Re: Implications of ICE crash

["M.B. and C.M.McDonald" <Michael.and.Colleen.Mcdonald@xtra.co.nz>]

If you use the standard formula    
     v^2 = u^2 + 2as
substituting
    v = 0
    u = 55.6  m/s  (= 200 km/hr)
    s = 2300
we get
    a = -0.67 m/s^2

This is only an average.
The braking efficiency is roughly 7%.

At high speeds there would be significant aerodynamic drag.
I believe the maximum tractive/braking is normally about 25%, with up to
33% available if anti-slip/slide systems used and the rail is clean and
dry.

If the wheels do slide, the braking efficiency is very much lower.

(If the braking effort was able to be held at 25%, the stopping distance
would be 620m.)

>From the formula, you can see that if the loco had been travelling 100km/hr
the braking distance would have been 575m, or if travelling at 300km/hr
then 5175m.

Talking of slip/slide, saw a rubber-tyred tractor shunting about 20 wagons
at one station in Queensland.  It was having considerable difficulty, and
the amount of slipping and sliding I observed made me wish I had shares in
Dunlop.

Also saw a not-too-small indentation in the rails at another station
obviously caused by a loco spinning its wheels.

Michael McDonald

Paul Jones <dj_nightshade@geocities.com> wrote in article
<357d1b7f.0@139.134.5.33>...
> hmmm let's see.
> 180,000kg's of train
> 55.55... metres per second
> 2300 metres in wich to stop
> but how much force is applied by the brakes? (in newtons)
> i'll assume a decceleration of .5m/s (can someone tell me that too?)
> ...
> ...
> = 166 seconds??
> guess i need more info...
> 
> --
> Paul Jones