Thanks, Lloyd.... It's taken a lot of time to distill all the knowledge down to a few relatively simple concepts.... Every time I go over it, it seems I learn more, and get another piece of the puzzle and it makes a bit more sense.... Here is another way to think about the issue of caliber and long range.... You need the best BC possible to fight the wind, and maintain velocity (trajectory) and energy.... BC is a combination of SD and FF, in fact BC = SD / FF.... so for the best BC, you need a high SD (heavy for caliber) and a low FF (low drag shape).... If you maintain the SD as a constant, as you increase the caliber, the bullet gets fatter, but stays about the same length, so the shape has greater drag, and the FF goes up, reducing the BC.... If you maintain a constant shape (FF), as you increase the caliber the SD goes up in direct proportion to the caliber.... which increases the BC.... However, since there is a limit to the SD that you can push at any given velocity, with a given pressure and barrel length, you will a point where you can no longer drive the bullet fast enough because it is too heavy (ie too high an SD).... This produces an optimum caliber for any SD, to produce the best BC.... The larger calibers become the best choice only if the power available is enough to allow very high SDs.... as was the case with the 200 gr. Whiteout in .308 cal because of the Helium at high pressure.... For most airguns, a smaller caliber will be a better choice, providing you can get a fast enough twist barrel to use the proportionately long bullets.... That is why I had a 7" twist barrel made in .257 for my Monocoque PCP....Shaun, I can't answer that question without knowing the reservoir size and the pressure range from fill to refill....Bob
The Form Factor can be calculated, as it is in bullet design programs, or the Drag Coefficient can be measured and then the FF derived from that.... Empirical results always trump theory, of course.... Most often the FF is calculated from the reverse formula.... FF = SD / BC .... This formula is the easiest to understand if you consider the G1 projectile.... It is 1" caliber, weighs 1 lb., and by definition has a BC of 1.000.... It also has an SD of 1.000.... and hence a FF of 1.000.... If you had a bullet with the same SD but twice the drag, the BC would be 0.500 and the FF would be 2.000.... If you had a projectile that had an SD of 1.000 but only 2/3rds the drag, it would have a BC of 1.500 and a FF of 0.667.... IIRC, a sphere has a FF (G1) of about 1.55.... so it's BC will be about 64% of it's SD.... making it easy to calculate for a lead ball of any caliber.... The problem is, that if you use a different projectile for your model.... for example a sphere (GS) or a long spritzer boattail (G7).... while they can have the same SD, both the FF and the BC will be different, because you are using a different reference point.... So, just like the BC, the FF depends on the Drag Model used.... NOE's bullet design program calculates the FF (I presume using the G1 Model) and also the BC (which is G1 based), and has done so for all the designs I submitted.... The thing that is obvious is that the longer the bullet (in calibers) the lower the FF.... However, it is virtually the same number for a .224 cal BBT that is 4.0 cal. long as it is for a .457 cal that is 4.0 cal long.... What IS different is the BC, because the SD is higher for the larger caliber, when you scale a design up or down....This puts the edge in BC towards the bigger calibers, if you hold the shape a constant.... The problem is that you can't drive a 4.0 cal long .457 cal bullet (~635 gr.) at a reasonable velocity with reasonable pressures and barrel lengths.... The Physics just won't allow it, because the SD affects the ACCELERATION inside the barrel just like it affects the DECELERATION outside the barrel.... Bob