Figuring Energy of a round

[Dave Waits]

A lot of times when I quote a Velocity from one of my tests I'll also quote the Energy. Some of you might wonder how I come up with these numbers( No Mike, there's no room for them in my posterior). Chuck and I use the Factory standard formula for figuring Energy in a round.

Velocity times velocity
450400
Times bullet-weight =FPE
For example; a 158 grain bullet at 1250 fps.
1250 x 1250 divided by 450400 times 158 equals 548FPE

[Lonestar grips.]

very interesting Dave! i'm going to write this down in my reloading manual.

[sdmc530]

that is great to know. I have a chrony and always messing with loads and this is an invaluable formula...now if I only knew how to do math....

[Lane]

Thanks, Dave. As an engineer, I'm always wanting to know the backing info. With that in mind - what does 450400 represent?

Lane

[Dave Waits]

Lane,The constant 450400 takes into account the proper unit conversions and the half mass in order to get the energy balanced correctly.

You can use different constants to figure it in Joules or Ergs. However, Americans prefer Pounds*feet or, more commonly, foot-pounds, for energy.

[ColMike]

A lot of times when I quote a Velocity from one of my tests I'll also quote the Energy. Some of you might wonder how I come up with these numbers( No Mike, there's no room for them in my posterior). Chuck and I use the Factory standard formula for figuring Energy in a round.

Velocity times velocity
450400
Times bullet-weight =FPE
For example; a 158 grain bullet at 1250 fps.
1250 x 1250 divided by 450400 times 158 equals 548FPE

Interesting.

The general formulae for Kinetic Energy is E=.5(mass)(velocity*velocity) European standard uses meters/second and yields Joules.

of course the bullet is in grains and needs to be converted to pounds for consistency. Many sources when publishing kinetic energy tables for small arms ammunition, use an acceleration due to gravity of 32.163 ft/s2 rather than the standard of 32.1739 ft/s2 which is used above.

The formula then becomes
KE=.5(mass)(velocity*velocity)/(7000*32.163) with the alternate gravity number. The ICAO standard accepted for the gravitational constant is 32.1742 which gives 450,437.4. Speer uses this standard number for acceleration due to gravity. Although several online sources use the alternate value, I can not find a 'why' to the use of it. I doubt the small variance will matter much at our level of specificity regardless of your choice.

Now for a completely different approach:
1
Divide the bullet weight by 100. For example, the average weight of a 9mm full metal jacket bullet is usually either 115 grains or 124 grains. When divided by 100, the result is 1.15 and 1.24, respectively.

2
Divide the velocity by 100. Again, using an average 9mm full metal jacket bullet at 115 grains, a common velocity is 1,145 feet per second. Divided by 100 to get 11.45.

3
Multiply the adjusted weight by the adjusted velocity, then by the adjusted velocity again, then by 2.22.

Using the figures from the example, the equation is 1.15 x 11.45 x 11.45 x 2.22 = 334.7046825. Rounded off, the muzzle energy ends up as 334.7 foot-pounds.

Compare this with the actual formulae you get 334.9... close enough I'd reckon.

Don't know what happened, I just got in the physics mode this morning.

[Dave Waits]

Colonel, mine is the same, just modified to use with a standard pocket calculator. Plugging your 115 grain 9mm at 1145 into mine I come up with 334.7 FPE. This way you don't have to break down the velocity to a managable size. also, for all practical purposes, .2 FPE is a non-existant problem. It's just a faster way to do it, unless you're a Math-Junkie who enjoys the minutia.

[TLE2]

I don't if any of you noticed, but this is cut from the same cloth as the famous equation for energy from mass E=Mc2

[ColMike]

I don't if any of you noticed, but this is cut from the same cloth as the famous equation for energy from mass E=Mc2
and C is the speed of light.

[TLE2]

Right, energy = mass times velocity squared.

[DrHenley]

The ICAO standard accepted for the gravitational constant is 32.1742 which gives 450,437.4. Speer uses this standard number for acceleration due to gravity. Although several online sources use the alternate value, I can not find a 'why' to the use of it.

