Muzzle Velocity Formula

[Logi]

I want to report that I seem to have found a rather good correlation between gun diameter, caliber, shell weight, gun year and muzzle velocity. That is; I, with a small modification to my Steam and Sails Ballistic method (which was derived from P3D's fictional gun formula), managed to bring the majority of the errors of the predicted muzzle velocity down to ±5% from 1820-2010. This extends the range of P3D's muzzle energy prediction from 1898-1945 by 143 years.

I would use the same method to extend the prediction range even further than 1820, but as you will see from the graphics, there isn't much information past 1850.

Like P3D, the method is empirical in nature, relying on hard data from various guns to produce. Unfortunately this also means unusual weapons are not well covered.



The formula is as follows:
         _______________________
V = √ ( d³ * L * (1/c₂) * 10⁶ ) / m

c₂ = c₁ * ( 0.1*( Year - 1870) + 14 )

c1	Year Range
6.00	<1860
1.40	1860-1880
1.00	1880-1900
0.68	1900-1930
0.65	1930-1950
0.60	1950-1980
0.42	1980-2000
0.40	2000<

How does it fare?



As you can see, the muzzle velocities rise slowly throughout. The sharp jump around 2010 is due to that data being from the CLGG demonstrations. The prediction seems to correlate with the data quite well, but how well?



The vast majority of the gun's predicted muzzle velocity have an error of between +10% and -5%.
Look closer and you see the vast majority of that majority is in the ±5% band. There are some outliers so the empirical formula is not perfect, but it's pretty darn good if I say so myself.

[Logi]

By a similar approach, I attempted to derive an empirical formula for gun range at elevation.
Unlike the usual methods which involve stepping through the air drag, etc. equations moment by moment, this approximates based only on the gun's static parameters, namely:
Diameter, shell weight, muzzle velocity, and elevation of gun.



x = x_parabolic * R
Note: x is in feet, divide by 3 to convert to the usual measure, yards

x_parabolic  = V₀²sinθcosθ / g
V₀ as determined by my muzzle velocity formula.
g = 32.174 ft/s²

R = ( c₃ * (0.4 + m/d³) + 0.135) ) / c₄
c₃ = 0.0021758 * θ + 0.082838
c₄ = cos(θ^0.78 * 1.3) * V₀/3000 * sin(θ + 9)




The error of the range as a function of elevation. The data is mostly within ±5% of the prediction, and 99% of the data is within ±10% of the prediction.

The USN 16"/50 Mark 7 firing at 45 degrees will have an error range of ±2117 yds. The formula actually predicts a range of 40,553.5 yds, a 1,791.5 yd difference with reality. An error of -4.23%.
At 10 degrees, the error range is ±882.5 yds. The formula predicts 16,948.5 yds, a difference of 701.5 yds and an error of -3.97%.

That seems decent enough for a first estimator.

[Logi]

I've derived another empirical formula. This one calculates the striking velocity of a gun based on it's muzzle velocity, range, diameter, and caliber.

I haven't tested this formula as extensively as the previous formulas. The formula has been checked against the 16"/50 Mark 7 and the error range is ±2%

The correction factors c₁ and c₂ are based on the data from several different guns.



V_f = c₁( 19.428(^x/₅₀₀₀)² - 267.71(^x/₅₀₀₀) + c₂*V₀ )

c₁ = (16*50 / (d*L))^1/3
c₂ = 1.05 if using my muzzle velocity formula

V₀ = muzzle velocity as predicted by my formula
x = shell fall distance as predicted by my formula (yards)
d = diameter of shell
L = caliber of gun

The corrective figure c₁ attempts to correct the strike velocity for differences in the muzzle energy.
The corrective figure c₂ attempts to correct for the error within my muzzle velocity formula.

[Nobody]

I'm hoping you're using metric units?

[Logi]

No, a lot of the data I'm using listed their data in imperial units. In addition, sources like NavWeaps have the bulk of their gun information on British and American guns, both which were measured originally in imperial. It stands, therefore, that the metric numbers are subject to rounding errors, so I opted not to use their metric equivalents.

Ideally, I would derive a metric equivalent for each, but each formula is a huge time and energy sink. I want to (hopefully) derive a similar formula for shell angle of fall by tomorrow and have a relatively quick series of methods to ultimately determine shell penetration and other details necessary for a battle simulation. Perhaps after a few weeks I will test and (if it doesn't match) develop metric equivalents.

Unfortunately for the time being, metric-oriented people will have to pre- and post-convert their units.

[Nobody]

No, a lot of the data I'm using listed their data in imperial units. In addition, sources like NavWeaps have the bulk of their gun information on British and American guns, both which were measured originally in imperial.

Understandable, however:

It stands, therefore, that the metric numbers are subject to rounding errors, so I opted not to use their metric equivalents.

That's wrong, it's the other way around! It's perfectly possible to convert Inch and feet to metric without error and a reasonable number of digits. Converting metric to imperial on the other hand almost always leads to rounding errors.
Pounds are another matter though. It's pretty much impossible to convert from or into them without rounding.

