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Old June 22, 2008, 06:50 PM   #1
Moloch
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Sectional density - Round ball

I was just running some numbers on beartooth bullets ballisticians corner.-the relative penetration calculator. With the calculator I've got 9.81'' of relative penetration with a 133 grain .45 round ball. Next I calculated the relative penetration of a 1500 grain 1.00'' ball, and guess what I've got?
21'' of penetration! How can that be? All round ball bullets have the same sectional density as the weight/size ratio is the same, the bigger ball needs the increased weight to make a larger wound channel as deep as a smaller ball. -but the penetration should be equal right?

I dont get it. #

Does round ball penetration go up with the same amount of size increase of ball? What I mean is, if I have a .45 round ball, If I take a 50% bigger ball with 50% more weight it penetrates is 50% more than the .45?

The link to the calculator.
http://www.beartoothbullets.com/resc...v1=133&v2=0.45
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Old June 22, 2008, 07:10 PM   #2
Sarge
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Well, no.

The heavier ball will penetrate deeper. It is exactly why the early hunters used 4 and 8 bores for African game, instead of Tennessee squirrel guns.

There are some things that simply don't follow a mathematical pattern, and this is one of them.
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Old June 22, 2008, 08:31 PM   #3
Moloch
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I am well aware that the heavier ball will penetrate deeper, but why? Everything is mathematically explainable, this case is no exception.
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Old June 22, 2008, 08:44 PM   #4
Hawg Haggen
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I think a thing called momentum plays a part in it. It takes more to stop a bigger bullet.

That should have been heavier bullet.

Last edited by Hawg Haggen; June 23, 2008 at 03:40 AM. Reason: correction
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Old June 22, 2008, 10:07 PM   #5
FL-Flinter
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I don't recall how the SD comes into play for the formulas when used for RB's but I can tell you these things:

0.440" RB driven from a 36" bore with 80gr of 2F will enter the chest of a whitetail at 40yds, cut the boiler almost in half, penetrate at an angle to the opposing hind quarter and come to rest against the bone. Same ball & load effectively enters the WT nose at 30yds emptying the brain box through a 2.5" diameter hole in the rear. Same ball & load at 60yds goes through a WT boiler room taking out the boiler and lays under the hide on the opposit side.

0.575" RB driven from a 34" bore with 100gr of 2F will effectively take out one WT front shoulder and penetrate at an angle exiting through a 1.5" hole in the rib cage on the opposit side on a WT at 85yds then go another 10yds or so before burying itself about 2" deep in a pine tree. Same ball & load at 65yds will enter the WT's neck from dead front center at spine level and open a hole a foot plus long like a zipper where the spine used to be over the shoulders, no idea where it went from there. Same ball & load on WT 70yds out that suspected I was there somewhere punched through both shoulder bones with ease and without destroying all the meat (was supposed to be a boiler room hit not a shoulder breaker - he started turning to bolt)

I don't care what the math says, RB's work just fine.
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Old June 23, 2008, 09:56 PM   #6
Dave Haven
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Sectional density = mass/dia^2; and mass of a round ball is proportional to dia^3, therefore, sectional density of a round ball is directly proportional to ball diameter. A larger ball has a higher sectional density.
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Old June 24, 2008, 10:33 PM   #7
FL-Flinter
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Dave,

I thought there was a longer formula to obtain a number for use in a penetration calculator program when dealing with round balls as opposed to conical bullets?
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