Originally Posted by
Hjl
Ill give it a try
Force = Perimeter of impact area x Thickness of material x material shear strength
T = 0.530m
S= 10~11MPa based on properly prepared white oak timber
Now to work out P
Diameter of a 12lb cannon ball is ~4.52" = 0.115 meters
The balls are spherical though and so there will not be a flat area hitting the hull. To account for this i will have to make a guesstimate. Lets say that the impact diameter is closer to 1" , 2.54cm, 0.0254m.
The perimeter of the shearing area will be pi*D = 0.0798m
So combine all of this
F=(0.0798m)(0.530m)(11Mpa)
F=465.2 KN
This is the minimum force required to break the hull on the USS Constitution.
I made a program a few weeks ago to determine final speed of a ball moving through the air given drag,
so with a 300m/s muzzle velocity from a 12lb carronade and a distance of 450m the impact speed should be ~ 58.6m/s
Force of impact
Mc = 3.78kg = Mass of cannon ball
Wc = 37.07 N = Weight of cannon ball
d = 0.02m = distance of deformation of wood due to deceleration of cannon ball
Rate of deceleration
a= v^2/2d
a= 85849 m/s2
Force of impact
F=W.a/g
F= 324.4 KN
Fi<Fw
324.4 KN < 465.2 KN
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So based on these back of the envelope calculations, it shouldn't be possible for a cannon ball fired from a 12lb carronade to penetrate the armor of the USS Constitution at 450m
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If we did the long gun the only difference would be the initial velocity of the shot, which should be around 440 m/s, impact velocity will be around 86m/s
This will give an impact Force of around 698KN which will be enough to punch through old iron sides.
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