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The stiffness of springs in gunsmithing catalogs is generally measured in pounds. How are the pound designations made? H. C. | ||
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One of Us |
Modern springs are designed using computer programs that automatically calculate the math to determine the resistance value of a spring using factors such as the material, diameter of material (in coil springs), width and thickness of material (for flat springs) and overall length or outside diameter of the finished spring. There are also spring compression test machines that give direct readings on a gauge for a finished spring. It’s all voodoo to most of us mortals. | |||
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One of Us |
It's voodoo, but that numerical designation in pounds means something. Let's say I have a spring. Let's say it's a longish one, like the recoil spring on a 1911. Let's say I compress that spring 1 centimeter and measure a resistance of 3 lb. If I compress that spring another 1 cm, I will measure a resistance of 6 lb. If there's room withot the coils touching to compress it another 2 cm, the measured resistance will be exactly 12 lb. That's Hooke's Law, and real-life springs obey it incredibly well. Is there some degree of compression at which the measured resistance equals the numerical poundage designation, or is it more vodoo than all that? H. C. | |||
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One of Us |
You’re way out of my league on this one...perhaps someone with an engineering background would have an answer for you. | |||
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new member |
It is usually a number of pounds for one inch of spring compression. A 15 lb. spring would require 15 lb. to compress it 1 inch. The only other designation I have seen is kg/mm which usually gives a very low number. | |||
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One of Us |
Henry, Why not call a company that manufactures springs and ask someone in their tech department for an answer. I’m sure they can give you more information than you probably even want. | |||
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One of Us |
scotty said it perfectly and Henry gave the criteria that showed the math. The force to compress a spring a unit length. | |||
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