by Thomas B. Bahder and John D. Bruno

Transient Response of anElectromagnetic Rail Gun:

A Pedagogical Model

ARL-TR-1663 May 1998

Approved for public release; distribution unlimited.

The findings in this report are not to be construed as an official Department ofthe Army position unless so designated by other authorized documents.

Citation of manufacturer’s or trade names does not constitute an officialendorsement or approval of the use thereof.

Destroy this report when it is no longer needed. Do not return it to the originator.

Army Research LaboratoryAdelphi, MD 20783-1197

ARL-TR-1663 May 1998

Transient Response of anElectromagnetic Rail Gun:A Pedagogical Model

Approved for public release; distribution unlimited.

Thomas B. Bahder and John D. BrunoSensors and Electron Devices Directorate

Abstract

We present a highly simplified model of an electromagnetic rail gun. Theenergy source for the gun consists of a rotating magnetic dipole momentpositioned at the center of a stator coil. The dipole moment is free to ro-tate about an axis normal to the stator coil’s axis, inducing a current in therail gun stator circuit, and thereby propelling the projectile. The system’sdynamical behavior is described by three nonlinear coupled ordinary differ-ential equations, involving time-dependent functions representing the arma-ture’s position, the stator circuit’s current, and the dipole moment’s angularposition. We numerically solve this system of differential equations and plotthe solutions versus time. The results exhibit a complicated dynamics due tothe interaction of the electrical and mechnical degrees of freedom.

ii

Contents

1 Introduction 1

2 The Model 2

3 Derivation of the Dynamical Equations 5

4 Numerical Solution of the Differential Equations 10

5 Conclusions 17

Acknowledgments 18

References 19

Distribution 20

Report Documentation Page 24

iii

Figures

1 Schematic of electromagnetic rail gun model . . . . . . . . . . 3

2 Plot of position x(t) versus t . . . . . . . . . . . . . . . . . . . 12

3 Plot of velocity x′(t) versus t . . . . . . . . . . . . . . . . . . . 13

4 Plot of acceleration x′′(t) versus t . . . . . . . . . . . . . . . . 14

5 Plot of voltage V and current I(t) in stator circuit versus t,for initial condition θ(0) = 0 . . . . . . . . . . . . . . . . . . . 14

6 Plot of voltage V and current I(t) in stator circuit versus t,for initial condition θ(0) = π/2 . . . . . . . . . . . . . . . . . . 15

7 Plot of rotor angular velocity θ′(t) versus t, for initial condi-tions θ(0) = 0 and θ(0) = π/2 . . . . . . . . . . . . . . . . . . 15

8 Plot of rotor angular acceleration θ′′(t) versus t, for initialconditions θ(0) = 0 and θ(0) = π/2 . . . . . . . . . . . . . . . 16

iv

Table

1 Values of dimensionless parameters α, β, γ, and κ . . . . . . . 11

v

1. Introduction

The electromagnetic rail gun (EMRG) has been proposed as a viable technol-ogy for launching projectiles at speeds over 2500 m/s. A small-scale EMRGwas reported a number of years ago [1]; however, the physics of the propul-sion mechanism was sufficiently complex that it was still being discussed inthe literature 10 years later [2–4].

For nearly a decade, the Army has pursued the development of a largerEMRG for applications in armored vehicles (for a review of this work, seeMcCorkle [5]). Recently, the technical feasibility of developing a full-scaleEMRG, capable of launching a 6-kg projectile, has been questioned at afundamental level [5] by the Technical Director of the Missile Command(MICOM), W. C. McCorkle.∗ We have attempted to assess the merits ofissues raised in McCorkle’s critique, and in so doing, have found it useful toconstruct a simplified model of the EMRG. The purpose of this report is todocument this work.

In the following, we formulate a pedagogical model of an EMRG. The purposeof the model is to provide a simple context within which the basic operatingprinciples of an EMRG can be studied, without many of the complicationsof the real system. Such a model is useful because it permits us to developan intuition for much of the physics involved in the real EMRG.

