While calculating the performance of SK20.3/C34 based on ADM 213/951, it seemed like there is a mistake of its Cg value.
On ADM 213/951, SK20.3/C34 was suggested to have a Cg value of 675 for Psgr against KC, however, comparing to Ch value of 650 for Psgr against KC, on graph against same thickness of armor, it seemed like there will be a crossing between heil line and grenze line if Cg=675 was true.
According to my calculation based on Awa A IA 100/40 g.Kdos, I personnally believe that 635 should be SK20.3/C34 Psgr's true Cg value against KC.
The rest Ch=650, Bh=2.5, Ch=1.7 works fine for SK20.3/C34 Psgr against KC
I'm also asking help for calculating SK20.3/C34's Spgr Bdz against WH, the C values and B values I found were very chaotic.
For WH from 180mm to 60mm (10mm, 1 result), the corresponding C values are:
750.6
744.0
735.6
730.2
715.5
711.5
693.3
683.7
669.7
654.0
633.0
619.6
605.0
The B values seemed indeed large for thicker WH plates' curves (180mm, 170mm ......), but fell to smaller values for thineer plates' curves, they were so disporportional I don't even think they should be placed here.
By the way, it seemed like UK MKI 38.1cm gun in German assumptions had Ch=680, Bh=3.13, Cg=675, Bg=2.25, for KC plate and Ch=Cg=660, Bh=Bg=3.8 for WH plates.
Any one know where to find Awa A IA 100/40 g.Kdos Heft b, c for German assumptions on foreign guns' performances in pdf?
ADM 213/951
German steel armour piercing projectiles and theory of penetration
https://web.archive.org/web/20200124093 ... chive.html
Awa A IA 100/40 g.Kdos
Unterlagen und Richtlinien zur Bestimmung der Hauptkampfentfernung und der Geschosswahl
Excel
https://www.kbismarck.org/forum/viewtop ... =36&t=3291
Questions regarding to the performance of German 20.3cm guns
Re: Questions regarding to the performance of German 20.3cm guns
I will post some personnal use of GerckeTierberg formula from ADM 213/951 here:
Original state of GerckeTierberg formula:
V = C * (D/G)^0.5 * S^0.8 / (sin a)^m
m = 0.5 + Bx
x = (D/G)^0.5 * S^0.8
Conversion to C:
C = V * (G / D)^0.5 * (sin a)^m / S^0.8
For normal attack, a = 90 degree:
C = V * (G / D)^0.5 / S^0.8
In ADM 213/951 for the Tierberg section, we can see although abbreviations were different, it showed same function as Gercke section, the only difference was that for normal attack (90 degree in formula), Tierberg neglected the influence of B (in formula, when angle equals 90 degrees, (sin a)^m = 1) and assumed that under normal attack only C value would show its effect.
Conversion to B:
B = ( log( C/V * sqrt(D/G) * S^0.8, sina) – 0.5) / (sqrt(D/G) * S^0.8)
Original state of GerckeTierberg formula:
V = C * (D/G)^0.5 * S^0.8 / (sin a)^m
m = 0.5 + Bx
x = (D/G)^0.5 * S^0.8
Conversion to C:
C = V * (G / D)^0.5 * (sin a)^m / S^0.8
For normal attack, a = 90 degree:
C = V * (G / D)^0.5 / S^0.8
In ADM 213/951 for the Tierberg section, we can see although abbreviations were different, it showed same function as Gercke section, the only difference was that for normal attack (90 degree in formula), Tierberg neglected the influence of B (in formula, when angle equals 90 degrees, (sin a)^m = 1) and assumed that under normal attack only C value would show its effect.
Conversion to B:
B = ( log( C/V * sqrt(D/G) * S^0.8, sina) – 0.5) / (sqrt(D/G) * S^0.8)

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Re: Questions regarding to the performance of German 20.3cm guns
@ Cobralion.
