Estimating Gun Propellant Charge
Weights

Sometimes all you have is the barest pieces of data on a gun, particularly for German “Paper” guns in World War II, where they're described as “7.5cm KWK 42 L/100” and that's it. No detailed data.
NOTE: The U.S. measured “calibre” as the length from the muzzle to the rear of the barrel (bore), while the Germans measured it from the muzzle to the rear of the breech. Thus, the US 90mm M3 gun would have a calibre of L50 under the US system and L52.5 under the German system.
There are a few ways to estimate the propellant charge amounts, and they're listed below.
The following rule of thumb is sort of workable:
Increase in Muzzle Velocity 
Increase in Propellant Charge 
0.05 
0.1625 
0.0625 
0.203125 
0.1 
0.325 
0.125 
0.40625 
0.20 
0.65 
0.25 
0.8125 
0.5 
1.625 
0.75 
2.4375 
1.0 
3.25 
1.5 
4.875 
2.0 
6.5 
2.5 
8.125 
3.0 
9.75 
3.5 
11.375 
4.0 
13 
If you use table this to scale up the 7.5cm L24 (0.37 kg propellant, 385 m/sec) to 770 m/sec (7.5 cm L48), – a doubling of muzzle velocity, you end up with a charge weight of 2.405 kg, very close to the actual 2.43 kg.
Likewise, scaling the 7.5 cm L48 from 770 m/sec up to 930 m/sec (7.5 cm L70 and 20% more velocity) using the table results in a charge weight of 4.0095 kg, a bit off, but still close to the actual 4.07 kg.
However, if you try to scale the 7.5 cm L24 up all the way to 930 m/sec (2.4x), you end up with 3.00625 kg of propellant, a decent amount less than the actual L70's 4.07; so a good rule of thumb would be to never scale up or down more than double the muzzle velocity in each step.
If you scale in multiple steps, e.g, from 385 to 770 m/sec, then from 770 to 930 m/sec, you end up with a propellant charge of 3.96 kg; which is within acceptable margins of error.
Richard M Ogorkiewicz put forth in Technology of Tanks (1991) a simple power law for muzzle velocity of projectiles against their projectile/charge mass ratio.
V_{0} = 1500 (M_{C} / M_{P})^{0.45}
Where:
V_{0} = Muzzle Velocity (m/sec).
M_{C} = Propellant mass (kg).
M_{P} = Projectile mass (kg).
Example: A large caliber gun fires a 25 kg projectile using 10 kg of propellant. What is it's likely muzzle velocity?
1500 * (10 / 25)^{0.45 }= 993.15~ m/sec.
The above formula “kind of sort of” works.
V_{N} = SQRT( M_{PN} / M_{PO} ) * V_{O}
Where:
V_{0} = Original Muzzle Velocity (m/sec).
V_{N} = New Muzzle Velocity (m/sec).
M_{PN} = New Propellant Mass (kg).
M_{PO} = Original Propellant Mass (kg).
Example: You have a gun that fires a projectile at 750 m/sec with a propellant charge of 2.43 kg. What would be the new velocity if you increased the propellant charge to 4.07 kg?
SQRT( 4.07 / 2.43 ) * 750 = 970.63 m/sec.
First, you figure out the muzzle energy in megajoules (MJ) of the gun/ammo combination.
Then you calculate the total energy in megajoules needed to achieve that muzzle energy through the following equation:
E_{T} = E_{M} / B_{E}
Where:
E_{T }= Total Propellant Energy (Megajoules)_{}E_{M} = Muzzle Energy (Megajoules)_{}B_{E} = Ballistic Efficiency of propellant/gun (%)
Ballistic Efficiency varies with the time period and nation. US WW2 Tank Guns are about 30.9%, while German WW2 Tank Guns are 22.2%.
Example:
We have a gun with a muzzle energy of 2 MJ and a ballistic efficiency of 22.2%
2 / 0.222 = 9.009 MJ total propellant energy
We then divide the total propellant energy by 4 MJ/kg to get the propellant weight in kilograms.
9.009 / 4 = 2.252 kg of propellant.
This is pretty close to the 7.5 cm L/48 PzGr 40 (APCR) data.
There are two factors which apparently affect propellant efficiency significantly in internal ballistics.
1.) FRICTION: As muzzle velocity increases; propellant gas friction losses increase, decreasing ballistic efficiency.
Ballistic Efficiency is given as:
Ballistic Efficiency = Muzzle Energy / Propellant Energy.
To calculate propellant energy is somewhat simple. Find out the weight of the propellant charge and then multiply it by the list below:
Single Base Propellants: 3 MJ/kg heat of explosion (Q_{ex}) on average. Pre WWI (?).
Double Base Propellants: 4 MJ/kg heat of explosion (Q_{ex}) on average. Used from WWI to WWII in artillery. Extremely erosive on barrels.
Triple Base Propellants: 4 MJ/kg heat of explosion (Q_{ex}) on average. Developed in 1930s, adopted post war in tankguns. Significantly reduced erosive effect on barrels.
Modern Double Base Propellants: 4.6 MJ/kg heat of explosion (Q_{ex}). Adopted from the 1970s onwards in tank guns. The use of modern additives to reduce barrel erosion is a significant reason for the return of double base to tank guns. JA2 (US 120mm) is one of the more famous examples of the new “modern” double base propellants.
