Mathematical model and technical solution for small

Mathematical model and technical solution for small multistage suborbital launchers Teodor-Viorel CHELARU, PhD

Cristian BARBU, PhD

University POLITEHNICA of Bucharest, ROMANIA e-mail: [email protected]

Military Technical Academy, Bucharest, ROMANIA e-mail: [email protected]

Adrian CHELARU SC Comfrac R&D Project Expert SRL, Bucharest, ROMANIA e-mail: [email protected] Abstract—The paper purpose is to dignify some aspects regarding the calculus model and technical solutions for small multistage suborbital launchers used to test spatial equipment and scientific measurements. The calculus methodology consists in numerical simulation of launcher evolution for different start conditions. The rocket model presented will be with six DOF and variable mass. The results analyzed will be the flight parameters and ballistic performances. The discussions area will focus around the technical possibility to realize small suborbital multistage launchers recycling military rocket engines. Key words: suborbital; model

launcher; multistage; simulation;

NOMENCLATURE α - Attack angle (tangent definition);

β - Sideslip angle (tangent definition);

β p - Azimuth angle; λ p - Geocentric latitude;

ψ - Azimuth angle; θ - Inclination angle;

ClT ; CmT ; CnT - Thrust momentum coefficients in the mobile frame. q1 , q2 , q3 , q4 Quaternion components; F0 = ρ

V2 S - Reference aerodynamic force; 2

H oA = F0 l - Reference aerodynamic couple;

T0 - Reference thrust force H oT = T0 l Reference couple thrust l - Reference length; m – Mass; mi - Initial mass

m f - Final mass p, q, r - Angular velocity components along the axes of mobile frame;

φ - Bank angle;

S - Reference area;

ρ - Air density;

T - Thrust;

Ω - Body angular velocity;

t - Time;

Ω p Earth spin:

V - Velocity;

A, B, C , E - Inertia moments;

C xA ; C yA ; C zA - Aerodynamic coefficients of force in the mobile frame;

ClA ; CmA ; CnA , - Aerodynamic coefficients of momentum in the mobile frame; C xT ; C Ty ; C zT - Thrust coefficients in the mobile frame;

u, v, w - Aircraft velocity components in a mobile frame; Vxp V yp Vzp Velocity components in Earth frame; OX pYp Z p - Normal Earth-fixed frame [2]; Oxyz – body frame (mobile frame);

x p y p z p - coordinates in Earth-fixed frame; r - The distance between rocket and Earth center:

R p - Earth radius:

I.

INTRODUCTION

It is indisputable that today, the spatial program involves many collateral activities, like preliminary tests for equipment and the qualifications. It is well known that those auxiliary activities suppose huge technical and financial effort and this increases the total cost for any space program. Starting from this idea, the paper proposes an economical solution for the tests by using small multistage suborbital launchers by recycling military rocket engines. This solution can be also used for quickly meteorological determination, taking into account that meteorological balloons are very slow. To solve these problems and in general for evaluating the launching capability is necessary to elaborate a complex mathematical model that ensures the rigorous and accuracy evaluation of the flight data and ballistic parameters. The mathematical model presented below, developed with maximum of accuracy, seeks to answer these needs. This model, allows us to evaluate two technical solutions, one of them based on three stage launcher using the engines of short rocket 122 mm (Fig. 3) and the second using four boosters, also from short 122 mm rocket around a central body obtained from long rocket 122 mm (Fig. 4). These two technical solutions will be evaluated and the flight parameters and ballistic performances will be analyzed.

THE GRAVITY ACCELERATION, COMPLEMENTARY II. ACCELERATION, CONNECTION BETWEEN EARTH FRAME AND BODY FRAME At the beginning we will start by analyzing the influence of the secondary parameters like the variation of the gravity acceleration with latitude and altitude and the influence of Earth spine about the launching rocket trajectory.

g zp = − g r

zp r

Ωp

(1)

where: x p ; y p ; z p are the rocket coordinates in the Earth frame and the Earth radius can be approximated by: R p ≅ a (1 − α sin 2 λ p ) , the gravity acceleration components in the Earth frame are:

g xp = − g r

xp r

− gω

Ω xp Ωp

; g yp = − g r

y p + Rp r

− gω

Ω yp Ωp

;

(2)

and Ω xp ; Ω yp ; Ω zp - the spin components are given by: Ω xp = Ω p cos λ p cos β p ; Ω yp = Ω p sin λ p ; Ω zp = −Ω p cos λ p sin β p ,

(4)

where the two angels used are: β p - azimuth angle and λ p geocentric latitude. On another hand, complementary acceleration is: a c = −2Ω p × V , (5) with the Earth frame components given by: acxp = 2(V yp Ω zp − V zp Ω yp ) ; acyp = 2(V zp Ω xp − Vxp Ω zp ) ;

aczp = 2(Vxp Ω yp − V yp Ω xp ) ,

(6)

where Vxp ;V yp ;V zp are the Earth frame velocity components.

