2—
Kinematics

In order to gain some quantitative insight into the elements of such a system, we propose to adopt a particular set of assumptions and to study numerically the effect of variation of parameters. The Eqs. (17) which follow are obviously highly tentative and subject to many questions here unresolved.

The vehicle is considered to be saucer-shaped, of diameter about 10 meters, sufficient at any rate to intercept all or most of the exploding propellant. Its final mass Mf is perhaps 12 tons, which must cover structure, payload, instruments, storage for propellant and bombs, and, if required, apparatus for maintaining the magnetic field. The initial mass M₀ of the vehicle exceeds this by the mass of bombs and propellant.

The bombs are ejected at something like one second intervals from the base of the rocket and are detonated at a distance of some 50 meters from the base. Synchronized with this, disk-shaped masses of propellant are ejected in such a way that the rocket-propellant distance is about 10 meters at the instant the exploding bomb hits it. The propellant is raised to high temperature, and, in expanding, transmits momentum to the vehicle. The final velocity Vf is attained after N (-50) such explosions.

We regard now the i-th stage of the process. From the rocket, traveling at velocity V^i-1 with respect to the earth, are ejected first the i-th bomb (mass mB) and then the i-th mass of propellant m? at some small velocity vo relative to the rocket. It is supposed that, upon detonation, a certain fraction cr of the mass of the bomb collides inelastically with the ejected propellant mass. This fraction could be made, in our case, perhaps as much as 1/10, which is considerably more than the factor given by the solid angle. This could probably be achieved by a suitable distribution of the mass of the tamper surrounding the core

* Part II never appeared.(Eds.)

of the bomb. In this way, a larger fraction of the mass of the bomb would hit the propellant. (It is easy to make the distribution of the mass involved in the bomb explosion nonisotropic; the energy distribution is probably essentially isotropic.) If vB is the average velocity of explosion of the bomb in the sector reaching the propellant, we have OamB(V^i- - Vo + VB) + mip(Vi-l - vo) =('amB + mip)V ,

where Vp is the velocity relative to the earth of the center of mass of the combined system (oam, mp). If we introduce a velocity vp by means of the relation Vp = V^i--vo + vp we obtain amBvB = (amB + mp)Vp . (1)

The excess kinetic energy in this transfer is supposed to appear initially as thermal energy Hi in the propellant Hi= lmB(v) (amB + mp)(vp)(2)2 2 (

It is assumed that about half of this heat Hⁱ reappears in kinetic energy of expansion of the propellant, with an expansion velocity VE relative to its own center I 1 Hi -(amB + mp)(v)2. (3) 2 2

We assume, arbitrarily, that in the expansion of the propellant, one half of its internal energy becomes converted to kinetic energy of expansion. This fraction depends obviously on the distance d and is, in our case, higher.

In our schematic computation we prefer to adopt this much too conservative value.

We may consider that the upper and lower halves of the exploding propellant travel with average velocities Vp ± VE, respectively. Now Eqs. (1), (2), (3) show that )2 i )2

and since, in all cases we consider, mi > ² oamB, we have vE > vp.Thus Vp- vi = V^i- - vo + vp - v < V^i-l , and the lower half of the exploding propellant will not reach the rocket.

The momentum conservation equation for the rocket and upper half of the propellant should read - (OamB + mi)(Vi- l-v + p + ) + MiV1(umB + mp)(^i- l - (-VO + vp + VE)) + mivi 2 or, simplifying, - (JmB + mp). 2(-vo + vp + v) = MiiV , where M ⁱ is the present mass of the rocket, and AiV is the i-th increment in its velocity relative to the earth. This assumes total reflection of the propellant. To allow for side effects and imperfect reflection, we use the equation - (amB + ) + v) = MiAiV . (4)

Finally, we assume the time Ait for the i-th acceleration to be 2d /it- (=,) (5) where d is the distance from propellant to rocket. The i-th acceleration is thus Ai_V(>i =/t (6) There are two cases of mathematical simplicity which we outline, and for which we include some numerical examples. (Tables 1 and 2 for the cases 1 and 2, respectively.)

Case 1—
Constant Acceleration

We take as independent parameters: Vf the final velocity Mf the final mass of the rocket N the number of stages (bombs) a the acceleration at each stage (assumed constant) d distance from propellant to rocket mB mass of each bomb a fraction of mB hitting propellant and show how all other parameters may be expressed in terms of these.

Thus each change in velocity will be AiV_Vf (7) over a time interval __v Vf/it== aN (8) The propelling velocity vp + v _- wⁱ is thus 2d 2aNd a=i- - (9) jt Vf We now consider Eq. (4), setting C = (1 - )mB (10) and mi = mB + m, the total ejected i-th mass. Thus (4) becomes mi-C=k M-mj (4*) j=l where k :2AiV 1 (Vf) (11) and M₀ is the initial mass of the rocket.

