Physics question

Discussion in 'General Slingshot Discussions' started by coyote-1, Nov 6, 2013.

  1. coyote-1

    coyote-1 Member

    I'm imagining some folks here have to have researched this: Wouldn't the momentum of a slingshot projectile come almost entirely from the initial moment of release? In other words, a shortened band stretched to its max should provide exactly the same speed leaving the sling as a longer band at equal tension - after that first release moment, the rubber instantly loses the power it was providing.

    To wit, a 10" rubber pulled back a foot might provide 100fps. Pull it back two feet, and it might provide 140fps. So when released from a two-foot pull, as it passes through the one-foot pull zone it cannot provide any more momentum than at release because, at that point, it is providing less momentum than the projectile already has.

    So the objective has to be to get maximum stretch from a short segment of rubber. Longer stretch, if not providing more tension, won't give any more speed.

    I'm thinking of this in terms of the various slingshot Xbows and rifles etc. The length of draw could certainly increase your accuracy, just as it does with rifles vs pistols. But unlike guns, where the burning powder keeps increasing pressure (and therefore projectile speed) as the bullet travels the length of the barrel, with slingshots the bands LOSE power long before the projectile has passed the 'muzzle'. So I was wondering why Jeorg's sling xbows do not have longer 'barrels'.... now I get it.
  2. dolomite

    dolomite Banned

    there's way more to it than that. you can make the bands wider and stretch them less and get the same results as full stretch with thinner bands. longer stretch with less tension does provide more speed, Joerg has shown us this.

  3. BeMahoney

    BeMahoney Builder of things

    re-post, too tired to translate- watch the youtube-vid


    v = a*t (+v0) (längere Bänder, theor. längere Beschl.-Dauer)

    .. da längere Bänder auch mehr Energie speichern können ..

    ach und: F equals m times a

    Ansonsten gilt immer:

    Viel hilft viel! Weil es aber eben NICHT einfach ist ne sehr gute Betrachtung des Themas vom Baumpabst:

    Die Mechanik der Schleuder: (Zur Frage ab 5:30min)

    Viel Spass,

  4. Flipgun

    Flipgun Well-Known Member

    Ah'm not a smaht man, but the acceleration of the load is consistent to the contraction range of the rubber regardless of its its length. When the rubber has contracted to its original length, momentum of mass takes over and the heavier load will leave the lighter pouch and bands behind If you have 2 sets of bands of the same width with the same load and pull them to the same elongation factor relative to their length, you will achieve similar velocities. I think.
  5. Moin moin,
    Experiment makes know ....
    Just try it out myself once!
    Otherwise, clips have other people already set it on youtube ... because you look around!!
    You can see the clip in slow motion shots, even beautiful, the ball when it has passed the forks, even still lies on the leather, or get a nudge that would have on your mind no longer be the case, yes ..... The fact that latex should not be so quickly, and would have its own mass, which is in the way by the higher air resistance (as the ball) and - to brake ....
    But we prove it to me, even by experiments conducted .....
    I'm waiting on it ...!

    Versuch macht wissen....
    Probiere es einfach selber mal aus!!!
    Ansonsten, haben andere Leute, schon Clips darüber bei youtube eingestellt... schaue dich da mal um!!!
    Man sieht auf den Clips bei Zeitlupen aufnahmen, sogar schön, das die Kugel, wenn sie die Gabeln passiert hat, sogar noch am Leder liegt, oder einen schups bekommt, das müsste nach deinen Gedanken ja nicht mehr der Fall sein..... Den eigentlich dürfte das Latex ja nicht mehr so schnell sein, und müsste durch den höheren Luftwiederstand (als die Kugel) und durch seine eigene Masse, die im Weg ist - mehr abgebremst werden....
    Aber beweise uns es mal, durch selbstdurchgeführte Versuche.....
    Ich warte drauf...!!!
  6. Achso_42

    Achso_42 Senior Member

    The resulting energy is: Ekin = Fdraw * s