For the body alone the resultant of the components of momentum W V -cos andW V sin 0 is W V -sec. lb, acting along 00', and so is unaltered.
If At seconds is the time during which the resistance of the air, R it), causes the velocity of the shot to fall Av(f/s), so that the velocity drops from v+2Av to v-2Av in passing through the mean velocity v, then (3) Rot = loss of momentum in second-pounds, =w(v-+ZAv)/g - w(v - 2 Av)/g = wAv/g so that with the value of R in (I), (4) At =wAv/nd2pg.
Equating the muzzle-energy and the work in foot-tons (2) E= w V 2 _XM.E.P.
2240 2g 12 (3) M.E.P. = w V 2 12 2240 2g B - C Working this out for the 6-in.
Then, by the principle of Archimedes, W = Vwo; or wo = W/V.
in the time At, during which the velocity falls from v+2Av to v-2Av, we have (12) RAs = loss of kinetic energy in foot-pounds =w(v+ZOv) 2 /g - w(v - ZOv) 2 /g=wvAv/g, so that (13) As =wvAv/nd 2 pg =CAS, where (14) AS = vAv/g p = vAT, and AS is the advance in feet of a shot for which C =1, while the velocity falls Av in passing through the average velocity v.
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