Velocity vs energy vs barrel length. Question

[1911man]

I am not real a math guy so maybe any of you can resolve a question for me. I want to compare, for example, gun 1 is a .243 with a 24" barrel, shooting a factory 80 gr. GMX hornady round. Gun 2 is a 6.5 creedmoor with a 16.5" barrel, shooting a factory hornady 120 gr. GMX. The velocity of the 6.5 drops about 200-250 fps from 24" to a 16.5" barrel. My question is energy transfer from a shorter barrel different than a 22" barrel at 300 yards? If so, how much difference?
Would gun #1 equal the energy from gun # 2 at 300 yards, I guess I need a balistic app.

[1911man]

I just looked at the hornady ballistics chart and the vel. matches at 300 yards. Both gun #1 80 gr. Gmx and gun #2. 120gr. Gmx. Both with the same length barrel. But does the energy suffer if gun #2 has a 16.5" barrel?

[Epaphroditus]

Estimates are one thing but a chronograph tells all.

50-100 per extra inch is a rule of thumb estimate.

[1911su16b870]

http://www.sadefensejournal.com/wp/?p=1093

...here you go.

Where is AR15barrels at? He used to have some of this data on his website, maybe my goog-fu is lame...

[Epaphroditus]

For the same velocity the energy is proportional to bullet weight. At the same velocity the 120 gr has 1.5 times the energy as the 80 gr.

At different velocities the energy is proportional to the square of velocity.

[1911man]

Another question is, if you had to, would you rather shoot a 250 lb. pig, at 300 yards, with gun #1 or gun # 2 with a 16.5 " barrel?

[Epaphroditus]

Hornady has a ballistic calc on their website.

I don't think I can reliably place a kill shot on a pig at 300 yards. If I could I think either choice would do the job. Placement is king.

[mooner760HD]

I am not real a math guy so maybe any of you can resolve a question for me. I want to compare, for example, gun 1 is a .243 with a 24" barrel, shooting a factory 80 gr. GMX hornady round. Gun 2 is a 6.5 creedmoor with a 16.5" barrel, shooting a factory hornady 120 gr. GMX. The velocity of the 6.5 drops about 200-250 fps from 24" to a 16.5" barrel. My question is energy transfer from a shorter barrel different than a 22" barrel at 300 yards? If so, how much difference?

Would gun #1 equal the energy from gun # 2 at 300 yards, I guess I need a balistic app.

The faster a bullet is moving, the more kinetic energy it carries. Longer barrels provide more time for the expanding gases from the gun powder to accelerate the bullet. This means that up to a point, longer barrel= higher velocity= more energy.
The mass of the bullet is the other factor that affects energy. So the larger bullet moving a few hundred fps slower could have similar energy to the smaller bullet going faster.

[NapalmCheese]

is energy transfer from a shorter barrel different than a 22" barrel at 300 yards?

Energy TRANSFER is a property of how quickly the bullet expands and/or fragments. The force equation at impact, assuming both bullets are hitting an immobile object, is 1/2*m*v^2.

A .50 cal FMJ may very well transfer less energy into a target than a 55 grain frangible .223.

[Jimi Jah]

Is this at sea level? I notice less bullet drop at 10,000 feet.

[Preston-CLB]

The force equation at impact, assuming both bullets are hitting an immobile object, is 1/2*m*v^2.

Actually, the equation you quoted is the Kinetic Energy of the bullet, not force of impact. The force would be F=m*a. "a" being the acceleration.

Both of these equations neglect the frictional force on the bullet due to air resistance. In the case of the kinetic energy, the resistive force acts to reduce the velocity.

There two forces then that act on the bullet: the resistive force due to air resistance and gravity, neglecting any force imparted by wind.

I notice less bullet drop at 10,000 feet.

Since the density of the air is less than that at sea level, and thus offering less air resistance, I believe your observation is likely correct.
--P

[NapalmCheese]

Actually, the equation you quoted is the Kinetic Energy of the bullet, not force of impact. The force would be F=m*a. "a" being the acceleration.

No, the force equation still holds and kinetic energy is just another representation of force. Impact with an immobile object implies rapid deceleration to the tune of instantaneous, thus the first integral.