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# new twist on b.c.'s

#15
09-22-2004, 05:32 PM
 Platinum Member Join Date: Sep 2004 Location: on the rifle range in Utah Posts: 2,704
Re: new twist on b.c.\'s

Jeff Rogers,
Yes, I did receive your email. Thank you. I have already added it to my documents so I can refer to it often. Sorry I didn't tell you sooner upon receiving.
goodgrouper
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#16
09-23-2004, 07:13 AM
 Silver Member Join Date: Jan 2004 Location: New Mexico Posts: 113
Re: new twist on b.c.\'s

<BLOCKQUOTE><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><HR>
One thing I didn't catch though was the "CD" you mentioned. What is that, and how do you calculate it?
<HR></BLOCKQUOTE>

CD is the drag coefficient. It's what everybody is trying to avoid by using BCs. The drag of the bullet is proportional to the cross sectional area of the bullet, the density of the air, CD and the velocity squared [there are some constants and the mass in there too]. A drag function, associated with a BC, is used to find the CD of the bullet as a function of mach number. So to find a trajectory, you can use a BC and a drag function or just the CD. The idea is that the curve of CD versus mach number is the same for like bullets, e.g. flat base, boattails, etc. The drag function models the shape of the curve and the BC adjusts it so that the magnitude is correct. It's not perfect -- just take a look at Sierra's lists of BCs for different bullets, they change with bullet velocity. If they used a G7 drag function which has a different curve, they could probably get away with fewer BCs. Take a look at my MPM program (it's free) -- is uses the CD and a few other coefficients to find the trajectory.

<BLOCKQUOTE><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><HR>I was wondering if you could maybe tell me the formula for figuring bc.<HR></BLOCKQUOTE>

I just do it iteratively. Find a trajectory and adjust it until it works. Not real elegant though. Unfortunately, there isn't a simple formula because the equations of motion are a differential equation. There is no simple solution to the equations. You must integrate them to get the trajectory.

<BLOCKQUOTE><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><HR>I called Art Pejsa and asked him for the formula, but he was such an arrogant bugger I couldn't stand being on the phone long enough with him to get a straight answer.<HR></BLOCKQUOTE>

Yeah, I talked to him ONCE too.

[ 09-23-2004: Message edited by: JBM ]
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#17
09-23-2004, 09:34 AM
 Silver Member Join Date: Jan 2004 Location: New Mexico Posts: 113
Re: new twist on b.c.\'s

I may not have answered this, so I'll follow up my own post...

You can pretty easily calculate the CD of the bullet for "small" changes in velocity and time using the formula:

CD = dv/dt*(8*SD)/(Pi*Density*V*V)

where dv is the change in velocity, dt is the change in time, SD is the sectional density, Pi is 3.14..., Density is the air density. Watch the units, the SD must be lb/ft^2. This formula is just the differential equation and we approximate the derivative (rate of change) of the velocity with the the ratio of the measured changes. This is why it is only valid for small changes.

To find the BC, you would have to interpolate into a table of CD for the the G1 projectile and then use:

BC = SD*CD(G1)/CD

where CD(G1) is the CD of the G1 projectile for the mach number (velocity over the speed of sound). The ratio of CDs is just the "form factor".

Calculated this way, you wouldn't have to do any integrals, but again, it's not really valid if you want to measure velocities and times many yards apart.
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JBM Small Arms Ballistics -- http://www.eskimo.com/~jbm
#18
09-23-2004, 08:55 PM
 Platinum Member Join Date: Sep 2004 Location: on the rifle range in Utah Posts: 2,704
Re: new twist on b.c.\'s

JBM,
Most fascinating! Thank you so much for taking the time to go into detail on that. There is still so much to learn about ballistics, I feel I'm still in Kindergarten sometimes!
Two other questions: You mentioned an MPM program that is free? Were do I go to try it and what is it?

[ 09-23-2004: Message edited by: goodgrouper ]
__________________
Find it
Range it
Click it
Pull it
Dump it

If it's not far, it's boring.
#19
09-24-2004, 05:52 AM
 Silver Member Join Date: Jan 2004 Location: New Mexico Posts: 113
Re: new twist on b.c.\'s

It's on my website, JBM. Just click on "MPM" at the top of the page. It is a command line program meaning it must from a "DOS box". It's not the typical ballistics program in that it takes bullet dimensions as input and calculates a modified point mass trajectory including drift, aerodynamic jump, etc.
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JBM Small Arms Ballistics -- http://www.eskimo.com/~jbm
#20
02-08-2007, 09:53 AM
 Silver Member Join Date: Jun 2007 Location: Buenos Aires, ARGENTINA Posts: 131
Re: new twist on b.c.\'s

[ QUOTE ]
It's on my website, &lt;A HREF="http://www.eskimo.com/~jbm" TARGET=_blank&gt;JBM&lt;/A&gt;. Just click on "MPM" at the top of the page. It is a command line program meaning it must from a "DOS box". It's not the typical ballistics program in that it takes bullet dimensions as input and calculates a modified point mass trajectory including drift, aerodynamic jump, etc.

[/ QUOTE ]

Just wondering, about measuring those velocities over 1000 yards, how do you account for the transonic/subsonic shift?

Regards,
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regards, Gus

http://www.patagoniaballistics.com
#21
02-08-2007, 07:38 PM
 Platinum Member Join Date: Aug 2003 Location: NC, oceanfront Posts: 3,974
Re: new twist on b.c.\'s

I calculate BC of a 140VLD @5Kft(std conditions &amp; 3000ft/sec) at .706
Since I seem so close, I'll say it's BC under standard conditions &amp; 3Kft/sec at 10,000ft is .824
If up to 50deg there BC would be .870 and @75deg = .912

It's because drag goes down with lower air density compared to the G1 curve under Std Metro conditions. There is nothing here to be surprised about really.

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