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Aviation History
1945
1945 - 0041.PDF
JANUARY 4TH, 1945 \ FLIGHT 2T CORRESPONDENCE is the mass of the rocket when the fuel is exhausted. Thisequation can be adopted for vertical flight and becomes, In the case of. a rocket projected vertically upwards, it canbe shown that the velocity required for it to escape completely from the attraction of the earth (velocity of liberation) is givenby, , v lib — v * 5 r where g is the gravitational acceleration at the surface of theearth, and r is the radius of the earth. Substitution of known values will give the velocity of liberation as II.I kilometres persecond (6.95 miles per second) slightly less than the 7.5 miles per sec. of Mr-. Spearman.Having recollected these equations it is possible to apply the known details concerning V.2 to them. First, however, itis necessary to adopt a figure for the efflux velocity, which is not difficult to estimate. Ethyl alcohol and liquid oxygenburning together should yield 1970 calories. The jet velocity- is according to the equation, vt = V2 H Jwhere H is the heat yield per gramme moL, and J is the mechanical equivalent of the calorie. The theoretical effluxvelocity (i>t) is thus found to be 4.0 kilometres per second. The thermal efficiency of a rocket engine is according to theexpression, , vE"' = U and in 1040 the American Rocket Society were obtaining effi-ciencies of 40 per cent. No doubt the Germans using, as they do, a turbine pressure pump for the fuel feed, exceed this figure.A conservative estimate may be 50 per cent so that the actual jet velocity will be 2.82 kms/sec.The official details of V.2 give 8J tons as the amount of fuel and 12 tons as the original weight of the rocket plus fuel. Theload ratio (R) will thus be 3.42. It is also reported that the time of burning is less than 60 seconds, and proceeding from thesefigures it is possible to calculate the final velocity, namely, V = (2.82 logc 3.42) — 0.456 kms/sec.giving 4.776 kms/sec. Allowing that some of the complex equipment necessary for the horizontal trajectory can be omittedin-vertical flight, and also for the effects of air resistance, this figure will be reduced to about 4.5 kms/sec.At a height of about 100 miles (160 kms.), there will thus be a projectile travelling vertically upwards at a velocity of4.5 kilometres per second. This projectile will weigh 3J tons, of which 1 ton comprises a smaller rocket taking the place of thenormal warhead. The mass ratio of this small rocket can be calculated in order that it may escape from the earth. Airsistance can be neglected at this height. Rearranging the Second equation gives, As the small rocket is already travelling at 4.5 kms/sec. due toits upward motion inside the large rocket, the additional velocity required for liberation is only 6.6 kms/sec. Assuming the sameefficiency as before, R = e -3t = 10 (approx.) If the total weight of the small rocket is I ton (2000 kgms.),then there must be 1800 kgms. of this fuel, and 200 kgms. •rocket and payload. 50 kgms. of the payload could be flashpowder arranged to explode upon impact with the surface of the Moon. The slight discrepancies between these figures and those ofMr. Gatland arise, of course, due to the fact that he assumed a total weight of 15 tons instead of the 12 tons now officiallyreleased. Regarding your correspondent's final remarks concerning theeffects of acceleration. Slight consideration will show that Mr. Spearman's observations are far from being correct. Con-sider first the acceleration of V.2 on this hypothetical vertical trajectory. The minimum acceleration will be experienced atthe beginning of the flight when the weight is greatest. The acceleration will then be, _ 26 — 12 __ ~ 12 ' "and the maximum acceleration will be when all the fuel has been nearly exhausted. The thrust at this stage will still be thesame, namely, 26 tons, so that, 26 ~ 3'5 -hi,a — 3-5Both these figures being nowhere near to the limit which the human body can withstand. Of interest it is to recall an articleby Mr. F. C. Sheffield (" G-men of the Air," Flight, 12th Feb., 1942), where it is stated that aeronautical research has provedthat the human body may withstand an acceleration of 14-16 g for 120-180 sees, if in the prone position. From the well-knownequation, v ~ at, it can be seen that in 150 seconds at 15 g's the human body can be safely accelerated to a velocity of 17 1kms/sec., much more than is required for a voyage tathe Moon. E. BURGESS,President, the Combined British Astronaulical Societies.[1 ton = 1,016 kgms, which affects the value of R.—ED.] AIRBORNE OUNCES Open Cage or Airtight Box ?I N response to Jonathan Peel re the perched and airbornecanary, I think the solution is simple. The cage and canary airborne equals 20 oz., i.e., the weight of the cage,because the air displaced by the canary flapping its wings would pass out through the bars of the cage. If, however, it were in an airtight box and flew around, theweight would be 21 oz. Possibly when the bird was "taking off" the scales would register over 21 oz. (presuming boxequals 20 oz. and bird 1 oz., of course) because the downward thrust transmitted through his feet would have to be takeninto account—also the weight of the volume of air, but this problem is more in Horace's line than mine.ROY E. FFRENCH DENITT. Newton's Third Law Again T^HE problem raised by your correspondent, Mr. Jonathan-*• Peel, in the December 14th issue of Flight, is a hardy old perennial that can usually be relied on to provoke anargument in any place where mathematicians congregate. Following is my analysis, which I believe to be correct. It involves an application of Newton's Third Law, whichmay be stated accurately thus: If a body A acts on a body B with a force F, then B reacts on A with an equal and oppositeforce — F; or, more usually, as: Action and Reaction are equal and opposite. Taking the weights of cage and bird to be 20 oz. and 1 oz.as suggested by Mr. Peel, then clearly when the bird is perched the total weight registered will be 21 oz. Now sup-pose the bird becomes airborne, and consider the forces acting at an instant when the vertical component of the accelerationof the bird is nil. For vertical equilibrium of, the bird, the air must be exerting upwards on the bird a force equal to theweight of the bird, namely, 1 oz. weight, and hence, by New- ton's Third Law, the bird must be exerting on the air anequal and opposite force of 1 oz. weight downwards. Now consider the forces acting on the mass of air in thecage. The air, considered a whole, is in equilibrium, and is being acted on by the bird with a force of 1 oz. weightdownwards, and thus there must be a balancing force of 1 oz. weight exerted upwards on the air by the floor of thecage; hence, again by New-ton's Third Law, the air must exert an equal and opposite force on the cage, that is, a forceof 1 oz. weight downwards. Thus the total force registered by the scales will be the weight of the cage (20 oz.) plus thisforce of 1 oz. weight—21 oz. in all, as before. If the bird is flying so that there is a vertical componentof acceleration upwards, the apparent weight will be increased by a force sufficient to produce this acceleration, and similarly,if a vertical component of acceleration downwards is present, the apparent weight will be decreased. I do not know how this result can be demonstrated experi-mentally for a bird in a cage, but a rough analogy is furnished by a well-known experiment in support of Archimedes' prin-ciple. If a body, say a metal ball, is weighed in air and then in a beaker of water, the beaker being supported on astand, the ball will suffer an apparent loss in weight equal to the weight of water displaced; this apparent loss in weightis a measure of the upthrust of the water on the ball. If now the ball is supported on a stand, so that the balance is notaffected by the weight of the ball, and the beaker and water weighed (1) by themselves, (2) with the ball immersed in thewater, it will be found that the apparent weight of the beaker and water is increased. This apparent increase in weight isa measure of the downthrust of the water on the beaker, and will be found to be equal to the weight of water displaced bythe ball. This demonstrates that the upthrust of the water on the ball is equal to the downthrust of the water on thebeaker. The analogy is not exact, but is a useful illustration of th-.principles involved. C. L. BROKENSHAW.
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