The essence of our EMRG model is a rotating magnetic dipole, which con-stitutes the system’s rotor. This magnetic dipole is inductively coupled tothe stator coil that feeds current into the rails of the EMRG. The model isdescribed by three coupled nonlinear ordinary differential equations in time.In section 1, we describe our simplified model of the EMRG. In section 2, wederive the dynamical equations that govern the time evolution of the EMRGsystem. In section 3, we numerically solve these equations, for two differentinitial conditions. Finally, we summarize our conclusions in section 4.

∗The delivered energy density is believed to decrease with increasing total machineweight, assuming that the rail gun length is constant. This constraint is a requirementfor applications in armored vehicles. The total machine weight is taken to be the powersupply, electronics, and all supporting mechanical equipment.

1

2. The Model

Figure 1 shows a schematic representation of a simplified EMRG. Our concep-tual model of the EMRG consists of a rotating hard magnet with a magneticmoment M located at the center of a fixed stator coil. The moment is free torotate about an axis normal to the stator coil’s symmetry axis. This rotat-ing magnetic moment plays the role of the field windings on the rotor thatis used to power the EMRG. The rotating magnetic moment M induces anelectromotive force (emf) in the surrounding stator coil (single turn), whichhas an area A = πa2, where a is the coil’s radius. We call the axis of thestator coil the z-axis, and assume that the magnetic moment M has a largemechanical moment of inertia I about an axis perpendicular to this z-axis.At times t ≤ 0, the rotor (magnetic moment) has a large angular velocitydθ/dt = ω = ω0, and we assume that the stator circuit is open so that there iszero current in the stator circuit, I = 0. At t = 0, we close the stator circuit,and the induced emf creates a current I for times t > 0. The current I in thestator circuit, as shown in figure 1, creates a magnetic field B perpendicularto the plane of the paper. The railgun armature, shown as the black barin figure 1, conducts the current I, and is free to move in the x-direction.This armature is in physical contact with the projectile, accelerating it fromits initial position at the breech of the gun’s barrel to its muzzle. By theLorentz force law, a current element of length dl in the armature experiencesa magnetic force

dF = Idl×B. (1)

The current I is, of course, coupled to the rotor with the magnetic momentM. The rotational kinetic energy of the rotor at t = 0 is the only energystored in the system for times t < 0. For times t > 0, after the stator circuitswitch is closed, some of this rotor kinetic energy goes into kinetic energy ofthe launch package (armature + projectile) of mass m, Joule heating of thestator circuit, and stored energy in the magnetic field of the stator circuit.∗

∗The rotor kinetic energy also goes into radiative energy loss, since the rotating mag-netic moment produces a time-varying magnetic field. However, the power associated withthis radiative loss is negligible for all rotational frequencies of interest.

2

Figure 1.Schematic ofelectromagneticrail gun model.

M

x

m

I L = L ′ x

R

θ

Ls

We now derive the differential equations that describe the dynamics of thissystem. We take the hard magnet to be a simple magnetic dipole moment M,located at the center of the stator coil. The BM field of this dipole momentis given by

BM =µ0

4π

[−M

r3+

3(M · r)r

r5

], (2)

where r is the vector from the point dipole M to the field point. The hardmagnetic moment is intended to represent the effects of real currents in thegenerator’s field windings. Since the field-winding circuit dynamics are notincluded in the present model, we assume that the hard magnet is turnedon at some time before t = 0, and retains a constant magnetic moment fort ≥ 0. Hence, for t ≥ 0, M has the functional form

M(t) = M0 (cos θ(t) z + sin θ(t) x), (3)

where M0 is a constant, z and x are unit vectors, and θ(t) is the angularposition of the dipole. At a given time t, the vector M makes an angle θ withthe z-axis, which coincides with the axis of the stator coil. The magnetic fieldflux density passing though the plane of the stator coil (z = 0) is given by

BM · z = −µ0

4π

M0 cos θ

ρ3, (4)

3

where ρ2 = x2 + y2, and the center of the stator coil is taken to coincide withz = 0, the origin of coordinates in the x-y plane. The flux passing through thearea occupied by the stator coil, given by ρ ≤ a, is the integral of equation(4) over this region. This integral is divergent, because there is a Dirac deltafunction contribution that is missing from the expression for the dipole fieldgiven in equation (2). However, the total magnetic flux passing through theplane of the stator coil is Φin+Φout = 0, where Φin is the flux passing throughthe area occupied by the stator coil, and Φout is the flux passing through theplane of the coil for ρ ≥ a. Consequently, we can find the flux passing throughthe stator coil by integrating the flux density in equation (4) over the regionρ ≥ a. We obtain