Writing the equation (4*) for i + 1 and subtracting shows that mi+1 = mip where 1 (12)

Thus mi = mlp^i- , i = 1,2,...,N. We determine M ₀ and ml as follows. Substituting -=mj = mi(1 - pi)/(l - p) into (4*) shows that mi( + k) =kMo + C, while, by definition, N Mo - Mf = E j = ml( -N)/(1 - p) =- (1 +k)(1 - pN) j=l

Eliminating Mo between these two relations yields nl = (kMf + C)(1+ k)^N- (13) and so Mo = [mi(1 + k)-C]/k . (14) Thus we have trivially the i-th mass: mi = mⁱ , (15) the mass ratio: M.R. = M, (16) the total expelled mass: T=Mo-Mf, (17) the total bomb mass: MB = NmB, (18) the total propellant mass: Mp = T- MB, (19) and the i-th mass of propellant: mp = mi- mB· (20) Now, solving equations (1), (2), and (3) for vp and vg in terms of v, we get vp = ornmBvB/mi - C (21)

and ·i1i-amBmp V Emi - C 2B Substitution into v+ + vp = w (22) yields v = (mi-C)/{l mB + a\MBm (23) whence the values of v, and vg may now be obtained, using (21) and (22), respectively.

Thus all parameters are determined in terms of the fundamental set Vf, Mf, N, a, d, mB, a. It is interesting to note that the mass ratio mi(k + 1) - C mi(k + )pN - C is (approximately in general and exactly when C = 0) (1 + k)N where k = (1/ad)(Vf/N)² , which indicates the extreme sensitivity of the mass ratio to a, N, d, and especially to Vf, in the constant acceleration case.

A rough indication of the energy of the i-th bomb is given by the kB = (1/2)mB(vB)² included in the tables. The actual yield of each bomb is several times greater since we assumed a special shaping of the tamper to concentrate as much as possible the mass, but not the energy of the exploding bomb, towards the propellant.

Table 1 is intended to show how the various factors in the problem depend on the initial parameters N, a, d and mg. None of the twelve "problems" is intended as an optimum case. It may be noted that problems 1 and 2 with Vf = .7 x 106 are included for the sake of comparison with various intercontinental ballistic missiles schemes. It should be noted that our mass ratios are considerably less than those contemplated in such cases, while the accelerations are very much more (w 10,000 g's), lasting for periods of about I millisecond each. One also notes that the bombs are rather "small" (1019 - 1020 ergs).

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Case 2—
Constant Mass

In this case, which closely corresponds to the usual rocket assumption, we take as independent parameters Mf, N,d, mB, , and now mp, VB (assumed constant) instead of a and Vf.

Thus we have for the mass expelled at each stage: m = mB + mp , (24) the total bomb mass: MB = NmB, (25) and the total propellant mass: Mp = Nmp , (26) the total mass expelled: T = MB + MP, (27) the initial rocket mass: Mo = Mf + T, (28) and the mass ratio: M.R. = Mo/Mf . (29) Since vB is given, we find from Eq. (1) that vp = crmBvB/(crmB + mp) , (30) while Eqs. (2) and (3) show that 1 2mp 2 omB + mP) (31) and VE = VH/(amB + mp) . (32) Hence we again have a constant propelling velocity given by J = VP + VE . (33) The "rocket equation" (4) now becomes L = - (amB + mp)w = MiiV , ~~172~~ 2

the left side being a known constant, and Mi = M^O - im being a known function of i = 1,. ., N. Hence we can compute the i-th increment of velocity AiV = L/M , (34) and the velocity after i stages: i Vi - E V . (35) j=l In particular the final velocity is N Vf = VN= V. (36) j=l The time Ait is given by the constant Ait =2d/w , (37) and hence we have the i-th acceleration ai = AiV/zAit. (38)

In particular, amin = Ol= ()/(Mo - m) , (39) 2d / and armax = oN = ( )/Mf (40) In analogy with the usual rocket equation, our Eq. (34) might be written (L) m = MiAi^V , or, letting 3= L/M= (B + mp-) w- dM=MdV 2 \ m whence dM M =-dV

and Mo In M = /3-lV or Mo- evf// which affords a rough estimate of Vf, namely Vf -Lin(M.R.) m

In Table 2, Problem #4' is intended to be an analogue of Problem #4 of Table 1, while Problem #12' is intended as a companion to Problem #12 of the former table. It may be noted that in order to duplicate the performance of a given rocket of constant acceleration a by the second method, one requires accelerations whose average is ~ a and which, therefore, individually greatly exceed a in the final stage. It may be that the method of Case 1, although unorthodox, has advantages in this sense which might justify the use of bombs of variable yield.