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Given two points and angles, how can I calculate the connection between the two points in the form of a straight line and a circular arc?


Context of the question: I am trying to make a program that converts a model railway layout created in XtrkCAD into a SCARM layout (scarm.info). Both design applications have a different approach, and a different way of saving the layout to a file. Luckily both use a human readable text file format.

XtrkCAD knows the concept of the easement or transition curve, but SCARM doesn't. When converting a layout I now get gaps in the track where XtrkCAD had an easement. When I close that gap using SCARM, it creates a piece of track consisting of a straight part and a circular curve.

The following image shows an example. A straight track (length 50) has to be connected to a curved track (radius 276, angular length 56.5 degrees). The red track created by SCARM is a straight track of length 56, followed by a curved track with radius 360 and an angular length of 16.8 degrees.

enter image description here

What I would like to know is how I can calculate the straight track and the curved track, based on the end points of the two pieces of track that should be connected and their angles.

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  • $\begingroup$ 1.Do you have to connect always a straight track to a curved track with another straight track and curved track? 2. Is the straight track to be joined always vertical or horizontal, or can it assume any direction? $\endgroup$ Commented Dec 12, 2015 at 16:18
  • $\begingroup$ Ad 1. It is possible that an easement is connecting two curved tracks. Ad 2. All tracks can assume any direction. $\endgroup$
    – Ge We
    Commented Dec 12, 2015 at 16:53

1 Answer 1

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Suppose you must connect two straight segments $AB$ and $CD$, or two arcs, or a segment and an arc (colored blue in the picture), ending at points $B$ and $C$.

Extend the segments (or draw the lines tangent to the arcs at $B$ and $C$) so that they meet at $F$. If $BF>CF$ (as in the picture) construct point $G$ on $BF$ such that $FG=FC$ and draw the perpendicular lines to $GF$ and $CF$ at points $G$ and $C$.

Let $K$ be the point where these perpendicular lines meet and draw arc $GC$ centered at $K$: this arc, together with segment $BG$, provides the needed connection (red in the picture).

enter image description here

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  • $\begingroup$ Thanks, that is exactly what I need! $\endgroup$
    – Ge We
    Commented Dec 15, 2015 at 18:34

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