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Civilian Space Launch Imminent

rossjudson writes: "Looks like the Civilian Space Exploration Team has gotten clearance from the FAA and the Bureau of Land Management to attempt the first amateur flight into space. That's pretty cool. Maybe one of you space-mathematics types out there can educate us on just what 0-Mach 5 in 15 seconds really means! Is this thing gonna just blow up?"

4 of 53 comments (clear)

  1. X-Prize? (or was that something else?) by dschuetz · · Score: 4, Interesting

    Wasn't there a prize for the first team to loft 1 kg to 100 km? Would this qualify?

    (no, it wasn't X-prize, that was for bringing people into space and re-launching in, what, a week?)

    Been a while.

    I'm disappointed, though, that they're being asked to keep the exact launch date a secret. If they can't protect a rocket in the middle of the Nevada desert from terrorists (or tourists), then we're all screwed.

  2. Where is going to land? by ManDude · · Score: 2, Interesting

    I wonder what their insurance is like? They have the rocket landing somewhere near Quebec City. I don't think Quebec knows or ready. This could be more interesting then I thought.

  3. Speed of sound w/ altitude... by paploo · · Score: 2, Interesting
    Don't forget that the speed of sound is a function wich, among other things, is a function of ALTITUDE.

    Now, the speed of sound at sea level is around 340m/s, which gives us around 11.7g's of acceleration, and an altitude of 42,000 feet in 15 seconds.

    However, the speed of sound at that altitude is significantly slower, around 290m/s-300m/s (according to my information), so our numbers for the acceleration should be a bit HIGH.

    Of course, without a nice function for me to integrate against, it is a little trickier to figure out what the acceleration really is, but we can put some bounds on it. As determined, the upper limit is 11.7g's of acceleration. The lower limit will be around 10g's. (This is found by using the slowest speed of sound number that we could possibly run across).

    So, with that range in mind, we can see that a reasonable estimate for the acceleration would be about 11g's +/- 1g.

    There, I feel more rigorous. I feel better now. :) (This is what I get for being a physicist.)

    -Jeff

  4. Re:Here's the numbers... by Louis_Wu · · Score: 4, Interesting
    I'm heading under the bridge, hold this rope and pull me out if the troll gets his hands on me.

    :)

    Part of the reason for using Mach as a reference is that it is so flexible. Many aspects of fluid mechanics depend on knowing if the fluid flow is supersonic or sub-sonic. The behavior of subsonic fluid is fairly familiar to most people, but that changes radically when the flow becomes supersonic. Knowing the flow speed in relation to the shock wave speed (the speed of sound in that fluid) tells you which equations to use, and what to expect. Handy.

    The problem we have here is that we don't care about the fluid mechanics. All we care about is the speed in relation to the ground, and we can't get that easily because the reporter & the PR flack thought that spouting off Mach numbers sounded cooler or more scientific. (My idea of a scientific speed reference in this case would be two-fold: the raw speed number and the velocity vector broken down into components: vertical, North/South, & East/West. But I'm a mechanical engineer, and I want useful information.) We can get a good idea about the bounds of the acceleration involved by using speeds for Mach 1 at various altitudes (10 - 11 gravities as posted already), but a more accurate calc would account for the variation in Mach number with altitude.

    Of course, for a trully rigourous theoretical treatment, what we really want is the mass of the craft, the mass loss rate (fuel burn rate), and whatever measure of the craft's rocket power we can get (it might be the force the rocket produces [which could be a function of time], it might be the power of the rocket [which I think I could translate to a force if I had a few books in front of me], it might be the mean velocity of the rocket's exhaust stream [which I know that I could translate into a force, if I had the proper information about that flow stream]).

    With that {potential} boatload of information, we could apply the modern incarnation of Newton's Second Law, F=d(p)/dt ; force equals the derivative of momentum (p) with respect to time (t). That ends up being F=ma + v*d(m)/dt , the first part being very familiar to anyone who's ever taken physics, the second part much less so. F=ma : force (F) equals mass (m) times acceleration (a). The second term isn't very familiar, because most people don't think of mass changing over time. The classic example of this is a rocket - the topic of today's lecture. {I have no idea why I'm saying so much. Work must be more boring than normal.} F=v*d(m)/dt : force (F) equals velocity (v) times the change in mass (m) over time (t).

    [BTW, all of the "d"s are NOT variables, they are part of the notation of derivatives in calculus. (I was falling asleep during a lecture once, and I wondered why the teacher didn't cancel the extra "d"s from the top and bottom of the equation; then I woke up and almost died laughing at myself.:) ]

    Anyway, the upshot of the math is that if we knew how fast that fuel was being burned [ d(m)/dt ] how fast the craft was traveling at any time (v) and what the mass of the craft was at any time (m), we could back-calculate to get the acceleration. All that work to find out how heavy you'd feel. :)