Larger Calibers For Fun And Profit

[Genghis]

In this thread discussing power factors in single stack division [http://www.brianenos.com/forums/index.php?showtopic=32606], one of the participants said that the larger bullet diameter of a .45 is a “huge advantage” over the 9 mm/.38 Super. I’ve heard this argument before, and I wondered how much of a factor this is.

The diameter of a .45 ACP bullet is .452, and the diameter of a 9 mm bullet is .357. So the diameter of the .45 is .095 larger than the 9 mm.

USPSA Metric Target

If you hit on the edge of the A zone, measuring from the center of the bullet, you will be able to hit up to half of that width (.047) farther out with a .45, on either side, and still score an A. The A zone of a standard (metric) target is 28 cm (11.024 in) high, and 15 cm (5.955 in) wide, so the area is 65.1 sq inches. In effect going from .45 to 9mm adds a band .047 inches wide around the outside of the A zone, making it .095 wider and .095 taller.

My geometry skills are just sufficient to figure the area of a rectangle. Since the C and D zones are irregular octagons, I’ll let someone else figure the increased chance of having your .45 turn your D’s into C’s, and your M’s into D’s. After all, we all shoot A’s and C’s, right?

So with the added width your effective A zone is 6.05 X 11.119. The raw surface of the A zone is 65.648 sq inches. The “expanded” A zone with a .45 is 67.270. Therefore the area has increased by 2.5%. So the .45 should generate about 2.5% more fives than fours, for body shots. So if someone shoots 100 rounds in a match with a 9 mm, with 80 A’s and 20 C’s, upgrading to a .45 would have generated about two more A’s. Assuming major power factor, that would be two more points (2 X (5 - 4).

I did perform the same calculations for the (rectangular) A zone in the head portion of the target. Actual size is 7.748 sq inches, “expanded” size with a .45 is 8.318 sq inches. So using a .45 enlarges the A zone by 7.35% for head shots.

IDPA Target

The -0 zone of the body of the IDPA target is 8 inches in diameter, or 201.06 sq inches. A .45 would increase the effective radius by .047, increasing the area to 203.43 sq inches, or about 1.2%. So a bigger bullet makes less of a difference with IDPA body shots.

For head shots, the whole head is -0. The head is six inches square, or 36 sq in. The “expanded” head is 6.095 inches square, or 37.15 sq in. This is 3.2% bigger, but each shot upgraded from a Mike to a -0 saves the shooter 5 points, or 2 1/2 seconds.

______________

Of course all these calculations ignore any change from minor to major power factor, and the additional recoil of the larger round.

What does everyone think?

[Flexmoney]

The diameter of a .45 ACP bullet is .452, and the diameter of a 9 mm bullet is .357. So the diameter of the .45 is .095 larger than the 9 mm.

USPSA Metric Target

If you hit on the edge of the A zone, measuring from the center of the bullet, you will be able to hit up to half of that width (.047) farther out with a .45, on either side, and still score an A. The A zone of a standard (metric) target is 28 cm (11.024 in) high, and 15 cm (5.955 in) wide, so the area is 65.1 sq inches. In effect going from .45 to 9mm adds a band .047 inches wide around the outside of the A zone, making it .095 wider and .095 taller.

You need to double the difference, right? You get a bullet width on both the left and the right side of the A-zone. And, a bullet height on both the top and the bottom.

So, you increase the width and the height by 0.190 (.425 - .357 = .095, .095 * 2 = .190)

A-zone = 5.955 x 11.024

9mm A-zone = 6.669 x 11.738 = 78.28722

45 A-zone = 6.859 x 11.928 = 81.814152

For an increase in the A-zone surface area of ~ 3.5 square inches. (81.8 - 78.3 = 3.5), which is ~ a 4.5% increase over the 9mm area.

[Loves2Shoot]

I think it is better to learn to hit the center of the target.

The "major" advantage is on partials/movers where it makes more sense to accept a C.

