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I Cannonball - time required to stop its horizontal motion

  1. Jun 24, 2017 #1
    Let's imagine that we fire a cannon located on a platform high above the ground and we want to compute the time after which the horizontal velocity of the cannonball will be zero. Let's assume that we shoot at 0 degree angle.

    1. Air resistance force
    [tex]F = C_d \frac{\rho v^2}{2}S[/tex]

    2. Assuming that the force from point 1 is constant (terrible assumption but I hope it does not affect the answer too much) we have
    [tex]F = \frac{mv}{t}[/tex]
    [tex]C_d \rho v^2 \pi R^2=\frac{mv}{t}[/tex] assuming [tex]S=2\pi R^2[/tex]
    [tex]t = \frac{m}{C_d \rho v \pi R^2}[/tex]

    Now, let's make up some rational parameters
    initial speed: 800 m/s
    drag coefficient: 0.47
    mass of the cannonball: 300 kg
    Radius: 0.5 m
    air density: 1.2 kg / m^3

    Now, as I plug in my data to the equation, I get the answer - roughly 0.8 s. I know that this answer is not rational. My question is, is this significant error caused by the assumption that the force is constant?
    Last edited: Jun 24, 2017
  2. jcsd
  3. Jun 24, 2017 #2


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    That's a good bet. You need a differential equation since the force of air resistance depends on the velocity of the object, which is changing over time.
  4. Jun 24, 2017 #3
    Could you help me to write this equation? I've never done that before. Physics is not part of my college course, I'm simply an enthusiast.
  5. Jun 24, 2017 #4


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    Not tonight, I'm getting ready for bed. I might be able to help you tomorrow though.
  6. Jun 24, 2017 #5


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    I seriously think that, if you want the equation to be at all meaningful to you, you need to have some basic knowledge of the Maths involved. It is not much use diving in to a subject like this without working upwards from scratch. Yes, you can get a 'feel' for many of these things in Physics but that's as far as you can hope to go without the Maths.
  7. Jun 24, 2017 #6
    It seems the velocity will be of the form
    where ##t_{0}## is probably something like the ##t## you found. In any case, its horizontal motion will never stop.
  8. Jun 24, 2017 #7


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    Well, I come up with two possible equations.

    1. Assuming the force scales with the square of the velocity: ##V(t)=(0.0014758t+0.00125)^{-1}##

    2. Assuming the force scales proportionally with velocity (##v## instead of ##v^2##): ##V(t)=800e^{-0.0014758t}##

    Looking at the graphs of both equations, equation 2 looks to be a bit more realistic. Equation 1 results in the cannonball losing more than half of its velocity in one second, whereas equation 2 takes 527 seconds to drop to the same velocity. That 2nd equation is probably off by a significant amount, but equation 1 is entirely wrong for some reason. Not sure if I've solved my differential equation incorrectly or if we've used the wrong initial equations.
  9. Jun 24, 2017 #8
    Sounds OK to me. Even if by "radius" OP means "diameter" the ball is still pretty light. BTW should not the area be ##\pi R^2## rather than ##2\pi R^2##?
  10. Jun 24, 2017 #9


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    Taking 0.5 meters to be the diameter instead of the radius makes the equation seem much more realistic. A cannonball with a radius of 0.5 meters and a mass of 300kg has a density of 577 kg/m3. But if the diameter is 0.5 meters the density increases to 4,615 kg/m3, almost 9 times as much. The new equation becomes: ##V(t)=(0.000184t+0.00125)^{-1}##

    The largest naval artillery ever fielded by the U.S. was the 16"/50 caliber Mark 7 guns on the Iowa-class battleships. These guns fired shells 16 inches in diameter (0.406 meters) with a mass of up to 1,225 kg at 820 m/s. With a diameter of 0.406 m and a length of 1.829 m the density of these shells was approximately 5,000 kg/m3. However, the mass vs cross-sectional-area of these shells was much higher, 9400kg/m2 vs 1500kg/m2 for the cannonball of diameter 0.5 m.

