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  • #76
    Originally posted by ArmySGT. View Post
    I was going with 0.1 gram for a modern grenade.
    Based on data available on the M67, 0.1 gram is about right. The fragmentation coil is about 140 g and tests at the Armament Research, Development & Engineering Center in Picantinny, NJ showed about 1335 fragments.

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    • #77
      so how about this.....

      M67 Fragmentation grenade..

      using the spherical charge equation.

      (7620/(7620/0.333)=((0.1/0.18)+3/5)-1/2 <--- that in negative one half as an exponent.



      22882 = 0.65555555555555555555555555555556

      Fragment is moving a 0.66 meters per second

      But, then E-factor is based on speed and diameter of the projectile..... without weight as a factor.

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      • #78
        I think your algebra is off. I solved the equation for V and got the equation below.

        Substituting the values I get V = 2730 m/s
        Attached Files

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        • #79
          Originally posted by mmartin798 View Post
          I think your algebra is off. I solved the equation for V and got the equation below.

          Substituting the values I get V = 2730 m/s
          Wouldn't surprise me a bit. Algebra is not one of my best subjects. I was trying to use the calculator built into Windows too. My scientific calculator is packed away.

          So, with a velocity.... we need a diameter for the e-factor formula.

          Modern grenade 1mm or 0.5 mm for a fragment.......

          WW2 grenade, 3mm 5mm

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          • #80
            We might not need to know the size, but use a modified E-Factor calculation. The E-Factor is a measure of the wound cavity. As wound cavity is caused mostly by momentum conservation and transfer, it is the momentum of the fragment that matters. Momentum is just mass x velocity, which we have for fragmentation. We just need to find how the bullet diameter is related to mass and make the substitution.

            Edit

            Been giving this some more thought. We really don't need to mess with the formula if we have an idea of the fragment size, and that size can be calculated if we make some assumptions. If we assume the fragments are roughly cylindrical with diameter approximately the same as the height, we can use the density of steel and the mass we know to calculate the volume and extrapolate the dimensions. Sounds like a lunch time project.
            Last edited by mmartin798; 02-17-2016, 08:52 AM. Reason: Second cup of coffee

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            • #81
              Originally posted by ArmySGT. View Post
              So, with a velocity.... we need a diameter for the e-factor formula.

              Modern grenade 1mm or 0.5 mm for a fragment.......
              Ok, density of steel = 8.05 g/cm^3

              Mass of an M67 fragment = 0.1 g

              Volume of an M67 fragment = 0.1/8.05 = .01242 cm^3

              Volume of a cylinder is given by:
              V= pi * r^2 * h

              Assuming height equals diameter gives:
              V = pi * r^2 * 2 * r
              V = 2 * pi * r^3

              Solving for r gives:
              r = (V/(2*pi))^(1/3)

              Doing the substitutions and doubling the radius to get the diameter and height we get 0.251 cm or 2.51 mm

              If we assume a softer steel with a density of 7.75 g/cm^3, the diameter comes to 2.54 mm
              Last edited by mmartin798; 02-17-2016, 11:41 AM.

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              • #82
                Damn, that gives an M67 fragmentation grenade an Efactor of 18 per fragment and you roll 1d20 to determine the number of fragments that hit the target.

                Guess, players aren't going to shrug that off like they did with resistweave and the old 4 points per fragment.

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                • #83
                  Originally posted by ArmySGT. View Post
                  Damn, that gives an M67 fragmentation grenade an Efactor of 18 per fragment and you roll 1d20 to determine the number of fragments that hit the target.

                  Guess, players aren't going to shrug that off like they did with resistweave and the old 4 points per fragment.
                  I need to fix a couple things. First, I made an error with my equation illustration earlier and did a unit conversion wrong for the fragment velocity. The correct equation is below and the new M67 fragment velocity is 3278 m/s. So that makes the second thing to correct the EFactor for each fragment to 21. Sorry for the confusion there Sgt.