It appears that German WW2 powder just may have had a poor ballistic efficiency, with an average BE of 22.2% by my calculations (7.5 cm L24, L43, L48 and L70; 8.8 cm L56 and L70) against US WW2 powder with an average BE of 30.9% (US 75mm M2/M3, 76mm M1, 90mm M3), and this significantly skews propellant weight calculations.
2.) EXPANSION RATIO: A higher expansion ratio allows the propellant more time to do it's work, increasing ballistic efficiency.
Expansion ratio is given as:
Expansion Ratio = (Bore Volume + Chamber Volume) / Chamber Volume.
The effect that expansion ratios and muzzle velocity can have on Ballistic Efficiency is dramatic, as shown by the statistics below:
7.5 cm KWK 37 L24 (385 m/sec) (6.24 to
1 Exp. Ratio) (1.362 MJ/kg impetus) (34.1% Ballistic Efficiency)
21
cm Paris Gun L171 (1,640 m/sec) (14.55 to 1 Exp. Ratio) (0.896 MJ/kg
impetus) (22.4% Ballistic Efficiency)
7.5 cm KWK 40 L48 (750
m/sec) (3.76 to 1 Exp. Ratio) (0.873 MJ/kg impetus) (19.7% Ballistic
Efficiency)
7.5 cm KWK 42 L70 (920 m/sec) (4.10 to 1 Exp. Ratio)
(0.773 MJ/kg impetus) (19.1% Ballistic Efficiency)
Unfortunately, I can't puzzle out currently the way they interlock; but it appears that doubling the expansion ratio can have a significant effect on ballistic efficiency; enough to cancel out a significant amount of frictional loss from high muzzle velocities, as shown by the Paris Gun example.
V_{B} = [ Pi * (D/2)^{2} * (D * C) ] * 0.000001
Where:
V_{B} = Bore Volume (liters)
D = Shell diameter
(millimeters)
C = Caliber of Weapon in L/x
We have a 80mm gun with caliber of L20. What is it's bore volume?
(Pi * (80/2)^{2} * (80 * 20) * 0.000001 = 8.042 Liters
V_{C} = [ Pi * (D/2)^{2} * L ] * 0.000001
Where:
V_{C} = Chamber Volume (liters)
D = Shell diameter
(millimeters)
L = Useable Length of Case (millimeters)
Note: To get useable length of cartridge cases, subtract the Bore divided by 1.5 from the total length of the case to take into account the protrusion of a seated round into the cartridge case.
We have a 80mm gun with a straightwalled case length of 250mm. What is it's bore volume?
[ Pi * (80/2)^{2} * ( 250 – (80 / 1.5) ) ] * 0.000001 = 0.988 Liters
Terminology
of Cartridge Case Locations
V_{B} = [ π * (D/2)^{2} * L ] * 0.000001
V_{S} = (1/3) * π * L * ( (D_{T}/2)^{2} + (D_{B}/2)^{2} + [ (D_{T}/2) * (D_{B}/2) ] ) * 0.000001
V_{C} = V_{B} + V_{S}
Note: Throat volume not included due to this space being taken up by the seated round in most cartridges.
Where:
V_{B} = Cartridge Base Volume (liters)
V_{S} =
Cartridge Shoulder Volume (liters)
V_{C} = Chamber Volume
(liters)
L = Length of that Segment (millimeters)
D = Diameter
of that Segment (millimeters)
D_{T} = Top Diameter of that
Segment (millimeters)
D_{B} = Bottom Diameter of that
Segment (millimeters)

7.5 cm KwK 37 
7.5 cm KwK 40 L43/48 
7.5 cm KwK 42 L70 
Base Diameter 
81.7mm 
102.2~ mm 
112.7mm 
Base Length + (% of total) 
243 mm? 
392 mm (79.2%) 
529mm (83.7%) 
Shoulder Length + (% of total) 
N/A 
54mm (10.9%) 
54mm (8.5%) 
Throat Length + (% of total) 
N/A 
49.1mm (9.9%) 
49.1mm (7.8%) 
What would be the chamber volume of a 80 mm gun with a 250mm cartridge case length with the same rough proportions as the KWK 42 round (same rough angle of necking, etc)?
80 mm * 1.5 = 120mm Base Diameter
55mm Shoulder Length
250
55  50 = 145mm Base Length
V_{B} = [ π * (120/2)^{2} * 120 ] * 0.000001 = 1.357 liters
V_{S} = (1/3) * π * 55 * ( (80/2)^{2} + (120/2)^{2} + [ (80/2) * (120/2) ] ) * 0.000001 = 0.438 liters
V_{C} = 1.357 + 0.438 = 1.795 liters
Thus, the chamber volume would be 1.795 L.
With USStyle Caliber Measurements
E_{R} = (V_{B} + V_{C}) / V_{C}
With GermanStyle Caliber Measurements
E_{R} = ( (V_{B} – V_{C}) + V_{C }) / V_{C}
Notes: You have to subtract the chamber from the bore in this to take into account that the German system of “caliber” measures a gun from the muzzle end to the rear of the breech; messing up measurements if you are calculating bore volume via caliber.
Where:
E_{R} = Expansion Ratio
V_{B} = Bore Volume
V_{C}
= Chamber Volume
Using the US Style caliber measurements, what would be the expansion ratios of the aforementioned 80x250 StraightWall and 80x250 NeckedDown cases in the previously mentioned L20 gun?
80x250 StraightWall:
E_{R} = (8.042 + 0.988) / 0.988 = 8.14 Expansion Ratio
80x250 Necked Down:
E_{R} = (8.042 + 1.795) / 1.795 = 4.48 Expansion Ratio