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2009 Module 2 solved Mainline calcs
#9
I don't understand where '1/2 Deceleration time' or '1/2 Acceleration time' equates to 33.3 m/s?

By adding the above time to non stopping headway we get stopping headway.

3 aspect stopping headway = 3 aspect non stopping headway + ½ Deceleration time + Station dwell time + ½ Acceleration time

3 aspect stopping headway = 112+33.33+30+33.33


Can anyone clarify, as I was working it out to be: 112+66.6+30+66.6
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Messages In This Thread
RE: 2009 Module 2 solved paper Mainline - by PJW - 25-08-2011, 08:26 PM
RE: 2009 Module 2 solved paper Mainline - by PJW - 26-08-2011, 08:12 AM
RE: 2009 Module 2 solved Mainline calcs - by michaelmcnulty - 21-09-2016, 02:40 PM
RE: Module 2-2009 Calculations - by PJW - 21-09-2011, 12:46 PM

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