Φin =µ0M0 cos θ

2a= kM0A cos θ, (5)

wherekA =

µ0

2a. (6)

We assume that the stator coil has a self-inductance Ls and that there isa small resistance R in the stator circuit. The rails of the EMRG have aself-inductance

L = L′x, (7)

which depends on the position x of the launch package mass m. The param-eter L′ is the self-inductance per unit length of the rails. For two straightparallel rails of circular cross section, each of radius b, and separated by adistance h, we have the approximate form [6]

L′ =µ0

4π

(1 + 4 log

h

b

). (8)

Note that L′ is a geometrical constant that is independent of the launchpackage position x and time t. We ignore in this model the small increase instator circuit resistance as x increases.

4

3. Derivation of the Dynamical Equations

The EMRG consists of a magnetic moment M coupled inductively to thestator coil. The stator coil is coupled to the rails of the EMRG by the currentI. We assume that the flux of the magnetic field due to the current I in therails does not couple to the stator coil or to the magnetic moment M. Hence,the rails form an isolated magnetic system.

Within the quasi-static approximation, we consider the free energy of themagnetic system [6]

δF = −SδT +∫

H · δB dV, (9)

where S is the total entropy of the system, T is the temperature (assumeduniform), and the integration is over the volume where the fields are nonzero.The free energy F is the work that must be done on the system by externalelectromotive forces to set up the magnetic fields H and B. Assuming that allconductors and the surrounding medium in the EMRG have a linear relationbetween B and H,

B = µH, (10)

equation (9) can be integrated from an initial thermodynamic stateR0 = (T,B = 0) to a final state R = (T,B). We obtain∗

F = F0(T ) +1

2LMI

2M + LMSIMI +

1

2LSI

2 +1

2L′xI2, (11)

where F0(T ) is the free energy F in state R0, which depends only ontemperature.

∗When the linear assumption in equation (10) is made, with a temperature-independentpermeability, the free energy F can be written as a sum of the work done to set up themagnetic system, plus a temperature-dependent constant F0(T ). In this case, the currentsin the system do not depend on the thermodynamic state; i.e., they do not depend ontemperature. We can then neglect the constant F0(T ) and speak of the energy of thesystem. We do this in the ensuing discussion. The resulting quantity, F − F0(T ), is thenequal to the coenergy, W =

∫V

∫ RR0

B · δH dV . See, for example, Fitzgerald et al [7].

5

The second term represents the energy required to set up the hard magneticmoment M, in the absence of the other parts of the system. Here we treat thehard magnetic moment as a simple self-inductance LM carrying a current IM .Since we treat the magnetic moment as constant, the second term in equation(11) is a constant. The fourth term in equation (11) is the energy requiredto set up the magnetic field due to the current in the stator coil, I, in theabsence of the rest of the system.

The third term in equation (11) is the energy of interaction between themagnetic moment M and the stator coil. The interaction is given in termsof the mutual inductance LMS, which depends on the relative orientation ofthe magnetic moment and the stator coil. Consideration of the interaction oftwo circuits (one circuit representing the magnetic moment M and the othercircuit representing the stator coil) allows us to write the interaction term as

LMSIMI = +M ·Bs = M0Bs cos θ = M0µ0I

2acos θ. (12)

The fifth term in equation (11) is the magnetic energy required to set upthe magnetic field of the rails. Note that we have assumed that the rails areisolated magnetically from other parts of the system, so that there are nointeraction terms in F between the rails and other parts of the system: i.e.,the flux of the magnetic field from the rails does not couple to the stator coilor to the magnetic moment.

