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Old 12-19-2014, 08:47 PM   #1
BraselC5048
 
Join Date: Oct 2012
Default Linear shaped charge to cut through DR 1900?

Anyway, in the field of weapon design I need a linear shaped charge (manually placed) to cut through starship armor, which could be as much as DR 1900. It can't just make a hole, it has to cut through it along its length. The idea is to string a bunch of them together (rigid sections) in a circle to blast a large opening through an armored airlock hatch. (The pressure door gets the armor's full DR to protect it, as it's on the other side.) Obviously it's going to be heavy, rigid and short, but it doesn't need any more portability than somebody being able to carry it from the storeroom to the airlock. It'll take a bunch of sections to make a man sized hole, of course.

What I'm wondering is how much it would weigh per foot, so I can get some idea of the size, weight and number of sections. It will do maximum damage automatically, and if it gets the (10) armor divisor, instead of the (5) armor divisor the flexible but much smaller cord of the same basic thing from High Tech gets, (which weighs 0.5 lb per foot), it would need to do 6dx6(10). If it's stuck with the (5) divisor, then it would be closer to 6dx11 (5). Which going by the "Blowing stuff up" section of Basic and scaling (assuming the charge from High Tech is the equivalent of .11 lb TNT), would be 137 lb or so for a 1ft section. Which is outrageous. If it gets to do 6dx6 (10), again scaling, it would weigh 41 lb for a 1ft section. Which is actually pretty close to what I had imagined for a 1 or 2 foot section.

So use the 6dx6(10) scaling, and 41 lb per foot long section?

Or would it be better to use one smaller charge to cut partway through, then place another one in the same sport to finish the job, or would that not work with shaped charges at all?
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