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Acceleration of space shuttle at liftoff

  1. Mar 25, 2008 #1
    Well, first off I finally got around to making an account here.
    Enough with the introduction, lets get to business.

    I have been trying to find to no avail the acceleration that the shuttle experiences. I know that the shuttle is designed to go up to about 3G's but what I am really looking for is the ΔV/ΔT around the area where V>=0 ( the time shortly after liftoff )
     
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  3. Mar 25, 2008 #2

    DaveC426913

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    "At some point about one minute after launch and at an altitude of about 35,000 ft (10,675 m), conditions are such that the dynamic pressure has reached 'maximum q.'"
    http://www.aerospaceweb.org/question/aerodynamics/q0025.shtml

    Nice time/velocity graph there.

    And a chart here:
    http://www.aerospaceweb.org/question/spacecraft/q0183.shtml
     
    Last edited: Mar 25, 2008
  4. Mar 25, 2008 #3
    Would it be possible to get probably the first 15 or 30 seconds in increments from about 500 to 1000 ms?
     
  5. Mar 25, 2008 #4

    DaveC426913

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    All I did was google terms compiled from your original question.

    Is that milliseconds??? Yike.

    Your best bet might be to email the right person at NASA, explain why you want those numbers and appeal to their good nature.

    But why would you need real numbers instead of just some reasonably plausible numbers?
     
    Last edited: Mar 25, 2008
  6. Mar 25, 2008 #5
    I had read somewhere, don't have where at the moment, that at lower speeds the efficiency of the rockets are much less efficient. But as the velocity of the craft approaches the exhaust velocity it become extremely efficient.
     
  7. Mar 25, 2008 #6

    FredGarvin

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    The propulsive efficiency is maximum at the condition where the vehicle velocity is exactly equal to the exhaust velocity.
     
  8. Mar 25, 2008 #7

    russ_watters

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    It is also more efficient in space...

    I read somewhere once that the max g-force was right at liftoff and is 3 g's (which would be 2 g's of acceleration + gravity). The graph Dave found shows a little less than 1 g of acceleration for the first minute, though I guess that could be because the shuttle throttles back its engines soon after it clears the tower.

    I seriously doubt NASA even records acceleration numbers in ms, as the shuttle experiences strong low frequency vibrations that would make acceleration readings over such a short time slice very unstable and largely useless.
     
  9. Mar 25, 2008 #8
    On the same lines of space rockets, anyone know where I can get the volume of exhaust gases created by oh say the solid rocket motors, or the like?
     
  10. Mar 25, 2008 #9

    D H

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    Figgy, the detailed data you want is most likely unavailable to the public due to ITAR restrictions.

    No. The rocket's acceleration is the same in all inertial reference frames. The propulsive efficiency (in terms of force) is maximum in vacuum. In terms of acceleration, the propulsive efficiency reaches maximum right before the vehicle runs out of fuel. A simple model of thrust from a rocket is (see http://exploration.grc.nasa.gov/education/rocket/thrsteq.html)

    [tex]F = \dot m \, v_e + (p_e -p_0) A_e[/tex]

    All other things being equal, the thrust (see above) will be at maximum when the ambient pressure ([itex]p_0[/itex]) is zero.

    Maximum acceleration occurs shortly after Max Q. The increases thrust after Max Q. You can see this in the graph Dave found. During the first 60 seconds the Shuttle accelerates to 1400 ft/sec, and adds another 2600 ft/sec in the next 60 seconds. The Shuttle throttles up to 104% thrust just after Max Q. After this, Shuttle throttles down to keep acceleration at a constant. If it instead maintained a constant force, the acceleration would grow as the fuel becomes depleted.
     
  11. Mar 26, 2008 #10

    FredGarvin

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    I didn't say anything about maximum thrust. I said that max propulsive efficiency is a maximum when the exhaust jet velocity is equal to the vehicle velocity. It's a basic definition.

    [tex]\eta_p=\frac{Tu}{Tu+\frac{\overdot{w}}{2 g_o}(c-u)^2}[/tex]

    It has to be at that point because that means that all of the kinetic energy is used, i.e. there is no residual kinetic energy in the exhaust stream.

    ref: Sutton, Rocket Propulsion Elements.
    http://en.wikipedia.org/wiki/Image:PropulsiveEfficiency.GIF
     
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