MATHEMATICS FOR THE WARGAMER
Paul A Allcock paul.a.allcock@ntlworld.com
This is a simplified discussion (with mathematics!) on armour thicknesses and armour penetration.
Firstly, because it’s the easier of the two, armour thickness, and, specifically, sloping to increase the value. If you do a calculation to see exactly how much weight is saved by sloping the armour, compared to just increasing the thickness, you’ll find that the answer is NOTHING (in fact there may be a weight INCREASE due to a requirement for extra structural members). See the first drawing, DIAGRAM 1, giving sectional areas for vertical and sloped armour with the same HORIZONTAL thickness, the two areas are identical, so why bother sloping? Easy, because the effective increase is greater than the actual horizontal distance as, the greater the slope, the greater the chance of the armour-piercing round skidding off. This, believe it or not, is quantifiable. The working value is found in the graph, DIAGRAM 2, or the formula Teff =T(1/Cosa)1.67. Where T is the plate thickness, Cosa is the cosine of the slope angle a from VERTICAL, and Teff is the effective thickness.
To give two worked examples showing the saving in weight for thickness increase, if we consider 25o and 55o of slope respectively. The first example gives an effective thickness increase of 1.18 (to 3 significant figures), compared to an actual thickness increase of 1.10, -a 7% overall increase and hardly worth bothering with. But looking at 55o tells a different story; the actual increase is 1.74 times the plate thickness and the effective thickness works out at 2.53 times, a massive 92% greater than pure trigonometry would indicate and certainly worth incorporating into your next MBT design.
The other article gave the formula WV2=Kd3(t/d)n (actually it said MV2.. etc, but the original states W). This is a 19th century formula credited to DeMarre, and is, to understate matters, slightly unwieldy in this form, particularly as, being wargamers, we want to get at the meat as quickly as possible. Therefore, after some re-arrangement we arrive at
t=d(((WV2)/(Kd3))1/n)
which at least puts the known values together. As before, the units are imperial, I.e.
W is projectile weight in pounds
V is impact speed in feet per second
K is usually taken as 1000000 (see later)
d is penetrator diameter in inches
t is thickness penetrated in inches
n is 1.414 i.e. square root of 2

OK, we’ve got somewhere useful now but surely it’s easier to work in metric units? Sure it is, so we get:-
t=(20d/39)(((WV2)/(d3))1/n)
using kilograms for weight, millimetres for thickness and diameter and metres per second for velocity. [Strangely enough, I’ve been using this formula for nearly twenty years and I always remember the original, i.e. double-you-vee-squared-equals-kay-dee-cubed-into-tee-over-dee-to-the-en, and every time I want to use it I either have re-work from first principles or dig out my reference note book, still, that’s life). Back to the subject at hand.
We now have two formulae, one to calculate armour effective thickness and the second to work out how much of the tank we could punch holes in if it was sat on the end of the gun barrel, yeah that’s right dudes, all the reference books give MUZZLE VELOCITY. So we need a way of calculating the projectile velocity at any given range. There’s a good number of works & treatises on this subject, but: anybody into second and third order calculus? Me neither, but we do have a workable first-order approximation available. If you pick a sensible range increment, say in the region of 150 to 300 yards/ metres then a projectile will lose velocity over that range increment fairly constantly throughout its trajectory. The percentage loss is (roughly) proportional to projectile diameter, (strictly speaking its areal density and drag coefficient) so if we divide the range increment in yards or meters by the projectile diameter in millimetres (note I say projectile diameter, because some weapons use discarding sabot rounds), we come up with a figure that we can use. Example 250 metre increments; 25, 75 and 150 mm shot gives a velocity drop of 10%, 3.33% and 1.66% respectively see also the graph, DIAGRAM 3. These losses apply to the velocity as LAST CALCULATED, i.e. from the muzzle, after 250 metres, after 500, after 750 etc etc. (If you plot velocity against range the curve is asymptotic, i.e. it never actually reaches zero speed, OK real ones do but we’re ignore form drag, base drag, gravity drop and a dozen other things), example again using the 25mm round with say V0 of 1000 metres/sec:-
RANGE VELOCITY

0 1000
250 900
500 810
750 729
1000 656
1250 531
1500 478
1750 430
2000 387 (to zero decimal places)

i.e. if the loss is 10% then the velocity drops to 90% of the value it had at the last calculation. This does not work in real life!! i.e. if we take a 1000 metre increment for a 25mm round then the velocity drop is (1000/25)=40% so at 1000 metres the velocity should be 60% of 1000 metres/sec =600m/s, this is only 90% of the 1000 metre value in the chart above. This is why I say this is a first-order approximation only and that increments of 150 to 300 should be used, anything larger decays small-calibre rounds too rapidly, anything smaller lets large-calibre rounds travel forever.
It is not only possible, but relatively simple to combine these two equations into a BASIC program for computer or programmable calculator, which should reduce arithmetical errors at the very least. So outlined below is a BASIC program to do the calculations for you. Note that the BASIC that I use most is actually a language called BACIS and is from the mainframe at work, so some conversion may well be necessary. Conversion notes are given later.
100 PROMPT ‘MUZZLE VELOCITY METRE/SEC’ VEL REA
110 PROMPT ‘PROJECTILE DIA MILLIMETRES’DIA REA
120 PROMPT ‘PROJECTILE WT KILOGRAMS’ WT REA 
130 PROMPT ‘RANGE INCREMENT METRES ‘ INC INT
140 PROMPT ‘CALCULATE TO RANGE METRES’ RNG INT
150 LET DROP =<(100-(INC/DIA))/100>
160 LET RANGE=0
170 LOOP LOOP
180 LET PEN1=<W*V*V/(DIA**3)>
190 LET PEN2=<PEN1**(1/SQRT(2))>
200 LET PEN =<PEN2*20*DIA/39>
210 WRITE ‘RANGE IS @VRANGE VELOCITY IS @VVEL PENETRATION
IS @VPEN’
220 LET VEL=<VEL*DROP>
230 LET RANGE=<RANGE+INC>
240 BREAK LOOP IF<RANGE.GT.RNG>
250 ENDLOOP LOOP
260 ENDRUN

Okay, a couple of remarks on the above, the PROMPT is the direct equivalent of INPUT, i.e. you’re expected to give the infernal machine some answers, the REA and INT suffixes tell it that the number is REAL (e.g. may contain decimal places) or an INTEGER (who wants to use range increments of 110.678 metres?). The <> brackets are because that’s the only way this computer will accept expressions for calculation. ** is, of course, RAISED TO THE POWER OF, i.e. **2 is squared etc. LOOP LOOP is this machine’s equivalent of a FOR...NEXT loop statement, you have to tell it that a LOOP is a LOOP, you also have to give it a name (with stunning originality mine are either LOOP or FROG -personal foible). BREAK LOOP ends the loop when the range reaches or exceeds the limiting value you’ve entered and ENDLOOP LOOP is the NEXT equivalent in this language with the loop name added because the dumb thing can’t tell that the loop you want to stop just happens to be the one its running and not a different loop that’s in some other program. The only other strange thing is the @V prefix to the variable names in the print (WRITE) statement. WRITE tells it to write to the screen and the @V makes that particular variable the currently-looked-at one. Any other differences should be simple to work out.
One last point, earlier I made the remark ‘see later’ against the K value, this is because K works out at 106 for ‘standard’ armour and ‘standard’ armour-piercing rounds, the standard being established around WW1 or earlier, and later armour and projectiles, projectiles particularly, gave a ‘FIGURE OF MERIT’ value for K which could vary wildly. (And are very hard to get hold of, for example at FARNBOROUGH ‘88 I was having a chat with the guys from ROYAL ORDNANCE FACTORIES about APFSDS rounds, and made the comment that the samples on display didn’t look quite right. ‘Correct’, they said, ‘these are display versions and dimensionally inaccurate, so as not to give too much information away’. ‘OK’, I said, ‘what’s the real stuff?’ The answer was ‘Piss off’.) But, personally, I consider the above formula to be sufficiently accurate IF consideration is given to how the results are applied in a set of rules, bearing in mind that, according to some sources, how well the target suspension works has an effect on how easily it may be penetrated, and that no matter how many decimal places you calculate the armour thickness to, it will be a very rare shot that hits exactly in the horizontal plane or at 900 to the vehicle axis giving yet another angle to consider. My method is to use the theoretical penetration value and each vehicle has its armour thickness calculated for each face shown on the data card for game use. But in the armour section there are three columns, the first with the calculated thickness, the second with 150% of this value and the third with twice the calculated thickness. If the penetration of the weapon at the impact range is higher than the third (x2 number) I consider the tank automatically penetrated, (but see below), possibly straight through and out the other side, if less than twice but higher than x1.5 then a roll of 4, 5 or 6 is needed, if above the lower value and less than x1.5 then a roll of 2 or 3 is required. Note that a dice is ALWAYS rolled, a 1 indicating a dud round that has skidded off, tumbled, broken up or whatever.

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