A few notes on ammonition.

(Probable) Contents:

Chapter 1: Calculate Kinetic energy (Joules, footpounds). (done)
Chapter 2: Speed: meters/second, or feet/second, Energy: Joules, or foot-pounds. (done)
Chapter 3: The anatomy of cartridges, powder, primers and a few words on Calibers. (done)
Chapter 4: Listing of rather common (small-arms) ammo and associated guns.
Chapter 5: A few words on air-resistence, gravity, trajectories, and stuff.
Chapter 6: Larger guns, and large ammo, > .50
Chapter 7: Self propelled projectiles.

Chapter 1: Calculate Kinetic energy (Joules, footpounds).

In a handgun, very short after the firing pin hits the primer (or cap), centered at the bottom of the
cartridge, a few things will happen. Inside the cartridge, at the bottom, the primer uses a small explosive
substance, which is sensitive for shock or friction. This will result in a small explosion, which ignites
the main charge inside the cartridge. The main charge explodes, and immediately gasses are released resulting in a high pressure
which immediately starts to force the bullet to accelerate in the barrel.
For a simple handgun, a pressure of around 2000 bar is quite typical. Some guns have even a much higher pressure.
Ofcourse, it all depends on the cartridge, amount of charge, sort of charge, resistence in the barrel,
mass of the projectile, length of barrel.

The above, is a bit of Jip and Janneke language, but A correct analysis will follow later this note.

Now, suppose the bullet has a mass of 11 g (11 grams). And suppose immediately after leaving the barrel, it's
velocity is 400 m/s (meter per second). What can you say about it's energy?
The energy of a bullet is an important parameter, since it can tell you all about the impact it can deliver
on a target.

First, "mass" and "weight" are not the same. However, usually very clean physics is not used and
"mass" and "weight" are used interchangedly in many discussions. It's all not that exciting.
So, folks may say that this bullet "weights" 11 grams, but a Physicist will say that it has a mass of 11 grams,
or even better, it has a mass of 0.011 kg (kilograms).

In formulas, you need to use the correct "units" of the parameters, like velocity = v = 400 m/s,
and mass (weight) = m = 0.011 kg. Otherwise, it all goes wrong. So here, the units are m/s and kg.

Now, the kinetic energy E_kin of a bullet is:

E_kin = ½ x m x v²

This formula, can be derived, but that is not neccesary ! It sort of speaks for itself.

- Because, if the "mass" m is higher, then you can imagine that it's energy increases too.
- The same for velocity v. If the velocity is higher, then it's energy must have increased too.

Example 1:

In our example above, we have a bullet of 0.011kg, with a speed of 400m/s.
It's kinetic energy then must be:

E_kin = ½ x 0.011 x (400)² = ½ x 0.011 x 160000 = 880 Joule.

Yes, energy is expressed (or measured) in "Joules".

This actually is quite a lot of Energy. But, it was quite a heavy bullet (11 grams).

Note 1:

If you found the energy in Joules, and want to know how much that is in foot-pounds (ft-lbs) units, then you
should know that 1 foot-pound (ft-lbs) is about 1.355 Joules. The digits go further, but as a good
approximation, that is good enough. So in example 1, we found 880 Joules. In foot-pounds that corresponds to:

Simply always start with this equation: 1 ft-lbs = 1.355 Joules

Since that is an equation, both sides of the "=" sign may be multiplied by the same number, in this case (880/1.355), so:

(880/1.355) x 1 ft-lbs = (880/1.355) x 1.355 Joules

Note that on the right side, the 1.355 cancels out, since one is above, and the other below the division symbol.

Thus:

649 ft-lbs = 880 Joules. The same example expressed in foot-pounds then becomes (about) 649 foot-pounds.

In some countries, you are not allowed to hunt for Big Game, with less than 1000 Joules.
Ofcourse, it all depends on local regulations and laws.

So, the gun (and ammo) in this example, is not allowed for big game hunting.
(I would not mind, since I do not hunt anyway).

Some examples will be repeated in later chapters with US standard units, like "feet/second", "foot-pounds".

Example 2:

The formula for calculation of Energy of a projectile, can be used in another way too.
This time, suppose you know the Energy, and the velocity, but NOT it's mass.
Then it's quite easy to learn the mass in the following way.