B. The connection between Earth frame and body frame The Earth frame is a geocentric frame with the origin in the mass center of the Earth, which can be considerate an ellipsoid of revolution [2],[3], [7]. In order to obtain body frame we are passing through three intermediary frames. The first frame is the starting frame (Ox s y s z s ) , which has y s axis normal to the tangent plane at the Earth’s surface (Fig.1). Booths frames participant in Earth rotation are not inertial frames. But, if we introduce as corrections the Earth spin influence by gravity and complementary acceleration, we can consider them as inertial frame. Ωp

N xp

ys

yp Q

γp

xs

βp

Op λg

r = x 2p + ( R p + y p ) 2 + z 2p

,

where the radials and polar components of the gravity acceleration [7], [3] are: g r = g Ar − Ω 2p r ; g ω = g Aω + Ω 2p r sin λ p . (3)

A. The gravity acceleration and complementary acceleration In order to write the movement equations we will use a geodesic frame [7] connected to the Earth. Due to diurnal spin, beside attraction force, we must consider two supplementary accelerations: carrying acceleration: −Ω p × (Ω p × r ) and complementary acceleration (Coriolis acceleration) given by: −2Ω p × V , where Earth spin has the value:

Ω p = 7,2921⋅10 −5 s −1 . If we designate r - the distance between rocket and Earth center :

Ω zp

− gω

R λ pp

P

zs

zp

G

Q

Figure 1. Connection between Earth frame and starting frame

If taken γ p the angle between geodesic normal and geocentric normal: γ p = λg −λp , the connection between frames is:

(7)

[xs

[

z s ] = A γp x p T

ys

yp

zp

]

T

The connection between these frames is given by:

(8)

[X 0

Overlapping the Earth frame above starting frame can be done by the rotation matrix:

Y0

Z 0 ] = A Ωp [x s T

ys

zs ] , T

(12)

where the rotation matrix elements are:

A γp

⎡ sb 2 + cb 2 ⋅ cg − sb ⋅ cb(1 − cg ) − cb ⋅ sg ⎤ ⎢ ⎥ = ⎢− sb ⋅ cb(1 − cg ) cb 2 + sb 2 ⋅ cg − sb ⋅ sg ⎥ ⎢ cb ⋅ sg sb ⋅ sg cg ⎥⎦ ⎣

a11 = cos 2 β p cos 2 λ g (1 − cos Ω p t ) + cos Ω p t

; a12 = sin β p cos β p cos λ g (1 − cos Ω p t ) − sin λ g sin Ω p t

(9)

2

; a13 = cos β p sin λ g cos λ g (1 − cos Ω p t ) + sin β p cos λ g sin Ω p t

where:

a 21 = sin β p cos β p cos 2 λ g (1 − cos Ω p t ) + sin λ g sin Ω p t

sb = sin β p ; cb = cos β p sg = sin γ p ; cg = cos γ p In order to obtain the angle γ p between geocentric and geodesic normal we must start from the relation [7]: 2

2

a tg λ p = b tg λ g ,

(10)

;

;

a 22 = sin 2 β p cos 2 λ g (1 − cos Ω p t ) + cos Ω p t

; a 23 = sin β p sin λ g cos λ g (1 − cos Ω p t ) − cos β p cos λ g sin Ω p t a31 = cos β p sin λ g cos λ g (1 − cos Ω p t ) − sin β p cos λ g sin Ω p t a32 = sin β p sin λ g cos λ g (1 − cos Ω p t ) + cos β p cos λ g sin Ω p t

; ; ;

a33 = sin 2 λ g (1 − cos Ω p t ) + cos Ω p t

where a, b are the Earth semi-axis. From relation: α = (a − b) a = 1 298,25 which defines Earth flattering we can obtain the relation:

. (13) Obviously, if the flight time is short, the matrix A Ωp becomes unitary matrix. Finally, with Euler angle (ψ, θ, φ) or Euler modified angle ∗

tg γ p =

α(1 − α / 2) sin 2λ p 1 − 2α(1 − α / 2) cos 2 λ p

.

