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B_Mc
Member
USA
577 Posts |
 Posted - 11/04/2009 : 04:09:40 AM
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Are they directly related? Does velocity come into play with the BC like energy? And finally how are both calculated?
Thanks for the quick answers.
Nononsense, thanks for the articles and I dont always want a quick simple answer. I would not ask if I did not want to know to the fullest extent. |
Edited by - B_Mc on 11/04/2009 12:13:04 PM |
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Tailgunner1954
Advanced Member
5760 Posts |
Posted - 11/04/2009 : 05:31:56 AM
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BC is a measure of a bullets ability to move through the air. The number you see is based on a standard of 1.000. If it loses velocity twice as fast as the standard, than it has a BC of .500, 4 times as ast and it has a BC of .250.
There are several "form standards" (G #) but the number you see advertised is G1 which is a 1" diameter solid with a 2d radius for the nose ogive. IF the matching G# is used, than BC dosn't change with velocity, if the wrong G# is used than it does change with velocity. They like using the G1 standard because, quite simply, it gives the highest number for advertising purposes.
Energy is simply a Mass/Velocity calculation. Without looking it up, IIRC Energy = Mass x velocity x velocity.
For most hunting purposes, BC dosn't matter as the range isn't long enough for the air to have slowed the bullet enough for the real world difference to show up. For those that hunt in the "long range" zone, than it does start to have an effect |
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 Some guys like a mag full of lead, I still prefer one round to the head.
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nononsense
Moderator
6840 Posts |
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tsr1965
Advanced Member
USA
3163 Posts |
Posted - 11/04/2009 : 09:05:02 AM
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B_Mc,
In short, the higher the ballistic coefficient, the less velocity loss, or higher retained velocity at any given distance. The higher the retained velocity, the higher the energy...so yes, they do relate to each other. however, if I were you, I would wait for the elongated reply of nononsense. I am sure it will be in great detail, and a superb explaination indeed.
Best |
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heavyiron
Senior Member
USA
1293 Posts |
Posted - 11/04/2009 : 6:34:14 PM
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I like these type of questions and feel compelled to respond.
The ballistic coefficient is an abstract number. Its definition is a mathematical expression of a bullets ability to overcome atmospheric resistance. In other words, it is a numerical value derived from a mathematical formula which predicts the bullets performance to overcome friction from the air or "drag". The higher the BC the more streamlined the bullet and the farther it will retain velocity. As Tailgunner stated so well, the BC is partially based on a standard bullet design.
To the best of my knowledge, today most BCs are determined by testing and measurements in ballistics labs because bullet design deviates so much from the standard bullet. In fact, Sierra messed up their BCs a couple of years ago when they were in California and had to recalculate them.
BCs also vary by velocity, so there isn't just one BC associated with a bullet. For example, given a 0.223 diameter, 40 grain bullet, it can have a BC of 0.117 @ 2800 fps and above, 0.123 between 2000 and 2800 fps and 0.136 @ 2000 fps and below - as determined by the manufacturer. It can be very complicated.
Now, the second part of the question is, how are the two related. The more velocity a bullet can retain, the more energy it will deliver.
Just as Tailgunner stated, it is a Newtonian physics formula. The formula is:
Ke = 0.5x((Wb/7000)/32.2)x(Vb^2)
Ke=Energy (kinetic) (ft/lbs)
Wb=Bullet weight (grains)
Vb^2=Velocity of bullet squared (feet/second)
The weight of the bullet in pounds is divided by 32.2 ft/sec/sec to convert bullet weight to mass. As you can tell from the formula the velocity is squared; the more velocity that a bullet can retain (due to a higher BC), the higher the energy. The relationship between energy and velocity is therefore exponential to the limit where drag dominates.
Sorry, I have gone on too long.
Hope this helps.
Heavyiron |
"If I don't see you nomore on this world, I'll meet you on the next one, and don't be late!" - Jimi Hendrix |
Edited by - heavyiron on 11/04/2009 10:14:36 PM |
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beantownshootah
Advanced Member
USA
8595 Posts |
Posted - 11/04/2009 : 11:38:08 PM
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Making this simple, ballistic coefficient is just away to compare how much resistance a bullet experiences in moving through the air.
Imagine a car being tested for aerodynamic drag in a windtunnel, and you'll get the idea. A well designed car (eg a Corvette) should be more aerodynamic (slippery) than a poorly designed one (eg a Yugo).
The same is true of bullets. Some offer less air resistance than others.
The better (higher) the BC, the more "streamlined" the bullet is, and the less velocity it will tend to lose over a given distance.
Does velocity come into play? Well. . .yes.
The amount of drag a bullet experiences is related to velocity, and the relationship is non-linear. But more velocity always means more drag.
I'll pass on how ballistic coefficient is calculated, since the answer is far too complicated for a brief response here.
Suffice it to say, that its a function of the bullets weight divided by its cross-sectional area (the so called sectional density) multiplied by a correction factor that takes into account the bullets aerodynamic "slipperiness".
To make this easier to understand, for a given bullet diameter all else being equal, heavier bullets should maintain their velocity better. Given two bullets of identical weight and diameter the one with the "slipperier" shape/design should maintain its velocity better.
In terms of kinetic energy, that is a specific physical measure representing the amount of work that a moving body can do.
The formula from simple Newtonian physics is that kinetic energy is equal to the one half the mass of the moving object multiplied by the square of its velocity. Or more briefly, 1/2mv^2. So this is easy to calculate in the real world, so long as you know the weight of the bullet and its velocity.
Note that kinetic energy is more dependent on velocity than mass.
Also note that the "real world" effectiveness of projectiles is often HIGHLY dependent on their mass, so that just looking at kinetic energy as a measure of effectiveness can be deceiving.
For example, if you look at JUST kinetic energy, a standard 9mm luger handgun round and a .45 ACP actually have identical (or nearly identical) kinetic energies. That's because the lighter 115 grain 9mm is flying a lot faster than the heavier 230 grain .45.
Still, the .45 ACP fires a bullet with a larger diameter and twice the mass, and it is generally considered to have better terminal ballistics (ie effectiveness).
The same principle can hold with rifle bullets, with small lightweight bullets generating tremendous energies because of their speeds.
The problem is that such bullets tend to give up their energy quickly due to air resistance (see above, light bullets tend to have poor ballistic coefficients). Also, lightweight bullets with high energy tend to give up their energy quickly on impact, often by total destruction of the bullet! This can have an absolutely devastating effect on small game, but can result in poor penetration and a shallow wound in larger targets.
Is there a relationship between kinetic energy and ballistic coefficient? Well. . .not a direct one. Two bullets of the same weight and velocity will necessarily have identical kinetic energies, though the ballistic coefficients can be wildy different.
The difference is the bullet with the better BC will tend to retain more energy over distance than the one with the poorer BC.
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BC & Energy question