Suppose we have a projectile with an Energy of 70 Joules.
Also, suppose we know that it's velocity is 190 m/s.
So what is it's mass?

Using the formula, we have:

70 = ½ x m x (190)² => 70 = ½ x m x 36100 => 70 = m x 18050 => m = 70 / 18050 = (about) 0.0038 kg.

Remember, in most countries we use the units of Joule (for energy), m/s (for velocity), and kilogram (for mass).

So, our projectile has a mass of 0.0038 kg, which is (about) 3.8 gram.

This seems not to be a firearm, since with firearms, the energy is often at least a couple of hundreds of Joule.
Or much higer, for example with a "high-power" rifle, it can even be over a few thousends of Joules.

Note 2:

If you wonder how to get "m" if you have something like this: 70 = m x 18050

Well, then take a look at 8 = 2 x 4. Then we know that 2 = 8/4.

So, if C = A x B, then we know for sure that A = C / B, and we know that B = C / A.

You can try that we some easy numbers, like 16 = 2 x 8, or 30 = 5 x 6 etc...

Chapter 2: Speed: meters/second, or feet/second, Energy: Joules, or foot-pounds.

Most countries use the "units" as defined by the "International System of Units (SI)".
Here we find units as Joule, meter, second, kilogram, and many more, and derived units like "m/s" (meter per second)
for speed. And many, many others.

Unfortunately, many Countries, like the USA, have their own Standard system (U.S. Standard Units). So here we may find
feet, yard, pound, ounces, second etc.., and derived units like "feet/second" to measure speed.
Also, some historical units are in use, especially in ballistics, like the "grain" (gr) for bullet weight (mass).

It's a tiny bit of a pain, that not the whole world uses one set of standards.
But..., it is... what it is...

There are many sites and documents, with the precise definitions. However, we stay "practical" here, and only want
to know how to go from the one system to the other. So we need a conversion table.

Some conversions can be found in the table below. Values are approximated (rounded).

US:	SI:
1 pound	0.4535 kilogram (Kg)
1 grain = 1/7000 pound = 0.000143 pound	0.00006476 Kg
1 feet	0.305 m
1 yard	0.914 m
1 ft-lbs	1.355 Joules
1 bar = 0.987 atm	same units used
1 bar = 100000 Pascal	same units used
1 psi = 6895 Pascal pound per square inch	0.689N/cm2 6894.7 N/m2

The units which has to do with pressures, can be a bit cumbersome. However, it immediately
warns us too, that one should always 100% use the correct ammo for a certain gun.
Sometimes, cartridges "look" very similar, but in detail they are not.
Fortunately, mistakes in taking the wrong ammo is very rare, but in principle, it might happen.

For example, in a (rare) highpower handgun, the peak pressure may even reach 50000 psi, which
corresponds to 3447 bar, which is very, very, high.
At the moment, the bullet is about to leave the case, the rim of the case is a tiny bit
lifted, and given the high pressure, is why always 100% correct ammo must be used.
Otherwise the strain on the gun (e.g. chamber, breech) may be too high, which may result in nasty accidents.
Receiving ammo from certified shops for that particular gun, practically eliminate any risk.

Pressures will be discussed in Chapter 3.

Let's now replay Example 1, from chapter 1, but this time in US units.

-In SI units, we had:

A bullet of 0.011kg, with a speed of 400m/s. It's kinetic energy then must be:

E_kin = ½ x 0.011 x (400)² = ½ x 0.011 x 160000 = 880 Joule.

-In US units, we can do the following:

Method 1:

Once you have the energy in Joules (and did a calculation as above, using kg, m/s), then you can simply
convert the number of Joules into ft-lbs. However, many folks accustomed to US units, do not like to start
a calculation with SI units to start with. However, suppose you have the number of Joules anyway.
Then simply use the basic equality:

1 ft-lbs = 1.355 Joules   (equation 1).

Now, if "a" and "b" are some numbers, so that "a = b", then for example "5 x a = 5 x b" or "120.7 x a = "120.7 x b" etc...
Indeed, you can simply multiply both sides of any equation with the same number.
So, on the right side of equation, I want to have 880, so I multiply it with (880/1.355),
which cancels out the 1.355, since 1.355 is then present above and below the division symbol.
Ofcourse, I need to multiply the leftside too, with that number (880/1.355):

1 ft-lbs = 1.355 Joules => (880/1.355) x 1 ft-lbs = (880/1.355) x 1.355 Joules =>

(880/1.355) x 1 ft-lbs = 880 Joules =>

880 Joules correspond to 649 ft-lbs.

Method 2:

This time, we will use the US units.

