billsnuff
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The formula is actually Diameter in inches = 0.005*92^(36-n/39), where n is the AWG gauge number.Lxnxjxhx said:...I wasn't there, of course, but I can't imagine the people who designed this gauge system to base it on 2.00503151950125:1. They based it on 2...
While I believe significant figure rules have their place, I don't believe the rules should be used in all cases. I consider them more of a guideline than a rule. After all, you must consider those rules were put into practice back when people had to use a slide rule or such to calculate results.I'd think if, in the answer, you need more than 3 "significant figures" (not the same as decimal places) you've gone off the track somewhere.
Intermediate calculations are supposed to carry two more than the final answer (so, 5, in this case).
That looks handy. Where did it come from?Smart $ said:The formula is actually Diameter in inches = 0.005*92^(36-n/39), where n is the AWG gauge number.
I remember one professor had us to use 3 significant digits unless the number started with a one, and then we were to use 4 significant digits. I don't know why. I would have thought you might want more digits if the number started with a five.Smart $ said:While I believe significant figure rules have their place, I don't believe the rules should be used in all cases. I consider them more of a guideline than a rule. After all, you must consider those rules were put into practice back when people had to use a slide rule or such to calculate results.
Can't remember exactly where I got it from, but this "reference" explains it... http://en.wikipedia.org/wiki/American_wire_gaugemivey said:That looks handy. Where did it come from?