Improvement of conical shaped charge system and comparison o

Acta Astronautica61(2007)617–

625

719813d5360cba1aa811daef/locate/actaastro

Theory of a multistage light gas gun

J.G.Linhart?,F.Cattani

Department of Physics,University of Ferrara,Via Saragat1,Ferrara,Italy

Received22December2004;accepted11December2006

Available online30March2007

Abstract

The theory of multistage kinetic energy transfer is applied to the acceleration of a light projectile in a light gas gun(LGG). It is argued that a better overall gun ef?ciency can be obtained if the kinetic energy1/2M1V21of the?rst piston is not directly transferred to the light projectile but proceeds via an intermediate stage transfer mediated by a second piston M2. Application of such a gun to satellite or robot craft launching is outlined as well as some new technological features in the gun design.

?2007Elsevier Ltd.All rights reserved.

0.Introduction

There are two possible methods for launching pay-loads into a high earth orbit or even out of the gravita-tional?eld of the earth.The?rst and the only one used today employ a rocket propulsion where the rocket is driven by combustion of advanced chemical jet fuels.

It is well known that this implies a very large fuel to payload ratio.This unfortunate state of affair is due to

the jet speed V j obtainable by the combustion of chem-ical fuels(V j 2.5×105cm/s)being much smaller than the speed of the rocket V required to reach a high orbit or the escape speed(V esc≈11km/s).

In order to obtain a suitable high jet speed it has been suggested to use,as a driver,nuclear explosion(project

?Corresponding author.

E-mail address:linhart@fe.infn.it(J.G.Linhart).

0094-5765/$-see front matter?2007Elsevier Ltd.All rights reserved. doi:10.1016/j.actaastro.2006.12.008Orion,[1]).Unfortunately it can be shown[2]that in order to transmit the explosion impulse to the payload, one must provide a suitably robust structure whose mass is proportional to the energy of the explosion. This unavoidable mass ballast implies that in order to reach V 10(km/s)the number of nuclear explosions must be very high.It seems that,at least for an escape from the Earth,the Orion concept is not suitable. The second method for space propulsion is the use of a super-gun.However,even in this case the mean thermal speed V T in the gas pushing the projectile must be at least as high as the?nal speed V of the projectile.This implies gas temperatures approaching T=104(K).Such temperatures cannot be obtained by chemical combustion.It has been suggested that for large payloads(M p 10ton)the driver could be a small nuclear explosion(W n 1kton TNT).This may

be dif?cult to obtain apart from being politically and ecologically inadvisable[2,3].

618J.G.Linhart,F.Cattani/Acta Astronautica61(2007)617–625 However,there is another way in which T～104

(K)can be obtained and this is a large light gas gun

(LGG).In such a gun,a heavy piston(M1)acquires

a velocity V1～1(km/s)driven by a chemical explo-

sion(e.g.methane and air).The piston compresses as

a volume of light gas(H2).Very high pressure,tem-

perature and thermal speed V T can thus be achieved

[4]and consequently a large projectile speed is ob-

tainable(V>5km/s).However,it can be shown that

this one stage transfer is extremely inef?cient when

the mass m of the projectile is a very small fraction

of M1,in fact some features of such a gun make one

doubt whether it is at all feasible.

We shall show that in order to obtain a good ef?-

ciency in a high speed LGG it will be necessary to

use an intermediate piston M2,achieving thus a much

better kinetic match M1→M2→m.We shall?rst

outline a basic theory of such velocity staging[5].

Subsequently,we shall describe the main concept for

optimum matching of M k→M k+1,and?nally we

shall sketch a simple design of a three stage LGG aim-

ing at launching satellites whose mass is of the order

of1ton to V>6(km/s).

1.A strategy for optimum energy transfer

Let us consider the following simple mechanical

model.A heavy slug(mass m1)moving with a speed,

v1,compresses a spring, .When m1loses all its

kinetic energy,the spring is blocked by a latch S.The

slug m1is removed and replaced by a much lighter one

(m2).The latch is then released and m2is accelerated

to a speed v2.In the ideal case when m1and m2are

incompressible,the spring is mass-less and the latch is

free of any friction;the law of conservation of energy

gives

V1=V2

m1 m2

.