To play according to the example, I must make sure that the bullet mass and speed are expressed
in US units.

-I must first convert the 11g, or 0.011Kg, into grains:

1 grain = 0.00006476 Kg.

(0.011/0.00006476) x 1 grain = (0.011/0.00006476) x 0.00006476 Kg

So, about 170 gr corresponds to 0.011 Kg.

So, my bullet is about 170 grains.

-Now I must convert the 400m/s into feet/second:

1 feet=0.305m => (400/0.305) x 1 feet = (400/0.305) x 0.305 m =>

So, now I have the speed in ft/s, namely 1312 feet/sec = 400 meter/sec

Now that I have the correct values, I can start the math:

The correct formula to use this time is:

E_kin = ½ x Mass_{in grains} x (v_{in feet/s})² / (7000 x 32.174)

The strange factor "1/(7000 x 32.174)" is partly due to the fact that we want to use grains for mass.
Actually, there exists a few of such formula's, which are "very close" in the expression, but there
are some differences, for example if applied to small arms, or slightly more heavy arms.

So, let's use the formula:

E_kin = ½ x 170 x (1311)² / (7000 x 32.174)

= 146091285 / (7000 x 32.174) = about 648 foot-pounds.

In the former section we found 649 foot-pounds. The small deviation is due to the approximations I made.

If you would fire testrounds, say 10 or so, and measure the muzzle energy or speed, you will find small
deviations anyway, in those testresults. Don't worry too much about 1% variation (or a few %) in calculations.

Just for comparison:

-A .22lr pistol, usually sits somewhere around 250 Joule muzzle energy.
-A 9mm pistol, usually sits somewhere around 500 Joule muzzle energy.
-Rifles usually are wel above 1000 Joule energy.
-Real High-Power rifles usually go beyond 3000 Joule energy.

It's just for an impression. Many parameters (later more on that) may have a large effect on muzzle energy.

Kinetic energy is the main energy of an inert projectile (common bullet). However, some torque effects
are present too in a fast moving object. Such effects are neglected in this note.

When the object impacts a target, the composition of the bullet is very important.
Suppose the bullet is extremely tough, it may go through the target, and not loose much energy
in that process.
On the other hand, if the bullet is relatively soft (e.g. lead), or has a hollow point
or something, it may loose all energy (or most energy) in the process.
It's a bit gruesome, but some specialized bullets will completely fragment on impact
and such special rounds are certainly not available to the general public.

Chapter 3: The anatomy of cartridges, and a few words on Calibers.

3.1 Structure of some well-known types of cartridges:

Sorry for the somewhat silly Jip and Janneke figure below.
You can find detailed images on the internet, but I can't use them. That's why my own,
rather "sad", Jip and Janneke figure.
Indeed, the figure below depicts 4 types of cartridges. Note that these are all "center fire",
meaning that the primer, is centered in the middle of the bottom of the cartridge.
However, One the the most popular rounds, .22lr, uses rimfire.



Anyway, some characteristics can be seen in that figure. For example, look at the pistol cartridge.
The "case" is clearly shown. So, for a common 9mm round, we often can see the designation "9 x 19",
the 19mm is the case length. The are a few rare exeptions in case length (historically),
but 19mm is very common (known as parabellum, Nato, Luger, para).

Note the third picture, which is supposed to illustrate a rifle round. Not all rifles use
such a bottleshape-like cartridge, but many do (or most do).
Make sure that you see that there are multiple diameters here. Like the "neck", and the "base",
and the slope of the "shoulder". There exists significant "ratio's" in those dimensions,
between the types of rounds.

Most pistol rounds, and certainly most rifle rounds, are slightly "tapered" (slightly broader at the base),
in order for easier (automatic) extraction of the empty case.
However, many rounds are straight if (automatic) extraction of the empty case is no feature of the gun,
or they are simply just straight by design.

For more realistic designs, take a look at the bullet database
which shows you close to 180 different cartridges and their properties.