The magnetic energy in equation (11) is a function of two currents, I andIM , and two coordinates, θ and x. Derivatives of the energy with respectto the coordinates, at constant currents, give the generalized forces on thesystem [6]. Here, these derivatives give the force on the mass m and thetorque on the magnetic moment:

mx =

(∂F∂x

)I,IM

, (13)

I θ =

(∂F∂θ

)I,IM

. (14)

The derivatives given in equations (13) and (14) lead to two dynamicalequations,

mx =1

2L′I2, (15)

I θ = −µ0M0

2aI sin θ. (16)

6

The above two equations could be obtained directly from Newton’s laws ofmotion and the Lorentz force, equation (1).

The third dynamical equation comes from consideration of the electric fieldsin the stator circuit, in the presence of magnetic flux passing through thecircuit. We derive Ohm’s law for this circuit in the presence of a magneticfield. The externally induced emf in the stator cicuit, which does work onthe circuit, is due to the external magnetic flux passing through this circuit,caused by the rotating magnetic moment M. The power dissipated in thestator circuit, EI, is given in terms of the externally induced emf E :

EI = I2R +d

dt

(1

2LSI

2)

+d

dt

(1

2LI2

)+d

dt

(1

2mx2

), (17)

where L is given by equation (7). From equation (17), this externally inducedemf is given by∗

E = IR + LS I + L′(xI +

1

2xI)

+mxx

I. (18)

On the other hand, the externally induced emf in the stator circuit is associ-ated with a nonconservative electric field EM , due to the changing field BM ,where BM is the field of the magnetic moment given by equation (4). Thesefields are related by the Maxwell equation

curl EM = −∂BM

∂t. (19)

The externally induced emf in the stator circuit can be written in terms ofthe line integral of EM around the stator circuit,

E =∮

EM · dl = − d

dt

∫BM · dS = −dΦin

dt(20)

= kM0 A θ sin θ, (21)

where Φin is given by equation (5).

∗The externally induced emf is not the total emf. The total emf is given by the lineintegral around the stator circuit,

∮Etotal · dl, where Etotal is the total nonconservative

electric field in the stator circuit. The externally induced emf is the line integral of theelectric field EM associated with the external magnetic flux through the stator circuit,due to the magnetic moment M. The electric field EM is related to the magnetic field BM

of the magnetic moment by equation (19).

7

Using the emf in equation (21) in equation (18), we obtain the third dynam-ical equation:

IR + LS I + L′(xI +

1

2xI)

+mxx

I− kM0 A θ sin θ = 0. (22)

We can put equation (22) into a more symmetrical form by using equation(15) to eliminate x, giving the third dynamical equation as

IR + LS I + L′(xI + xI)− kM0 A θ sin θ = 0. (23)

Equations (15), (16), and (23) are a complete set of dynamical equations forthe quantities x(t), I(t), and θ(t). These equations are nonlinear differentialequations of second order in x(t) and θ(t) and first order in I(t). Conse-quently, as discussed above, we take the initial conditions to be

I(0) = 0, (24)

x(0) = 0, (25)

x(0) = 0, (26)

θ(0) = 0, (27)

θ(0) = ω0, (28)

where ω0 is the initial angular velocity of the rotor at t = 0.

The goal of the EMRG is to launch a projectile with as high a kinetic energyas possible, subject to certain constraints on the acceleration to prevent ma-terial failures. To this end, we want to maximize the transfer of the initialrotor energy,

U0 =1

2I ω2

0, (29)

into launch package kinetic energy,

K0 =1

2mx2(tf ), (30)

where x2(tf ) is the speed of the launch package at the gun’s muzzle, wherethe time tf is given by x(tf ) = Lg. Consequently, we want to maximize thequantity

η =K0

U0

=κ

γx′(tf ), (31)

8

where the dimensionless quantities x and t are given by x = x/Lg, andt = ω0t/(2π), with x′ = dx/dt. Equation (31) gives the fraction of rotorkinetic energy that is transferred into launch package kinetic energy. Solutionof the system of differential equations in equations (15), (16), and (23) allowscalculation of the dependence of η on system parameters.

9

4. Numerical Solution of the Differential Equations

The nonlinear nature of the dynamical equations (15), (16), and (23) makes itunlikely that we can find a closed-form analytic solution. Therefore, we solvethese equations numerically, subject to the initial conditions in equation (28).

