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how about this moosey...mach 5/50you have one cartridge...

  1. 2,793 Posts.
    how about this moosey...

    mach 5/50

    you have one cartridge containg...

    containing 2 projectiles + 1 inner cartridge that sits behind the 2nd projectile

    the inner cartridge contains no projectile

    the propellant load for each of the 2 projectiles is ~ equal

    (where the propellant load determines the muzzle velocity)

    let each projectile's propellant load = A

    the propellant load of the inner cartridge is twice that of of each of the projectiles, therefore = 2A

    when firing..

    before the first projectile leaves the barrel, the second projectile is fired -- and almost instantaneously after the second projectile is fired, the inner cartridge is fired..

    the backward force from the first projectile, would be maintained at a high value, due to the second projectile maintaining the space the first projectile's propellant exerts its force.. i.e. the space between the rear of the first projectile and the front of the second projectile is approximately maintained due to the forward speed of the second project, as a consequence of the backward force exerted by the propellant of the second projectile, for a short, yet critically effective space moment within the barrel...

    but this above action at this point would still produce recoil, and would have the effect of stalling the second projectile after a moment

    so...

    almost instantaneously after firing the second projectile, the inner cartridge (containing no projectile), fires with a propellant force of 2A

    this inner cartridge propellant for a very small space moment meets no virtual resistance as it expells forward, but then meets the propellant from the second projectile that is moving with an effective backward acting force (recoil)..

    as the inner cartridge forward acting propellant force is stronger (2A), and as this propellant force is twice that of the backward acting propellant force coming from the second projectile then, the inner cartridge force being stronger neutralises over a very short space moment the backward travelling propellant force (recoil) from the second projectile, until the resultant propellant force from the inner cartridge equals A, at which time, it acts as a forward force to the rear of the second projectile, thereby increasing the forward velocity/momentum/force of the second projectile to approximately equal that of the first projectile..

    the value of this doubling of the forward velocity/momentum/force of the second projectile is approximately 2A...

    since the backward acting (recoil) force of the first projectile's propellant is approximately A, then the result of the first projectile's backward acting recoil force will reduce the velocity/momentum/force of the second projectile to approximately A, therefore, the forward speed/muzzle velocity of both the first and second projectile will be approximately equal...

    critically, as there will be an moment of time when the second projectile is moving faster than the first projectile, and as there results in a reduction in the forward force of the second projectile due to the backward force of the first projectile, then result is both the first and second projectiles will be extremely close to each other when leaving the barrel...

    the consequnce of the above train of events are...

    the recoil force of the first projectile is nearly reduced to zero by the second projectile, where the recoil force from the second projectile is nearly reduced to zero by the forward force from the inner cartridge, where the latter allows, after resultant forces settle, for the first and second projectile to leave the barrel almost at the same time, and with almost the same muzzle velocity...

    in the process, overall recoil from each of the first and second projectiles is substantially delayed and substantially reduced...

    this means, where there is a substantial reduction in the backward acting recoil force, then, as the system is a nearly closed barrel, then, as energy is difficult to lose in a near closed system, then the resultant energy loss to the substantial reduction in the backward acting recoil force is given to the forward energy of the first and second projectile, AND, the substantial delay in the release of the recoil force, which is a critical time issue, is also picked up and made up by the first and second projectiles, both of which transferences of energy to the forward moving first and second projectiles, is observed in the very very high muzzle velocity of the two projectiles that exit this near closed system...



    anyway, i think it may all be explained along the above lines...


    : )

    go the mach 5/50!!


    good luck

 
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