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AR15.COM
3/30/2005 3:49:44 PM EDT
Ok. I am interested in ballistic coefficients and the effects of air friction/drag on a bullet. Yet I understand very little about ballistic coefficients and drag coefficients, and how to incorporate them into the standard range equations. Could somone please explain the difference between BC/DC if there is one? And how do i use it to find the drag force.

Here's what im talking about:

Say I fire a 230gr .45ACP round out of a glock 21, it leaves the barrel with a velocity of about 830 feet per second. I am about 5 feet 9 inches, so if the gun is fired at eye level it has an initial height of about 5 feet 4 inches or 5.333 feet. To make things simple we will assume that the angle of the bore with respect to the ground is 0( They are parallel).

Anyone who has had a highschool physics class should recognize the following equations.



Now we said that the initial height and muzzle velocity are 5.33 ft and 830 ft/s. We know gravity is -32.2 ft/s/s. We will also say the initial x-position is 0.  Since the bullet is not falling until after it exits the barrel, the initial y-velocity is 0.

that makes two functions of time:


These are the parametric equations for the position of the bullet. Notice that we do not know a.
a will be the negative acceleration caused by the drag force of the bullet going through the air.

For the simple case we usually neglect air friction/drag.

Neglecting the drag on the bullet makes the a term drop out(a is 0):




By graphing those parametric eqqations this is what the trajectory of the bullet should look like



Now in reality there are forces that act against the bullet slowing it down, it is always directly opposite of the bullet.

My question is what are Ballistic Coefficients's, what are Drag coefficients. How the hell do I use them to find the force of drag on the bullet, and then can i simply set it equal to ma to find the negative acceleration for that x equation?. Im gonna neglect the force of drag of the bullet falling because that is much more negligible.





3/30/2005 3:56:08 PM EDT
[#1]
There is also some lift created by the nose of the bullet, especially if it is pointed.

Horizontal and vertical components of the trajectory are independent.

E=mc^2


3/30/2005 4:00:59 PM EDT
[#2]
All I can tell you is when I shot steel plates the gong at 50 yards was about
a 3 foot above aim to hit it.
3/30/2005 4:11:04 PM EDT
[#3]
I'm reminded of what a friend of mine, who has attended McMillan Sniper School for his Instructor's ticket, said after his return.

Seems there was a REAL rocket scientist, also in attendance at the class, for whatever reason.  He started to ask about the earth's rotation and how it would affect POI after the shot.

The instructor's response had something to do with "paralysis through over-analysis."
3/30/2005 4:17:59 PM EDT
[#4]

Quoted:
I'm reminded of what a friend of mine, who has attended McMillan Sniper School for his Instructor's ticket, said after his return.

Seems there was a REAL rocket scientist, also in attendance at the class, for whatever reason.  He started to ask about the earth's rotation and how it would affect POI after the shot.

The instructor's response had something to do with "paralysis through over-analysis."



Bah, thats different. This actually has a measureable difference.
3/30/2005 4:26:50 PM EDT
[#5]
The BC can be estimated from bullet weight, shape, and diameter.  2/3BL x (Density of Pb) = Weight.  Where B = cs area of bullet and L = bullet length.  Remember to respect the units.  The rest is fudging with a constant. *  

drag force = drag coeff * x 1/2 x density of air x velocity x velocity x B

You have all the equations you need with your post.   Have your program plot x and y and integrate with respect to time.

I could never get the program to produce good results without another decay function with another fudge factor.

Have fun.

* Compare your fudge factor's output to published ballistic tables.