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broken_arrow
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diameter/weight/length formula?
Dec 18th, 2013 at 10:52am
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Hi all,
Does anyone know of a mathematical formula for the following theoretical situation....
Can you determine the length of an unknown lead bullet if you have the diameter and the weight of said bullet?
Thanks,
Dave
  

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gunlaker
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Re: diameter/weight/length formula?
Reply #1 - Dec 18th, 2013 at 11:16am
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It'll depend a lot on the nose profile of the bullet. 

Chris.
  
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Re: diameter/weight/length formula?
Reply #2 - Dec 18th, 2013 at 11:30am
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WAG is the best without a lot more info!

Number and size of GG's
Nose shape and length
Then youre pretty much there.

But then again alloy is an issue.
Lead = 11,34gram/cc~2843,2grain/cui
Antimony = 6,7gram/cc~1680grain/cui
Tin = 7,3gram/cc~1830grain/cui

For a true wadcutter you can come pretty close:
Weight gives you volume
Diameter gives you area of a slice of bullet
And that leads you to length

Volume = radius x radius x PI x length
Length = volume /(radius squared x 3.1415)

For lead:
example 148grain .357" caliber
Volume = 148/2843,2 = 0.052 cui
Length = 0.052/(0.1875 squared x 3.1415) = 0.052/0.1 = 0.52inch
  
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Re: diameter/weight/length formula?
Reply #3 - Dec 18th, 2013 at 1:35pm
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This is a good example of the difference between two bullets of almost the same length. If the 320366 was cast of linotype and the PJ cast of 30/1, there would be even more difference in weight.

  

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joeb33050
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Re: diameter/weight/length formula?
Reply #4 - Dec 18th, 2013 at 3:39pm
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Yes. GREENHILL FORMULA WORKBOOK.xls" will estimate bullet length from weight and caliber. PM me your email address and I'll send it.


broken_arrow wrote on Dec 18th, 2013 at 10:52am:
Hi all,
Does anyone know of a mathematical formula for the following theoretical situation....
Can you determine the length of an unknown lead bullet if you have the diameter and the weight of said bullet?
Thanks,
Dave

  
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Cat_Whisperer
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Re: diameter/weight/length formula?
Reply #5 - Dec 18th, 2013 at 4:31pm
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For just an estimate, probably not a bad idea.

Greenhill's formula was for a given bullet shape that likely would be right-much heavier than other bullets - due to nose shape and grooves - but that's a great place to start.

  

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Re: diameter/weight/length formula?
Reply #6 - Dec 18th, 2013 at 9:51pm
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gunlaker wrote on Dec 18th, 2013 at 11:16am:
It'll depend a lot on the nose profile of the bullet. 
Chris.

Indeed. My 38cal Brooks mould casts a 340gr 'Money' bullet and the PJ mould casts a 368gr bullet. Both are 1.423" in length. 
  

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Re: diameter/weight/length formula?
Reply #7 - Dec 19th, 2013 at 1:30am
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Or, the same without Excel:

Bullet length ~ bullet weight/Factor

Bullet weight ~ bullet length * Factor

DIAMETER      FACTOR
0.257      122
0.286      151
0.308      175
0.323      192
0.357      235
0.375      259
0.410      310
0.429      339
0.457      385

The precise relationship depends on bullet shape, number and volume of grease grooves and alloy composition. Close-to-exact relationships can be found using other, more involved arithmetic that is seldom worth the effort.. 
  
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Re: diameter/weight/length formula?
Reply #8 - Dec 19th, 2013 at 8:16am
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I doubt both weights-both look too low, and the difference doesn't seem correct.
Does anyone have similar bullets to check weight?



frnkeore wrote on Dec 18th, 2013 at 1:35pm:
This is a good example of the difference between two bullets of almost the same length. If the 320366 was cast of linotype and the PJ cast of 30/1, there would be even more difference in weight.


  
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Re: diameter/weight/length formula?
Reply #9 - Dec 19th, 2013 at 8:53am
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Thanks guys,
Your responses have me a lot closer than my WAG were getting me.
And the simple formula that Joe posted will get me close enough for starters.
Thanks again,
Dave
  

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Re: diameter/weight/length formula?
Reply #10 - Dec 19th, 2013 at 1:02pm
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joeb33050 wrote on Dec 19th, 2013 at 8:16am:
I doubt both weights-both look too low, and the difference doesn't seem correct.
Does anyone have similar bullets to check weight?



frnkeore wrote on Dec 18th, 2013 at 1:35pm:
This is a good example of the difference between two bullets of almost the same length. If the 320366 was cast of linotype and the PJ cast of 30/1, there would be even more difference in weight.




Joe, I also own the bullet molds and the bullets.

Frank
  

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Re: diameter/weight/length formula?
Reply #11 - Dec 19th, 2013 at 2:26pm
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frnkeore wrote on Dec 19th, 2013 at 1:02pm:
joeb33050 wrote on Dec 19th, 2013 at 8:16am:
I doubt both weights-both look too low, and the difference doesn't seem correct.
Does anyone have similar bullets to check weight?



frnkeore wrote on Dec 18th, 2013 at 1:35pm:
This is a good example of the difference between two bullets of almost the same length. If the 320366 was cast of linotype and the PJ cast of 30/1, there would be even more difference in weight.




Joe, I also own the bullet molds and the bullets.

Frank


Frank;
Would you tell us about sample weights of the two bullets?

  
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Re: diameter/weight/length formula?
Reply #12 - Dec 19th, 2013 at 7:03pm
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I haven't looked at how those factors were generated, but obviously with a reference bullet of. Particular shape.  The "money" bullets are a relatively slender tangential ogive with a small hemispherical tip.  I imagine that they'll always come out a bit longer than the formula predicts.  But that's ok, it's still likely useful.  The only way to get a really accurate answer is to calculate the volume by integration.

For reference, I just weighed a Buffalo Arms 324gr "money bullet" in 20:1 alloy.  It is 1.322" long and the formula predicts 1.251".  Still probably good enough for whatever you want it for Smiley

Chris.
  
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Re: diameter/weight/length formula?
Reply #13 - Dec 19th, 2013 at 7:15pm
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gunlaker wrote on Dec 19th, 2013 at 7:03pm:
I haven't looked at how those factors were generated, but obviously with a reference bullet of. Particular shape.  The "money" bullets are a relatively slender tangential ogive with a small hemispherical tip.  I imagine that they'll always come out a bit longer than the formula predicts.  But that's ok, it's still likely useful. 
For reference, I just weighed a Buffalo Arms 324gr "money bullet" in 20:1 alloy.  It is 1.322" long and the formula predicts 1.251".  Still The only way to get a really accurate answer is to calculate the volume by integration.
probably good enough for whatever you want it for Smiley

Chris.



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Re: diameter/weight/length formula?
Reply #14 - Dec 19th, 2013 at 7:57pm
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Cat_Whisperer wrote on Dec 19th, 2013 at 7:15pm:

Just a few keystrokes when drawn in 3d cad!


Indeed!
  
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