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Physics behind a cannon

  1. Jun 19, 2005 #1
    Ok now im new to the whole physics thing so thats y im here.

    Ok well I know that there is a fomula that the US army has developed to figure out the elevation in degrees that a cannon should be pointed to land a round on a specefic target, But im having a problem finding this formula. I dont know If this has been asked before and i have done searches.

    Im almost shure that the formula inculded the varbiles:
    Distance from target to cannon
    muzzle velocity
    projectile weight
    wind direction
    elevation of cannon
    elevation of target

    there may be others but thats all i can think of right now.

    Basicly once you figure out the varibles it will produce a number that is between 0deg and 90deg.

    Any help is thanked

  2. jcsd
  3. Jun 19, 2005 #2
    Whenever I imagine a cannonballs trajectory I imagine a sin(x) graph, though one must take into account air resistance and other things. If you are looking for a mathematical equation I would look into calculus.
  4. Jun 19, 2005 #3
    Trajectories are parabolic, not sin waves.
  5. Jun 19, 2005 #4
    Gaaa big words, im sorry but im already a lil lost
    and yes i think i forgot wind resistance
  6. Jun 19, 2005 #5
    Very true, I should have specified that the sin wave I imagine is restricted to [itex]0°\leq x \leq180°[/itex], which is, in essence, parabolic.
    Last edited: Jun 19, 2005
  7. Jun 19, 2005 #6
    Those are the ones you're looking for. If you have a basic mechanics book or even access to google you can find models describing freefall with air resistance.

    edit: missed the first one
    Last edited: Jun 20, 2005
  8. Jun 20, 2005 #7


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    What? you need distance to target to figure out where to fire

    You also need a whole lot more to the equation... plus its most likely not really any simple equation. You must look at aerodynamics since artillery shells come in all sorts of sizes and shapes.
  9. Jun 20, 2005 #8


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    This may be more advanced than what you are looking for, but

    http://www.evac.ou.edu/jmpbac/appl.html [Broken]

    has some information, and some references. Interestingly enough, the army apparently has a weather division to support the artillery division, by providing meteorological information. They have portable balloons (sondes) that they can launch to get on-site data.
    Last edited by a moderator: May 2, 2017
  10. Jun 20, 2005 #9


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    If the range is long enough, you must take into account the Coriolis acceleration too.
    The air resistance is highly dependent of the shape of the projectile, meanly it's cross section. For small angles of incidence, the drag provided by the air is given by:
    [tex]D = C_{D0} \rho S v^2[/tex]
    [tex]C_{D0}[/tex] is the drag coefficient, dependent of the shape of the projectile and the Mach number.
    [tex]\rho[/tex] is air density.
    [tex]S[/tex] is the cross section area.
    [tex]v[/tex] is the velocity of the projectile relative to air.
  11. Jun 20, 2005 #10
    the range is neer going to be over 400 feet at sheer most.

    But what im getting at is i find these varibles and plug them into the equation and it should produce a number wich is between 0 and 90deg.
  12. Jun 20, 2005 #11
    Because wind resistance is very complex, there is no "standard" formula.

    This formula is correct if there is no wind and the launch site is at the same elevation as the target:

    [tex] Range= \frac{v_0 ^2}{2} Sin {2 \theta)[/tex]

    v in this equation is the velocity at the moment of launch.

    This will overestimate the actual range, but for an aerodynamic object (say a spear) it should be fairly accurate.
    Last edited: Jun 20, 2005
  13. Jun 20, 2005 #12
    here is a cruddy pic to try to explain what im saying

    for the sake of simplifacation we can eliminate wind direction and wind speed.

    Attached Files:

    Last edited: Jun 20, 2005
  14. Jun 20, 2005 #13


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    Even if the air is still, your projectile is moving relative to it, so you must take into account the expression I gave you in my previous post.
    Obtaining the drag coefficient [tex]C_{D0}[/tex] requires a wind tunnel. If you are doing this for a school project I suggest you do some experiments to see how much your range differs from the simplified one given by Crosson 's formula in order to estimate the drag.
    The forces acting on your projectile are the weight Mg directed downwards and the drag D directed in the inverse direction of velocity. You can divide the drag in two components: one vertical that sums to or subtracts from the weight and one horizontal that will shorten the range.
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