(11)

Obviously, if the start frame origin is on the Equator or on γ A North or South Pole the angle p is null and the matrix γp became unitary matrix. The next intermediary frame is the initial starting frame (O0 X 0Y0 Z 0 ) , which overlap the starting frame in the launching moment. The starting frame which is attached to the Earth, is rotating around the polar axe related to the initial starting frame, considerate fix, with an angle equal with rotation angle of the Earth in this time.

(θ , ψ ∗ , φ ∗ ) or Hamilton’s quaternion (q1 , q2 , q3 , q4 ) , we can overlap the initial start frame with the body frame. The rotation matrix A i , which makes this transformation, will be shown afterwards. In the issue, in order to pass some elements from Earth frame to the body frame we need three rotations, which can be concentrated in a single matrix: A p = A i A Ωp A γp ,

which is the rotation matrix between Earth frame and body frame. III.

Ωp

X βp 0

Y0

xs ys

zs Z0 P

λg Ω pt

G

Figure 2. Connection between starting frame and initial starting frame

(14)

GENERAL MOVEMENT EQUATIONS

A. Kinematical equations Unlike paper [4], which covers the regular ballistic Rockets, where the kinematical equations used Euler angles, in our case, including launching rocket for the satellite, with the almost vertical initial launching direction, the papers [3], [5], [6], [7] recommend modified Euler angles, which have first rotation in vertical plane. Using kinematical equations written with Euler angles, in addition to benefits related to the significance of physical measurable sizes, the following drawback is involved: the use of trigonometric functions and difficulties that these place in the building program algorithms. Therefore, in some works [3], it is proposed to build rotation matrices and a connection matrix related to rotation speed by using an algebraic operator called Hamilton's quaternion.

Quaternion of Hamilton is an operator that expresses the rotation. If the rotation is done in connection with the angle σ around an axis with the unit vector:

eσ = Il + Jm + Kn ,

(15)

σ σ σ σ ; q2 = m sin ; q3 = n sin ; q4 = cos 2 2 2 2

(16)

φ θ ψ φ θ ψ q1 = − cos sin sin + sin cos cos ; 2 2 2 2 2 2 φ θ ψ φ θ ψ q2 = cos sin cos + sin cos sin ; 2 2 2 2 2 2 , φ θ ψ φ θ ψ q3 = cos cos sin − sin sin cos ; 2 2 2 2 2 2 φ θ ψ φ θ ψ q4 = cos cos cos + sin sin sin , 2 2 2 2 2 2

the sizes:

q1 = l sin

are called quaternion components It is well known that a sequence of rotations of a rigid body with a fixed point can be replaced by a single rotation around an axis through the fixed point.

and reciprocal : ⎛ 2( q q + q q ) ⎛ 2( q q + q q ) ⎞ φ = arctan⎜⎜ 2 2 23 24 1 2 ⎟⎟ ; ψ = arctan⎜⎜ 2 1 22 23 4 2 ⎝ q4 + q1 − q2 − q3 ⎝ q4 + q3 − q1 − q2 ⎠

Using quaternion components the rotation we can show [3] that the rotation matrix has the following form: ⎡q42 + q12 − q22 − q32 ⎢ A i = ⎢ 2(q1q2 − q3q4 ) ⎢ 2( q3q1 + q2 q4 ) ⎣

− 2(q1q2 + q3q4 ) − q42 − q22 + q32 + q12 − 2( q2 q3 − q4 q1 )

− 2(q3q1 − q2 q4 ) ⎤ ⎥ − 2(q2 q3 + q4 q1 ) ⎥ − q42 − q32 + q12 + q22 ⎥⎦

(17) Using the inverse of this matrix B i = A i−1 = A T we can obtain the components of acceleration in the Earth frame from components of acceleration in the body frame, used in the dynamical equations. Like the kinematical equations we can write the relation:

[x [

p

y p

where Vxp V yp Vzp the Earth frame.