For convenience, we first rewrite equations (15), (16), and (23) in terms ofdimensionless variables:

x =x

Lg, (32)

I =R

kM0Aω0I, (33)

θ =θ

2π, (34)

t =ω0

2πt, (35)

where Lg is the length of the railgun, and time is measured in units of 2π/ω0,which is the period of rotation of the rotor at t = 0.

In terms of the dimensionless variables x, I, θ, and t, the dynamical equations(15), (16), and (23) become

2κ x′′ − βI2 = 0, (36)

γ θ′′ + I sin(2πθ) = 0, (37)

I + α I ′ + β(x′ I + x I ′

)− θ′ sin(2πθ) = 0, (38)

where the primes denote differentiation with respect to the dimensionlesstime, and the dimensionless parameters α, β, γ, and κ are defined as

α =LSR

ω0

2π, (39)

β =L′LgR

ω0

2π, (40)

γ =2U0

E0

, (41)

κ =2K0

E0

, (42)

10

where U0 is the initial rotational kinetic energy of the rotor,

U0 =1

2Iω2

0, (43)

and K0 is a characteristic kinetic energy associated with the linear translationof the launch package mass m:

K0 =1

2m[Lgω0

2π

]2

. (44)

In equation (42), E0 is a characteristic energy scale in the problem given by

E0 =V 2

0

R

2π

ω0

, (45)

where V0 is a characteristic emf induced in the stator circuit

V0 = kM0Aω0. (46)

Each of the dimensionless parameters α, β, γ, and κ has a simple physicalinterpretation. The parameter α is a measure of the ratio of the inductivetime constant, LS/R, to the initial (mechanical) rotation period of the rotor,2π/ω0. The parameter β is a measure of the inductive time constant of therails, L′Lg/R, to the initial rotation period of the rotor. The parameter γis a measure of the initial rotational kinetic energy of the rotor, and theparameter κ is proportional to the characteristic kinetic energy K0.

We integrate equations (36) to (38) numerically using the values of α, β, γ,and κ given in table 1, and subject to the initial conditions in equation (28).Initially, the angular velocity of the rotor is ω0.

The (dimensionless) launch package position x is plotted as a function of(dimensionless) time t = ω0t/(2π) in figure 2. The t axis corresponds to

Table 1. Valuesof dimensionlessparameters.

Parameter Value

α 0.36

β 0.36

γ 0.307

κ 0.0176

11

Figure 2. Plot ofposition x(t)versus t forinitialconditionsθ(0) = 0 (solidcurve) andθ(0) = π/2(dashed curve).

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.51

1.52

2.53

3.54

4.55

x~

~t

measuring time in periods of revolution at t = 0. The position x is generallyan increasing function of time t, as expected. However, the plot starts outrather flat near t = 0, and at approximately t = 0.3, there is an upturn. Thefigure shows x versus t for two initial conditions: θ(0) = 0 (solid curve) andθ(0) = π/2 (dashed curve). At any time t, for θ(0) = π/2, the armature hasnot traveled as far as for θ(0) = 0.

For t > 1.5, the dimensionless launch package position x appears to be astraight line on this scale; however, there is in fact a slight ripple, which showsup clearly in figure 3, which shows the dimensionless velocity of the launchpackage x′(t) plotted versus time t, for the initial conditions θ(0) = 0 (solidcurve) and θ(0) = π/2 (dashed curve). This velocity plot clearly shows ripplesassociated with the acceleration of the launch package, which is proportionalto I2 (see eq (15)).

Unlike in some actual EMRG designs, in this model there is an alternatingcurrent in the stator coil induced by the magnetic moment. However, the(dimensionless) acceleration x(t)′′ is proportional to the square of the currentI2, and hence there are no essential differences in the operating principlesbetween ac and dc designs. Note, however, that if we want to maximizethe velocity of the armature, we would choose to close the switch on thestator coil at a time that coincides with the θ = 0 position of the rotor. Thedifference in the velocities of the armature for the initial conditions θ(0) = 0and θ(0) = π/2 depends in a complicated way on the parameters in theproblem. Furthermore, the armature containing the projectile cannot leavethe gun muzzle (break the stator circuit) at an arbitrary time, since there

12

Figure 3. Plot ofvelocity x′(t)versus t forinitialconditionsθ(0) = 0 (solidcurve) andθ(0) = π/2(dashed curve).

