I have noticed somewhere (but can't find the document) the following theorem:

By widening as an example the neck of a 30/06 case to 338 (338/06) thus increasing the diameter by ~ 10 pct I can theoretically achieve a higher velocity of 2 to 2,5 pct with a bullet of identical weight and with identical chamber pressure. More generally, the increase of diameter by a certain percentage is followed by a potential increase of velocity by a quarter or a fifth of this percentage.

Is my memory working correctly. Who has the precise numbers for this theorem.

There was an article about 10 years ago in the "American Rifleman" about this. I don't recall the formula however or that one was proposed.

However if you look at a .243 Win at 2000 foot pounds and a .308 Win at say 2600 fpe and a .358 Win at almost 3,000 fpe the comparison is there.

This is a reason why I favor the larger calibers.

na, glaubst du mir nicht?

Don Martin29, thank you for drawing my attention to the comparison 308Win and 358Win. Indeed, in this case the difference in velocity and therewith even more in energy is very significant.

Previous to my post I completed a check by comparing the 30/06 with the .35 whelen (similar pressures as in your case). Strikingly, in this comparison the muzzle velocities and with that the energy are very close, turning it down range in a slight advantage for the more aerodynamic 30/06 - bullet.

esox,

Assuming equivalent loads and equally suitable powder, the increase in kinetic energy will be half the percentage increase in bore are.

For example, 338 has a 20.43% greater bore area than 308. Thus a 338/06 would deliver a 10.21% increase in kinetic energy.

This gives extremely accurate predictions. If calibers are tested in bench rifles and loaded to the point where accuracy falls off, the actual increase in energy from the larger bore size is slightly less than predicted and there are two reasons for this.

Firstly, smaller case capacities for the bore (which of course is what happens when we neck up) seem to have accuracy fall off at lower pressures. This if we compared a 6mm/06 and 30/06 in bench guns and loaded up until accuracy fell off, then 6mm/06 would probably be running at higher pressure.

Secondly, necking up often means we finish with a slightly smaller usable case capacity.

An extreme example would be comparing the 264 Winchester with 100 grain bullets and the 458 Winchester with 500 Grain Hornadies.

A general rule will be if we move up "one caliber" through the range of 270, 30, 338, 375, 416, 458 and 510, we will drop about 150 f/s for bullets of equal secdtional density.

As an example, a 270 might do 3150 with 130s and a 30/06 with top loads might be around 3000 f/s with 165 grain bullets.

Mike

esox...the last time I saw that it print it was an article by John Barness in either Rifle or Handloader magazine. I don't have it handy but basically it worked out to about 2.5% of the difference in the areas of the bores in question.

Necking up does not reduce the case capacity with equal weight bullets. It increases it and with modern partition bullets and their improved ballistic coeffients the larger caliber is always the most flexable choice.

Nor are larger calibers less accurate. The right powder can be found.

Larger calibers are more likely to hit the target. A .458 bullet radius reaches 20% closer to the center of the point of aim than a .264 bullet. And hiting the target is what it's all about.

Bringing up sectional densities really moves the discussion away from the greater energy from the larger caliber. You can't compare the effectivness of a .338 to a .264 at any range!

Then there is the loss of energy from the expansion of the bullet and also it's slower intial expansion as it hits the quarry. The larger caliber wastes less energy than the smaller in expanding it's bullet to an effective size.

Gentlemen,
the math for this phenomenon goes like this:

Double the diameter and you get:

3.14 x the amount of surface (actually circumfrance)

four times the area.

Because the area increases faster than the external surface you get more velocity from the same WEIGHT bullet with the same pressure. (assuming the same bearing length and bore fit) More area gives more force with the same pressure.

PaulS

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stay safe and live long!

[This message has been edited by PaulS (edited 04-19-2002).]

Paul...the area is "pi"(3.14) times the square of the radius.

If you have pressure of 35,000 psi on a work surface it will accomplish a given value of "work" So if you increase the the area of work surface and keep the pressure the same then simple math tells us the given value of "work" will increase with it.

Example:

35,000 psi on a 1" area = 35,000 lbs of "Work"

35,000 psi on a 2" area = 70,000 lbs of "work"

This example is not an actual demonstration of the formula involved but you get the idea.

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From my cold, dead hands!
Thanks Chuck!