On Jan 25, 2:50 pm, Tom Anderson wrote:
On Fri, 25 Jan 2008, wrote:
Presumably for trains with few stops the power consumption is
approximately constant regardless of the length of the train because
the main loss will be air drag.

As long as the train never wants to climb a hill, perhaps. If it does, the
old mgh term rears its head.

And you get it back again on the downhills. If we assume a 1000kg car
takes 10kW to maintain 25m/s (about 50mph) on the flat. On a 1 in 50
it will take 15kW to maintain that speed uphill but only 5kW to
maintain that speed downhill. As the speed is constant the time is the
same up and down so the total energy is the average, i.e. same as on
the flat over the same distance.

It's only if you are power limited on the climbs and descend faster
that you lose out because your power requirements go up as cube of
speed to overcome airdrag. This is typically the case for cyclists -
going uphill the speed is low enough that airdrag is negligible and
all the energy goes in mgh. A cyclist coasting down a 1 in 10 at 20m/s
will be using 2kW to overcome the airdrag.

(You also lose out if you have to use braking on the descent and it's
non-regenerative. On a tandem where the power doubles but the drag is
roughly the same the terminal velocity will be about 25% higher and
the tandem will have to dissipate 2kW via the brakes to keep the speed
at 20m/s - which is why tandems often have a hub brake as well as rim
brakes or, nowadays, disk brakes)

Do you think that drag is overwhelmingly greater than rolling resistance,
losses in the bearings, etc?

I'm not sure quite how many orders of magnitude it will differ by but
yes. Beyond any doubt air drag dominates everything else at any sort
of reasonable speed.

Power to overcome airdrag goes up as cube of speed. Friction losses
are linear.

Tim.