Home Forums Hunting & Shooting Hunting Rifles Why is .338-06 with 200 grain bullets faster than .30-06?

I was wondering why a .338-06 would drive a 200 grain bullet so much faster than a .30-06 could. Anyone know? They have the same powder capacity. My thought was that perhaps because the 200 grain .30 caliber bullet was longer, and had more contact with the barrel, this could explain it, but that's a lot reduced velocity. Is this the whole reason?
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<br>The 200 grain .338 even has the .30/200 beat by 100 fps out to 300 yards. Superior sectional density allows the .30 to catch up a bit, but not all the way till beyond 400 yards.

Hawkeye
<br> Not trying to be a smart a%$# here but you answered your own question perfectly ! Folks could go into great detail giving scientific data and all sorts of things but basically you hit the nail on the head. I have a 338-06 Ackley Improved and love it. Its on a Win 70 action, # 5 barrel at 23 inches including brake, High Tech stock. Took my first elk with this rifle.
<br>Charlie

That's a good efficiency gain, moving up to .338-06, because you increase velocity AND bullet size, while not increasing powder charge or recoil (assuming 200 grain bullets). Plus you have the added advantage of being able to step up to the 225 and 250 grainers for elk and moose (or grizzly). Nice deal. I am not sure which rifle of mine is going to become a .338-06, but one of them sure is. It's between my Dakota 76 and one of my pre-64 Winchester 70s.

Expansion ratio:
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<br>Within a given bullet weight and given case size, when loaded to equivelant pressure, a larger dia. bullet will achieve a higher velocity.
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<br>My 338-08 will shoot a 180 gr. bullet faster than a 308, but slower than a 358.
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<br>JimF

I guess that's because the .358, based also on the .308, has an even larger diameter than the .338-08. Is this strictly because of less surface area in contact with the barrel, or are there some other forces of physics involved? Could it also be because, at the base of the bullet, the larger diameter bullets provide more of a surface for the expanding gasses to act on? At first thought, that seems to make sense. I don't know, myself. Never took an advanced physics class, so this is all murky water for me. Curious though.

It's the expansion ratio. The ratio of the case capacity to the bore volume.

Think I got it. So, basically what I said above is correct, in different words?

Hawkeye
<br>Like my dad used to say, " You got er by the balls on a down hill drag".
<br>Charlie

I believe I can answer this one.....and it's not too technical. The diameter of 30 cal. is 0.308" and 33 cal. is 0.338". Now find the area of the base of the bullet.......where the pressure of the burning gases would be pushing. The rear of the bullet is just a circle and the area of a circle is PI times the radius squared. Using this formula you will find the area of a 30 cal. bullet to be 0.0745 sq.in. Similarly, for the 33 cal. the area is 0.0897sq.in.
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<br> If you took the 33 cal area and divided it by the 30 cal area you will see that the 33 cal has about 20.4% more area for the pressure of the burning gases to act on. Therefore, if the same pressure is applied to the rear of the bullet (from the burning gases) the force exerted on the 33 cal bullet will be 20.4% greater. More force on the rear of the bullet pushes it to higher velocity
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<br> Here is an example: Assume 60,000 psi (which is 60,000 lb/sq.in. ....... or said another way, 60,000 pounds of force that will act on each inch of area) If there were 1sq.in. of area at the rear of the bullet, 60,000 psi would produce 60,000 pounds of force on the bullet. It is calculated like this
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<br>(60,000 lb/sq.in.) x (1sq.in.) = 60,000 lb
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<br> If the area was 1/2 sq in, the force would be 30,000 lb. Here is the calculation for the 308 and 338 bullets using the areas calculated above:
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<br>(60,000 lb/sq.in.) x (0.0745sq.in.) = 4,470 lb (308)
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<br>(60,000 lb/sq.in.) x (0.0897sq.in.) = 5,382 lb (338)
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<br> The additional force acting on the bullet is responsible for the increased velocity. This does not mean there will be a 20.4% increase in velocity. There are other factors that must be considered. Many have already been mentioned such as friction between bullet and bore, expansion ratio, efficiency/burn rate/pressure curve of the chosen powder, etc.
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<br> The principle is the same as is used in hydraulics. A larger diameter cylinder can lift more weight than a smaller diameter cylinder even though the pressure of the hydraulic fluid acting on the base of the piston is the same.

Hogwild, thanks for that detailed explanation. That makes perfect sense. I had an intuitive feeling this was a big part of the explanation, but you've really put it in words so it is really clear. That steam piston example made it perfectly clear. I can visualize how a narrow piston would lift less weight from the same source of steam pressure than would a broader one, because there are fewer square inches being acted on by the pounds per square inch, even though the pressure remains constant for both pistons. Good job.

Hogwild is right on the money.
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<br> It's the same reason why a .45Long Colt operating at 25,000 PSI can equal the velocity of a .44 Mag at 35,000 psi when both are pushing 300gr bullets.
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<br>MM

and the recoil feels less with the 338-06, too, doesn't it. i think it has something to do with the expansion ratio, also. i noticed it with equivalent rifles shooting equivalent loads in 8x57 and .30-06. the 8x57 somehow is easier on the shoulder to shoot. seems to me the 338-06 more so since the diameter is slightly larger than the .323. but there is no science to my opinion. just plain old shoulder experience.

Fish:
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<br>The 8X57 delivers about 12% less recoil than an '06 because it uses about 20% less powder to get similar velocities with 180 gr. bullets.
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<br>You can credit expansion ratio for the velocities, but not the reduction in recoil.
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<br>JimF

ah, jim f, glad to know. that certainly makes sense to me.
<br>recoil also is subjective, as well, but no way to measure that except with opinion.
<br>thanks.

"I was wondering why a .338-06 would drive a 200 grain bullet so much faster than a .30-06 could. Anyone know? They have the same powder capacity."
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<br>No, they don't. The .30-06 holds more powder than the .338-06 when both are loaded to the base of the neck. When you neck any cartridge OUT (larger), the neck becomes longer (and the shoulder correspondingly shorter), reducing the net capacity of the body and shoulder.
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<br>The .25-06, OTOH, holds MORE powder than the .30-06 when both are loaded to the base of the neck. When you neck any cartridge DOWN (smaller), the neck becomes shorter (and the shoulder correspondingly longer), increasing the net capacity.

Ken, be that as it may, is it true that if loaded so as to produce the same pressure levels, the .338 would be faster?

Ken, I always learn something from your posts. I had never given this any thought before. Thanks.

TRH:
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<br>I would defer to the estimable Mr. Howell on this if he disagrees but you are correct. Theoretically, the slightly larger capacity of the '06 might give an advantage if both were loaded at maximum loading density, but they practically never are.
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<br>JimF

I have never sat down with close observation to investigate the advantages of the 338/06 over the 30/06. This is an interesting thread.

Actually, if you take the same case, say the 30-06 case, and neck it to different calibers, say .30 cal., 8mm cal, .338 cal and .35 cal, and load them with the same bullet weight to the same pressure, say all at 60,000 psi, then the larger diameter bullet will always go faster at the muzzle. This is because the base of the bullet is bigger, has more surface area for the pressure to push against, and thus will accelerate faster, since more foot lbs of energy are pushing against it. Any good physics text book will explain this somewhere in it.