Unfortunately the real situation corresponds to Fig.1, in which the slugs have an internal spring( 1and 2), the spring, ,has an effective mass,m ,and friction is associated with any change in the system.

To start with,the role of will be ignored—assuming the masses,m1and m2,to be incompressible.A friction-less system will also be assumed.

Energy Fig.1.Simple mechanical model for energy density ampli?cation.

conservation then gives

V2=V1

m1

m2

.

The ef?ciency of the transfer is

=

m2V22

m1V21

=

1+

m

m2

?1

.(1)

Let us suppose that m is related to m1.This is quite natural,as only a large and robust spring can effec-tively stop a large and heavy slug(for considerations of volume and pressure matching see Section2).It may be suspected now that the energy transfer from m1to m2>m1can be effected more ef?ciently if sev-eral stages of energy transfer are used in series,rather than if only one direct transfer is employed. Consider thus a chain of y=n?1stages,

W1(m1)→W2(m2)···→W k(m k)→···W n(m n), involving masses m1,m2,...,m n.

In order to simplify the analysis it will be assumed that m k/m k+1=cost= and m k/m k=const= , where m k is the effective mass of the k th spring.The ef?ciency per stage is

k=

1+

m k

m K+1

?1

=(1+ )?1.(2)

J.G.Linhart,F .Cattani /Acta Astronautica 61(2007)617–625

619

00.10.20.30.40.50.60.70.80.911

7654

3

2

89

10

11

number of stages

e f f i c i e n c y

γ = 100 β = 0.01

γ = 100 β = 0.1

γ = 100 β = 0.1 λ = 0.1

Fig.2.Total ef?ciency T depending on the number of stages y .

The total ef?ciency T of the energy transfer from m 1to m n is

T = y

k =(1+ )?y .

The parameter is a function of the number of stages,since m 1

m n

= y = or = 1/y .Consequently, T is the following function of y : T =(1+ 1/y )?y .

(2a)

If optimum staging existed then T (y)must have a

maximum in the range 1

ln

1+12

x ,(3)

the ?rst interaction giving y ln /(1+1.3 ).It follows that,for cases of practical interest ( >100, >0.1),the optimum y is in the range of 3–6.

The losses due to the compressibility of the slug will now be considered.The moment at which the latch must be inserted,i.e.at the moment of maxi-mum compression of and ,the pressure distribu-tion in m 1and 1will resemble that shown in Fig.3.In most cases [6]the compressional energy

density,

Fig.3.Pressure distribution in M 1.

u in m 1,is u ∝p

and consequently the total compressional energy,U ,will be

U = 1

2m 1v 21,

(4)

where is a constant depending on the material of

m 1.Assuming now that,for an instant,m =0and calculating v 2and 1give

V 2V 1= m 1(1? )m 2and 1=1? .

This result corrects signi?cantly the y opt obtained for m =719813d5360cba1aa811daefbining the and m departures from an ideal loss-less machine gives 1=

1?

1+(m /m 2)

.

When staging is considered the total ef?ciency of y stages becomes T =(1? )y

(1+ )y

.

(5)

The consideration of compressional losses results in the shift of y opt obtained for =0towards smaller values and smaller ef?ciencies (see Fig.2).

In order to be able to apply results to an LGG,let us ?rst discuss the dynamics of one stage in an LGG.

620J.G.Linhart,F .Cattani /Acta Astronautica 61(2007)617–625

2.A simple dynamics of one stage LGG

Let us consider following simple scenario.A pis-ton whose mass in M 1accelerated by gas (1)whose pressure is p 1compresses a gas (2)which in turn ac-celerates a second piston M 2(Fig.4)

The equations of motion of M 1and M 2are M 1¨X =A 1(p 1?p 2),(6)M 2¨Y

=A 2p 2.(7)

For a monoatomic gas

p 1=p 10

1

1+A 1X 5/3

,(8)p 2=p 20

A 1l

A 1(l ?X)+A 2Y

5/3

.(9)

Let us ?rst assume that A 2=0,i.e.M 2does not move.