3.2 Main charge: smokeless powder:

The "main charge" inside a cartridge, is not a socalled "high explosive", with a detonation
that characterizes such a high explosive. It's more a fast form of deflagration (fast burning).
This charge is often called "smokeless powder", but manufacturers most often call them "propellants".
Most types contain nitroclycerine / nitrocellulose, but often with mixtures added, and the total composition
does not give rise to a detonation. Some additions even are added to reduce wear on barrel, chambers etc..

It is important to know, that "nitrocellulose" can be manufactured with different "degrees" (amounts) of
nitro-groups. A very high degree, in pure form, might be called a High explosive.
This is why smokeless powder is a mixture of several compounds, next to nitroclycerine / nitrocellulose.
Many explosives contain molecular nitro groups.

This is all quite different from the older "black powder" charges, since that is mainly sulfur, charcoal,
and potassium nitrate.

Note: For older or historical guns, small marks on the gun, could show that it is also nitro capable.
If such a stamp was absent, it's probably for "black powder" only.
Ofcourse, really old guns, who "lived" in the black powder age only ("smokeless powder" was not available yet,
or not mainstream yet) do not have such markings. By the way, many historical guns often have several markings,
for several reasons.

Do not jump out of your socks, but there are well over 900 forms/substances (some only differ slightly)
of propellants around. The manufacturers give their range of powders names like for example "WC860"
and then in their documentation you can see the physical shape of their small balls, or flattened balls,
or small flocks, small cylinders etc..

These small flocks or balls ot cylinders, are often called "granules". It is generally so, that the smaller the granules are,
the faster the burning rate is.
Ofcourse, faster-burning propellants generate higher temperatures, and higher pressures, which ofcourse makes
it important to carefully select the amount of propellant, and purpose of the cartridge, and type of guns
which can use it.

Folks who reload cartridges should never "experiment" with the casing and powder grains.

If I just look up a few samples, often it shows to have as contents "Nitroglycerin, Diphenylamine, Dibutyl phthalate,
Ethyl centralite, 2-nitrodiphenylamine".
For some types of powder, additional components are added. And sometimes some components are absent.
Indeed, there is a large variety over the different types.

3.3 Primers:

Since the explosive substances of primers are interesting enough for a more thorough discussion,
please see the document simple overview conventional explosives
where I will place it shortly. The theory on Primers should be adequately described in such dedicated note.

3.4 Calibers:

Calibers Section I:

Europe and other countries, use the millimeter to measure (or express) the diameter of a bullet,
while folks in the USA (and other countries) use the inch to express the diameter of a bullet.
Yes, but does that represent the "caliber"...?

Formally, the caliber is actually defined as "internal diameter of the gun's barrel".
Or even better formulated: Caliber is the diameter of the bore at the "lands".
So, it's not really the bullet's diameter.

Intestingly, note that the upper formulation makes the Caliber a property of the Gun,
or barrel of that gun. Nowhere the bullet is mentioned, in that definition.

Is it entirely correct? No, it has some sort of "dual" character. Take for example the .30-06 cartridge.

Actually, the term ".30-06" represent the whole of the specifications of the cartridge, like
case length, all of the various diameters (neck, base etc..), allowable pressures, tolerances etc.. etc...
So, both the barrel diameter, and in general, the properties of the cartridge, must be taken into account.
Usually, folks look at the bullet's diameter.

Calibers can be a bit confusing, especially comparing rounds in inches and mm's.

Barrels might be "smooth" or "rifled". Often, smooth bore is often used with shot (shotshells), while
"rifled" barrels are most often used with guns that shoot bullets or slugs.

Rifling are the helical grooves inside a barrel (grooves and ridges/lands), with the purpose to impart
a spin to a projectile so that the final angular momentum makes sure that the orientation of the bullet in space,
after leaving the barrel, stays constant (as much as possible).
It's the same principle of a moving wheel of a bike, gyroscope etc..

Ofcourse, bullet (ammo) and Gun must match. Manufactures will make sure it matches. Certified supply (quality gunstores)
makes sure everyting will be compatible (gun and possible ammo).

There exists several mathematical formula's, which relate the Twist rate, to parameters as the bullet's
length and diameter. For this note, I do not consider that to be very relevant.

Calibers Section II:

Back to inches and mm's:

Fortunately: 1 inch = 2.54 cm = 25.4 mm (exactly).

