And putting this equation into more standard quadratic form I get
d_norm(v) = 0.072(v^2) + 0.229v + 0.449
which yields 0.75 as expected for v = 1, so it looks as though I have the right
equation here.
Graham
________________________________
From: GRAHAM ASHER <graham.as...@btinternet.com>
To: David Bevan <david.be...@pb.com>
Sent: Tuesday, 31 August, 2010 17:07:52
Subject: Re: [ft-devel] a satisfactory fix for the cubic spline bug
Can you give the workings? The equation I think you mean is this one:
0.07200(v + 3.180556)v + 0.449000,
so when I substitute 0.1 for v I get
0.07200(0.1 + 3.180556)0.1 + 0.449000
which gives 0.47262, and multiplying that by 50, the maximum deviation of a
control point, gives 23.631, not your y value, so obviously I'm missing
something;
Thanks,
Graham
________________________________
From: David Bevan <david.be...@pb.com>
To: GRAHAM ASHER <graham.as...@btinternet.com>
Sent: Tuesday, 31 August, 2010 16:20:08
Subject: RE: [ft-devel] a satisfactory fix for the cubic spline bug
________________________________
From:GRAHAM ASHER [mailto:graham.as...@btinternet.com]
Sent: 31 August 2010 15:56
To: David Bevan
Subject: Re: [ft-devel] a satisfactory fix for the cubic spline bug
Midpoint as defined by bisection is the point you get when you split a cubic
spline into two by the bisection method (as used by FreeType in
gray_split_cubic). It is defined as the midpoint of the line between the two
midpoints of the lines between the three midpoints of the lines between the
four
control points, taken in order. It is of course a point on the curve.
So that’s the point t = 0.5 in the parametric equation — found using de
Casteljau’s algorithm.
Your example gives a midpoint of (31.25, 20.625). Applying the midpoint method
again (hastily) to the left-hand curve seems to yield a y coordinate of
21.84375, so if my arithmetic is correct your example is a good one.
Using the equation in Hain’s paper (p.4; with v=0.1), the maximum deviation
tmax
in my example is at 0.3503928884, which is the point (16.26529900, 23.37564709).
David %^>
Thanks,
Graham
________________________________
From:David Bevan <david.be...@pb.com>
To: GRAHAM ASHER <graham.as...@btinternet.com>
Sent: Tuesday, 31 August, 2010 15:20:59
Subject: RE: [ft-devel] a satisfactory fix for the cubic spline bug
________________________________
From:GRAHAM ASHER [mailto:graham.as...@btinternet.com]
Sent: 31 August 2010 15:01
To: David Bevan; freetype-devel
Subject: Re: [ft-devel] a satisfactory fix for the cubic spline bug
David,
(vi) Out of interest, please give an example, with coordinates, of a cubic
spline with both control points on the same side of the straight line from
start
to end, for which the midpoint of the curve, as defined by bisection, is not
the
point of maximum deviation.
I expect that something like the following would be an example:
P0: 0,0
P1: 0,50
P2: 50,5
P3, 100,0
But to be sure I’d need to know what you mean by “midpoint as defined by
bisection”? Is it t = 0.5 in the parametric equation, the point perpendicular
to
the midpoint of the chord, or something else?
David %^>
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