Then

¨X

=A 1p 10M 1 1+A 11X ?5/3

?p 20

p 10 1?X l ?5/3 .Put X/l = and p 20/p 10= and get

¨

=A 1p 10M 1 1+A 1l 1

?5/3

? (1?˙ )?5/3 .(10)

Let us denote A 1l

1

= = and

A 1p 10l l 2M 1

=23 W 1l 2M 1=13 v 1l 2 ?1

1=a and because 12M 1v 21= 1W 1

we have ¨

=a {(1+ )?5/3? (1? )?5/3}.(10a)

Integrating we obtain 1

˙ 2=3a

1(1+ )

2/3?1 ? 1

(1? )2/3

?1

.

(11)

1When =0,Eq.(11)can be further

integrated.

Fig.4.The energy transfer from M 1to M 2.The cross sections A 1and A 2refer to the tubes 1and 2.We shall assume here that the piston M 2is free to move as soon as M 1begins to move.

The piston stops when ˙

=0and therefore when (1? )?2/3+ (1? )?2/3=1+ .(12)

For l ? = >1,Eq.(12)becomes (1+ )?2/3+ ?2/3

min = +1(12a)

and

min =( )3/2[1+ ?(1+ )?2/3]?3/2.

(13)

As we expect >1and >1we may approximate min as min ( )3/2.

(13a)

Example. =4, =0.1:we obtain min 0.0316.If l =100(m )then l ?X max 3.16(m ).

In the vicinity of min the equation of motion (Eq.(6))can be written as M 1¨X

=?A 1p 2and consequently

¨ =?a p 2m p 20 0 5/3.(14)

Integrating we obtain

˙ = 3a p 2m p 1m 0 1?

2/3 .(15)

If v 1=105cm /s then the characteristic compression time c (l ?X max )/V 1=6.3(m s ).

J.G.Linhart,F.Cattani/Acta Astronautica61(2007)617–625621 3.Optimum matching of M k to M k+1

In the case of an LGG,where the springs are

gas cushions,the ratio m k/m k= cannot be made

arbitrarily small for,at least,three reasons.

The?rst is that when tends to zero the ratio of

the compressional energy U to the stored energy in the

spring tends to in?nity and,therefore,the ef?ciency

of energy transfer T tends to zero.

The second reason is that the thermal losses from

too small a gas cushion squander most of the energy

transferred to the .

Both these effects limit the minimum of m k.But,

in order that the gun ef?ciency for the propulsion of

m k+1is not too small,m k+1must be larger than m k

and therefore,m k/m k+1= cannot be arbitrarily large.

Moreover,we shall see later,that is also limited

by the dynamics of m k+1;a too light m k+1will be

expelled before the piston m k is able to squash fully

the spring(gas cushion) k.

(a)Let us?rst address the effect of U on the m k

to m k+1matching.As we mentioned in the previous

paragraph U k= k12m k v2k.

According to Ref.[6]U k 0.112 k p max where k

is the volume of the k th piston.We shall require that

U k> W k?U k=(1?x max)A k32p max.Limiting our-

selves to the?rst piston we have

U1>32(l?X max)A1p,U1=0.05A1l1p,

where is the length of the piston M1.We obtain for

p max

l?X max

?1

30

.(16)

Example.If M1=200(ton)and A1=40(m2), = 5(ton/m3)then =1(m)and l?X max?0.033(m), say l1 min=2(m).

The mass m2of the gas cushion in case of H2?ling is

m2=A1(l?X max)p2max

kT2

m H

2

3

1W1

kT2

m H

and for hydrogen

m2=1.6×10?8 1W1

T2

.(17)

Example. 1W1=121018(J),T2=2500(K)then m2=

3.2(ton)and obviously M2?3.2(ton),say10(ton).

One notices that the limits on m2and M2are

dictated by the maximum temperature obtained in

the compression,the limitations imposed by U,i.e.

Eq.(16)is not serious.

(b)Let us now calculate the heat loss from the com-

pressed volume 2.For l?X max r1we can assume

that the heat?ow is mainly to the piston P and the

wall S(Fig.4).If most of the heat loss Q occurs dur-

ing the characteristic compression time c we require

that Q> 1W1.

But Q≈

?

Q c and

?