-So, if we have .32, it means .32 of an inch, or 0.32 x 1 inch = 0.32 x 2.54 = 8.128 millimeters.

-So, if we have .36, it means .36 of an inch, or 0.36 x 1 inch = 0.36 x 2.54 = 9.144 millimeters.

-So, if we have .38, it means .38 of an inch, or 0.38 x 1 inch = 0.38 x 2.54 = 9.652 millimeters.

Yes, but something is not right. You probably know that, for example, .32 (for pistol) is called a 7.65 mm round,
in European language.
But, .32 and 7.65 mm do not match. 8.128 is not equal 7.65. What is this?

If you look at a .32 ACP round (or .32 auto, .32 Browning auto etc..), then:

Bullet diam.	.313 inch	7.94 mm
Neck diam.	.3365 inch	8.55 mm
Base diam.	.337 inch	8.6 mm
Case length.	.680 inch	17.3 mm

One thing is for sure: 32 caliber ammunition, is fit for firearms with a bore diameter of 0.32 inches = 8.1 mm (8.128).
So, if someone calls it 7.65mm, it's fine, but the above will hold.

Also, do not forget, that the bullet must be forced along the grooves and ridges in the barrel (rifling),
and the high-pressure will, although very little, expand the diameter bullet as well.

Not withstanding all those large calibers, .32 CF is still reasonably popular today, notably for compact guns.

By the way, you notice the small difference in diameter with the base- and neck diameters (bottom/top of case)?

There are quite a few rounds, called .32, with small deviations.
7.65 mm Browning or Auto, is the European "label" for the .32 ACP or .32 Auto cartridge.

=>You will see such an effect, with many calibers in inches (USA), compared in mm's (European).

=>You simply have to be careful, in comparing rounds in inches (USA) and rounds in mm (e.g. European).

=>It is indeed so, that the diameter of the bore at the "lands", best defines the caliber, and that
the diameter of a projectile is less informative.


-As another example, 9 mm Short and the .380 ACP are compatible. But not with 9mm x 19mm Parabellum.
In such cases, the lenght of the "case", now is the primary difference.
-It can also be found, that there exists small differences in the ejector gap, at the bottom of the round.

A compatibility matrix of rounds, would be a good thing. Just a simple sheet with rows and columns,
would be good enough. Or better: a website with inputfields (maybe gun and/or ammo), and which
outputs compatible rounds. However, all in all there are so many (slightly) different rounds
and filearms, that would be daunting task.

Quickly something about one obsolete caliber and a bit of fun.
At the date around 1850, plus or minus several decades, percussion guns (e.g. revolvers), were mainstream.
One famous caliber was ofcourse the .36. Many guns used it, like for example the Colt Navy 1851 or 1861.
That caliber is not around any more. Ofcourse, a collector can still find such objects.
And ofcourse, the caps, blackpowder, and balls can be purchased, fortunately.

A Colt Navy 1851 or 1861, original, in very good condition (some minor pitting, minor scratches, will always be present),
in Europe, will cost easily 3500,- euro's. An original in mint condition (I mean flawless), then there is
probably no pricetag possible. I took this example, because it's a fairly well-known gun.
I know, Uberti produces quality replica's of famous guns. But..., there is "something" with those originals...
By the way, .36 ammo as cartridges, actually used a slug that had a 0.357-inch diameter,
(or there about) which was simply rounded to .36.

Now you know one of my favourites: "Colt Navy 1851 or 1861".

Calibers Section III:

SAAMI, C.I.P., Deva, Epvat Testing and Standards:

One leading principle of manufactures is, that the smallest chamber for a given cartridge, will always
accept the largest allowed cartridge. It leads to certain tolerances. Ofcourse, if the tolerances
are too large, then the kinetic energy, and accuracy will suffer.

There are organizations which will test rounds (e.g. for safety, compatibility), like those listed
in the title. But it seems to me that a "true" Global organisation is missing.

SAAMI and C.I.P, also define the standards for cartridges, like base- and neck widths, all diameters,
lengths of case and cartridge, max pressure allowed etc..
It does not mean that the whole world stricktly adheres to the protocols of SAAMI and C.I.P.

SAAMI is more US oriented, while CIP is more European oriented, it seems.
Especially in the approach and methodology around testing.

Calibers Section IV: Shotgun calibers.

