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In answer to an inquiry on "Ask Ken Howell," I wrote the posts copied below. The original poster isn't interested to this extent, but I still want to know whether my equation is sound -- so would appreciate confirmation or correction from someone who knows how to figure such things. STIA
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Pure lead weighs (nominally) 0.4096 pound per cubic inch. That's 0.4096 � 7,000 = 2,867.2 grains.

But your bullets -- cast or jacketed -- aren't all lead, so the workable weight per cubic inch is something less -- and I can't guess how much less. There's an equation for finding the weight of an alloy or heterogeneous bullet relative to the weight of the same volume of water -- by weighing the bullet suspended (by an essentially weightless thread or wire) in the air, then in water. Some of the better scales used to have a double hook on the beam and a beaker platform on the base for making these two weight measurements.

But I haven't seen that equation in decades, don't remember it, and don't know where to find it or how to derive it. I'm sure that some of the better educated guys here can supply or derive it and will post it here. (I hope so -- I've wanted to recover that equation, for lo! these many years!) If I find it, I'll post it. It may be more dependable than hardness for comparing cast-bullet alloys for their relative consistency.
-------------------------------------------

Well, FWIW, I think I've fingered it out. I'll give my reasoning, and somebody can check it for us.

Old Archimedes first figured-out that an object in a liquid is buoyed-up, supported, or "lightened" by a force equal to the weight of the liquid that it displaces. A bullet suspended in water, therefore, must weigh its weight in air minus the weight of the water that it displaces. Its relative density -- its weight in air divided by the weight of the water it displaces -- is therefore its weight in air divided by the difference in its two test weights.

If I'm right, then the "lost" equation is

x = a/(a-b)

if
a = weight of the bullet suspended in air
b = weight of the bullet suspended in water
x = specific gravity

The weight of the bullet per cubic inch, then, is the weight of a cubic inch of water (about 0.0361 lb or about 252 to 253 grains) times the bullet's specific gravity as determined above.


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Ken,

You are exactly correct! Archimedes ruminated the same thing in the King's quest to determine if his crown was made of pure gold. Since water is roughly 1g/ml the displaced volume (in mL) is very nearly the difference in mass.


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Damn you beat me to it 'Tikka LOL! <img src="/ubbthreads/images/graemlins/blush.gif" alt="" /> ~ any more questions Ken point them my way <img src="/ubbthreads/images/graemlins/smirk.gif" alt="" /> .... but don't hold your breath for the reply...


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THANKS, 'Tikka and 'Stove!

I'd hate to have to decide which'd be more welcome -- confirmation or correction. Whichever, I'm glad to have the correct equation again -- have made it into a Mathcad work sheet so that all I have to do is enter the values of a and b for any tested bullet and get its x automatically -- then enter the value of x so derived and get the density (gr/cu in.) of the tested bullet from another equation. Nice and handy.

THANKS again!


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I knew that ! Darn, I wish I could have answered quicker.


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For anyone who caught my error in this response pre-edit... sorry about that <img src="/ubbthreads/images/graemlins/blush.gif" alt="" /> I almost said someone made a mistake <img src="/ubbthreads/images/graemlins/blush.gif" alt="" /> <img src="/ubbthreads/images/graemlins/blush.gif" alt="" /> <img src="/ubbthreads/images/graemlins/blush.gif" alt="" /> WOnder who that could have been, making a mistake? <img src="/ubbthreads/images/graemlins/blush.gif" alt="" />

To extend the playing around a little

x=.58y+1(100-y)

where y is the % of alloying antimony or tin in the bullet making huge assumptions that those are the only alloying metals...
art

Last edited by Sitka deer; 04/02/03.

Mark Begich, Joaquin Jackson, and Heller resistance... Three huge reasons to worry about the NRA.
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Sitka,

Mmm. As I originally understood Ken, he was interested in the formula to determine the density (or SG) of a bullet with unknown composition (alloy, jacketing, etc) I assume that he wanted that value in absolute terms and therefore the displaced water volume (i.e. volume of bullet) is simply the difference between the two mass measurements. Therefore, if the bullet mass is simply divided by the volume the density is obtained directly.

If we perform the exercise in SI units, then the calculation of SG is moot (since the reference density of water at 4 C is 1 kg/L) and the density is simply equal to the SG. However, working in units other than SI one would have to ensure proper conversion factors...


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Sitka dear, .... <img src="/ubbthreads/images/graemlins/blush.gif" alt="" />


I would double check your math but the battery has just gone in my abacus damn bad luck....

<img src="/ubbthreads/images/graemlins/wink.gif" alt="" />


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Wannatika
Sorry, I did not realize you had posted when I edited my post... you are quite correct... I hit the equation and disengaged brain for some reason <img src="/ubbthreads/images/graemlins/blush.gif" alt="" /> I edited my response, but would not have done that had I seen your response first...
art


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Sitka,

No problem... I was just worried that I had missed something. <img src="/ubbthreads/images/graemlins/smile.gif" alt="" /> BTW, I've been known to disengage the old brain on occasion as well. Have a good one.


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'tikka
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Why not just fill a small graduated cylinder half-way with water and mark the volume, drop the bullet in and measure the difference in volumes? This negates any conversions needed to compensate for the density of water.


George
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Originally Posted by GOD
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Think it over a bit more, my friend.
-- The bullet would sink to the bottom of the graduate.
-- It could be aluminum, iron, brass, lead, or ...
-- You'd never be able to see any difference in their composition.
-- The displacement would be the same for all.
So your technique couldn't determine its specific gravity (relative density), the aim and purpose of the equation.


"Good enough" isn't.

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Ken
Actually Avagadro is correct... the increase in volume would easily convert to the weight difference between weight in water and weight in air.

To use your example, the different metals would have equal displacement, but that displacement would be applied to a different starting mass and easily calculated.

I think the key is the weight would be more accurate because the graduated cylinder would have inaccuracies built into the less exact scale and the difficulty in determining exactly where the level was (miniscus problems)
art


Mark Begich, Joaquin Jackson, and Heller resistance... Three huge reasons to worry about the NRA.
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Yes, Art -- requiring separate calculations nonetheless, not in a single calculation -- and probably less precisely, especially with a sample as small as a bullet.


"Good enough" isn't.

Always take your responsibilities seriously but never yourself.



















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Ok, I sort of understand this but...


How does this tell you the percent of lead you have in the bullet you just cast?


Whatever you are willing to put up with, is exactly what you will have.

When your ship comes in. ... make sure you are willing to unload it.

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It doesn't.

How much ice cream do you expect for a nickel? <img src="/ubbthreads/images/graemlins/wink.gif" alt="" />


"Good enough" isn't.

Always take your responsibilities seriously but never yourself.



















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It gives only a rough approximation when you are using pure lead and alloying it with a lead/antimony alloy. Commonly done when you want to change the hardness or run out of a particular batch of lead.

While it is not perfect, the SG of Antimony is about .58. Assuming only two alloying metals (not likely to be exact..) is not pure science, but it does give a working approximation.
art


Mark Begich, Joaquin Jackson, and Heller resistance... Three huge reasons to worry about the NRA.

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