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I Resistance drag

  1. Feb 25, 2017 #1
    Hi
    This example of KEnergy and resistance seems incorrect.
    Sabot. An artillery projectile with narrow core and wide plastic rings to give energy to smaller drag .
    .. "This, in combination with the sub-projectiles’ higher sectional density, gives the resulting sub-projectile vastly reduced aerodynamic drag in comparison to the APCR. Both the higher initial velocity and the reduced drag result in high velocity at impact."

    For concrete "The penetration depth increases almost linearly with impact velocity in that range . "

    Ian V Hogg - 2013 -
    "This was the Röchling anti-concrete shell (Rö-Geschoss), ... passing through 3m of earth cover, 36m of concrete, a layer of broken stone, gun casement and 5m into the earth .."
    ----------
    Resistance drag in air in given conditions is Area. Velocity sq. A modern sabot is about 5 /18 for core / calibre, core A is 7.5%. As KEnergy is ~ .5 M Vsq then the Area being the size of Mass the velocity sq is the inverse at 1300% . So the sq`/ 13 gives 3.6 times increased Velocity matching the reduced Area giving same drag.

    Artillery shells penetrate around .6 m into concrete with solid resistance as ~ *. Mass . Velocity. Area. Penetration rises by about .6 with triple the velocity so raising velocity by 3.6 should double penetration to 1m.
    7.5% Area should give 13m. penetration but the advantage of 7.5% Mass in the gun is the same disadvantage in concrete .
    How then did the Germans penetrate 36m concrete + 8m earth?
     
    Last edited: Feb 25, 2017
  2. jcsd
  3. Feb 25, 2017 #2

    boneh3ad

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    I am going to be honest, I don't understand what you are asking. It sure seems like your question has nothing to do with drag, and I am having a really hard time deciphering the rest of it.
     
  4. Feb 25, 2017 #3
    As I don't have expertise in this , I gave all the info connected with it. The sabot velocity of 5500 ft /sec shows the increase over normal artillery at 2500-3500 ft/ sec. The KE and drag equations are similar ,~ M.Vsq / ~A.Vsq. suggesting that Velocity is the inverse and can be derived from the sabot proportions. Concrete resistance is the issue . I don't know how Mass relates to penetration of concrete , maybe it's a hemispherical increase. But cutting 36m of concrete seems false. Meteorites make craters rather than tubes. So I'm asking if you agree it's false.
     
  5. Feb 25, 2017 #4

    mfb

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    For high-velocity objects, a good approximation for their penetration depth is given by comparing their area densities: An object will penetrate as deep as its own length in a material of the same density, through twice that if it has twice the density, and so on. A formula from Newton.

    At 2.5 g/cm3 for concrete, an uranium projectile (19 g/cm3) would need a length of 4.7 meters to penetrate 36 meters of rock. That sounds implausible, even for things like a shaped charge.
     
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