4th Ed Whole Body Damage from Explosives

[nuke11]

Those numbers do look more reasonable than the first set. I've been mulling over how to figure out what 1 dPW equals in psi.

I just found these calculators that will help figure out some of this stuff I'm looking for : https://www.un.org/disarmament/un-sa...d/calculators/

[mmartin798]

I had some time to kill today, so I went ahead and just calculated out the whole body damage for just about everything that explodes in the 4th edition and put them all into one PDF, one item per page. There are just a couple things that I need to point out.

I could not find any warhead specs for the Armbrust 300. But since the game has their damage effects virtually identical to the LAW, I assumed the same size warhead.

The M397 40mm HE grenade page is for a successful bounce back into the air. If it fails, just use the M381 40mm HE chart instead.

Hope you find this somewhat useful.

[nuke11]

Originally Posted by mmartin798

I had some time to kill today, so I went ahead and just calculated out the whole body damage for just about everything that explodes in the 4th edition and put them all into one PDF, one item per page. There are just a couple things that I need to point out.

I could not find any warhead specs for the Armbrust 300. But since the game has their damage effects virtually identical to the LAW, I assumed the same size warhead.

The M397 40mm HE grenade page is for a successful bounce back into the air. If it fails, just use the M381 40mm HE chart instead.

Hope you find this somewhat useful.

Very nice and helpful! I hope you plan on releasing the excel formula at some point.

For the specs of the Armbrust, I had to dig really deep into the archives, but the projectile is 0.99 kg and the filler is 0.16 kg of RDX (MP RE of 1.20) (it could also be some what closer to 0.19 kg) and it can penetrate 300 mm of rha.

[mmartin798]

Originally Posted by nuke11

Very nice and helpful! I hope you plan on releasing the excel formula at some point.

For the specs of the Armbrust, I had to dig really deep into the archives, but the projectile is 0.99 kg and the filler is 0.16 kg of RDX (MP RE of 1.20) (it could also be some what closer to 0.19 kg) and it can penetrate 300 mm of rha.

I do most of my work at home on my MacBook Pro using Numbers. However, I did export my master as an Excel spreadsheet. There are two sheets. The Explosives Data sheet and the Whole Body Damage sheet. I will explain the assumptions made and the formula used so people can understand where everything came from. If you just want the calculations, you can download the spreadsheet, goto the Explosive Data sheet, enter the weight of the explosive filler in Kg into cell B3, enter the explosive damage number from the game into cell B5 and then goto the Whole Body Damage sheet to see the results.

If you are still reading, I assume you want the math and such behind the sheet. I make two major assumptions in these calculations. The first is that the damage drops off exponentially with distance. The second, the point at which the Department of the Army Pamphlet 385-63 assumes a less than 1% chance of a ruptured eardrum from the over pressure generated by an explosive device, the minimum stand-off distance, is equal to 1 DPW. Using these assumptions, we have two known points with which to solve for the constants a and b in the general exponential function: f(x) = a(b)^x, where f(x) gives the damage at range x from the explosion. The values a and b are calculated on the Explosive Data sheet. When x=0, f(0)=a(b)^0. The term (b)^0 = 1, therefore at range 0 the value of a can be determined to be the full force of the explosion, Ex. To solve for b, we need our second known point. From our assumptions, we know at at the minimum stand-off distance (Msd), damage equals one. The Msd is given by D=23.4*(mass of explosive filler in Kg)^1/3. Therefore, F(Msd)=1. Substituting values we have, 1=Ex(b)^Msd. Dividing both sides by Ex and then taking the Msd-th root of both sides, we end up with b=(1/Ex)^(1/Msd). So our exponential equation is, f(x)=Ex*((1/Ex)^(1/Msd))^x. Easy peasy!

If there are any error in the math or questions about the assumptions, let me know.

[mmartin798]

Originally Posted by nuke11

Those numbers do look more reasonable than the first set. I've been mulling over how to figure out what 1 dPW equals in psi.

I just found these calculators that will help figure out some of this stuff I'm looking for : https://www.un.org/disarmament/un-sa...d/calculators/

I just found a CDC paper on the effects of overpressure on the human body. The rupture of ear drums in 1% of subjects happens at 5 psi.

Here is the whole paper:
https://www.cdc.gov/niosh/docket/arc...geChambers.pdf