02-01-2022, 10:25 PM
OK, between consulting meetings I have racked my brain and finally it seems as though this can be solved by conservation of momentum, but you have to make some assumptions. So anybody who could do some reality-verification of loads that don't work as well as loads that do, then we can tweak and come up with a good answer.
Basically, conservation of momentum ('55 you were on the trail but needed to go a bit further) says the M*V product of the bullet transfers to an M'*V' of the ram, no loss of momentum.
I'm assuming that the mass of the bullet is much much less than the ram so even if the 2 objects stuck together the increase in mass would be negligible (and a bullet bouncing would transfer more so I left that out too). And assuming that the Ram does not get deformed by impact. In real life the bullet may bounce and the ram would deform, but we can adjust the problem to allow for that later. Also the Ram dimensions are approx 33x27.5 as given and I called a guy that sells and ships them, he said Rams are pretty close to 50 lb weight on the money (52 lb with cardboard shipping box).
M*V of the bullet (with V the unknown), and we need the delta-distance and delta time of the Ram's center of gravity (Ds/Dt for the V of the Ram).
Looking at a mock up picture of the ram I'm estimating the center of gravity is 13.5" off the ground and it has all of the ram's "weight", and the bullet hits the ctr of grav. The feet, as per NRA rules are to be such that the cog has to travel 1.25" rearward to tip over (thanks '55 for pointing to the NRA rule book). So the c.o.g has to move 1.25", that's Delta-distance, Ds, =1.25"/12"-per-foot.
How long does it take for the ram to tip when it gets hit? That will be the Delta-time for us, Dt. I'm assuming 0.15 seconds. We need then only to calculate for V1, the impact velocity of the bullet.
So M1V1=M2(D2s/D2t), add rearranging for the unknown, V1, V1= (m2/m1)*(Ds/Dt).
For a 130 grn bullet (0.008424 kg), and 50 lb (22.67 kg) ram, the needed impact velocity comes to 1868 ft/sec.
This also assumes that the ram's c.o.g moves in a straight line... in fact it moves slightly radially but the initial angle involved is of the order of 4* (4 degrees) and for such a short distance straight-line movement is probably an ok assumption.
1868 ft/sec at 550 yds for a 130 gr bullet, you'd need, depending on the BC... somewhere in the order of 2700 ft/sec Muzzle Velocity to start with (this from my 6.5 Creed, 20" barrel, I found a bullet I have that would work). Other bullets/MV/BC combos would work, at whatever xyz elevation/humidity you have, but this gives a somewhat decent approximation.
Apparently the NRA competition can be done at either 500 m or 500 yds. For 500m I just used 550 yds, which is again, estimate I can use in my BC calc.
Basically, conservation of momentum ('55 you were on the trail but needed to go a bit further) says the M*V product of the bullet transfers to an M'*V' of the ram, no loss of momentum.
I'm assuming that the mass of the bullet is much much less than the ram so even if the 2 objects stuck together the increase in mass would be negligible (and a bullet bouncing would transfer more so I left that out too). And assuming that the Ram does not get deformed by impact. In real life the bullet may bounce and the ram would deform, but we can adjust the problem to allow for that later. Also the Ram dimensions are approx 33x27.5 as given and I called a guy that sells and ships them, he said Rams are pretty close to 50 lb weight on the money (52 lb with cardboard shipping box).
M*V of the bullet (with V the unknown), and we need the delta-distance and delta time of the Ram's center of gravity (Ds/Dt for the V of the Ram).
Looking at a mock up picture of the ram I'm estimating the center of gravity is 13.5" off the ground and it has all of the ram's "weight", and the bullet hits the ctr of grav. The feet, as per NRA rules are to be such that the cog has to travel 1.25" rearward to tip over (thanks '55 for pointing to the NRA rule book). So the c.o.g has to move 1.25", that's Delta-distance, Ds, =1.25"/12"-per-foot.
How long does it take for the ram to tip when it gets hit? That will be the Delta-time for us, Dt. I'm assuming 0.15 seconds. We need then only to calculate for V1, the impact velocity of the bullet.
So M1V1=M2(D2s/D2t), add rearranging for the unknown, V1, V1= (m2/m1)*(Ds/Dt).
For a 130 grn bullet (0.008424 kg), and 50 lb (22.67 kg) ram, the needed impact velocity comes to 1868 ft/sec.
This also assumes that the ram's c.o.g moves in a straight line... in fact it moves slightly radially but the initial angle involved is of the order of 4* (4 degrees) and for such a short distance straight-line movement is probably an ok assumption.
1868 ft/sec at 550 yds for a 130 gr bullet, you'd need, depending on the BC... somewhere in the order of 2700 ft/sec Muzzle Velocity to start with (this from my 6.5 Creed, 20" barrel, I found a bullet I have that would work). Other bullets/MV/BC combos would work, at whatever xyz elevation/humidity you have, but this gives a somewhat decent approximation.
Apparently the NRA competition can be done at either 500 m or 500 yds. For 500m I just used 550 yds, which is again, estimate I can use in my BC calc.
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