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Affect of air resistance on a projectile

  1. Aug 1, 2013 #1
    Ok, so I have calculated an ideal speed of a projectile, but now I would like to know how it is affected by air resistance, although I'm only interested in the air resistance when it's outside the barrel, hopefully that'll simplify it a bit.

    So the formula is: FD = [itex]\frac{1}{2}[/itex]CDρAv2

    Can I find the affect on the projectile with a = F/m?

    The projectile is an 8mm rod that is 290mm long, so how can I find the coefficient of drag? Also, I'm assuming 20°C at 0m above sea level, so the density of air is about 1.2kg/m3. Is that in the correct units for the formula?

    Thanks for any help.
     
  2. jcsd
  3. Aug 1, 2013 #2

    rcgldr

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    If the rod travels at mach 0.5 or more, then the function for aerodynamic drag becomes complex and usually a table for coefficient of drag versus speed is generated and then used via interpolation to calculate drag. Related wiki article:

    http://en.wikipedia.org/wiki/External_ballistics
     
  4. Aug 1, 2013 #3
    I have read the article, however I'm not sure what to use out of the stuff there. The velocity of te projectile in this case is 46 m/s, and for any reasonable-sized projectile (>10 grams) it will not exceed 100m/s. Does this mean that it will not require the complex equations listed on that webpage?
     
  5. Aug 1, 2013 #4

    rcgldr

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    100 m/s is about mach 0.291, so you can use a constant for Cd instead of a table as shown in that wiki article. You'll need to search for the Cd of a rod or cylinder with the stated dimensions.
     
  6. Aug 1, 2013 #5

    Baluncore

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    46 m/s is very very slow.
    For a simple rod, model the front and rear as disks, plus the sides as a wetted surface. The front will be subject to wind pressure proportional to density and speed squared. The rear will have cavitation, so a constant force due to pulling a vacuum.

    Air resistance will depend on both the front and rear profiles and on the axial stability. Give the nose a spherical or elliptical section. Taper the tail to a point. Spin the projectile rod to stabilise it in flight. It should now look like an aircraft's fuselage. Roughen the transition from the nose to the shaft so as to create a lower drag boundary layer along the sides and tail. Model the front as a sphere, the sides as a wetted surface and the tail as a cone.
     
  7. Aug 1, 2013 #6
    I have tried to search for it to no avail. It mostly says that the CD is found experimentally, but does anyone know a rought approximation of the value? Even a very rough guess?

    The 46 m/s is for a 71g projectile, it travels significantly slower than my other projectiles, but I was wondering if it would be better at knocking things over. It is just a length of an insulated metal rod from and old clotheshorse, I haven't tried to shape it or anything like that, and I lack the skill or equipment to make a decent projectile anyway... Thanks for the detailed response, though.
     
  8. Aug 1, 2013 #7
    Drag coefficients for various shapes are given on Wiki
    for a sphere it is 0.47, for a half sphere 0.42, for a cone 0.5 for a flat face it is about 1
     
  9. Aug 1, 2013 #8
    Well, seeing as it is a very long cyclinder compared to the diameter, and it has a slightly rounded end, I would guess it to be somewhere between the long cylinder and the stream-lined body. It says a bullet of non-ogive shape at sub-sonic speeds is about 0.295, so that's probably pretty close.

    So if CD = ~0.3, then the formula should be as follows:

    FD = [itex]\frac{1}{2}[/itex]CDρAv2

    FD = 0.5*0.3*1.2*5x10-5*462

    FD = 0.019044 = ~0.02N

    So then does that mean that the acceleration due to friction will be:

    a = F/m

    a = 0.02/0.071 = 1.42x10-3m/s2?

    That doesn't seem like much at all, have I done something wrong?
     
    Last edited: Aug 1, 2013
  10. Aug 1, 2013 #9

    cjl

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    I would think that your coefficient of drag would be quite a bit higher than 0.3.
     
  11. Aug 1, 2013 #10

    rcgldr

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    AlphaZero caught an error in your final divide calculation in the next post (.02 / .071 = .28). Otherwise the math looks OK assuming the rod manages to always be aligned into the direction of travel. The Cd for sports cars is around .3 so a long thin rod should be similar. The speed isn't all that high at 46 m/s, and the frontal area is small 5 x 10^(-5) m^2 (.004 m radius).
     
    Last edited: Aug 1, 2013
  12. Aug 1, 2013 #11

    AlephZero

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    Check your arithmetic. I get 0.28 m/s2

    That's still small, but not ridiculously small.
     
  13. Aug 2, 2013 #12
    Yeah, I accidentally multiplied instead of dividing. So 0.28m/s2 sounds reasonable?

    What formula could I use to calculate the velocity at any given time assuming it's initial velocity is 46m/s?
     
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