Christopher_Perkins
08/10/08 01:33 AM
24.125.201.167
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Quote:
What's more relevant to laser performance was 100,000 joules per second per square centimeter.
Thats good... That is what i have been driving at in a way..
The Laser is delivering the energy to a square centimeter of the targets armour to burn through or otherwise render Hors De Combat an unknown thickness of the magical clarkeian armour... or a determinable thickness of standard steel plate armour (unadulterated with things like carbon fibers, reactive blocks, etc...)
1 cm * 1 cm = 3.14 * x 0.56 cm * x 0.56 cm quiaff?
a plate of Standard Armour Steel of thickness X with a surface area of Y weighing Z
Z is the mass of armour defeated by a small laser
Given
62.5 kg per Tech B Bar 5 Armour Point
Number of Tech B Bar 5 Armour Points required to be roughly equal to Tech D Bar 10 is 2
Small Laser renders Hors De Combat 3 Points of Standard Tech D Bar 10 Armour.
Z = 62.5 kg * 2 * 3 = 375 kg of standard plate steel armour...
X = the thickness of the Armour with a Mass of Z and a surface area of Y
Solve For: a 1.12 cm diameter beam cutting through a thickness X plate of armour massing 375 kg of armour would only have to defeat what mass of armour?
And key holing & Penetrating that thickness of steel with an entry point 0.56 cm in diameter would require what amount of energy to melt a 0.56 diameter hole through the armour in approximately 0.2 seconds?
Alternately...
if Tech B Bar 5 Armour requires 4 points (250 kg Tech B Bar 10 / 62.5 Tech B Bar 5 = 4) to be equivalent to Tech D Bar 10 Armour, then 750 kg of Standard Steel Plate Armour would be the equivalent to 187 kg of Standard Tech D Bar 10 armour
And key holing & Penetrating that thickness of steel with an entry point 0.56 cm in diameter would require what amount of energy to melt a 0.56 diameter hole through the armour in approximately 0.2 seconds?
Quote:
**187 megajoules is the amount of energy needed to melt half of 187.5 kilograms of cheap carbon steel with a low melting point. The small laser destroys 187.5kg of armor, which is most definitely not cheap carbon steel. Depending on your assumptions about beam power (rate of energy delivery) and various inefficiencies (reflection from the target, energy wasted trying to hose the beam onto target, etc.), you could very well need much more energy from a small laser to destroy 3 points of armor.
humm, interesting...
What level of energy is needed to do the same to 187.5 kg of homogenous plate armour steel?
Christopher Robin Perkins
It is my opinion that all statements should be questioned, digested, disected, tasted, and then either spit out or adopted... RHIP is not a god given shield
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