Question about straight-line ballistics

  1. This article in the Economist magazine sparked a physics debate in my pub, on the topic of straight-line trajectories:

    My protagonist was arguing that any body in a gravitational field will be subject to downward forces, no matter how fast it was going horizontally. His beermat calculation was (doubling the speed of the 30k shot) 4000kph = 1,111.1 m/s and therefore a travel time of 27 seconds over 30km. With a gravitational constant of 9.81 vertical falling distance ought to be 9.81*27*27*0.5 or 3,575m. Rather a lot.

    But alas, the earth is not vacuous. So I plugged some numbers in to this terminal velocity calculator here and a 10kg mass, 0.02 cross-section, 0.3 drag co-efficient gives me around 140m/s. Lets assume 5 seconds to reach terminal velocity and decent is additive after, 140*22 still gives me over 3km vertical decent.

    So, my question is, how can our beermat estimates (and assumptions) be so at odds with the statement in the article? This thread https://www.physicsforums.com/showthread.php?t=281635 says
    - accepted, and I'm not asking for anything exact, something else is missing here.

    Could Mr Finkenaur be referring to a non-Euclidean spherical geometry straight line that meant the slug was always 20ft (say) off the ground and in effect in orbit? I'd go with that kind of 'straight-line'. But orbital velocity is way over atmospheric speeds.

    Side note: A Distance to the Horizon Calculator tells me that I'd need to be at least 18m above the ground in order to fire 30km tangentially at the earth and the target would also have to be 18m above the ground. Doesn't suggest its useful against tanks if truly Euclidean.
     
  2. jcsd
  3. mfb

    Staff: Mentor

    The projectile could have lift, depending on its orientation. At those speeds, even small effects can get significant.

    Those things don't reach terminal velocity over 30km, not even speeds close to it. I remember speed estimates as high as 3km/s. That would give 10 seconds flight time, and 500m corresponding drop. Too much for a straight line (still just 1 degree deviation), but add a bit of aerodynamics (and the curvature of earth as a smaller effect) and it can fit.
     
  4. SteamKing

    SteamKing 9,565
    Staff Emeritus
    Science Advisor
    Homework Helper

    If the author is indeed referring to non-Euclidean spherical geometry in an article for the Economist, it's probably the first time such a reference has been made in that august, albeit non-physics, publication.
     
  5. ok I buy that, it seems to be a good candidate for my missing factor...

    ..so as the slug drops as a result of gravity it we get an aerodynamic lift component via an angle of attack, so it effectively becomes a symmetrical airfoil. I was imagining the projectile was bullet-like plus fins (for spin stability and accuracy) so the forces would be balanced. Seems like a tiny force against the constancy of gravity but magnified by its speed?

    @steamking; indeed. Its true the conversation quoted in the Economist is inadequate for not qualifying the speakers geometry in his assertion about a "straight line". There is a great deal of information about the forces required to move the slug horizontally but little about the forces balancing it vertically to produce the straight-line effect. Hence my interest.
     
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