[PRCo] Fw: Acceleration of real trains [tm]

[Jim Holland]

wrote in message 
news:<1194839531.042430.164560 at v65g2000hsc.googlegroups.com>...
On Nov 11, 1:31 pm, Peter Niessen wrote:
 > Hans-Joachim Zierke writes:
 > > The Tokaido Shinkansen will see service by the replacement generation
 > > of the 700 type, the N-700. It features a 1° tilting system for
 > > eliminating some 250 km/h curve speed restrictions on the 270 km/h
 > > Tokaido Shinkansen.
 >
 > > For really using the 270 km/h speed, acceleration has been 
improved. The
 > > trains can now reach these 270 km/h (168 mph) within 3 minutes. For 
this,
 > > power was increased to 17080 kW (22905 hp). Despite the tilting 
system,
 > > and despite the power increase, the new generation is few tons lighter
 > > again, more than half a ton has been shaved from each carriage.
 >
 > Hm. I'm puzzled now. I pulled up the numbers for the ICE3 (wikipedia),
 > which has 8000 kW. Yet it seems to have a better acceleration:
 >
 > a = F / m = 300 kN / (32 axles * 16 * 10^3 kg/axles)
 > = 300 * 10^3 N / (512 * 10^3 kg)
 > = .6 m/s^2
 >
 > The Shinkansen N-700 gets
 >
 > a = dv / dt
 > = 270 10^3m / (3600 s) / (180 s)
 > = .4 m/s^2
 >
 > Am I reasonfoundationing averages and peak values here?

Something like that. Your calculation for the ICE3 is only the
_initial_ acceleration, starting at v = 0, while the Shinkansen figure
is the average acceleration from v = 0 to v = 270.

As speed increases, two things happen that result in slower
acceleration. The first is that you have to subtract train resistance
from the "F" term. If you have 300 kN of tractive effort, but it takes
50kN just to maintain your current speed, then you only have 250kN
left over for acceleration. The second thing to consider is that 300kN
is the _maximum_ tractive effort and is not available at all speeds.
In fact, the tractive effort available at speed v will be the maximum
of 300kN or (P/v), where P is the power that can be applied by the
motors - in this case, I guess, it would be 8000 kW. So at v = 150
km / h, the actual tractive effort will be

P / v = (8000 kW) / (150 km / h)
= (8000 N km / s) / (100 km / 3600 s)
= 192 kN

For the purpose of the example, if we pretend that the train
resistance is 50 kN when v = 150 km/h, then the acceleration would be
((192 - 50) kN / (512 * 10^3 kg)) = 0.28 m / s^2 when the train is
traveling at this speed.

Since acceleration is a function of speed, you have to integrate the
acceleration function from 0 to 270 in order to find out how long the
ICE takes to reach 270 km/h (or, even better, run a test :-)

Dan

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Jim  Holland
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Studying Pittsburgh Railways Company (PRCo)
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..............................From 1930 -- 1950
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Pennsylvania  Trolley  Museum  (PTM)
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http://www.pa-trolley.org/
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N.M.R.A.
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http://www.nmra.org/