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
Rifles, Reloading, Optics, Equipment
Rifles, Bullets, Barrels & Ballistics
Velocity Magnatude Questions
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="Kevin Thomas" data-source="post: 384226" data-attributes="member: 15748"><p>Eddybo,</p><p> </p><p>after taking a peek at the link you posted to Later's comments, it seems to me that things were just taken a bit out of context. I didn't get that he was saying anything about identical bullets, or that the slower bullet would go further because of air drag, etc.. He's correct about the faster bullet suffering more resistance. This resistance is roughly proportionate to the square of the velocity, sort of like how kinetic energy is calculated. Double the velocity and you've roughly quadrupled the atmospheric resistance. However, once you compare two identical bullets that have arrived at the same velocity, the trajectory, energy, drag, etc., is going to be identical. </p><p> </p><p>Here's a simple illustration; Maximum range is acheived at a departure angle of 45 degrees in a vacuum. No atmospheric resistance involved in that case, so the departure angle is the same for the max range of any projectile. In the real world, that departure angle needs to be about 30-33 degrees for a typical centerfire rifle, due to the atmospheric resistance, i.e., drag. However, a typical mortar usually hits its maximum range when the departure angle is close to 45 degrees, just like the theoretical vacuum trajectory. The difference here is velocity. The mortars are typically fairly low velocity, and as a result suffer much less drag compared to a higher velocity projo. Any clearer now?</p><p> </p><p>Hope this helps,</p><p> </p><p>Kevin Thomas</p><p>Lapua USA</p></blockquote><p></p>
[QUOTE="Kevin Thomas, post: 384226, member: 15748"] Eddybo, after taking a peek at the link you posted to Later's comments, it seems to me that things were just taken a bit out of context. I didn't get that he was saying anything about identical bullets, or that the slower bullet would go further because of air drag, etc.. He's correct about the faster bullet suffering more resistance. This resistance is roughly proportionate to the square of the velocity, sort of like how kinetic energy is calculated. Double the velocity and you've roughly quadrupled the atmospheric resistance. However, once you compare two identical bullets that have arrived at the same velocity, the trajectory, energy, drag, etc., is going to be identical. Here's a simple illustration; Maximum range is acheived at a departure angle of 45 degrees in a vacuum. No atmospheric resistance involved in that case, so the departure angle is the same for the max range of any projectile. In the real world, that departure angle needs to be about 30-33 degrees for a typical centerfire rifle, due to the atmospheric resistance, i.e., drag. However, a typical mortar usually hits its maximum range when the departure angle is close to 45 degrees, just like the theoretical vacuum trajectory. The difference here is velocity. The mortars are typically fairly low velocity, and as a result suffer much less drag compared to a higher velocity projo. Any clearer now? Hope this helps, Kevin Thomas Lapua USA [/QUOTE]
Insert quotes…
Verification
Post reply
Forums
Rifles, Reloading, Optics, Equipment
Rifles, Bullets, Barrels & Ballistics
Velocity Magnatude Questions
Top