Rate of rail elevation change during train transition curve

[tomtomtom1]

Hi everyone

Firstly this is not a homework question, i work in transport engineering. ( i do study part time which is why i sometimes post in the homework section)

Now to the problem.

I am trying to understand how to solve this problem, i already know the answer.

A train is traveling on straight & level track at 136.794kph.
It approaches a transition. The transition is 60m long.
Along the transition the Left Rail rises linearly over the length of the transition relative to the right Rail by 25mm.
Calculate the rate of change the left rail raises over the right in mm per sec over the 60m transition at 136.794kph.

i know that the answer is (25mm * 136.794kph) / (3.6 * 60m)

what i need help with is where does the 3.6 some from? and why do you times 25mm by 136.794kph?

can anyone explain

Thanks

 

[SteamKing]

The transition length is 60 m long. In this distance, the rails rise by 25 mm. Doesn't this suggest a slope?
The train is traveling at 136.794 km/hr. How do you convert 136.794 km/hr to a speed measured in m/s?

 

[tomtomtom1]

The transition length is 60 m long. In this distance, the rails rise by 25 mm. Doesn't this suggest a slope?

Yes there is a slope and the slope is 25mm/60m = 1m per 0.416mm which means for every 1m the train travels along the transition the left rail rises by 0.416mm.

The train is traveling at 136.794 km/hr. How do you convert 136.794 km/hr to a speed measured in m/s?

1kph = 0.277777778 m/s so 136.794km/hr in m/s is 136.794*0.277777778 = 37.998m/s.

 

[jacket]

rate of change the left rail raises = 25mm / (time needed to cross 60m)
= 25 mm / (60 m / velocity)
= 25 mm / (60 m / 136.794 km per hour)
Now, 1 km per hour = 1000 m / 3600 sec = 1/3.6 meter per sec
So,
rate of raise = 25 mm / (60 m / ((136.794 / 3.6) meter per sec))
= 25 mm / (3.6 x 60 / 136.794) sec
= (25 mm x 136.794) / (3.6 x 60) sec