Why does BC go UP with increasing speed?

[Syncrowave]

This has always puzzled me: When you look at a bullet's ballistic coefficient, at supersonic velocities, the higher the speed, the higher the BC.

With things like cars and air drag, or boats with water/air interface drag, if I recall correctly, drag increases proportionally to the square of the speed. In other words, if you double the speed, the drag increases by 4X (two squared) and if you triple the speed, the drag increases 9X (three squared).

Is there a simple explanation of why the BC of supersonic bullets seems to go UP with increasing speed, instead of DOWN, when it seems that the drag should be increasing?

Thanks in advance.

 

[Bryan Litz Ballistics]

BC is comprised of a bullets sectional density and its form factor. Form factor describes the drag of the bullet in comparison to a standard. The old G1 standard is not representative of modern bullets in terms of how its drag changes with speed.

So, as velocity increases, the drag of modern bullets is lower than the standard, so the form factor decreases at higher speeds, which means the BC increases.

In other words, the BC of a bullet is higher at high velocity because it's being compared to a non-representative standard, whose drag changes differently with velocity.

The variation in BC with velocity is a big problem when you use the G1 model as the standard. But if you look at your bullets G7 BC, the velocity variation is much less. This is because the G7 standard is more representative of modern long range bullets in shape and it's drag curve.

Check out this article which goes into more detail: http://www.bergerbullets.com/a-better-ballistic-coefficient/

Berger, Lapua, and Nosler among others have started to provide G7 BC's for all or parts of their bullet lines which is a big step forward.

Hope this helps,
-Bryan

 

[CatShooter]

This has always puzzled me: When you look at a bullet's ballistic coefficient, at supersonic velocities, the higher the speed, the higher the BC.

With things like cars and air drag, or boats with water/air interface drag, if I recall correctly, drag increases proportionally to the square of the speed. In other words, if you double the speed, the drag increases by 4X (two squared) and if you triple the speed, the drag increases 9X (three squared).

Is there a simple explanation of why the BC of supersonic bullets seems to go UP with increasing speed, instead of DOWN, when it seems that the drag should be increasing?

Thanks in advance.

In less words.

BC is not drag - it is the ratio of velocity loss compaired to a "standard" bullet (which is a somewhat clunky design)... so at higher velocities, the modern bullet is more efficent, so the ratio of loss is less, so the BC numbers are higher.

Easy Peasy.

 

[CaptainMal]

Heh?

Now I have seen what "smart" is. Thanks.

Actually, Brian, your article and drawing is quite simple informative. Now steer my 105 Hybrids to the target.

 

[Syncrowave]

OK, thanks, all. That article helped especially.

So can I assume that for a given bullet shape, at supersonic velocities, the drag component alone does indeed increase with increasing velocity – just as it does with cars or boats?

I can't imagine how drag could decrease at increasing speeds above the speed of sound...

 

[CatShooter]

OK, thanks, all. That article helped especially.

So can I assume that for a given bullet shape, at supersonic velocities, the drag component alone does indeed increase with increasing velocity – just as it does with cars or boats?

I can't imagine how drag could decrease at increasing speeds above the speed of sound...

Yes... with the square of the change in velocity.

 

[dcali]

To be extra clear, the drag FORCE increases as velocity increases. The drag COEFFICIENT decreases as velocity increases. I think that is the source of confusion. The drag force is proportional to the drag coefficient and to the square of the velocity. Generally speaking, the square of the velocity rises faster than the drag coefficient drops, resulting in a net increase in drag force with increasing velocity.

Note that the drag coefficient is not the same as the ballistic coefficient. It's just a representation of a object's shape's impact on drag. A higher drag coefficient means more drag force, all else equal.

 

[Syncrowave]

Excellent. Just the information I was looking for. Thanks, fellas.