(8 April 2012, 23:30)Stokey Wrote: (8 April 2012, 23:24)00111001 Wrote: (8 April 2012, 10:08)Weyoun Wrote: I'll be keeping a motorcycle helmet put to one side. that'll take an air rifle round with out any problems. Less visibility, and you can't hear as well, but it'll definitely save a thwack to the old noggin.
I'm gonna get really boring here with my defence of air rifles, an uprated air rifle will shoot through a steel sheet let alone a motorcycle helmet :-)
The one I have will kick out 23lb/ft2 apparently. I doubt very much that'll go through a steel sheet?
A motorcycle helmet is built to withstand a huge impact. Admittedly, the surface area of the impact is greater, but the forces involved are higher.
Quote:Airgun energy generally is expressed in foot pounds. It is well worth considering, in practical terms, what a foot pound represents. One foot pound is the energy that a one pound object releases when falling one foot (ignoring the friction of air, which, from a very rough practical sense over very short distances, we may do for very dense objects). Thus, one can roughly consider that a gun firing with a muzzle energy of 12 ft./lbs. is about the same as dropping a 16 oz. hammer head from about 12 feet. Extend that to a 30 ft. drop for a 30 ft./lb. gun. Imagine being hit by that falling hammer head and you are doing a very crude visualization of the gun’s potential hitting strength.
(9 April 2012, 00:14)00111001 Wrote: (8 April 2012, 23:30)Stokey Wrote: [quote='00111001' pid='12044' dateline='1333923890']
[quote='Weyoun' pid='11927' dateline='1333876083']
I'll be keeping a motorcycle helmet put to one side. that'll take an air rifle round with out any problems. Less visibility, and you can't hear as well, but it'll definitely save a thwack to the old noggin.
I'm gonna get really boring here with my defence of air rifles, an uprated air rifle will shoot through a steel sheet let alone a motorcycle helmet :-)
The one I have will kick out 23lb/ft2 apparently. I doubt very much that'll go through a steel sheet?
A motorcycle helmet is built to withstand a huge impact. Admittedly, the surface area of the impact is greater, but the forces involved are higher.
Quote:Airgun energy generally is expressed in foot pounds. It is well worth considering, in practical terms, what a foot pound represents. One foot pound is the energy that a one pound object releases when falling one foot (ignoring the friction of air, which, from a very rough practical sense over very short distances, we may do for very dense objects). Thus, one can roughly consider that a gun firing with a muzzle energy of 12 ft./lbs. is about the same as dropping a 16 oz. hammer head from about 12 feet. Extend that to a 30 ft. drop for a 30 ft./lb. gun. Imagine being hit by that falling hammer head and you are doing a very crude visualization of the gun’s potential hitting strength.
How to Calculate Metal Piercing Force
By Vivek Saxena, eHow Contributor | updated July 12, 2011
To pierce metal, you must first calculate the force required.
The only way to calculate metal piercing force, which is the force required to pierce sheet metal, is by using the mathematical equation F = P * T * PSI. F equals the force required to pierce sheet metal in pounds, P equals the perimeter of the area to be pierced in inches, T equals the sheet metal thickness in inches and PSI equals the sheer strength rating of the sheet metal in pounds per square inch.
1 Measure the perimeter (P) of the area on the sheet metal that you intend to pierce by using a ruler.
2 Determine the thickness (T) of the sheet metal you want to pierce by using a ruler.
3 Obtain the sheet metal's PSI. Below are the PSI(s) of several common metals:
Brass: 20,000 lbs/in^2
Gun-Metal: 12,000 lbs/in^2
Cast Copper: 20,000 lbs/in^2
Bronze Aluminum: 50,000 lbs/in^2
Muntz Metal: 40,000 lbs/in^2
4
Plug the values obtained in Steps One, Two and Three into the equation from the Introduction and solve. Thus, given a perimeter of 1 inch, a thickness of 3 inches and a PSI of 20,000 lbs/in^2, the equation would be solved as follows:
F = 1 * 3 * 20,000
F = 60,000
Thus, the piercing force required to shear this metal is 60,000 pounds.
Read more: How to Calculate Metal Piercing Force | eHow.co.uk
http://www.ehow.co.uk/how_8727459_calcul...z1rUiRxMZh
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