] = [V

]

are components of the velocity in

T

]

T

xp

V yp Vzp ,

− q3 q4 q1 − q2

q2 ⎤ ⎡ p⎤ − q1 ⎥⎥ ⎢ ⎥ q q4 ⎥ ⎢ ⎥ ⎢ ⎥ r⎥ − q3 ⎦ ⎣ ⎦

(21)

B. Dynamical equations Developing cross products from paper [3], with the supplementary notations from [3], [4] we can obtain matrix representation of the dynamical equations: 1) Force equations in the Earth frame ⎧ ⎡C xA ⎤ ⎡Vxp ⎤ ⎡C xT ⎤ ⎫ ⎡ g xp ⎤ ⎡acxp ⎤ ⎪ ⎢ A⎥ ⎢ ⎥ 1 ⎢ T ⎥⎪ ⎢ ⎥ ⎢ ⎥ ⎢V yp ⎥ = m B p ⎨ F0 ⎢C y ⎥ + T0 ⎢C y ⎥ ⎬ + ⎢ g yp ⎥ + ⎢acyp ⎥ ; (22) ⎪ ⎢C A ⎥ ⎢Vzp ⎥ ⎢C zT ⎥ ⎪ ⎢ g zp ⎥ ⎢ aczp ⎥ ⎦ ⎣ ⎦ ⎣ ⎦⎭ ⎣ ⎦ ⎣ ⎩ ⎣ z⎦

(18)

In order to obtain the connection between the derivatives of quaternion components and components of rotation velocity in the body frame we use [3] the following relation:

⎡ q4 ⎡ q1 ⎤ ⎢ ⎢q ⎥ ⎢ 2 ⎥ = 1 ⎢ q3 ⎢ q3 ⎥ 2 ⎢− q2 ⎢ ⎢ ⎥ ⎣q4 ⎦ ⎣ − q1

θ = arcsin(2(q2 q4 − q3 q1 ) ) .

⎞ ⎟; ⎟ ⎠

2) Moment equations in the body frame

z p

T

(20)

⎧ ⎡ClT ⎤ ⎫ ⎡ClA ⎤ ⎡ ( B − C )qr ⎤ ⎡ p ⎤ ⎢ q ⎥ = J −1 ⎪ H A ⎢C A ⎥ + H T ⎢C T ⎥ ⎪ + J −1 ⎢ (C − A)rp ⎥ , ⎨ ⎬ 0 ⎢ m⎥ 0 ⎢ m⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎢CnT ⎥ ⎪ ⎢CnA ⎥ ⎢⎣( A − B) pq ⎥⎦ ⎢⎣ r ⎥⎦ ⎣ ⎦⎭ ⎣ ⎦ ⎩

where we denoted:

0 ⎤ ⎡1 / A 0 ⎢ 0 ⎥⎥ . J = ⎢ 0 1/ B ⎢⎣ 0 0 1 / C ⎥⎦ −1

(19)

which, together with the relation (18) represent kinematical equation. Because the rotation matrix (17) is the same regardless of the variables used, in work [3] are obtained the following relationships between different variables (Euler angles, quaternion):

(23)

(24)

The inertial moment inverse matrix, where the inertial moments are given by:

∫

∫

∫

A = ( y 2 + z 2 ) d m; B = ( z 2 + x 2 ) d m; C = ( x 2 + y 2 ) d m .

(25)

and the matrix B p = A Tp is the transpose matrix given by (17) For the aerodynamic coefficients calculus we used the method indicated in [4] and [8], based on polynomial series expanding:

Figure 4 shows the second model, called „VLS B4”, and main characteristics are included in Table II. TABLE II.

VLS B4 CENTRAL BODY FROM LONG ROCKET 122 MM AND FOUR BOOSTERS, FROM SHORT ROCKET AROUND

VLS B4

C xA = a1 + a2 (α 2 + β 2 ) + a3 (α 4 + β 4 ) + a4 α 2β 2 + a13 ( zˆ p − zˆ pc );

mi [kg] m f [kg] Length [m]

C yA = −b1α − b2 α 3 − b3β 2 α − b4 rˆ; C zA = b1β + b2β3 + b3 α 2β + b4 qˆ; ClA = c2 (α 2 + β 2 )αβ + c3 pˆ + c5 (αqˆ − β rˆ); (26)

I Stage (with boosters)

164

120

2,4

(27)

II Stage

57

31

2,4

Final stage

16

16

0,6

CmA = d1β + d 2β 3 + d 3βα 2 + d 4 qˆ; CnA = d1α + d 2 α 3 + d 3 αβ 2 + d 4 rˆ, where, by definition [8] α = − arctan(v / u ) , β = arctan( w / u ) .

The following notations are used for the non dimensional angular velocities: pˆ = p l V ; qˆ = q l V ; rˆ = r l V .

(28)

T y

C ; C ; C

T z

T l

T m

T n

C ; C ; C .

(29)

INPUT DATA, CALCULUS ALGORITHM AND RESULTS

IV.

A. Input data for the model Figure 3 shows the first model, called „VLS T3”, and main characteristics are included in Table 1. TABLE I.