                  This means one of two things. We may need to adjust the number of fragments that strike a person to something lower or we just start carrying a big bag of M67s and lots of body armor.
                  Attached Files

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                  • #84
                    It turns out that short cylinders have a high drag coefficient and lose velocity very quickly. This chart shows the effect of range on velocity and EFactor of M67 fragments from 0 to 50m.
                    Attached Files

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                    • #85
                      Probably why they hammer into you that 5 meter spread... lethal radius is 10 meters and no two soldiers should get caught in it.

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                      • #86
                        another job well done by the way!

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                        • #87
                          For those who want to calculate values for other grenades, here is a spreadsheet you can use. Remember the assumptions used in these calculations are that the fragments are generated from a wire coil and the fragments are all uniform and the same length as the diameter of the wire. The fragments are therefore short cylinders and have a drag coefficient of 1.15.
                          Attached Files

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                          • #88
                            Originally posted by nuke11 View Post
                            Yes this is still my current work. I've been digging around for Soviet information, but it seems to be a bit harder to find for some reason.
                            Did I send you the Defense Intelligence Agency munitions guide I have

                            I got it off of Scribd.... Which has been a good source for these.

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                            • #89
                              Ok, the problem you have at the moment is the document is based on TNT equivalent of an RE of 1.00. The game uses C4 as it's equivalent for an RE of 1.00 and TNT has an RE of 0.75 in the game.

                              You will need to adjust all of the formulas from the document accordingly.

                              Doesn't the M67 have a solid steel case that also needs to be taken into account for the fragments Here is a good website that shows what some of the fragments look like: http://machinesforwar.blogspot.ca/2012/03/m67.html

                              Some more images of the inside of the case and what the fragments look like: http://www.big-ordnance.com/grenades...CutawayM67.jpg and http://www.big-ordnance.com/grenades/loworderM67.JPG

                              The M26 is the grenade with a spiral wound wire core: http://img.photobucket.com/albums/v1...ut/OR-034A.jpg
                              Last edited by nuke11; 02-17-2016, 07:39 PM.

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                              • #90
                                Originally posted by nuke11 View Post
                                Ok, the problem you have at the moment is the document is based on TNT equivalent of an RE of 1.00. The game uses C4 as it's equivalent for an RE of 1.00 and TNT has an RE of 0.75 in the game.

                                You will need to adjust all of the formulas from the document accordingly.
                                The Gurney equations don't care about the RE at all. It only uses the uses the Gurney constant which does not necessarily correlate with RE. Taking the example of making C4 1.00 and TNT 0.75, we would expect TNT to perform at 75% of C4. But that is not the case. C4 has a Gurney constant of 2530 m/s and TNT is 2438 m/s, which makes TNT perform at 96% of C4. The equation in my spreadsheet uses the approximation of the Gurney constant, which is one third of the detonation velocity. For C4 that works out to 8200/3 or 2733 and TNT is 6900/3 or 2300. That makes TNT 84% of C4, which is closer to the 75% you are looking for already.

                                Originally posted by nuke11 View Post
                                Doesn't the M67 have a solid steel case that also needs to be taken into account for the fragments Here is a good website that shows what some of the fragments look like: http://machinesforwar.blogspot.ca/2012/03/m67.html

                                Some more images of the inside of the case and what the fragments look like: http://www.big-ordnance.com/grenades...CutawayM67.jpg and http://www.big-ordnance.com/grenades/loworderM67.JPG

                                The M26 is the grenade with a spiral wound wire core: http://img.photobucket.com/albums/v1...ut/OR-034A.jpg
                                The method of determining fragment size was an approximation intended to simplify the math and make fragments of uniform dimensions that can be directly plugged into the EFactor equation. While we could take the mass of the case and the diameter of the grenade to determine an approximate thickness and then divide the surface area up into the number of fragments to get the area of the face which is more accurate as to what happens with the case of the M67, we are then stuck with fragments that are essentially flat tiles. There is no diameter that we can reasonably put through the EFactor equation as it is shown in the rules. Even though we can calculate the kinetic energy of these flat tile fragments, that is still not enough information to determine the wound generating capability of the fragments. KE alone has been shown time and time again to bot be reliable in wound cavity calculation. The EFactor used in the rules is similar to wound approximation models for significant number to ballistic rounds. So while the wire core is not what happens with an actual M67, it works better for incorporating with the EFactor of other weapons in the game.

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