16.6 (42.2 cm)


Weight:1.6oz (45 g)


Fins:T3 plastic molded


Recommended Engines:A8-3(First Flight), B4-4, B6-4, B6-6, C6-5, C6-7

m

Bandit AltitudeBandit Velocity

400

140

A8 Motor

A8 Motor

B4 Motor

B4 Motor

B6 Motor

350120B6 Motor

C6 MotorC6 Motor

100

300

80

25060

200140

ms

15020

0

100

20

5040

02468101214161820602468101214161820

00

ss


Bandit Acceleration

200

A8 Motor

B4 Motor

B6 Motor

150C6 Motor

100

50

2

ms

0

50

100

15002468101214161820

s

(c) Acceleration

Figure 4: Bandit Kinematics

C Estes Banshee17

Estes Banshee

Length:16.6 (42.2 cm)


Weight:1.6oz (45 g)


Fins:T3 plastic molded


Recommended Engines:A8-3(First Flight), B4-4, B6-4, B6-6, C6-5, C6-7

m

Banshee AltitudeBanshee Velocity

400

140

A8 Motor

A8 Motor

B4 Motor

B4 Motor

B6 Motor

350120B6 Motor

C6 MotorC6 Motor

100

300

80

25060

200140

ms

15020

0

100

20

5040

02468101214161820602468101214161820

00

ss


Banshee Acceleration

200

A8 Motor

B4 Motor

B6 Motor

150C6 Motor

100

50

2

ms

0

50

100

15002468101214161820

s

(c) Acceleration

Figure 5: Banshee Kinematics

D Estes Echostar18

Estes Echostar

Length:29 (73.7 cm)


Weight:3.6 oz (102.1 g)


Fins:Die cut balsa


Recommended Engines: 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 AltitudeEchostar Velocity

450

120

B4 MotorB4 Motor

B6 MotorB6 Motor

400C6 Motor100C6 Motor

C6+C6 Motor

C6+C6 Motor

35080

30060

25040

m1

ms

20020

1500

10020

5040

051015202560510152025

00

ss


ms2

Echostar Acceleration

60

B4 Motor

B6 Motor

C6 Motor C6+C6 Motor

40

20

0

20

40

60 0 5 10 15 20 25

s

(c) Acceleration

Figure 6: Echostar Kinematics

E Estes Engines19

Estes Engines

Length:1.73 (4.4 cm)


No.TypeStageI (N s)td (s)ml (g)Tmax (N)tb (s)me (g)mp (g)

15031 A3-2TSingle1.25256.67.80.365.61.75

15072Single2.50456.67.80.867.63.50

A3-4T

1511A10-3TSingle2.503141.513.30.267.93.78

150421 A3-4TUpper1.25428.37.80.366.01.75

1510A10-0TBooster2.500141.513.30.266.73.70

Table 2: Estes Blackpowder Mini Motors




15931 A6-2Single1.25270.812.80.2015.01.56

15982Single2.503113.213.30.3216.23.12

A8-3

1601B4-2Single5.002113.213.31.2019.88.33

1602B4-4Single5.00499.113.31.2021.08.33

1605B6-2Single5.002127.413.30.8319.36.24

1606B6-4Single5.004113.213.30.8320.16.24

1620B8-5Single5.005141.522.20.6019.36.24

1617C5-3Single10.003226.422.22.1025.512.70

1613C6-3Single10.003113.313.31.7024.912.48

1614C6-5Single10.005113.213.31.7025.812.48

1599A8-5Upper2.50556.613.30.3217.63.12

1604B4-6Upper5.00642.513.31.2022.18.33

1607B6-6Upper5.00656.613.30.8322.16.24

1615C6-7Upper10.00770.813.31.7026.912.48

1608B6-0Booster5.000113.213.30.8016.46.24

1616C6-0Booster10.000113.313.31.6822.712.48

Table 3: Estes Blackpowder Motors




1666D12-3Single20.003396.228.51.7042.224.93

1667D12-5Single20.005283.028.51.7043.124.93

1668D12-7Upper20.007226.428.51.7044.024.93

1665D12-0Booster20.000396.228.51.7040.924.93

1669D11-PPlugged20.000453.127.61.8244.024.93

Table 4: Estes Blackpowder size D Motors

F MATLAB Code20

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 4) % check correct number of arguments help rocketelseif (nargin

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