    The equation for this shell: ##V(t)=(2.67*10^{-5}t+0.00122)^{-1}##

    Graphing these equations shows that it takes 8 seconds for the cannonball to lose half its velocity compared to 0.847 seconds for the original equation.
    However, the shell takes 45 seconds to lose half of its velocity and 136 seconds to reach a quarter, which seem like very reasonable numbers considering that they had a maximum flight time of about 90 seconds when fired at maximum range.
    Last edited: Jun 24, 2017
  11. Jun 24, 2017 #10
    What is the source of these numbers, your equation or actual measurement? Don't forget that artillery shells are not balls. Their drag coefficient is probably much smaller than that of a sphere.
  12. Jun 24, 2017 #11


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    Oh, @ChessEnthusiast, here's your equation with just the variables: ##V=(\frac{C_dρS}{2m}t+c)^{-1}##
    To find little ##c##, set ##V## to the initial velocity, ##t## to zero, and then solve for ##c##. So ##V_0=\frac{1}{c}##
    Then just plug in your value for ##c## into the equation.

    I used the drag coefficient for a half-sphere, which is 0.42. I looked again before making this post, but had a difficult time quickly finding a drag coefficient for an artillery shell. The best I could do was use the ~0.23 value from section 2.4 of this article. That certainly reduces the drag on the projectile by a good amount.
    The rest of the values I found on wikipedia or a few sites on naval guns. I calculated the approximate density myself using these values.
  13. Jun 25, 2017 #12


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    Using a drag coefficient of 0.23 as per my link in the previous post, along with a projectile velocity of 762 m/s, the new equation is: ##V(t)=(0.0000146x+\frac{1}{762})^{-1}##

    Looking at some information on this naval weapons website, the striking velocity of a shell launched at a target at 10,000 yards is 632 m/s and the flight time is 13.2 seconds (look under the "Range" and "Armor Piercing with AP Mark 8" sections). Using our equation, the predicted velocity after 13.2 seconds of flight is 664 m/s. That's within 5% of the real value, so I'd call our little exercise in using the drag equation a resounding success any day.
  14. Jun 25, 2017 #13
    Let's say that assuming a constant drag is very erroneous, then we come in with calculus.
    Now, from the equation
    F=½CρAv² —(i)
    F=(mv–mu)/t —(ii)
    To an infinitesimal point eqn (ii) becomes F=m(δv/δt) and eqn(i) becomes δF/δv=CρAv ∴ δF=CρAvδv, say eqn(iii)
    From eqn(ii) δv=Fδt/m, say eqn(iv)
    Put eqn(iv) in eqn(iii), eqn(iii) becomes δF=CρAv(Fδt/m).
    Now F=∫ FCρAvδt/m with upper and lower limit being tfinal and tinitial respectively
    F=[(FfinalCρAvfinaltfinal)/m]+k] –[(FinitialCρAvinitialtinitial)/m]+k] say eqn(v). But for initial velocity in this case, tinitial=0
    m=CρAvt ∴ t=m/(CρAv), luckily same as you wrote. But note that the v is final velocity, not initial. If its initial that implies that the faster its start the quicker it stops. That's the error in your analysis which caused a very wrong value, and not because drag force reduces.
    It makes sense, since it says the ball will never stop(by replacing zero as v says time is infinite). If the velocity reduces the force reduces "squarely". As it is reducing the force is not sufficient to bring it to rest since it also reduces, hence continues forever.
    However, this may be untrue because if its get slow enough it instead obeys stokes law which is F=6πrηv.

    Some of the parameters are still erroneous.
    0.5m in radius? that's more that 12 inches as radius only ie diameter is 1m that's too large. 300 kg is thrice the human weight, are you saying 6 people needs to load the cannon? (Average human carries half his weight). And the cross-sectional area is πr² not 2πr².

    Using your parameters, the eqn only that for the ball to get to 1m/s it will take about 677 seconds, which make sense. The only problem to this eqn I noticed is that it does not depend on initial velocity. Nevertheless there seem to be a remedy, we could fine the velocity after some short time rather than the instantaneous initial velocity where t tends to 0. After which we compute the drag force for that velocity, and insert it in eqn(v).

    With brotherly love, let others comment on this method and eqn to be certain of its credibility. Thank you
    Zaya Bell
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