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## RE: Converting Outlines to postscript Vector

**From**: |
Satpal Chander |

**Subject**: |
RE: Converting Outlines to postscript Vector |

**Date**: |
Tue, 29 Feb 2000 09:07:16 -0000 |

>* >Hi,*
>* > Has anyone tried doing the above from the outline structure*
>* as returned*
>* >buy*
>* >TT_Load_Glyph().*
>* >I have found a page*
>* >http://www.icce.rug.nl/erikjan/bluefuzz/beziers/beziers/beziers.html*
>* >It does give a good definition there of howto do it but I'm new to fonts/*
>* >beziers /postscript and the like.*
>* >Can anyone tell me if I can/should get at the curve data*
>* directly of through*
>* >the*
>* >api, I'm using 1.3.1.*
>* >Also if the cruve data split into segements or is it just one long curve.*
>
>* I'm not 100% sure, but it seems FT1.x doesn't have the decompose outline*
>* API like FT2 has, so I guess you'll have to walk trhough the*
>* outline object yourself.*
If this is the case then how stable is the FT2 stuff considering that I'm
only
interested in the outline data for each glyph.
>* Even so, it should be fairly straightforward, though:*
>
>* - There are on-curve and off-curve points in TT (and therefore in FT1.x)*
>* - Between any two consecutive off-curve points there is an*
>* implied on-curve*
>* point (exactly in the middle between them).*
>* - A quadratic segment (onCurve, offCurve, onCurve) can be converted to*
>* cubic beziers (as used by PS) like so:*
>* let (from, control, to) be the three points making up de*
>* quadratic curve*
>* you need to convert these three points to four points for*
>* cubics: (from,*
>* control1, control2, to)*
>* control1 = from + (control - from) * 2.0/3.0*
>* control2 = to + (control - to) * 2.0/3.0*
>* from and to remain the same.*
Does the algorithm above meant that I don't have to actually
do anything with the curves using the curve maths, just manipulating the
control points and outputting them to postscript. ;-))
Cheers
Satpal