Home · Maps · About
Home > SubChat
 

[ Read Responses | Post a New Response | Return to the Index ]
[ First in Thread | Next in Thread ]

 

view flat

Re: Calculation

Posted by Jeff H. on Wed Jan 5 03:46:44 2005, in response to Re: Calculation, posted by Stepney on Tue Jan 4 22:06:26 2005.

edf40wrjww2msgDetail:detailStr
The issue of train dynamics on a curve is a very complicated one.

I assure you, on a 90 foot radius curve, the radius of the point of
contact of the outer wheel does not shift by 1.82 inches!!! This
would mean the wheel climbs to nearly to end of the flange.

There are three ways that wheelsets can steer around a curve:

1) Through the action of the flange and fillet of the inner wheel.

2) Through the action of the guard rail against the outer wheel.

3) Through "coning"

Let's start with (3).

The typical AAR wheel profile has a tread which is tapered 1-in-20.
On very mild curves (one guideline is less than 5 degree, i.e.
a radius of approx 1200' or more), the wheelset shifts its point
of contact on both rails. The wheels move towards the outside of
the curve, so the inner wheel makes contact further out on the tread,
where the diameter is slightly smaller, and the outer wheel sees
a slightly larger diameter.

This coning is sufficient to steer around mild curves with no
assistance from the flanges.

(1) Clearly the flange, either the rounded fillet between the
tread and flange, or the straight diagonal face of the flange itself,
can steer the wheelset. However, this sort of action is very
destructive to the flange profile and is to be avoided. The tighter
the curve, the worse it is on the flange. Flange steering happens
at a switch, between the point and the heel.

(2) For tight curves, a guard rail is used on the inner rail. This
steers the wheelset by a sliding contact against the back face
of the flange. Since this surface is not used for traction, the
guard rail can be greased heavily, cutting down on flange wear.

Unless the guard rail is installed wrong, the flange of the inner
wheel never makes contact with the rail, and thus is not worn down.

Now, how do we reconcile this with the apparent need to match the
linear speed of the two wheels? Undoubtedly you've derived the
formula r2/r1=(R+g/2) / (R-g/2), where r2 is the radius of the
outer wheel, r1 the inner wheel, R the track curve radius, and g
the track gauge.

However, in doing so, you've made the assumption that the axle of
the wheelset is normal to the rail. Clearly, this can not be,
because the two axles of the truck are rigidly held parallel.
Can two parallel lines be normal to the same circle? Of course
not. The leading axle is pointing "out" of the curve, and the
trailing axle is "digging in".

So what is actually happening in a sharp curve is that the force
exerted by the point of contact of the wheel with the rail is
not quite tangent to the curve. There is a radial component as
well. That latter force is resisted by the guard rail reaction,
which causes the wheelset to pivot (steer) around the contact point.

The interface between the wheel tread and the rail is not strictly
static friction. In fact, exactly what is happening here is subject
to some debate among rail engineers. It is a slip-grip-slide
contact and some believe that it oscillates among these states
rapidly

Responses

Post a New Response

Your Handle:

Your Password:

E-Mail Address:

Subject:

Message:



Before posting.. think twice!


[ Return to the Message Index ]