Figure 3. VLS T3

VLS B4

Total impulse of the long engine: I Σ = 6.2 kNs

Also, we have the thrust coefficients: T x

Figure 4.

VLS T3 THREE STAGE LAUNCHER USING THE ENGINES OF SHORT ROCKET 122 MM

VLS T3

mi [kg]

m f [kg]

Length [m]

I Stage

96

85

4.2

II Stage

69

58

3.0

III Stage

42

31

1.8

Final stage

16

16

0.6

Total impulse of the short engine: I Σ = 2.5 kNs

B. Calculus algorithm The calculus algorithm consists in multi-step method Adams' predictor-corrector with variable step integration method: [1] [10]. Absolute numerical error was 1.e-12, and relative error was 1.e-10. C. Test calculus For the test calculus the following initial conditions were used: Geographic orientation: Azimuth angle β p = 90° (towards the East); Geocentric latitude λ p = 45° (Romania latitude); Altitude: y0 = 1[m] ; Initial velocity V0 = 40[m / s ] ; Initial inclination angle θ 0 = 85° D. Rezults In the figures 5-7 are comparatively shown the flight parameters and the ballistic performances of the two models. In Fig 5 is presented the velocity diagram. It can be observed the difference between the models, the velocity of the second being greater.

V. 1.8 VLS T3 VLS B4

1.6

V[km/s]

1.4 1.2 1 0.8 0.6 0.4 0.2 0

50

100

150

t[s]

Figure 5. Velocity diagram

Consequently, due to the velocity difference the trajectory of the second is higher (Fig. 6). 50

The paper presents synthesis aspects of the simulate model developed for the calculation of an operational rocket - which will launch carrier equipment to be tested with the aim of integrating them into complex spatial systems. The application is made for two variants of the small launchers, whence will be tested in the national projects. We considered more possible solutions and we presented and analyzed the flight parameters for two launchers that can be used. Since the major objective of the launch vehicle carrier consists of testing solution for assembling and detachment of more stages rockets and to launch at large angles, to eliminate the additional risk, it will be used the rocket motors from current production connected in parallel or in tandem, solution what was presented during the work. In conclusion, the main sub-assemblies of the operating launcher, will be provided from current production, the engine being developed in Romania and being in significant amounts in deposits. Assembly and testing of the ground will be made at the plant where usually such systems are manufactured, and finally the launchers will be tested by firing in an area from the Black Sea maybe next year.

45

REFERENCES

VLS T3 VLS B4

40

[1]

BAKHVALOV, N. Methodes Numeriques – Analyse, algebre, equations differentielles ordinaires Ed. Mir Moscou , 1976. [2] BOIFFIER, J.L. The Dynamics of Flight – The Equations, John Wiley & Sons , Chichester, New York, Weinheim, Brisbane, Singapore, Toronto , ISBN 0-471-94237-5, 1998. [3] CHELARU T.V. Dinamica Zborului–Racheta dirijată, Ediţia a-II-a revizuită şi adaugită ,Ed.Printech, Bucureşti, ISBN 973-718-013-5 , mai 2004. [4] CHELARU T.V. Dinamica Zborului–Racheta nedirijată, ,Ed. Printech, Bucureşti, ISBN 973-718-428-9, 155 pag. ,martie 2006. [5] CHELARU, T.V., COMAN, A, Some theoretical aspects regarding the mathematical model for launching a small satellite on low orbit International Workshop on SMALL SATELLITES, NEW MISSIONS AND NEW TECHNOLOGIES 05-07 June 2008, Istanbul, TURKEY [6] KУЗОВКОВ, н.T., Системы стабилизащии летателных аппаратов, балистических и зенитных ракет, Издателъство Высшая школа , Moskva, 1976. [7] ЛЕБЕДЕВ, A.A., ГЕРАСЮТА, Н.Ф., Балистика ракет, Издателъство Машиностроение, Moskva, 1970. [8] Nielsen, J.N., Missile Aerodynamics, McGraw-Hill Book Company, Inc., New-York, Toronto, London, 1960. [9] ISO 1151 -1:1988; -2:1985-3:1989 [10] SLATEC Common Mathematical Library, Version 4.1, July 1993

35

y [km]

30 25 20 15 10 5 0

0

5

10

15

20

25

30

35

40

45

50

x [km]

Figure 6. Ballistic trajectory 80 60 VLS T3 VLS B4

40

Te [G]

20 0 -20 -40 -60 -80 -100 0

50

100

150

t[s]

Figure 7. Inclination angle diagram

CONCLUSIONS

200