Shotguns come in various formats, like the main versions:

=> Traditional 2 barrel shotguns, like:

-Juxtaposed (juxapose), where the barrels were side-by-side (horizontally parallel).
-Superposed (superpose), where the second barrel is below the first one (vertically parallel.

Due to the fact, that originally, the hammers were outside (on top), the "juxtaposed"
was technically more easily to manufacture.
Once hammerless technology was matured, variants like the "superposed" was easier
to implement as well.

Ofcourse, the single shot, single barrel shotguns were along, almost "forever"...

One other main feature ofcourse, is that the barrel is smooth, and so does not have any
rifling, which ofcourse is not needed at all, with small bb-like shotgun pellets.

=> (More modern) Pump action shotguns:

These more modern shotguns, most often use a tubular magazine below the barrel, and shotgun rounds
can be loaded by pumpaction.
Such guns are ofcourse geared towards "defense", and not for hunting.

In general, shotguns use shotshells. However, larger "one-piece" projectiles are possible too, like
the Brenneker cartridge, or many other variants, having all sorts of slugs.

In a later section, we will go somewhat deeper in Shotguns. However, for the "traditional"
shotguns, it is true that quite often a "choke" is present (near the end of the barrel),
which is essentially a narrowing inside the barrel. So, the internal barrel is not straigt, since various
degrees of narrowing can be implemented. It's true, that a choke will lower the scatter
of hail, so increasing the effective distance of the shot (reduced spread).

The above, more or less, represents the main types.

Let's take a look at some well known calibers. It may come as a surprise, but the larger
the caliber (gauge) is, the lower the diameter of a shotshell (or barrel).
So, caliber 12 is quite a bit larger than caliber 24, while you might have assumed
that it is the other way around (24 seems bigger than 12, but not with shotshells).
The calibers are "historically determined" like for example the ability to produce 12 identical
balls from one pound, or, the diameter of the barrel on a 12 gauge shotgun is the size
of a 1/12 pound lead ball (caliber 12, 12 gauge).

Nowadays, the shotshell diameter, or diameter of the bore, leads to the following table
of the most popular gauges:

Gauge:	Bore diam. inch:	Bore diam. mm:
10	.775	19.69
12	.725	18.42
16	.665	16.89
20	.615	15.62
28	.545	13.84

Nowadays, certain calibers are considered to be obsolete like cal. 4, and cal 24, and a few others.
It also depends a bit in which country you are.
Cal. 12 is very popular, in most countries.
It's further very important that shotshells of a certain caliber, may come in various lengths.
Shorter lengths, in general, is allowed, but longer shells certainly not. Here too, only use the correct ammo
for your specific gun.
Generally, the proper usable length of the shell, is often noted on the barrel or chamber.

Calibers Section V: Rifle calibers.

Again, here too we have some "special" denotions, like for example .30-06
What does it mean? Or something like 308?

There exists an enormous ammount of different rifle ammo, and in chapter 4, I will try to make
a rather comprehensive table of famous small-arms rounds (pistol/revolver, rifle).

Here we are only going to take a look at a few well-known examples.

First, take a look at the silly drawing in figure 1, namely the rifle cartridge (figure 3).
Any such cartridge, has a neck width, base width, length of the case/no bullet,
and length of the cartridge/with bullet, and slope of the shoulders.
However, There are more characterizing dimensions (SAAMI, C.I.P specifications).

In such case, it's even more critical that the correct ammo is used, for the correct rifle (sorry!).

Example 1: .30-06 (Springfield):

In the example of caliber ".30-06" you may immediately see that we are talking about caliber .30.
The "06" refers to the year of first production of the Springfield round, namely 1906.

Actually, the term ".30-06" represent the whole of the specifications of the cartridge, like
case length, all of the various diameters, allowable pressures, tolerances etc.. etc...

As always, calibers (like .30), are a feature of the gun, and here we have .30 x 2.54 = 7.62 mm.

The round has a long military history (e.g M1 Garant rifle, and many more), and is still used
as a hunting/sporting round, for hunting/sport rifles.
While the .30-06 is still in use, the US Military choose for a slightly shorter round, with different
dimensions. The shorter round fits machines guns better.

