It’s quite uncommon for spin-stabilized bullets and shells to be 10 times longer or more than they are wide.
I remember reading that dreadnoughts and early battleships increase their caliber as a way of increasing their range. It was specifically the range which was aimed to be increased. I understand why one would want to increase the length of the projectile: with the same volumetric density, increasing the length increases the frontal area density which means that the projectile will not slow down as quickly. By increasing the diameter as well as the length, the increase in projectile weight and recoil is much greater than its range increase.
The chainguns used in navy CIWS also increase the projectiles across all 3 dimensions as a way of increasing the range. One would think that increasing the length of the projectile and the weight in proportion to the length would be sufficient to increase range.
So, it seems that all 3 dimensions tend to be increased when designers want to increase range although this comes at significant weight and recoil cost. Either that’s consistently bad design or there is a reason that having a projectile 10 times longer than it’s wide is a problem.
I note that fin-stabilized projectiles do not exhibit the same phenomenon and are commonly much longer than they are wide.
I will be following this as I work with flight arrows, we shoot strictly for maximum distance. Our thoughts have always been smallest diameter overall with widest at the front. I am thinking for projectiles it might have something to do with the propellant exposure surface area.
Maybe its the lack of any point to making artillery shells in the shape of a very large arrow ?
Also the arrow shape has larger cross-wind profile , for the same mass. - drifts off course more… rotation doesn’t stop drift.
Doesn’t penetrate better, doesn’t make a bigger bang at the end.
and yeah high acceleration means high forces… well to get the same force from the smaller area, the pressure would be higher… So the shell’s tail would have to be made to withstand higher forces which means sacrificing something at the front…
I know why you would want the heaviest at the front; because the center of gravity should be in front of the center of pressure. But why the widest at the front? Is it because it’s the simplest way to put weight at the front? Couldn’t a lead/iron ball at the front combined with light fins at the back be sufficient to stabilize the arrow?
FYI - Caliber refers to both the bore diameter of the gun barrel (or the diameter of the projectile) and to the length of the gun barrel in multiples of the bore diameter. For example, U.S. battleships used the USN 16"/50 Mark 7 naval gun. It had a bore of 16 inches and a barrel length of 50 calibers (16" x 50) or 800 inches. The 16"/45 caliber Mark 6 gun had a bore of 16" and a barrel length of (45 x 16) 720 inches. Assuming that the powder charge and projectile weight are the same, the longer 50 caliber length will allow gas pressure to push the projectile for a longer period of time than a 45 caliber length barrel will, resulting in higher muzzle velocity. Higher muzzle velocity means increased range and also higher impact velocities at the same distances.
Here, I am using “caliber” to refer to he diameter of the projectile. Nothing in my post refers to barrel length and my question refers strictly to the relative sizes of the round across its 3 dimensions.
My w.a.g. is that making a projectile too long and skinny makes it too susceptible to tumbling; the slightest perturbation would cause the nose of the projectile to wobble wildly, eliminating the advantage of spinning. A lot of modern projectiles that are long and skinny use stabilizing fins, but that introduces a trade-off of extra drag.
Sure seems like this is the answer. Note that kinetic energy penetrators (like those used in anti-tank rounds) have a very high aspect ratios, but use a sabot to increase the effective bore. They also allow the use of stabilizing fins to avoid the problems associated with rifling.
Another way of looking at it is that using a sabot allows using a relatively light (and therefore fast) round in a barrel of a given size. If the round were the full diameter, it would either have poor aerodynamic performance (short and stubby to stay light), or be slow (too much mass for the given barrel length to handle).
This is not true. Many different shapes of arrows have been used to set flight records, non with an arrow that was “widest at the front”, but I’m sure this mixup is more about semantics than anything else. Typically, a flight arrow has the largest diameter somewhere along the length of the arrow shaft, with both ends smaller in diameter, a type called a barreled shaft. Some of the shots made in the 1930’s and '40’s, still unbeaten in their analogous classes, were made with arrows that have the maximum diameter quite close to the rear end of the arrow. But even cylindrical arrow shafts have been used to set records, occasionally.
Your correct, I should have said lately our thoughts have been toward wider in the front half. The tear drop stretched out into an arrow is my goal and several others. This will throw the weight preferably front of center, even though some records were set with rear of center ballance. Too much taper in the front they fly horrible. But thats another thread, this thread is bullets.
My post was an FYI that addressed the term caliber in more detail. You were talking about increased range. Increasing the caliber (length) of the barrel increases muzzle velocity which results in increased range. I apologize if that was confusing.
You also stated, “I note that fin-stabilized projectiles do not exhibit the same phenomenon and are commonly much longer than they are wide”. I’m not aware of any firearm projectile that is not longer than they are wide. It’s the nature of the beast.
FYI - There is a balancing act between what can be shot out of a barrel and what will result in either a catastrophic failure of the breach and barrel or result in the projectile being stuck in the barrel. The longer a bullet is, the more surface area it has. Increasing the surface area results in more friction between the bullets surface area and the barrels rifling or bore. It takes more gas pressure to push a longer bullet down the barrel. The propellant (gun powder) is selected based on it’s ability to rapidly reach a pressure that will start a bullet down a barrel and push it all of the way thru but not so quickly that the gas pressure exceeds the strength of barrel before the bullet exits the muzzle.
It depends. Short of actually going onto the Abrams assembly line in Lima, Ohio, and measuring it (and if I could, I still wouldn’t be able to say anything here), we public-source people have to go by the figures that are released. Wiki lists the various Abrams tank models at having anywhere from 450 to 1300mm RHA equivalent thickness, depending on what’s being shot at it.
What does that mean? In theory, the armor composition and configuration is such that it would take a sheet of Rolled Homogeneous Armor that thick to duplicate its resistant to being penetrated. The current USA RHA standard is MIL-DTL-12560J, and the public spec .pdf document is [here](file:///C:/Users/pots/Downloads/MIL-DTL-12560J.pdf). One of the manufacturers of it, with a bit more info on its physical characteristics. Here’s an interesting World of Tanks discussion thread, with photos allegedly of M1s and Challenger tanks suffering battle damage. In one of the photos, the width of the turret ring is easily visible.
Length of a spin-stabilized projectile is almost always less than 5x its diameter. The larger the length/diameter ratio, the faster it needs to spin to remain stable. Also, faster spin amplifies the effect of any small asymmetry (which is unavoidable in mass produced bullets).
Very little energy goes toward overcoming the skin friction. If the friction is too much, the surface of the bullet may partially melt, but the bullet will get out of the barrel.