Physics behind a cannon

  • Thread starter Decker
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  • #1
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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
gravity

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

Decker-
 

Answers and Replies

  • #2
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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.
 
  • #3
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Trajectories are parabolic, not sin waves.
 
  • #4
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Gaaa big words, im sorry but im already a lil lost
and yes i think i forgot wind resistance
 
  • #5
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whozum said:
Trajectories are parabolic, not sin waves.

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.
 
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  • #6
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Decker said:
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
gravity

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
 
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  • #7
Pengwuino
Gold Member
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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.
 
  • #8
pervect
Staff Emeritus
Science Advisor
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Decker said:
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
gravity

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

Decker-


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.
 
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  • #9
SGT
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]
where
[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.
 
  • #10
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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.
 
  • #11
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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.
 
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  • #12
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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.
 

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  • #13
SGT
Decker said:
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.
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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