Because of the different centrifugal force, the gravity at the equator is smaller than that at the poles. So the precise gravitational acceleration is a function of latitude. They just took the gravitational acceleration at 45° Latitude (halfway between the poles and the equator) as the standard. Paris is at 48.86°, New York City is at 40.67°.

Poles: 32.26 ft/s²
Equator: 32.09 ft/s²
45° Latitude: 32.17 ft/s²

[ColMike]

[QUOTE=DrHenley;74487]Because of the different centrifugal force, the gravity at the equator is smaller than that at the poles. So the precise gravitational acceleration is a function of latitude. They just took the gravitational acceleration at 45° Latitude (halfway between the poles and the equator) as the standard. Paris is at 48.86°, New York City is at 40.67°.

Poles: 32.26 ft/s²
Equator: 32.09 ft/s²
45° Latitude: 32.17 ft/s²[/QUOTE

Excellent, thanks!

[Lonestar grips.]

now my Brain hurts:faint:

[DrHenley]

now my Brain hurts:faint:

Here, let me help :D

Here is how you find the acceleration due to gravity at any particular latitude (at sea level of course):

http://www.cotep.org/forum/picture.php?albumid=22&pictureid=740

g45: 32.17 ft/s²
gpoles: 32.26 ft/s²
gequator: 32.09 ft/s²
π: 3.1415927

[Lonestar grips.]

Here, let me help :D

Here is how you find the acceleration due to gravity at any particular latitude (at sea level of course):

http://www.cotep.org/forum/picture.php?albumid=22&pictureid=740

g45: 32.17 ft/s²
gpoles: 32.26 ft/s²
gequator: 32.09 ft/s²
π: 3.1415927


well now it makes sense:D

[TLE2]

I love math! is latitude in degrees or radians?

[TLE2]

Found this on Yahoo Answers:

Altitude
Acceleration due to gravity decreases as the inverse of the square of the distance from the center of mass of the body imparting the gravitational acceleration; g=k/x2
Because the earth is not perfectly spherical, the distance from the center of the earth for any person standing on the surface depends upon the latitude L.
The equatorial radius of the earth is approximately 6,378,140 meters.
The polar radius is approximately 6,356,755 meters.
If we model the earth as an ellipsoid, this means that the radius of the earth at latitude L is given by R= 6,356,755* sqrt( 1 + .0067396*cos2(L) ).
At the latitude of Terre Haute, this is approximately 6,369,502 meters.
Thus, a person at altitude of H meters above sea level experiences and acceleration due to gravity of
a= g*R2/(R+H)2.
G310 is at approximately 580 feet above sea level. This is roughly 176.8 meters above sea level.
This gives an acceleration due to gravity of approximately 9.8095794 meters per second2.
Moving to the classroom ceiling, this constant becomes 9.8095701 meters per second2.
Moving to the circle in front of Hadley Hall (L-> .689103, H->180m), we get an acceleration due to gravity of approximately 9.80956953 meters per second2.

[AutoMag]

Way to technical for me!!!:D

[DrHenley]

I love math! is latitude in degrees or radians?

Degrees

The π/180 part converts degrees to radians. If you have the latitude in radians already, just leave out the π/180 part.

[Dave Waits]

Gentlemen, we're figuring energy for handgun and rifle bullets, not the Space Shuttle!

As the Good Colonel aptly put it, a few final inch*pounds at this level are unnecessary. It's getting taken way above most pay-grades on here.

All I was trying to do was to put a simple yet fairly accurate formula in front of the guys. Nobody on this forum is going to be taking one-mile .416Barrett shots on Terrorists.

You guys are making my head spin and I have a math background myself, though evidently not as heavy as some of you. LOL!

[Riverpigusmc]

Bump.

[reddoggm]

Ok I don,t know if I,m sneaking in on a Sticky (without permission ) OR for Dummys Like me that Chrono and want a Quick Figure I go Here--http://www.handloads.com/calc/quick.asp Hope I,m not over stepping some Site Rule here
By the way I looked for a RULES Sec here and I couldn't believe my Monitor
I couldn't find One
DOGG!!!

[Rick McC.]

Very handy, thanks!

[Riverpigusmc]

bump. Time to get some gun threads up