c-g-designs

ON THE QUESTION OF FIRING PINS DIAMETERS:  

          A 1,6mm(.0625”)  requires some 33% LESS than its 2mm (.080’’) equivalent. The small protrusion difference even increase slightly this difference. Assuming mechanical properties  and tthickness of the primer cup being equal,it would be easy to determine the impact energy necessary  for the indentation of the two firing pins diameters,taking also Into account the protrusion difference (1,4mm-.056’’ and 1,6mm-.0625’’ ) respectively.

          Not trying to bore anyone with other concept considerations, my firm belief  is that the firing tip diameter reduction should go pair with the reduction of the impact energy

          Indented  area                            1,6mm firing pin:

Cylindric éformation  (corresponding to protrusion 1:4mm– r =0,6 or:  0,8 x 0,8 x 3,14  x 0,6  =       1,21 mm².

Hemispheric deformation  (1,6 x 1,6 x 3,14 / 2 =  4,02 mm².

                              TOTAL (deformation at full mpact)  = 1,21 + 4,02 = 5,23mm².

Indented area firing pin  2mm:

          Cylindrical deformation: protrusion 1;6 –r 1mm= 0,6mm  ) or 1 x 1  x 3,14 x 0,6 = 1,90 mm².

          Hémispheric deformation :  2 x 2 x 3,14 / 2 = 6,28 mm².

                             TOTAL (défprmation at impact : 1,90  + 6,28 =              8,18mm².  

           Assuming the primer cup thickness a constant,it is easy to conclude that the smaller firing pin require 1/3 LESS energy than the larger one (5,23 /  8,18 = 0,64. Also, SR primers are thinner than LR

          I would have gone further but I have little information on primer cups material charactéristiques and thicknesses.

          An estimate, based on a 0,5mm thickness and CuZn30, condition H111 or H112 would give approximately:

          70 Newton for the  1,6mm firing pin.

          100 Newtons for the 2mm firing pin.

                    This is perhaps go quite far, but the difference deserve consideration and is to compare with S. Otteson conclusions on accuracy.

          R.G.C