Since you're discussing many weapons from before say the 1930s you might even want to consider that US and UK inch are of different length. I believe the UK was some 2,53something cm and the US 2,54something cm. Even more reason to calculate in metric in the first place.

[Logi]

What I meant is rather the number of significant digits. There are more on the imperial side than the metric side (imperial with 4, metric with 3), I would suppose their conversion number, which is not inaccurate, is cut off by 1 digit. Of course, I don't know the details of their reporting methodology so I may be entirely wrong in this line of thought.

However, I never knew about the UK/US inch difference, although given the origin of the unit it makes sense.

[Logi]

Obliquity was a difficult one to derive a formula for. I tested over a dozen methods, which were all accurate to the ±10% to ±20% range. The main problem was the scatter in the data, most likely resulting from the comparatively small numbers being worked on (0 to 50 degrees).

Unfortunately, I wasn't able to reduce much of the scatter, although in retrospect, a scatter of ±10% isn't bad. That's 0.5° out of 5° and 4.5° out of 45°. Since obliquity is not as important a term in most armor penetration formulas, it'll have to do.

I chose the simplest out of my derived formulas.



θ_f = c₁θ * (0.6592 / (m/d³))^0.3

c₁ = 1 + 0.15 * c₂(sinθ + 0.95)⁵

c₂ = cos⁵θ - 0.1      θ > 20
    = cosθ               θ ≤ 20



The error seems larger than it is primarily because of the graph axis range. It's mostly within ±10%.

[Logi]

The same as the Thompson F-Formula (US, 1930), the F value is derived from the extended version where F_STD is solved for.

T_penetration = 1728.04 * m/d³ * [ (V_f/F)Cos(θ_f) ]²

where F = 6 * (T_armor/D - 0.45)(θ_f² + 2000) + 40,000

T_armor = the thickness of the armor the shell impacts
T_penetration = the shell penetration, if it's less than the armor thickness the shell fails to penetrate

[Logi]

A Hit Location formula based on P3D and Miketr's hints towards their systems as well as the system in place in Seakrieg 4.

Hit Location Rolls (Seakrieg 4)

Short Range	Long Range	Location
01-08	01-30	Deck
09-40	31-40	Sidebelt
41-55	51-70	Conning Tower
56-65	71-82	Superstructure
86-00	83-00	Dud

H_{probability,belt} = A_belt/A_vertical * H_{probability,sidebelt}
A_vertical = (F_avg + 0.2*T_m) * L_pp



The probability of a hit on the vertical side of a ship is determined by rule in Seakrieg 4. However, where it hits on that vertical side is determined by breaking down the armor sections.

The vertical side consists, therefore, of the above-water main belt, underwater main belt, ends belt, upper belt, and the unarmored portions.
Their respective areas are the result of their length multiplied by their height. The unarmored portion is just whatever area of the vertical side is left over. That area is divided by the total area of the vertical side in question.

So to recall, you roll the dice to determine the hit location generally and then (if rolled belt) once more to determine where in the belt specifically.

A similar scheme could be put in place for oddies such as a hit on the turret roof vs turret face or barbette.

[Logi]

Ship Hitpoints

Some of the grievances Miketr mentioned about SeaKrieg was the low amount of hitpoints ships had (such as destroyers) and the generic nature (unable to tell the difference in a 1890 10,000 PDN and a 1925 10,000 CA).

For that a fusion with SS numbers might be the best way. It was discussed before that flotation (Survivability) could be used to simulate ship resistance to torpedo attack, but I believe it could be used in conjuction with the SeaKrieg DP calculation method (0.033 * Δ_standard).

My suggested method:

DP = 0.025 * Δ_light + F/50

where F = flotation in Kg


The special cases of merchant ships could be handled, as previously suggested, by dividing the displacement by 4.



Gun Damage
Gun Damage in Seakrieg is a function of shell diameter only, which while a great over-simplification is prudent given the lack of necessary information. For our purposes, I am inclined to think it ideal, since diameter is a major influence on the size of the bursting charge. We simply need a way to calculate the odd shell diameters people research that Seakrieg does not handle.

SeaKrieg's DP/diameter correlation follows closely that given by the equation

DP = 0.6445 * d^1.3294

d in inches

[Logi]

Torpedo Damage

Torpedo damage a bit more difficult than shell damage due to the underwater shock-waves and venting that happens on a torpedo hit.
I've taken the data from 4 different cases (the only ones I could find decent information on) and referenced it with SeaKrieg 4's system of calculating torpedo damage to produce the formula.

• The SMS Istvan was sunk in 1918 after getting hit by 2 x 45cm (18", ~176 kg TNT) Italian surface torpedoes abreast of her boiler rooms. She was struck 3:25 am and capsized 6:05 am.
• The MN Leon Gambetta was sunk in 1915 after getting hit by 1 x 45cm (18", 122.6 kg TNT) German/Austrian submarine torpedo. She sank in 10 minutes.
• The IJN Musashi was sunk in 1944 after getting hit by 22 x 56.9cm (22.4", 274kg TPX/HBX) American aerial torpedoes.
• The IJN Yamato was sunk in 1945 after getting hit by 10-12+ x 56.9cm (22.4", 274kg TPX/HBX) American aerial torpedoes.

