Ok now this is a little embrassing, now I consider myself to know a bit about ww2 History from politics, economies, tactics, units and weapons employed etc. But there is one thing that I just don't know the answer too and the is what makes an 88mm an 88mm, or a 6pdr a 6pr or even a 15 inch shell? I have heard that it is calulated by the middle of the bottom of the shell and the 88mm (for example) is 88mm from the middle of the bottom of the shell to the outer edge, or in other words the radius of the shell. Now I measuered out 88mm and the is only around a total 170mm wide and that seems pretty small, so it that how it is done or did I hear wrong.
You ARE complicated! In inches or mm it's the internal diameter of the gun tube! So a 15 inch shell gun will have a 15" internal diameter, the projectile will be a tad less. A 6lb is something else, it will be the nominal weight of the projectile - in this case the metrical calibre is 57mm. Have you ever heard of Wikipedia? http://en.wikipedia.org/wiki/Calibre P.S. http://en.wikipedia.org/wiki/Ordnance_QF_6_pounder
To tell you the truth Za, I don't really understand a word of that, lol. Is there a simplified answer you can give me, or is that it already?
88mm is the diameter of the projectile. For artillery, linear measurements (e.g. 75mm, 40mm, 105mm, 76.2mm) are diamters of the projectile and has no relevance to the length of the shell or projectile. For weight measurements (17pounder, 2 pounder) the projectile's diameter is the same as the diameter of a spherical steel ball that weighs the named amount. A 6 pounder gun is about 57mm, a 25 pounder is about 4in/105mm. If a spherical ball that weighs 25lbs will fit in the breach of a gun, then that gun was classified as a 25 pounder, even if the actual projectile that the gun fired weighed much more that 25lbs. This method of determining gun size goes back the early 19th century. Calibers in hand guns is a measure of the bore of the barrel in inches. A .50 caliber gun has a bore of 1/2 inch. A .30 caliber has a bore of nearly 1/3 in and so forth. In navy parlance, calibers mean something different. It is the length of the barrel as a mulitplication of the diameter of the breach. If a gun is 16"/45 caliber gun, it means that the breach is 16" in diameter and the barrel is 45 times as long as the the diameter of the breach - 16 inches x 45 = 720 inches or 60 feet. So that tells you the gun has a diameter of 16 inches and a lenght of 60 feet. Clear as mud?
About that, lol. Seriously it is making sense, except with the Naval guns, why are they measured differently, and why do the English measure it in weight and not in milimetres?
Old habits die hard. The metric system was not an accepted from of measure when the "pounder" form of gun measure came into being and I guess the British Army was loath to changed until forced, much like the US was when changing to metric after WWII for military hardware.
I think you made a slip with the 25lb as the WWII gun was slightly less than 88mm bore diameter. Avtually the naval guns are measured the same, you must distinguish between caliber (singular) and calibers (plural) the first is bore diameter while the second is barrel length expressed as a multiple of the gun's bore size.
I have two charts showing Pounder conversions. One shows 4 inches and the other 87.6mm. I guess I'll go with the 87.6mm. Gun Caliber by Weight 2 pounder (pdr) 40 mm 6 pounder (pdr) 57 mm 17 pounder (pdr) 76.2 mm 18 pounder (pdr) 83.8 mm 25 pounder (pdr) 87.6 mm I wonder why the small variation in diameter with the 7 pound difference for the 18:25 pounders, as compared with the 17:18 pounders?
Why does the different shell sizes or weights matter, I always thought that assuming you had a high velocity you could pierce virtually anything even when using a 37mm cannon?
Kinetic energy is 1/2 m Vsquared so while V (Velocity) is the most important m (mass) also matters. I may be completely off base but I think it there is a limit to what a small caliber can do because of hardness, assuming both shell and armour are the same if the shell is thinner than the armour it's the shell that's going to break first no matter how much energy (velocity and mass) it had. This is why there is a continuing search for always harder materials for penetrators and face hardened armour may stop an AP but not a harder APC shell with the same caliber, mass and velocity.
So its a balance between both weight and velocity? Since the heavier the shell the more power required to fire the shell, which would also mean it would need a higher velocity in order for the heavier shell to pierce armour and not simply bounce of?
If I remember my geometry the volume of a sphere varies with the cube of the radius. But the formula ((bore/2) * (bore/2) * (bore/2) * 4 / 3 * PHI) / weight_in_pounds gives values that range between 13627 fot the 17lb and 17117 for the 25lb while I think it should be a constant (weight of cast iron).
No, weight and velocity gives you the energy, but you also need your shell not to shatter on impact as that would disperse that energy all over the place and for that you need either a material that's harder than the armour or a shell that's thicher than the armour. That's why unless you have a much harder material available at one point you need a bigger shell . I'm not really an expert on this but you basically have two pieces of steel hitting each other, the thicker one wins no matter which one was moving and which was standing still.
Sounds reasonable to me, but since the sum total of my physics education consisted of walking by those particular classrooms on my way to biological sciences, I have no basis for arguement.
The calibre is the diameter of the shell -the main part where it is cylindrical -not the tapered part at front or back and not the driving bands which are slightly greater. It can be in inches (ie 3 inches) or Millimetres 88 mm -and just to make it tough -sme are designated in pounds -which is the typical weight of an average round. Hence a 25 pounder round would have a projectile weighing approximately 25 pounds -but as I remember it is close to 75 mm. The reason why this designation is used is that when developed there may have already been a shell for a different gun that was called (for example) 75 mm -so to make the difference obvious the "pounder" designation was used The 105 mm recoilless rifle is called a "106mm" because there were already 105 mm shells in production for a howitzer and eventually a totally different 105 mm shell for tank armament. These rounds were NOT interchangeable... the howitzer round was way smaller than the tank round -although the shell diameter itself was correct the tank shell CASING was probably twice as big -and the recoilless shell was perforated whereas neither of the other two were. Similarly a .22 calibre bullet has a diameter of 22/100 of an inch and a .50 cal is half an inch (.5 inch) across. 9 mm is approximately .38 calibre -but there are several types of 9 mm rounds and .38 calibre rounds and they are NOT interchangeable. It can be very dangerous if incorrect ammo is inserted in any firearm. That is why the markings are -or at least should be -very clear! Jock Williams Yogi 13
Indeed. That is why, when the British fitted a highly modified 17pdr gun into the late war Comet tank they called it a 77mm gun, even though its calibre was the same as a normal 17 pdr (76.2mm), to stop any confusion about ammo, as the rounds were not interchangerable.
Here's some useful information on caliber nomenclature, along with links to other basic topics. I believe that site is maintained by a member of this forum.