Q 2 (T/ )A1where is the

thermal conductivity and is the heat skin-depth.For

monoatomic gases(He,H)we have

4×103

√

(erg/cm s,K)and =

x c

3

2

kn2

.

Consequently,our condition can be written as

2A1T

3

2

kn2 c> 1W1.(18)

But3

2

kn2T=32p2= 1W1/(l?X max)A1and the equa-

tion can be expressed as

T>10?3

1W1v1

A1

2/3

.(18a)

Example. 1=12,v1=12×105,A1=4×105,W1=

1018(erg)→T>108(k).

This indicates that the heat loss due to the heat con-

ductivity will not limit the T obtained in gas compres-

sion.The limitation will be of technological nature

affecting the density n2:

n2>

p2

kT2

=2

3

1W1

2(kT)

?1.

Let 1W1=1/2×1018(erg)and 2=5×108(cm3),

T=5×103then n2 1021(H/cm3),the pressure p2=

700(bar).

This determines the mass M 2of the gas in 2:

M 2 A1(l?X max)n2·m= 2n2m.

622J.G.Linhart,F.Cattani/Acta Astronautica61(2007)617–625 For hydrogen(H2)

M 2 0·81 1W1

T1

×10?8.

Example.As above,then M

2 0.81(ton)and in or-

der that the propulsion of M2was ef?cient M2= 1M 2, where 1?1,and therefore M2 10(ton)and if M1= 100(ton)we have M1/M2=10.

Performing a similar analysis for M3we obtain

M2 M3=

1

2

A1

A2

2/3

v1

v2

2/3

?1/3

2

.(19)

Example.Put A1/A2=10,v2/v1=3, 3=12, 1/ 2≈1then M2/M3=12.

From all this it follows that M k/M k+1≈should not be larger than10.

(c)Let us now address ourselves to the limitation on M k/M k+1imposed by the requirement that M k+1 should not move too much during the compression time t c k.This can be treated by considering M1and M2.The compression time t c is

t c l?X max

1

2

v1

.

During this time M2will travel a distance S2:

S=1

2

A2p2

M2

t2c,where p2= p2, <1,(20)

or S=1

2

A2 p2

M2

x0

1

2

v1

2

,x0=l?X max,(20a)

but1

2

M1v21= 1W1and therefore

S=1

6

A2

A1

x0M1

M2

.

We require that

SA2>x0A1,

2The often suggested remedy using a membrane which rup-tures at t=t c is not applicable here.If p max is suitable for propul-sion to v?1(km/s)then the p at which a membrane ruptures will be much smaller than p max.which can be written as

M1

M2

>2

A1

A2

2

.(21)

Example.A1/A2=5then M1/M2>50, e.g. M1/M2=10and it is con?rmed that normally, M k/M k+1≈10.Ratios higher than20will result in an inef?cient energy transfer from M1to M2.

All this has not touched the fact that,even in a multistage LGG,the last stage M k must be propelled by a high pressure(p k>1000bar)and high tem-perature(T k>5000K)gas cushion.Admittedly,the advantage of staging is that it allows us to use a rela-tively smaller gas volume k subjected to high p k and T k.Consequently,the last gas chamber may use the techniques of inertial con?nement and sweating walls (Appendix A).

4.Example

As an example of our previous conclusion about the matching of M k+1to M k we shall consider a three stage LGG in which M1/M2=M2/M3=10.A pro-posed structure of such an LGG is shown in Fig.5 (not to scale).The two?rst stages,where the masses M1and M2are large,will consist of two almost ver-tical shafts in order to eliminate,as much as possible, the friction between the shafts and the projectiles M1 and M2.

Let us prescribe the masses M1,M2and M3as follows:M1=100(ton),M2=10(ton),M3=1(ton). Let us also aim at a?nal velocity v3～6(km/s). The kinetic energy of the projectile M3is18GJ, equivalent to3.6ton TNT.3In order to get an idea of the initial energy we obtained by burning some fuel in the sphere S,let us assume(based on considera-tions in Sections2and3)that the ef?ciencies of energy transfer from S1to S2,from S2to S3and?-nally the ef?ciency of the M3launch are 1= 2= 3=1

2

.

