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Hunting
Long Range Hunting & Shooting
I am very dissapointed with Berger bullets regarding the 338 hybrid bullet.
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<blockquote data-quote="BryanLitz" data-source="post: 403248" data-attributes="member: 7848"><p>Your approach is correct (f=ma; Newtons second law)</p><p></p><p>However,</p><p></p><p>.169*.169*3.14*65000 = 5829 lb, not 1856 lb </p><p>(1856 results from .169*.169*65000, in other words, you left out pi).</p><p></p><p>So then;</p><p></p><p>5829/.0429 = 135,881 G's.</p><p></p><p>This number is higher than the range I gave (80,000 to 100,000 G's) for two reasons:</p><p>1) I based my range on lower pressure.</p><p>2) The above calculation (f=ma) ignores friction. I've estimated (using Quickload) that you need to scale down the 'a' implied by the simple f=ma force balance by about .86 in order to get close to the actual bullet acceleration, accounting for friction. This is a rough estimate, but is better than ignoring it.</p><p></p><p>Thanks for running the numbers, always good to have someone checking the math.</p><p></p><p>-Bryan</p></blockquote><p></p>
[QUOTE="BryanLitz, post: 403248, member: 7848"] Your approach is correct (f=ma; Newtons second law) However, .169*.169*3.14*65000 = 5829 lb, not 1856 lb (1856 results from .169*.169*65000, in other words, you left out pi). So then; 5829/.0429 = 135,881 G's. This number is higher than the range I gave (80,000 to 100,000 G's) for two reasons: 1) I based my range on lower pressure. 2) The above calculation (f=ma) ignores friction. I've estimated (using Quickload) that you need to scale down the 'a' implied by the simple f=ma force balance by about .86 in order to get close to the actual bullet acceleration, accounting for friction. This is a rough estimate, but is better than ignoring it. Thanks for running the numbers, always good to have someone checking the math. -Bryan [/QUOTE]
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Long Range Hunting & Shooting
I am very dissapointed with Berger bullets regarding the 338 hybrid bullet.
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