[–]moeburn [score hidden] 28 minutes ago  Margin of safety. They calculate exactly what the minimum machining tolerance would be required to make the parts work perfectly in its most extreme circumstances (full load, takeoff, banking, etc), then apply a 1.5 factor on top of that. permalinksaveparentreportgive goldreply [–]jacksonian7 [score hidden] 26 minutes ago  To add to this: It's the minimum machining tolerances required for part function (what ^ s/he ^ said) + performance safety factor + machining tolerance factor + what would be easier to manufacture + which would cost less to make/repair/rebuild? permalinksaveparentreportgive goldreply [–]mikeash [score hidden] 23 minutes ago  I imagine it all comes down to a complicated tradeoff with costs. Reducing tolerances to X will have benefit Y but will cost an additional $Z, is Y worth at least $Z? Complicated because Y is going to involve long-term stuff like reliability after a couple of decades, minor changes in fuel burn, or noise, and $Z is going to involve delicate machining technology. permalinksaveparentreportgive goldreply [–]chocodilesupreme [score hidden] 13 minutes ago  A new car built by my company leaves somewhere traveling at 60 mph. The rear differential locks up. The car crashes and burns with everyone trapped inside. Now, should we initiate a recall? Take the number of vehicles in the field, A, multiply by the probable rate of failure, B, multiply by the average out-of-court settlement, C. A times B times C equals X. If X is less than the cost of a recall, we don't do one. permalinksaveparentreportgive goldreply [–]IllustratedMann [score hidden] 15 minutes ago  Most likely testing and equations. The engineers figure that let's say +/-.00006in might cause an imbalance, wear, etc, so they give it a bit of room, and put it at +/-.00004in. Now, it's also probably not the case that they can't get it to .00002, but it's an order of magnitude in price difference. What I mean is, for this part, if all we need is .00004, and the price between machining the part to .00004 at $20,000 and machining the part to .00002 is $100,000, then .00004 is good You see, when a mechanical engineer gets his degree, along with it they give him 1" of tolerance. Early in his career he is very miserly with it, and gives it out .0001 at a time, fearful that it won't last. As he gets towards retirement, he sees how much he has left and starts handing it out in bigger and bigger chunks. That's why when you see a tolerance that is excessively tight it is usually from a very young engineer.