[Rotwang]

In this thread discussing power factors in single stack division [http://www.brianenos.com/forums/index.php?showtopic=32606], one of the participants said that the larger bullet diameter of a .45 is a “huge advantage” over the 9 mm/.38 Super. I’ve heard this argument before, and I wondered how much of a factor this is.

The diameter of a .45 ACP bullet is .452, and the diameter of a 9 mm bullet is .357. So the diameter of the .45 is .095 larger than the 9 mm.

What does everyone think?

I think the real question you have to ask is -- "How many of my 9mm shots miss the A zone by .05 inches?" Because that's how many will get promoted into an A. Assuming you hit the same place.

[shred]

or, you could just say "I don't have to care where I hit the steel..".

To be excruciatingly pedantic , to figure out the extra surface area available, you have to measure the rectangle between the 9mm 'bonus zone' (the area 9mm or less away from the outside of the perf) and the outside of the .45 bonus zone, and the corners aren't quite square either.. ick.

[Carmoney]

Now which one of you pointy-heads wants to factor back out the hits that because of the bigger bullet diameter just became no-shoot penalties?

[MikeFoley]

In the time I read this, scratched my head, and responded, I could have been practicing with either caliber and gained more points.

[pskys2]

Dang, Carmoney you beat me to it. I guess great minds think alike! LOL!

Course, instead of checking out this website I was watching 24 and dryfiring Reloads on Pudgy the Bear. My wife's used to the click, click flurry of activity !expletives! click, click!

But, now I'm at work and there's time to sit back.

[Genghis]

You need to double the difference, right? You get a bullet width on both the left and the right side of the A-zone. And, a bullet height on both the top and the bottom.

Actually I already doubled the difference. The bullet diameter (or width) is .095 larger (.452 - .357 = .095).

Back to dry firing. Since I'm at work, I'll have to use the mouse instead of the heater.

[Flexmoney]

yeah, but for some reason you halved it to start with...

If you hit on the edge of the A zone, measuring from the center of the bullet, you will be able to hit up to half of that width (.047) farther out with a .45, on either side, and still score an A.

That is the part I am missing. ?? Why are you only using half? You should get the full 0.095 on each side, right?

Edited February 21, 2006 by Flexmoney

[Genghis]

If you hit the edge of the A zone with a 9 mm, then with a .45 right above it, the .45 hole will stick out .047 inches farther into the C zone. It will also stick out .047 farther into the A zone. So the A zone will effectively be .047 wider on the right, and .047 wider on the left, for a total of .095.

[Flexmoney]

A bullet only needs to touch the outside of the scoring line at one point, it doesn't need to stradle the line at all.

So, with a 45, you get the width of the A-zone...plus .452 on each side.

[Bigbadaboom]

That statement was made regarding major vs. minor vs. capacity. I was also compairing 9mm/.38 Super to .45 and didn't take .40 into consideration. In Limited I will shoot .40 over .45 due to how many I can get in a 140mm mag because the difference in bullet diameter when both make major P.F. doesn't justify the sacrifice in capacity. My point was that, when shooting single stack, not only does .45 have a huge advantage over 9mm/.38 Super in scoring major vs. minor it also gains a large advantage with bullet diameter when touching the line comes into effect. I R.O. a lot and I've noticed that tight to the line shots happen quite frequently, enough so that major .45 touching the outside line of the A-Zone vs. a 9mm being .03" away into the C-Zone means 5 points vs. 3 points and that's a "Huge" difference. In major .40 vs. major .45 it's only 1 point difference so I wouldn't call it a "Huge" advantage when you start giving up capacity.

Edited February 21, 2006 by Bigbadaboom

[Flexmoney]

All three of the hits in the drawing are Alpha hits.

[short_round]

That is the part I am missing. ??

I think you forgot to factor in the air speed velocity of a fully laden swallow

[Genghis]

So, with a 45, you get the width of the A-zone...plus .452 on each side.

Good point. So to calculate the actual difference you'd have to compute the effective size of the rectangles with each caliber, adjusted for the "stretch" added by the caliber, then subtract. But the original figures are close enough for gubmint work.