When talking about barrel and bullet, the bullet diameter is important, but more defining
is the "lands and grooves" of the barrel. By the way, if a bullet is fired, you will see the
land and groves imprinted on the bullet (unless it struck a hard object and was shattered).
Forensic personell often shoot a bullet into ballistic gel, resulting in a clean bullet with
perfect grooves and lands (if the gel is thick enough to stop the bullet).

So, the gun is .30, but the bullet diameter is .308 inch (7.62mm).
The ".30-06" cartridge, adheres to the SAAMI, C.I.P specifications like lengths, all diameters
involved with the cartridge, max pressures, and allowable tolerances.

It's an enormous powerful round. The max pressure when fired is somewhere near 60000 psi,
and that in combination with a longer rifle barrel (compared to handguns), makes that
we can find energies somewhere near 3800 Joules.

The .30-06 (7,62x63mm), was replaced (a noted before) by the shorter 7.62×51mm NATO cartridge.

A similar cartridge to the .30-06, exists, namely the .308 (Winchester).
The .308 is shorter (and a little bit less powerfull), and ofcourse not interchangble.

Example 2: Some other examples:

A large array of rifle ammo, all with complete specs, exists, like for example:

.223R, .30-06, .303Brit, .308, .30M1, .45-70, 6.5x55, 7.5x55, 7.62x54R, countless .300 rounds etc...

But ofcourse, for example the popular "sport" pistol-, revolver round, namely the .22lr,
exists for such suitable rifles too.
There are countless examples of rifles, which are suitable for pistol-, revolver rounds.

For some reason, I do not often refer to Wikipedia articles, but the following
listing is quite informative.

However, in the following chapter, I will do it myself.

Chapter 4: Listing of common (small-arms) ammo and associated guns .

Chapter 5. A few words on air-resistence, gravity, trajectories, and stuff.

What effects the path of a bullet, are at least the following factors:

(1). Bullet shape, mass + internal mass distribution, spin (helicity grooves), caliber, charge, amount charge.
(2). Muzzle velocity (also in relation to 1).
(3). Speed of sound barrier: Supersonic- or Subsonic bullet? Barrier important?
(4). Gravity.
(5). Air resistence.
(6). Angle with respect horizon.
(7). Wind, humidity, temperature.
(8). Rather Evident: Is the gun (barrel) cleaned?

There are more effects, like the "Coriolis drift", due to Earth's rotation, but for small-arms (pistol, revolver, rifle etc..),
it is not significant.
Let's take a look at the points 1 - 7. (8 might be doubted by some folks, but it is not to quantify anyway).

What are we asking ourselves here? It seems that we are after the reduction in velocity at time "t", after the bullet
left the muzzle at t=t₀ with speed v₀. If we want a formula, then the physical characteristics
of the bullet must be a parameter, as well as it's mass (?) and spin (?) and other parameters (?).

That might be too cumbersome. Anyway, you can find arguments to leave the charge, and amount of charge, out of the
discussion, since that determines the initial velocity, so you can stick with initial velocity v₀.
Furthermore, many bullet properties are wrapped up in some bullet coefficient (parameter), which will
be part of the calculations (reduction of speed, range/time, total range etc...).

The most important factors influencing the bullet's trajectory, are:

- air resistence, where the force is proportionate to v(t)²
- drop by gravitation, where the force is F = mg (m: mass, g:acceleration due to gravity)

For realistic simulations, you may take a look at the Ballistic Calculator (gundata.org)
at the site "gundata.org" and make a few tests with cartridges you are familiar with.

Looking at that site, you may immediately ask: "where is the barrel length", since that too determines the initial velocity.
Well, note that the initial velocity is one of the input parameters, so the ballistic calculator, in effect, eliminates any aspect
of the gun itself.

5.1 Bullet shape, mass + internal mass distribution, initial velocity.

Mass:

While accelerating in the barrel, in general, the less mass, the higher the obtained muzzle velocity.
Now, ofcourse, for impact, too little mass (probably) isn't good.

So, a bit of mass is good too, as we can see from two formula's from Classical Mechanics:

- (Impuls) momentum p = mv (SI units: m:mass in kg, v:velocity in m/s).  (1)

- Kinetic energy E_kin = ½mv².  (2)

- Δp = m Δv = FΔt  (3)

(As of now, in formula's like p = m x v, I will leave out the multiplication operator "x", and simply
write it as p = mv).

Note:
Wanna see more on Classical Mechanics (including momentum)? You can see my note here.