My thanks for posting these equations, but they are not of any practical use unless unless variables and units are specified, or unless one has ADM213/951 at hand, in which case the equations are actually redundant. It's nice to know, for example, that m = 0.5 +Bx, but not very handy unless the user knows exactly what Bx is.
In that regard, could you provide more detail, and perhaps one or two worked examples?
Bill Jurens
My thanks for posting these equations, but they are not of any practical use unless unless variables and units are specified, or unless one has ADM213/951 at hand, in which case the equations are actually redundant. It's nice to know, for example, that m = 0.5 +Bx, but not very handy unless the user knows exactly what Bx is.
In that regard, could you provide more detail, and perhaps one or two worked examples?
Bill Jurens

 Senior Member
 Posts: 864
 Joined: Mon Apr 27, 2009 4:17 pm
Re: Questions regarding to the performance of German 20.3cm guns
Adm 213... gave only general values from memory.
The individual values used in the performance charts differ slightly from these.
I have reproduced all values.
The individual values used in the performance charts differ slightly from these.
I have reproduced all values.
Meine Herren, es kann ein siebenjähriger, es kann ein dreißigjähriger Krieg werden – und wehe dem, der zuerst die Lunte in das Pulverfaß schleudert!
Re: Questions regarding to the performance of German 20.3cm guns
I have to apologize to everyone that due to schedule issue the work is still incomplete.
The complete GerckeTierberg formula from ADM 213/951:
D: shell caliber (cm)
G: shell weight (kg)
V: striking velocity (m/s)
a: angle of incidence between trajectory and plate (degree)
S: plate thickness (dm)
V = C * (D/G)^0.5 * S^0.8 / (sin a)^m
m = 0.5 + Bx
x = (D/G)^0.5 * S^0.8
Certain examples of data set with 203mm Psgr vs KC plates will be listed:
For D=20.3cm, G=122kg, a=90degree, S=2dm=200mm, to calculate velocity for penetration with intact shell:
Ch=650, Bh=2.5
V = 650 * (20.3/122)^0.5 * (2^0.8) / (sin(90))^(0.5 + 2.5 * ((20.3/122)^0.5 * (2^0.8)))
Vh = 650 * (20.3/122)^0.5 * (2^0.8), since sin(90) = 1, which means for normal attacks, the formula seems to think B values (angular sensitivity of penetration difficulty?) has no effect, only C values (direct penetration difficulty?) effects the outcome.
Vh = 461.6 m/s (approximately)
For incidence angle a=70degree, D=20.3cm, G=122kg, S=2dm=200mm, Ch=650, Bh=2.5
Vh = 650 * (20.3/122)^0.5 * (2^0.8) / (sin(70))^(0.5 + 2.5 * ((20.3/122)^0.5 * (2^0.8)))
Vh = 531.8 m/s (approximately)
To calculate velocity for penetration with broken shell
If Cg=675, Bg=1.7
For a=90degree, D=20.3cm, G=122kg, S=2dm=200mm,
Vg = 675 * (20.3/122)^0.5 * (2^0.8) / (sin(90))^(0.5 + 1.7 * ((20.3/122)^0.5 * (2^0.8)))
Vg = 479.4 m/s (approximately)
For a=70degree, D=20.3cm, G=122kg, S=2dm=200mm,
Vg = 675 * (20.3/122)^0.5 * (2^0.8) / (sin(70))^(0.5 + 1.7 * ((20.3/122)^0.5 * (2^0.8)))
Vg = 533.1 m/s (approximately)
A graph will be like this:
To calculate the cross point of red line and black line:
The result of cross point (69.347091 degrees, 536.96904 m/s) is still an approximation because:
Vh = 650 * (20.3/122)^0.5 * (2^0.8) / (sin(69.347091))^(0.5 + 2.5 * ((20.3/122)^0.5 * (2^0.8)))
Vh = 536.9690444 m/s approximately
Vg = 675 * (20.3/122)^0.5 * (2^0.8) / (sin(69.347091))^(0.5 + 1.7 * ((20.3/122)^0.5 * (2^0.8)))
Vg = 536.9690438 m/s approximately
The interesting issue is "Did German 203mm guns have Cg=675, Bg=1.7 for their performance of armor piercing shells (Psgr)?"