Torpedo DER = 4*m * 0.5^(B_TDS/3) * DER_mod

m = (d_torp/2)² * PI * L_torp * 0.012 * 0.00165

where:
B_TDS = beam (in meters) devoted to TDS
DER_mod = 150% if using TPX/HBX, else 100%
d_torp = diameter of torpedo (in cm)
L_torp = length of torpedo (in cm)
0.012 = % of torpedo devoted to warhead weight
0.00165 = density of TNT, TPX, HBX (kg/cm³)



An analysis of several different torpedo from various countries revealed that their warheads consumed 0.8% - 1.6% of the total volume of the torpedo. I averaged this as 1.2%, you multiply this with the warhead density to get the warhead weight.

As Miketr suggested in the previous iteration, later Hexanite/Torpex warheads vs the earlier TNT warheads should be modeled by giving Hexanite/Torpex warheads 50% more effectiveness.

-snip-
Using the seekrieg dud chance I suggest we start with a 40% for earlier torpedo's dropping to 30% for our current gen weapons and then going down to 20% for 1930's weapons.  Ditto for Magnetic triggers 40% for 1st generation, 30% for 2nd gen and 20% for 3rd gen (with 1940's).  We also will need a improved warhead tech for Hexanite / Torpex in late 30's that gives a 50% boost to warhead yields.

Michael

There is a large variation in the amount of torpedoes needed to sink a ship (such as between Leon Gambetta and Istvan) given similar displacement. I attributed this to the TDS system (or absence thereof). The use of a TDS was to give space to absorb the shock-waves generated by a torpedo hit. I abstracted the exponential rate the shock-waves die off given distance by introducing the variation coefficient, 0.5^(B_TDS/3). Simply put, for every 3m of ship beam devoted to TDS (you set this in SS3) the effectiveness of the torpedo (DER) halves.

Given this system, it takes ~0.9 hits to sink Leon Gambetta, ~2.1 hits to sink Istvan, and ~13.3 hits to sink a Yamato. We can include the special case of Musashi where torpedo hits on both sides helped the counter-flooding effort by multiplying the DER of a hit by 0.60 (60%) if it hits the side of the ship where there has been less torpedo hits. For the time being, that is more complexity than necessary, so I will refrain from pushing it along.

[Nobody]

How many hits in you system would be required to sink Bismarck/Tirpitz?


I should be able to calculate that myself, but the chance that I use the right values and use the right units is pretty low I assume.

[Logi]

Torpedo DER is in metric, but I'll calculate that for you:

The Bismarck was hit by 2 aerial torpedoes and 3 surface torpedoes.

Since I don't have a SS file of Bismarck, I'll use the SeaKrieg 4 rule for DP instead of my suggestion. kmbismarck.com says the tds depth is 4.5-5.5 meters amidships, let's average this as 5 meters.
The Swordfishes seemed to have carried the 45 cm (18") Mark XIV torpedo, which had a 170kg warhead. Unfortunately NavWeaps does not detail the length of the torpedo, so warhead weight can not be calculated via my method.
The HMS Dorsetshire carries 53.3cm (21") torpedoes, seemingly the 53.3 cm (21") Mark IX** with a warhead of 327 kg or later 365 kg Torpex. The torpedo length was 7.277 m (23' 10.5") so by my method the warhead was 321.5 kg. I will assume Torpex was used.

DP = 1376.1
B_TDS = 10
m_aerial = 170 kg
m_surface = 321.5 kg
DER_{mod, surface} = 1.5

So the DER per torpedo is,
DER_aerial = 67.5
DER_surface = 191.4

The total damage inflicted on Bismarck = 709.2, 52% of it's total DP. In reality, this was augmented by shellfire that inflicted damage as well.
If only the surface torpedoes were used, it would take ~7.2 hits to sink the Bismarck.

A TDS depth of 4.5 meters on each side would come to 893.4 DP damage and ~5.67 hits needed to sink the Bismarck. Similarly, a depth of 5.5 m would give 562.8 DP and ~9 hits.

To give a reference point, if I compare it to Miketr's calculated value:

If we plug in ww2 torpedo like the German G7 with a 300 kg Hexanite warhead that would be equal to a 450 kg /990 lb TNT warhead.  So 6.2 hits to kill.

By my system with the same warhead, it would take 7.7 hits to kill.

Edit: Corrected a typo, "Similarly, a depth of 5 m" to "Similarly, a depth of 5.5 m"

[Logi]

Shell Dispersion

Derived empirically from data from the:
USN 16"/50 Mark 7 firing in 1987
MN 14.96"/45 Model 1935/1936 firing in 1948
USN 5"/38 firing in 1925-1929.
USN 16"/45 Mark 5/8 firing in 1930-1931 Force Battle Practice.
RM 203mm/53 Model 1927 firing.
IJN 12cm/45 10th Year Type firing in 1936 and earlier.
USN Battleships true mean dispersion, 1918-1945.



Dispersion_{as % of range} = -4.71438E-4 * Year + 0.918483           Y < 1934.5
                                = 179,053,72,000.355 * e^{-0.016 * Year}