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
Hunting
The Basics, Starting Out
"Knock down" power
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="Warren Jensen" data-source="post: 51603" data-attributes="member: 21"><p>I knew somebody would want to break it down.</p><p></p><p>You must add the weight of all of the ejecta which includes the weight of the powder combusted. Your Free Recoil Impulse is 5.0, you Free Recoil Velocity is 16.1 f/s, and you Free Recoil Energy is 40.5 ft-lbs. at the butt of the rifle. When calulating perceived recoil most figure the weight of the rifle, which is incorrect. The weight and resistance of your shoulder has to also be inputted, if the rifle is being held against your shoulder. But this will vary from shooter to shooter and rifle to rifle. If you are calculating this in terms of "knockdown" then you would have to assume movement of the entire body in which case the entire weight of the shooter must be added to the weight of the rifle for recoil calulations.</p><p></p><p>One question is the deceleration energy dispursed more quickly than the acceleration energy is accumulated, thereby making the terminal end more abrupt. You have to make several assumptions. First is that the bullet reaches it's maximum velocity as it exits the barrel. I have seen it argued that the bullet may actually accelerate for a short distance after exiting the barrel as the compressed gasses are accelerating past the bullet and before undisturbed air is encountered by the front of the bullet. If it does it is not much. Anyway, the distance traveled by the bullet from it's point of rest to it's maximum velocity is around 20" to 24" on 22" to 26" barrels. This distance is traveled in around 3 milliseconds, depending on cartridge, load and rifle. It is very seldom less than 2 milliseconds or greater than 4 milliseconds. On the receiving end (the animal) the deceleration occurs in approximately the same time and distance. The deceleration time and distance can sometimes be twice or greater than that, which will serve to extend or dilute the energy dissipation. It is very seldom half the 3 millisecond and 20" to 24" value with big game rounds. The point is the acceleration time and deceleration time are relatively equal, ergo there is no disparity between these to separate energy impulse.</p><p></p><p>The point is if you and your rifle weigh approximately 200 lbs. and the animal you are shooting weighs approximately 200 lbs., then if the bullet "knocksdown" the animal it would have to "knockdown" you when you fired your rifle. If it "knocked down" a 600 lb. elk, it would pert near have to flatten you.</p><p></p><p>[ 11-30-2001: Message edited by: Warren Jensen ]</p><p></p><p>[ 11-30-2001: Message edited by: Warren Jensen ]</p></blockquote><p></p>
[QUOTE="Warren Jensen, post: 51603, member: 21"] I knew somebody would want to break it down. You must add the weight of all of the ejecta which includes the weight of the powder combusted. Your Free Recoil Impulse is 5.0, you Free Recoil Velocity is 16.1 f/s, and you Free Recoil Energy is 40.5 ft-lbs. at the butt of the rifle. When calulating perceived recoil most figure the weight of the rifle, which is incorrect. The weight and resistance of your shoulder has to also be inputted, if the rifle is being held against your shoulder. But this will vary from shooter to shooter and rifle to rifle. If you are calculating this in terms of "knockdown" then you would have to assume movement of the entire body in which case the entire weight of the shooter must be added to the weight of the rifle for recoil calulations. One question is the deceleration energy dispursed more quickly than the acceleration energy is accumulated, thereby making the terminal end more abrupt. You have to make several assumptions. First is that the bullet reaches it's maximum velocity as it exits the barrel. I have seen it argued that the bullet may actually accelerate for a short distance after exiting the barrel as the compressed gasses are accelerating past the bullet and before undisturbed air is encountered by the front of the bullet. If it does it is not much. Anyway, the distance traveled by the bullet from it's point of rest to it's maximum velocity is around 20" to 24" on 22" to 26" barrels. This distance is traveled in around 3 milliseconds, depending on cartridge, load and rifle. It is very seldom less than 2 milliseconds or greater than 4 milliseconds. On the receiving end (the animal) the deceleration occurs in approximately the same time and distance. The deceleration time and distance can sometimes be twice or greater than that, which will serve to extend or dilute the energy dissipation. It is very seldom half the 3 millisecond and 20" to 24" value with big game rounds. The point is the acceleration time and deceleration time are relatively equal, ergo there is no disparity between these to separate energy impulse. The point is if you and your rifle weigh approximately 200 lbs. and the animal you are shooting weighs approximately 200 lbs., then if the bullet "knocksdown" the animal it would have to "knockdown" you when you fired your rifle. If it "knocked down" a 600 lb. elk, it would pert near have to flatten you. [ 11-30-2001: Message edited by: Warren Jensen ] [ 11-30-2001: Message edited by: Warren Jensen ] [/QUOTE]
Insert quotes…
Verification
Post reply
Forums
Hunting
The Basics, Starting Out
"Knock down" power
Top