On 9/6/2012 3:00 AM, scott wrote:
>>> Won't speeding up and slowing down make the train swing out more or
>>> less? (Like going round a corner in a car, if you speed up it rolls
>>> more).
>>>
>>> In a rotating reference frame you'd be essentially turning even when on
>>> a straight piece of track, so I would not say that changes in forward
>>> motion won't have any effect.
>>>
>> Yes, to an extent, this is true. But, since the only angle its changing
>> at is side to side, forward motion isn't going to have an affect
>> *except* when making turns.
>
> But as I said, in a rotating reference frame the train will be "making
> turns" all the time, even when on a straight track. So any forward/back
> acceleration will, in general, affect the side to side swinging.
>
> You can easily see this by imagining a straight piece of track rotating
> and plotting out the course the train will take as it goes along the
> track - it will be a curve, so some sideways forces must be involved.
>
Ok, true. Maybe I am phrasing wrong then. What I meant is that, since 
the reference is going to be, in principle, constant, what ever forces 
it is applying are not going to have the same effect as say, 
acceleration or deceleration of the train **on** the track, if the 
direction is directly parallel to the existing force being applied by 
the reference. The result will, thus, only matter when the resulting 
vectors actually change, with respect to the angle it can swing in. I.e.

Possible angle = <0,1,0>
Forces = <4,2,1>
Result = <0,2,0>

This is not the case, obviously, when the track is not directly parallel 
to the force, or if the possible rotations where <1,1,0> (i.e., both X 
and Y, not just Y).

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