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 01-17-2013, 02:48 AM PM User | #2 Old Pedant Supreme Master coder!     Join Date: Feb 2009 Posts: 24,952 Thanks: 75 Thanked 4,308 Times in 4,275 Posts Ummm...I think you need to go back and take pre-algebra again. If you do: x1 = 0; x2 = 4; x3 = 0; x4 = 0; Then you do: y = (x2-x1) + (x3-x2) + (x4-x3); you will be doing: y = (4 - 0) + (0 - 4) + (0 - 0) which is y = (4) + (-4) + 0 which is, correctly: y = 0 ********* In any case, just going one step further into elementary algebra, why not SIMPLIFY this equation: y = (x2-x1) + (x3-x2) + (x4-x3); which becomes: y = x2-x1+x3-x2+x4-x3; which becomes y = x2-x2 + x3-x3 - x1 + x4 which becomes y = 0 + 0 - x1 + x4 which becomes y = x4 - x1 In other words, the values entered for x2 and x3 *DO NOT MATTER*. The end result will only depend on x1 and x4. Me thinks you have much much bigger problems than just your JavaScript. Which is actually working correctly. __________________ An optimist sees the glass as half full. A pessimist sees the glass as half empty. A realist drinks it no matter how much there is.