Table 5 AKB 9215 from "Heft e Awa A IA 100/40 g.Kdos"
As we can see in the GerckeTierberg formula, because sin(90 degrees)=1, no matter what overall m values have for (sin(90))^m, it always equals to 1
As the result, for 90 degree normal attack, only C values will have their effects, generally a higher C value will result in a corresponding higher velocity for penetration
2 examples of 2dm=200mm with "Ch=650, Cg=675" will result in "Vh = 461.6 m/s, Vg = 479.4 m/s", but this will not fit in the graph 5 AKB 9215 as the velocity for a 203mm Psgr broken penetration against 200mm KC only needs around 450 m/s of velocity under normal attack.
If we only change Cg from 675 to 635 while other things remain the same:
For Cg=635, Bg=1.7, a=90degree, D=20.3cm, G=122kg, S=2dm=200mm
Vg = 635 * (20.3/122)^0.5 * (2^0.8) / (sin(90))^(0.5 + 1.7 * ((20.3/122)^0.5 * (2^0.8)))
Vg = 451 m/s approximately
For Cg=635, Bg=1.7, a=70degree, D=20.3cm, G=122kg, S=2dm=200mm
Vg = 635 * (20.3/122)^0.5 * (2^0.8) / (sin(70))^(0.5 + 1.7 * ((20.3/122)^0.5 * (2^0.8)))
Vg = 501.5 m/s approximately
A graph of 203mm Psgr against 200mm KC with Cg=635 will be like this:
A graph of 203mm Psgr against 203mm KC with Cg=635 will be like this:
203mm Psgr against KC plates greater than 80mm:
V schutz = 0.92 * V grenze according to ADM 213/951, whether this empirical calculation considers the effect of explosives of armor piercing shells is still uncertain especially for homogeneous plates:
203mm Psgr against Wh plates from 200mm to 20mm, when
Ch=Cg=660, Bh=Bg=2.03, D=20.3cm, G=122kg
original table 5 AKB 9262
The complete GerckeTierberg formula from ADM 213/951:
D: shell caliber (cm)
G: shell weight (kg)
V: striking velocity (m/s)
a: angle of incidence between trajectory and plate (degree)
S: plate thickness (dm)
V = C * (D/G)^0.5 * S^0.8 / (sin a)^m
m = 0.5 + Bx
x = (D/G)^0.5 * S^0.8
Certain examples of data set with 203mm Psgr vs KC plates will be listed:
For D=20.3cm, G=122kg, a=90degree, S=2dm=200mm, to calculate velocity for penetration with intact shell:
Ch=650, Bh=2.5
V = 650 * (20.3/122)^0.5 * (2^0.8) / (sin(90))^(0.5 + 2.5 * ((20.3/122)^0.5 * (2^0.8)))
Vh = 650 * (20.3/122)^0.5 * (2^0.8), since sin(90) = 1, which means for normal attacks, the formula seems to think B values (angular sensitivity of penetration difficulty?) has no effect, only C values (direct penetration difficulty?) effects the outcome.
