1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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:

  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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor

    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


    User Avatar

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook