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PS: This answer have a few minor mistakes and incorrect input values I will have to change when I get the time.

If you know the armor thickness and fracture toughness you can calculate how much kinetic energy you’ll need for penetration to happen. For instance, you need 55 Joules in order to penetrate a 1 mm mild steel plate with a fracture toughness of 235 kJ/m2 with a 10*10 mm square shaped bodkin arrowhead 40 mm deep. Now you can calculate the kinetic energy you need in order to penetrate a 1.5 mm wrought iron plate. The fracture toughness of iron is around 150 kJ/m2.

55 * 1.5^1.6 * (150/235) = 67.16 Joules

If the arrow strikes at a 15 degree angle you do it like this:

55 * 1.5^1.6 * (150/235)/cos(15) = 69.53 Joules

The fracture toughness of other armor qualities are as follows:
Wrought iron 120–150 kJ/m2
Low carbon steel 180–210 kJ/m2
Medium carbon steel 240–260 kJ/m2
Tempered medium carbon steel 300–400 kJ/m2
Tempered medium carbon steel around 1480 400–500 kJ/m2

A perfect 1.8 mm breastplate with a fracture toughness of 500 kJ/m2 will take:

55 * 1.8^1.6 * (500/235) = 299.7 Joules

If the size of the arrowhead changes, to, say, 7*7 mm you do this and switch out 55 with the number you get.

((square root(7 * 7 * 2)) * 2) *1.767766952966 )+ 5)
((square root 98) * 2) *1.767766952966 ) + 5)
((9.89949 * 2) * 1.767766952966) + 5)
((19.79898 * 1.767766952966) +5) = 40 Joules

If the arrowhead is 8*8 mm, switch out those two 7s with 8s in the firs line.
We use the Pythagorean theorem to calculate the hypotenuse of the 10 *10 mm arrowhead. We multiply the number we get with 2 because the arrowhead looks like a cross from the front. The coefficient, 1,767, is derived from the fact that you need 30 Joules to leave a 5*5 mm hole and 55 Joule to leave a 10*10 mm hole. If you make a ladder of calculations to see how this changes with smaller holes, you’ll end up with 1.767 * the length of the cutting edges of the arrowhead + 5 Joule on top of this answer. It’s the same with lozenge arrowheads and triangular arrowheads. When I say length I mean as you see the arrowhead from the front. This holds true with the tests Alan Williams has done and the test results from The Warbow trials 2005.

This is how you calculate 1.767.

(30-10)/(((square root(5*5*2))*2)-((Square root(1*1*2))*2))

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