Vh = 461.6 m/s (approximately)
For incidence angle a=70degree, D=20.3cm, G=122kg, S=2dm=200mm, Ch=650, Bh=2.5
Vh = 650 * (20.3/122)^0.5 * (2^0.8) / (sin(70))^(0.5 + 2.5 * ((20.3/122)^0.5 * (2^0.8)))
Vh = 531.8 m/s (approximately)
To calculate velocity for penetration with broken shell
If Cg=675, Bg=1.7
For a=90degree, D=20.3cm, G=122kg, S=2dm=200mm,
Vg = 675 * (20.3/122)^0.5 * (2^0.8) / (sin(90))^(0.5 + 1.7 * ((20.3/122)^0.5 * (2^0.8)))
Vg = 479.4 m/s (approximately)
For a=70degree, D=20.3cm, G=122kg, S=2dm=200mm,
Vg = 675 * (20.3/122)^0.5 * (2^0.8) / (sin(70))^(0.5 + 1.7 * ((20.3/122)^0.5 * (2^0.8)))
Vg = 533.1 m/s (approximately)
A graph will be like this:
To calculate the cross point of red line and black line:
The result of cross point (69.347091 degrees, 536.96904 m/s) is still an approximation because:
Vh = 650 * (20.3/122)^0.5 * (2^0.8) / (sin(69.347091))^(0.5 + 2.5 * ((20.3/122)^0.5 * (2^0.8)))
Vh = 536.9690444 m/s approximately
Vg = 675 * (20.3/122)^0.5 * (2^0.8) / (sin(69.347091))^(0.5 + 1.7 * ((20.3/122)^0.5 * (2^0.8)))
Vg = 536.9690438 m/s approximately
The interesting issue is "Did German 203mm guns have Cg=675, Bg=1.7 for their performance of armor piercing shells (Psgr)?"
Table 5 AKB 9215 from "Heft e Awa A IA 100/40 g.Kdos"
As we can see in the GerckeTierberg formula, because sin(90 degrees)=1, no matter what overall m values have for (sin(90))^m, it always equals to 1
As the result, for 90 degree normal attack, only C values will have their effects, generally a higher C value will result in a corresponding higher velocity for penetration
2 examples of 2dm=200mm with "Ch=650, Cg=675" will result in "Vh = 461.6 m/s, Vg = 479.4 m/s", but this will not fit in the graph 5 AKB 9215 as the velocity for a 203mm Psgr broken penetration against 200mm KC only needs around 450 m/s of velocity under normal attack.
If we only change Cg from 675 to 635 while other things remain the same:
For Cg=635, Bg=1.7, a=90degree, D=20.3cm, G=122kg, S=2dm=200mm
Vg = 635 * (20.3/122)^0.5 * (2^0.8) / (sin(90))^(0.5 + 1.7 * ((20.3/122)^0.5 * (2^0.8)))
Vg = 451 m/s approximately
For Cg=635, Bg=1.7, a=70degree, D=20.3cm, G=122kg, S=2dm=200mm
Vg = 635 * (20.3/122)^0.5 * (2^0.8) / (sin(70))^(0.5 + 1.7 * ((20.3/122)^0.5 * (2^0.8)))
Vg = 501.5 m/s approximately
A graph of 203mm Psgr against 200mm KC with Cg=635 will be like this:
A graph of 203mm Psgr against 203mm KC with Cg=635 will be like this:
203mm Psgr against KC plates greater than 80mm:
V schutz = 0.92 * V grenze according to ADM 213/951, whether this empirical calculation considers the effect of explosives of armor piercing shells is still uncertain especially for homogeneous plates:
203mm Psgr against Wh plates from 200mm to 20mm, when
Ch=Cg=660, Bh=Bg=2.03, D=20.3cm, G=122kg
original table 5 AKB 9262

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Re: Questions regarding to the performance of German 20.3cm guns
(sin a)^mBill Jurens wrote: ↑Mon Sep 13, 2021 2:00 am It's nice to know, for example, that m = 0.5 +Bx, but not very handy unless the user knows exactly what Bx is.
Bill Jurens
m describes the obliquity term wich with to some o degree of accuracy described the general oblique performance of german L/ 4,44,6 APC projectiles with crh ~1,3 and about 15% cap
depending on material properties, projectile shape, APcap and projectile lenghth the exponent varies
value of m should be around 1,2 for german projectiles
by adjusting B one can adjust the angular prediction somewhat to fit a real "Sollkurve"
Meine Herren, es kann ein siebenjähriger, es kann ein dreißigjähriger Krieg werden – und wehe dem, der zuerst die Lunte in das Pulverfaß schleudert!

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Re: Questions regarding to the performance of German 20.3cm guns
My thanks and appreciation for the last few postings here, which have helped to considerably clarify some of the points under discussion.
Bill Jurens
Bill Jurens