Figuring Energy of a round

[DrHenley]

Originally Posted by ColMike

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

[DrHenley]

Originally Posted by Lonestar grips.

now my Brain hurts

Here, let me help

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



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

[Lonestar grips.]

Originally Posted by DrHenley

Here, let me help

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



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

well now it makes sense

[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!!!

[DrHenley]

Originally Posted by TLE2

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!