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

x'~

~t

is typically a significant current in the stator circuit, and this would lead toarcing problems in practice. The requirement of zero stator current beforethe stator circuit is broken (or the switching of stator circuit current) is anadditional constraint in a real EMRG system.

The fraction of rotor kinetic energy that is transferred to the launch package,given by η in equation (31), is η = 0.070 and η = 0.063, for the initialconditions θ(0) = 0 and θ(0) = π/2, respectively, and time tf = 3.0. However,the value of η depends in a complex way on the system parameters and onthe precise choice of time tf (see fig. 3).

The (dimensionless) acceleration x′′(t) versus t is shown in figure 4. Sincethe acceleration is proportional to the square of the current, the accelerationpeaks at the peak magnitude of the current (see eq (15)).

However, the peak in the stator current lags the emf induced in the statorcircuit by a variable amount of phase through the ac cycle, in contrast toa simple steady-state inductive resistive circuit, where the current lags thevoltage across the inductor by π/2. The emf induced in the stator circuit, E ,given in equation (18), is given in dimensionless form by

V (t) =EV0

= I + α I ′ + β(x′ I + x I ′

). (47)

Figure 5 shows the voltage-current phase relations for the initial conditionθ(0) = 0. Note that closing the stator circuit switch at θ = 0 corresponds toclosing the switch at minimum voltage. The variable phase through the accycle between the current and the induced voltage is due to the particular

13

Figure 4. Plot ofaccelerationx′′(t) versus tfor initialconditionsθ(0) = 0 (solidcurve) andθ(0) = π/2(dashed curve).

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.20.40.60.8

11.21.41.61.8

22.2

x''~

~t

Figure 5. Plot ofvoltage V andcurrent I(t) instator circuitversus t, forinitial conditionθ(0) = 0.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50.60.40.2

00.20.40.60.8

I~

V~

~t

initial conditions and to the variable coupling of the stator circuit to thelaunch package mass as the mass m moves down the rails. The detailedphase relationship between the current and voltage depends in a sensitiveway on the values of the parameters in the problem.

A similar complicated phase relationship exists for the initial condition θ(0) =π/2, which is shown in figure 6.

During the firing of the EMRG, the kinetic energy of the rotor goes intothe magnetic field energy of the stator circuit, Joule heating, and the kineticenergy of the launch package. Consequently, on average, the rotor kineticenergy decreases. However, this process is nontrivial. Figure 7 shows the

14

Figure 6. Plot ofvoltage V andcurrent I(t) instator circuitversus t, forinitial conditionθ(0) = π/2.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

0.80.60.40.2

00.20.40.60.8

1

I~

V~

~t

Figure 7. Plot ofrotor angularvelocity θ′(t)versus t, forinitialconditionsθ(0) = 0 (solidcurve) andθ(0) = π/2(dashed curve).

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

~t

~'θ

15

angular velocity of the rotor for the two initial conditions, θ(0) = 0 andθ(0) = π/2. Consider the solid curve for initial condition θ(0) = 0. Withincreasing time, the rotor angular velocity initially decreases, then it increases(near t = 0.9), and again decreases. The increase in rotor angular velocity is aconsequence of the coupling of the rotor to the stator circuit. During the shot,starting at t = 0, energy from the rotor is transferred to the stator circuit,thereby decreasing the rotor angular velocity. At approximately t = 0.6,some of the energy from the stator circuit is transferred back to the rotor,causing the rotor to speed up. The transfer of energy from the stator circuitback to the rotor occurs only for a short period of time, from approximatelyt = 0.6 to t = 0.9. At time t = 0.9, energy is again transferred from the rotorto the stator circuit. The process continues with a decreasing amplitude ofoscillation with time.

The behavior of the system for the initial condition θ(0) = π/2 is similar tothe behavior for the initial condition θ(0) = 0. However, for θ(0) = π/2, thephase relationships are different, and the rotor has not lost as much energyfor times later than t ≈ 0.25.