In that case W0=144(GJ).If this energy were to be obtained by burning methane and oxygen whose

31tTNT is equivalent to burning0.14ton of petrol.

J.G.Linhart,F.Cattani/Acta Astronautica61(2007)617–625

623

Fig.5. ’s are the ballast recoil masses.

total initial pressure is p0=9bar then the volume of S1must be4

1 1050(m3).

The radius of the sphere S1is then R1=6.3(m). From consideration of mechanical strength of the S1shell we need to satisfy(M s1=mass of the S1shell) M s1>100W0(ton,tTNT).(22) In which case the thickness of the shell S1is larger than0.1(m).Let us choose =30(cm).

The volume of the?rst shafts must be at least5times larger than 1in order that the cannon ef?ciency of

the W0transfer to1

2M1V21is satisfactory(e.g.70%).

This gives us

5 1= r21l1.

4CH

4+2O2=CO2+2H2O+1.7×10?14(erg)or about 0.46MJ/m3at NTP,the partial pressure of CH4being only3bar.The length should not be a large multiple of2R1—so as to ensure an ef?cient transmission of pressure from 1to M1.Choosing l1=20R1can obtain

L1=126(m)and r1=3.6(m).

This determines the dimensions of M1(steel+carbon ?bre)since

M1= r21 =100, ～3,

or 0.8(m).

It is clear that the shape of M1will resemble a saucepan.

The burning of CH4according to the equation in footnote4is not straight forward;the temperature T will rise to about2700(K)with much of C,O and H remaining dissociated.As T1drops,the reaction pro-ceeds further,maintaining T1above the value corre-sponding to the adiabatic drop in T1and pressure.If the cannon ef?ciency remains above60%,the speed v1acquired by M1at the end of its course will be 1.3(km/s).

Let us now consider that M1compresses H2;M1 will then pass the inlet of S2at a speed of v>v1and stop shortly afterwards.Thus,almost all the kinetic energy of M1will be transferred to H2in the sphere S2.As the?nal speed of M2cannot be larger than 2.9(km/s)we can operate with a temperature of hy-drogen in S2of only T2=.2300(K)corresponding to a speed of sound in dissociated hydrogen of5(km/s). Such a T2will not cause an appreciable loss of energy from the H gas in S2(see Eq.(18b))and consequently our guess of 2 12is a credible one.

The maximum pressure p2can be obtained from 1W1=3/2p2 2.

Let us allow p2=1000bar;then

2=2

3

1W1

p2

500(m3)(23)

corresponding to R2=5.9(m).The mass of H is m2= 2(p2/kT2)m H 1

2

(ton)much smaller than M2. Making the same considerations for the l2as were made for l1we obtain for r2=2.5,5 2= ·r22l2and hence l2=127(m).

624J.G.Linhart,F .Cattani /Acta Astronautica 61(2007)617–625

Let us now address ourselves to the design of the S 3,M 3hydrogen gun.

Let us start by assuming that for a short pulse of p 3,the inertial con?nement mentioned in Appendix A will help to ignore the static,mechanical strength of the condition of Eq.(22).Let us start by assuming that p 3=5000(bar )is admissible.Then 3=

23 1 2W 1

p 3

50(m 3),

or R 3 2.3(m ).

The radius r 3will be determined by the geometry of the projectile.For M 3=1ton,r 3cannot be larger than 1

2(m)considering the air resistance during its transit in the Earth’s atmosphere.Taking r 3=0.5(m )we get for l 3l 3

3

r 23

= ·64(m ).For M 1and M 2we have taken =5;here we shall be

content with =3and therefore l 3=192(m ).

Let us now check if l 3and p 3are large enough to be consistent with v 3=6(km /s ).From the equation of motion we get

l 3 12v 23M 3

r 23ˉp 3

where ˉp 3≈

12

p 3,M 3= r 2

3

3ˉ 3,which gives

l 3 v 23

3ˉ 3

p 3

.In our case 3=1.3(m )for ˉ 3=1and l 3 92(m )which is amply satis?ed.

