explain exaggeration

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Motiejus Jakštys 2021-05-14 20:24:24 +03:00
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@ -1178,8 +1178,6 @@ Combination operator was not implemented in this version.
\subsection{Exaggeration Operator}
\label{sec:exaggeration-operator}
% TODO: change for azimuth-based algorithm.
Exaggeration operator finds bends of which \textsc{adjusted size} is smaller
than the \textsc{diameter of the half-circle}. Once a target bend is found, it
will be exaggerated it in increments until either becomes true:
@ -1208,9 +1206,23 @@ implementation. A single exaggeration increment is done as follows:
\textsc{midbend} to the other baseline vertex.
\item Mark each bend's vertex with a number between $[1,s]$. The number is
derived with elements linearly interpolated between the start vertex
and \textsc{midbend}. The other half of the bend, from \textsc{midbend}
to the final vertex, is linearly interpolated between $[s,1]$.
derived with elements linearly between the start vertex and
\textsc{midbend}, with values somewhat proportional to the azimuth
between these lines:
\begin{itemize}
\item \textsc{midbend} and the point.
\item \textsc{midpoint} and the point.
\end{itemize}
The other half of the bend, from \textsc{midbend} to the final vertex,
is linearly interpolated between $[s,1]$, using the same rules as for
the first half.
First version of the algorithm used simple linear interpolation based
on the point's position in the line. The current version applies a few
coefficients, which were derived empirically, by observing the
resulting bend.
\item Each point (except the beginning and end vertices of the bend) will
be placed farther away from the baseline. The length of misplacement is