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You're going to make me dust off my engineering text books! This is going to be a ridiculously over-simplified analysis, so please bear with me and feel free to make additions or corrections. The governing theory here is the linear thermal expansion of a material. Note that by "linear expansion" we're already simplifying the problem to one dimension, when in fact the real world exists in three dimensions. The three dimensional problem is highly complex and well beyond my capacities, so fortunately the one dimensional problem yields a reasonable approximation. Here's the governing equation: (alpha) = (change in length) divided by (original length) times (change in temperature) We're curious about the (change in length), so we'll solve the equation for it: (change in length) = (alpha) times (original length) times (change in temperature) Now we insert our three known quantities. Original length is easy, that's 300 miles. Change in temperature can be determined from M636C's information. He gives us a range of 0 to 40 C. We'll assume the rail was welded (or installed) at a neutral temperatue of 20 C. Thus, the maximum change in temperature is 20 C in either direction (warming from 20 to 40 or cooling from 20 to 0). Alpha is the linear thermal expansion coefficient for the material. For steel, we'll use 12 x 10^-6 (that's 0.000012). The units here are inverse degrees C. This is only an approximation, as the exact number would depend on the specific type of steel. Plugging these numbers into our equation gives: (change in length) = 0.000012 x 300 miles x 20 C (change in length) = 0.072 miles = 380 feet So, if we have a solid, continuous piece of uniform steel that's 300 miles long, it would grow by 380 feet in the heat of a 40 C day or shrink by 380 feet on a cold 0 C night. Again, please bear in mind this is a grossly over-simplified approximation. A more precise answer would require a three dimensional calculation (since steel expands in all three dimensions, not just length), specifics on the rail itself, information on the weld and insulating joints and their expansion properties, etc. Hope this was a useful exercise, Scott Lothes Cleveland, Ohio
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