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| Re: Oft asked question Re: Breaking News MNRR Derailment (1262375) | |||
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Re: Oft asked question Re: Breaking News MNRR Derailment |
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Posted by AlM on Wed Dec 4 13:07:29 2013, in response to Re: Oft asked question Re: Breaking News MNRR Derailment, posted by Joe V on Wed Dec 4 12:22:45 2013. Actually not.Basic physics: Radial acceleration = v^2/r, where v = speed and r = radius of the curve. Leverage factor because of high center of gravity is approximately = (height of center of gravity above rails) / (1/2 the distance between the rails) = approx. 6/2.35 = 2.5. Speed = 82 mph = 120 feet/sec. Looking at an arial Google view, raidus = about 1000 feet. So radial acceleration times leverage factor = 120^2 x 2.5 /1000 = 36 ft/sec/sec. You need 1 g = 32 ft/sec/sec of upward acceleration to lift the outer wheel off the rail, so even at 82 mph the train barely exceeded that. I admit the 1000 foot curve radius is an estimate and the 2.5 leverage factor is too. And trains might not behave exactly the way simple physics says they do. But I truly believe that if the train had been going 60 mph it would have stayed on the rails. It's the square of the speed that matters, and 60^2 is a lot less than 82^2, even though it's still 4 times 30^2. |
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