Figure 8 shows the angular acceleration of the rotor, which is proportionalto the torque acting on the rotor. Clearly, the torques on the rotor havea complicated time dependence for both initial conditions considered. Suchcomplex phase relationships must be carefully considered in the design of anypractical EMRG.

Figure 8. Plot ofrotor angularaccelerationθ′′(t) versus t,for initialconditionsθ(0) = 0 (solidcurve) andθ(0) = π/2(dashed curve).

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

1.21

0.80.60.40.2

00.2

~t

~''θ

16

5. Conclusions

We have solved the dynamical equations (15), (16), and (23) for several dif-ferent initial conditions (in addition to the ones discussed above) and withdifferent values of the parameters. We have found that the dynamical behav-ior of the system can be very complex. In particular, the maximum currentdepends not only on the position of the rotor when the stator circuit switchis closed, but also on values of the stator circuit’s resistance and inductance.Furthermore, the velocity of the launch package depends in a complicatedway on the actual parameters in the problem. Consequently, it will also beprofoundly affected by the characteristics of the field winding circuit, whichwas not included in our simplified model. In order to approach more closelythe performance of a real EMRG, we are presently analyzing a model ofan EMRG that includes a primary circuit. We will report our results andconclusions from that analysis in a future report.

17

Acknowledgments

The authors would like to thank D. E. Wortman for many fruitful conversa-tions. One of the authors (T.B.B.) would like to thank W. C. McCorkle forbringing this problem to his attention and for numerous discussions.

18

References

1. S. C. Rashleigh and R. A. Marshall, “Electromagnetic Acceleration ofMacroparticles to High Velocities,” J. Appl. Phys. 49, 2540 (1978).

2. P. Graneau, “Electromagnetic Jet-Propulsion in the Direction of CurrentFlow,” Nature 295, 311 (1982).

3. P. Graneau, “Compatibility of the Ampere and Lorentz Force Laws withthe Virtual-Work Concept,” Nuovo Cimento 78B, 213 (1983).

4. P. T. Papas and P. G. Moyssides, “On the Fundamental Laws of Elec-trodynamics,” Phys. Lett. 111A, 193 (1985).

5. W. C. McCorkle, “Electromagnetic Gun (EMG) Report to the DeputyUndersecretary of the Army for Operations Research” (1997, manuscriptin preparation).

6. L. D. Landau and E. M. Lifshitz, Electrodynamics of Continuous Media,Pergamon Press, New York (1960).

7. A. E. Fitzgerald, C. Kingsley, Jr., and S. D. Umans, Electric Machinery,fifth edition, McGraw-Hill, New York (1990).

do not print me

19

21

Distribution

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NSN 7540-01-280-5500 Standard Form 298 (Rev. 2-89)Prescribed by ANSI Std. Z39-18298-102

Unclassified

ARL-TR-1663

Interim, from Dec 1 1997 to Feb 27 1998May 1998

PE: 61102ADA PR: AH47

Transient Response of an Electromagnetic Rail Gun:A Pedagogical Model

Thomas B. Bahder and John D. Bruno

U.S. Army Research LaboratoryAttn: AMSRL-SE-EP (bahder@arl.mil)2800 Powder Mill RoadAdelphi, MD 20783-1197

Approved for public release; distribution unlimited.

AMS code: 611102.H47ARL PR: 8NENFF

Electric gun, electromagnetic gun, rail gun, magnetic propulsion, physics,model

Unclassified Unclassified

25

UL

U.S. Army Research Laboratory2800 Powder Mill RoadAdelphi, MD 20783-1197

We present a highly simplified model of an electromagnetic rail gun. The energy source forthe gun consists of a rotating magnetic dipole moment positioned at the center of a stator coil.The dipole moment is free to rotate about an axis normal to the stator coil’s axis, inducing acurrent in the rail gun stator circuit, and thereby propelling the projectile. The system’s dynamicalbehavior is described by three nonlinear coupled ordinary differential equations, involving time-dependent functions representing the armature’s position, the stator circuit’s current, and thedipole moment's angular position. We numerically solve this system of differential equations andplot the solutions versus time. The results exhibit a complicated dynamics due to the interaction ofthe electrical and mechnical degrees of freedom.

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