The Rohrbaugh Forum
Miscellaneous => The Water Cooler -- General Discussions => Topic started by: GeorgeH on July 03, 2004, 12:51:47 PM
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In determining the foot pounds of energy of a bullet, they divide the product of part of the formula by 450,000--sometime a more exact number. What does this number mean?
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The main equation George is E=MC^2 .... or based on that. C tho is speed of light ... 186,000 miles/sec. Then we have the question of units ... metric, imperial - whatever.
Using as we do, imperial ... feet/sec ... and Avoirdupois ... grains ... the whole deal when worked out the ''long way'', yields a constant for use in the simplified version.
I am fudging here cos - I forget the whole deal to reach that figure but .... that's all it is - a constant to enable us to use the ''simple'' method!
So all we do - is square velocity ... multiply by grains of bullet weight - and then divide by that constant. I use a more exact version which is 450,356 ... even that is fractionly off the exact but I happen to have memorized it! (It's actually 450,436 - just checked - oh well, I was close!).
My math gets real rusty - at least all the stuff I used to know and use - cerebral atrophy they call it!
So - explanation of sorts ... but fudged! ;D
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Chris:
Thanks for clearing that up for me.
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Ok here goes...... using Newtonian physics...a bit more of a challenge to derive from E = Mc2 :o
The enegry of a moving object is typically represented as a combination of its potential and kinetic energies. So E=PE+KE.
Potential energy is 'stored' energy due to gravity:
PE = Mass * (gravitational constant) * height
Kinetic Energy is 'moving' energy:
KE = (Forward Velocity) + (Rotational Velocity)
KE = [1/2 * (Mass) * (Veloctiy)2] + [1/2 * (moment of intertia) * (angular veloctiy)2]
Total Energy:
E = PE + KE
E = [Mass*(gravitational constant)*height] + [1/2 * (Mass) * (Veloctiy)2] + [1/2 * (moment of intertia) * (angular veloctiy)2]
For the sake of simplicity we can neglect Potential Energy altogether. Do the math, and you'll find that it is less than 1 ft-lb for most grains. The rotational velocity can be calculated to be less than 5 ft-lb depending on grain and geometry. So in essence Energy ~ Kinetic Energy
We now have something resembling the equation you mentioned:
E = 1/2 * (Mass) * (Veloctiy)2 or
E = 1/2 MV2
Now for the units game (ft - lb's in this case)
Since bullet weight is typically represented in grains...
Convert to mass
M = weight / gravity
M = weight / ((7000 grains/lb) * (32.2 ft/sec^2))
Substituting......
E = 1/2 * (weight / 7000 gr/lb)(32.2 ft/sec^2) * V^2
E = wieght * V2 / 450800
Add in all the specifics for rotational energy and you'll come out with a slightly different constant, but I'll leave that to the more ambitious! :D
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Shelby - thank you Sir - for doing more than my lazy ol' brain was prepared to dig out!! :D
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You're welcome :)
I was up doing some number crunching anyway, so it was nice to take a break and wrap my mind around something fun.
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Thank's guys, but I'm now more confused. R9S used 450,436, and Shelb used 450,800. Why the difference? Remember when it comes to this, I'm a moron...
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George .... Shelby may well come back on it - I am short on time right now.
Consider tho ..... an example .....
230 grain bullet, 800 fps .....
{(800^2 = 640,000)x230}=147,200,000
147,200,000/450,436 = 326.8 ft lbs
147,200,000/450,800 = 326.5 ft lbs
So - you can see that even if we are lax and use an approximation, the result is probably only affected to the right of decimal point ...... for our purposes, insignificant IMO!
N'est pas?? :)
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Ahhhh, the devil is in the details! ;D
I am speculating, but the descrepency is likely to be attributed to the value that is used for g.
If we use a more precise value for gravity: 32.1734 ft/sec˛, we get a constant of 450427.6
Now for some real fun:
The value for g, and thus an object's weight varies slightly depending on latitude and altitude anyway! ;D
An approximate value for g, at a given latitude (L) and height above sea level (H in meters), may be calculated from the formula:
g = 9.7803184 * (1 + 0.0053024 * sin2 L - 0.0000059 * sin2 2L) - 3.086×10-6 * H
*don't forget to convert this number back to English units ;)
I think that you'd find that however precise of a theoretical constant you get, the true emperical data would vary widely for various bullet shapes and experimental conditions.
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i am sooo comfused :P
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Hi Shelb:
Back to the class. You wrote:
Ahhhh, the devil is in the details! ;D
I am speculating, but the descrepency is likely to be attributed to the value that is used for g.
If we use a more precise value for gravity: 32.1734 ft/sec˛, we get a constant of 450427.6
What is multiplied and divided to get a constant of 450427.6?
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Revisiting the mass & energy equations with the new constant
Mass:
M = weight / gravity
M = weight / ((7000 grains/lb) * (32.1734 ft/sec2))
Plug Mass into kinectic energy equation: E = 1/2 M * V2
E = 1/2 * (weight / 7000 gr/lb)(32.1734 ft/sec2) * V2 = (weight * V2) / (2 * 7000 * 32.1734)
2 * 7000 * 32.1734 = 450427.6
E = wieght * V2 / 450427.6
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Shelby - great stuff ... you saved me having to spray brain cells with ''Liquid Wrench'' .... way too much corrosion there ... i think they call it ''disuse atrophy''!! ;D ;D
Or - could it be ..... ''Misuse atrophy'' ... haha! :P
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Lord have mercy, I am impressed! All that math makes my head swim and reminds me why I decided it was best for me to major in history with a minor in English.
RS :D
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Daaaaaaaammnnn! ;D We'uns shure doose have some smart uns here! ;D