There is one more check to be carried out,normally using the condition that the sound speed in the hy-drogen in S 3must be higher than v 3.This forces us to consider T 3>5000(K ).From the point of view of thermal loss (Eq.(18b))this is tolerable;it remains to be seen if the technology of sweating walls can cope with this.Remember the duration T of this extreme temperature will be only T

2l 3

v 3

≈60(ms ). 5.Conclusion

A case has been made for an LGG in which the energy momentum of the piston M 1is not transmitted directly to a small projectile m but passes through an intermediate stage represented by a second piston M 2.The advantage of using a three stage LGG is a much better ef?ciency in energy transfer from M 1to M 3as well as a possibility to use a much higher pressure and temperature in the third stage of the gun.We have not treated the many technological dif?culties inherent in the realization of such a gun,some of which will determine the repetition rate of the operation.

Even if an LGG functions satisfactorily there are two major problems to be resolved related to the ?ight of the projectile M 3.The ?rst is the heating of the M 3during its passage through the Earth’s atmosphere,and the second is its passage to a circular orbit.The solutions to both these problems were indicated in [2,3].

It was suggested that M 3’s contact with the atmo-sphere be substantially reduced if a portion of the gas pushing the projectile in the l 3shaft is left to pre-cede M 3,forming thus a jet in which the projectile is imbedded during its passage across the denser layers of the Earth’s atmosphere.

The second problem,i.e.of circuiting the M 3tra-jectory,must be resolved by additional rocket drive entering operation near the zenith of the projectile’s elliptic trajectory.The additional necessary velocity increase required to do this depends on the launch an-gle (elevation of the l 3shaft)and is of the order of v =2(km /s ).This implies that about one half of the M 3mass is that of the jet fuel required to augment the payload’s speed by v .

The only solution that appears suitable in order to launch a projectile whose m >10(ton )is the use of a nuclear explosion driver [2,3].

Appendix A.Limits on the inertial con?nement of large pulsed pressures in a cylindrical cavity Provided the internal pressure p r is below the me-chanical strength p m of the wall material,a suf?ciently robust vessel will contain the gas pressure statically.However,even if p r >p m but the duration of p r is short enough the vessel will not rupture;this is the

J.G.Linhart,F.Cattani/Acta Astronautica61(2007)617–625625 principle of elastic,inertial con?nement.Let us derive

the limits for p r and permitting a vessel stressed in

the pulsed manner to return to its original form after

time .The impulse I=2 R·p r· due to a pressure

p r applied for time equals to M·v r,the momentum

acquired by M.

It follows that v r=(2 Rp r/M) ,but M=2 R· · ,

and therefore

v r=p r

.(A.1)

This radial movement has to be stopped by the az-imuthal forces p · ,where

p =F e· R/R(Hook s law).(A.2) From energy conservation we have a condition for stopping the v r displacement:

1 2Mv2r=2

R

p ( R)d(2 R),(A.3)

giving a R for which the expansion will stop.We have

( R)= R

R+ R = 1

1+ R/R

and therefore Eq.(A.3)is

1 2Mv2r= ·F e

R

R

R

·2 ·d( R)

1+ R/R

.(A.4)

Using the expression for M we get

1 4 v2r

F e

=

x

x(1+x)?1d x,x=

R

R

,

1 4 v2r

F e

=

R

R

?ln

R

R

+1

.(A.4)

We wish that x>1and therefore

x (2F e )?1/2

p r

.(A.4a)

Example.p r=1010(dyn/cm2), =10, =50and

F e=2×1012then x=31.6 .

In order that x<0,1the <0.001(s),indicating

that only very short p r pulse can be inertially con?ned.

A similar expression can be obtained for a spherical

vessel.

The relative elastic extension x= R/R for suit-

able materials(steel,carbon?bre,etc.)is often below

0.1.If the product p r· is too high for achieving an

elastic expansion,the vessel will either not return to

its original R or will rupture(e.g.x>1).

This kind of inertial con?nement can be improved if

the vessel is imbedded in a medium capable of taking

on and transmitting a part of the impulse I and also

one which,when the system is at rest,exercises a

compression on the vessel(underground cavities).

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[2]J.G.Linhart,Orion revisited,Proceedings of IFSA,paper

FP o5.53,2000.

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