Forums
New posts
Search forums
What's new
Articles
Latest reviews
Author list
Classifieds
Log in
Register
What's new
Search
Search
Search titles and first posts only
Search titles only
By:
New posts
Search forums
Menu
Log in
Register
Install the app
Install
Forums
Chatting and General Stuff
General Discussion
Stubby--Lapua
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Reply to thread
Message
<blockquote data-quote="Pdvdh" data-source="post: 158266" data-attributes="member: 4191"><p>Dave, </p><p></p><p>I understand and agree with what you are saying. But I happen to be an engineer and studied the physics that pressure acting over an area determines the force exerted over that area. In the example we are discussing here, if equal pressures (say 60,000 psi) are generated in two different cartridges over an equal duration of time, the force exerted on the base of the bullet with the larger diameter will be greater than the force exerted on the smaller bullet diameter. The area across the base of a 7mm bullet is about 0.0634 square inches. The area across the base of a .338 bullet is about 0.0897 square inches. The pressure generated by the cartridge acts against the base of each bullet. In the case of the 7mm bullet, the 60,000 psi creates a force of (60,000 lbs/square inch) X (0.0634 square inch) = <em>3804</em> lbs of force. With the .338 bullet the 60,000 psi pressure creates a force of (60,000 lbs/square inch) X (0.0897 square inch) = <em>5382</em> lbs of force.</p><p></p><p>We can see that even though the pressure is equal in each bore, much more force is applied against the base of the .338 bullet than the 7mm bullet, because the force exerted is a function of the area that the pressure is able to act on. And I think we can all agree that the greater the force acting on two different bullets of equal weight, the faster the bullet will be accelerated down the bore. Which is why bullets of <u>equal</u> weight will always be driven faster in a larger caliber, provided that the same pressures are maintained over the same length of time. So this simple example has a straightforward explanation.</p><p></p><p>What Kirby describes involving sectional density involves a greater number of variables (for example different bullet weights and even greater differences in bearing surfaces), and I find it interesting that in his emperical testing, he's come up with a general rule of thumb. I'm sure there is a mathematical explanation for it also, but it's more complicated than determining the forces that pressures exert on the base of bullets of different diameters. So I'll leave that one for the professors!</p><p></p><p>FWIW /ubbthreads/images/graemlins/crazy.gif</p></blockquote><p></p>
[QUOTE="Pdvdh, post: 158266, member: 4191"] Dave, I understand and agree with what you are saying. But I happen to be an engineer and studied the physics that pressure acting over an area determines the force exerted over that area. In the example we are discussing here, if equal pressures (say 60,000 psi) are generated in two different cartridges over an equal duration of time, the force exerted on the base of the bullet with the larger diameter will be greater than the force exerted on the smaller bullet diameter. The area across the base of a 7mm bullet is about 0.0634 square inches. The area across the base of a .338 bullet is about 0.0897 square inches. The pressure generated by the cartridge acts against the base of each bullet. In the case of the 7mm bullet, the 60,000 psi creates a force of (60,000 lbs/square inch) X (0.0634 square inch) = [i]3804[/i] lbs of force. With the .338 bullet the 60,000 psi pressure creates a force of (60,000 lbs/square inch) X (0.0897 square inch) = [i]5382[/i] lbs of force. We can see that even though the pressure is equal in each bore, much more force is applied against the base of the .338 bullet than the 7mm bullet, because the force exerted is a function of the area that the pressure is able to act on. And I think we can all agree that the greater the force acting on two different bullets of equal weight, the faster the bullet will be accelerated down the bore. Which is why bullets of <u>equal</u> weight will always be driven faster in a larger caliber, provided that the same pressures are maintained over the same length of time. So this simple example has a straightforward explanation. What Kirby describes involving sectional density involves a greater number of variables (for example different bullet weights and even greater differences in bearing surfaces), and I find it interesting that in his emperical testing, he's come up with a general rule of thumb. I'm sure there is a mathematical explanation for it also, but it's more complicated than determining the forces that pressures exert on the base of bullets of different diameters. So I'll leave that one for the professors! FWIW [img]/ubbthreads/images/graemlins/crazy.gif[/img] [/QUOTE]
Insert quotes…
Verification
Post reply
Forums
Chatting and General Stuff
General Discussion
Stubby--Lapua
Top