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Alex Terrell · August 30th 04, 10:39 PM posted to uk.transport.london,uk.railway

(Neil Williams) wrote in message ...
On 30 Aug 2004 03:11:06 -0700, (Alex Terrell)
wrote:

Which is why you don't see too many of them.

AFAIAA, all non-DMU new stock builds for local services in Germany
have been DDs - or certainly most of them. (The PumA-Modus stock
doesn't count as it's a Networker Classic-style rebuild on old
underframes). They are very common in Northern Germany.

I think on the Freiburg -
Titisee route long trains are difficult because of the curves and
gradient.

Why should curves restrict train (not vehicle) length? Gradient I'll
give you, because if weight is an issue DDs may win by a small margin.

Neil

From a knowledge of engineering, rather than trains:

Every time a carriage pulls on the following carriage, there is a loss
of force (to the rails) dependent on the angle between the two
carriages. So to pull the last carriage with force F, I need an
engine force

= F/((cos(X))^n),

where n is the number of joints, and x is the angle of curve between
each carriage.

Lets say radius = 150m, carriage length = 25m, each carriage occupies
7.2 degrees. 8 carriages = 7 joints, I lose 10% of the pull on the
last carriage perpendicular to the rails. This also increases the wear
on the rails.

As said, I know little about railways, but this would seem a logical
explanation.

Piccadilly Pilot · August 31st 04, 12:41 AM posted to uk.transport.london,uk.railway

"Alex Terrell" wrote in message
m...
(Neil Williams) wrote in message
...
On 30 Aug 2004 03:11:06 -0700, (Alex Terrell)
wrote:

Which is why you don't see too many of them.

AFAIAA, all non-DMU new stock builds for local services in Germany
have been DDs - or certainly most of them. (The PumA-Modus stock
doesn't count as it's a Networker Classic-style rebuild on old
underframes). They are very common in Northern Germany.

I think on the Freiburg -
Titisee route long trains are difficult because of the curves and
gradient.

Why should curves restrict train (not vehicle) length? Gradient I'll
give you, because if weight is an issue DDs may win by a small margin.

Neil

From a knowledge of engineering, rather than trains:

Every time a carriage pulls on the following carriage, there is a loss
of force (to the rails) dependent on the angle between the two
carriages. So to pull the last carriage with force F, I need an
engine force

= F/((cos(X))^n),

where n is the number of joints, and x is the angle of curve between
each carriage.

Lets say radius = 150m, carriage length = 25m, each carriage occupies
7.2 degrees. 8 carriages = 7 joints, I lose 10% of the pull on the
last carriage perpendicular to the rails. This also increases the wear
on the rails.

As said, I know little about railways, but this would seem a logical
explanation.

The maths are Greek to me, but the argument is correct. The more vehicles
(more accurately the more wheelsets) the more force needed to get round a
bend.

Ian Johnston · August 31st 04, 08:18 AM posted to uk.railway,uk.transport.london

On Mon, 30 Aug 2004 22:39:39 UTC, (Alex Terrell)
wrote:

: From a knowledge of engineering, rather than trains:
:
: Every time a carriage pulls on the following carriage, there is a loss
: of force (to the rails) dependent on the angle between the two
: carriages. So to pull the last carriage with force F, I need an
: engine force
:
: = F/((cos(X))^n),

Resolve along a tangent at the coupling.

Ian

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