From the Energy relation, "v" seems to be the main factor, for the energy associated with a projectile.
However, from the momentum relation, "m" may not be underestimated. You can simply see that from "mv",
"m" and "v" are (sort of) equal footing.

Actually, for penetration, "m" is very important, as might be made likely by the following reasoning.

The "delta" impulse (Δp) over a time interval (Δt) (e.g. a bullet enters material like wood,
or bangs against metal, loosing Δp, taking time Δt) is defined to be Δp = FΔt (with F as Force)

So, viewed from momentum side, "m" and "v" are equally important for "delivering a punch" to a target.

Example:

It's instructive, to use the relations in another way:

Suppose a bullet is accelerated down the barrel of a gun (by the hot gases produced in the fast combustion of the powder).
What is the average force exerted on a 0.02 kg bullet (in the barrel) to accelerate it to a speed of 650 m/s,
in a time of 1.50 ms (milliseconds) = 0.0015 seconds? (time in barrel since start of acceleration).

Ofcourse, the Force F when the bullet goes through the barrel, is not really constant during the Δt of 1.5 ms.
But the average force can be considered to be a constant. As we will see later on, we will use the trick
in other occasions as well. So:

The average force exerted is F_avg = Δp/Δt.

Since the momentum of the bullet went from p=0 (bullet not fired yet) to p=0.02 x 650 = 13 (kg m/s),

we have:

F_avg = Δp/Δt = (13 kg m/s)/(1.5x10^-3 s) = 8667 Newton (= 8667 kgm/s²).

Mass distribution inside a bullet:

Some trivial remarks:
Many bullets have a thin mantle (e.g. brass, copper), surrounding lead, or lead-antimony, or other stuff.
The projectile often has some aerodynamic shape. But a wide variety exists in shapes and consistuents.

Very large rounds may even contain depleted Uranium (heavy mass) in order to maximize penetration.

Now, different types of bullets, may be meant to use for (slightly) different purposes.

Many rifle bullets often have some resemblence to the (longer) "spitzer bullet", but ofcourse many other
shapes exist too. Handguns often have shorter bullets.

Some have a soft tip, some a hard tip, some have a hollow cavity in front (hollow point),
some are extremely tough in the inside, some have explosive substances inside (hopefully not for sale),
some make use of plastics for some purpose, some have a "self-sharpening" mantle (on impact),
some have multiple (mostly 2) partitions, etc.. etc..

Ofcourse that all is not new to you.

There is a lot to say as well about the shape.

Long projectiles:

For example, for a certain caliber gun, how can you make a bullet heavier? One easy answer is applying
heavier constituents in the bullet.

However, given a certain fixed caliber, making a bullet longer will make the projectile
heavier compared to shorter ones. This also explains why you might encounter rather slim, but quite long projectiles,
most notably, for high-power rifles.

The longer types may also achieve a higher angular momentum, which increases stability in flight.

But apart from the total mass, and shape, and inner constituents, does the Mass distribution
inside a bullet, has any effect?

A bullet can be seen as a cilinder, with a central axis. Non-symmetrical distributions of mass, will give
rise to wobbling effects, which is no good ofcourse. Any Non-symmetrical distributions of mass with respect
to the central axis can be considered to be negative for the trajectory.
However, I have noticed some articles experimentlly studying various mass distributions, although symmetrical, but having more mass
in the back, or contrary at the front of the bullet. More mass at the back seems to favour stability and deeper penetration.
Ofcourse, that practise had been applied many years by now.

Partitioning of bullets into compartments (like the Nosler partitioning):

Some bullets have a copper mantle, but a copper layer (inside) also divides the bullet in 2 parts, which contain
the heavy parts like lead, or lead-antimony. So, the bullet has two compartments. It's a sort of dual core partitioning.
The bullet remains symmetrical along the central axis.
The objective is often, to have a mushroom effect on the front compartment, while the other heavy part remains
intact for deeper penetration. One example is the Nosler type partitioning.

Bullets: shapes, weights, and primary purpose.

I would not be surprised if well over 5000 different types exist, but ofcourse many of which are
only slightly different.

It's not usefull to build a database here, while actually informative websites exist.

Chapter 6. Larger guns, and large ammo, > .50.

Chapter 7. Self propelled projectiles.

Just started this note....