XPost: rec.models.rockets   
   From: niall@nospam.oswald.ndo.co.co.uk.retro.com   
      
    wrote in message   
   news:pan.2004.07.08.11.24.50.829436@null.com...   
   > I wonder if someone on this list can help me with something that's   
   > confusing me:   
   >   
   > In rocketry we want to get the maximum amount of thrust from the minimum   
   > mass of propellant, so we need to accelerate the minimum of propellant to   
   > the maximum velocity possible in order to achieve this aim.   
   >   
   > In some other applications (e.g. if the mass of propellant is small   
   > compared to the mass of the rocket - as in some rocket-powered aircraft)   
   > we don't need to worry about using propellant mass so efficiently.   
   >   
   > This is what's confusing me:   
   >   
   > It should take 200 J to accelerate 1kg of exhaust to 20m/s   
   >   
   > It should take 400 J to accelerate 0.5kg of exhaust to 40m/s   
   >   
   > Both scenarios provide the same amount of thrust (1 x 20; 0.5 x 40), but   
      
   Same amount of thrust, or same amount of impulse (change in momentum)?   
   Thrust will be related to the time in which the mass is expelled from the   
   motor. Both scenarios provide the same total impulse, if thats what you   
   mean.   
      
   > scenario 1 seems to be more efficient than scenario 2. I don't see where   
   > any energy is lost, so I'm confused as to why that should be.   
      
   I think the issue is that momentum (i.e. thrust, impulse, time, all that   
   stuff) is related to energy, but not the same. For a particular amount of   
   momentum, say in a moving object, the amount of energy required to acheive   
   this momentum depends on the mass, and the relationship is not linear. P=mv,   
   KE=1/2mv^2   
      
   Thus for the same momentum at half the mass, v must double, meaning that KE   
   will be 4 times as high. Momentum and KE are not the same thing. It makes   
   more sense to talk about rocket motors in terms of momentum (i.e. impulse),   
   since that tells you what the motor does, and in terms of specific impulse   
   (which tells you effectively how much 'woosh' you get from a unit amount of   
   propellant).   
      
   > I'm sure there must be some flaw in my reasoning but I don't see it. I'd   
   > be very grateful if someone could point it out to me.   
   >   
   > Thanks in advance,   
   >   
   > vne   
   >   
   >   
      
      
   Just going back to simple physics:   
      
   Kinetic Energy = 1/2mv^2   
   Momentum = mv   
   (Impulse = ft = Delta mv)   
   Average Velocity - Impulse/propellant mass (mv/m)   
      
      
   So double the velocity, you have 4 times the KE, but only twice the momentum   
   (and thus for the same time, say one second, double the thrust). I was   
   thinking about the exhaust velocity/propellant mass stuff the other day, but   
   I seem to remember there being another term relating to pressure and nozzle   
   area in there.   
      
   HTH   
      
   --   
   Niall Oswald   
   =========   
   UKRA 1345 L0   
   EARS 1151   
   MARS   
      
   "Gravity assisted pieces of the rocket raining from the sky should be   
   avoided. It is also financially undesirable."   
   -Portland State Aerospace Society   
      
   --- SoupGate-Win32 v1.05   
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