On 09/07/2008, at 5:24 PM, Graham Cox wrote:
You will need to also take into account of the line width (as I'm
sure you do) so I think the combined offset would be best expressed
as:
max(w / 2, ml * w / 2)
where ml is the mitre limit.
That's right, and precisely what I'
On 9 Jul 2008, at 5:11 pm, Chris Suter wrote:
You could ignore the mitre limit and assume it was always mitre
joint and it would give you a worst case bounds. However, it
wouldn't be sensible to do that since the bounds would stretch out
excessively for very acute angles.
Without knowing
Hi Graham,
On 09/07/2008, at 4:37 PM, Graham Cox wrote:
On 9 Jul 2008, at 4:02 pm, Chris Suter wrote:
so you can get a worst case bounds by ignoring the mitre limit.
No you can't. If the stroke width is 'w' then you can outset the
bounds by w/2 to enclose the path for straight edges and a
On 9 Jul 2008, at 4:37 pm, Graham Cox wrote:
Thus the angle 'a' of the corner is equal to 2*sin(w/m)
That of course should be 'arcsin', not 'sin'
Graham
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Indeed, I think this is the same as m = 2w/sin(a). But for worst case
bounds, the angle itself doesn't need to be measured or known, only
the miter limit.
Thanks for your help!
cheers, Graham
On 9 Jul 2008, at 4:34 pm, Peter Zegelin wrote:
Cool I got it right - finally! Its on page 154
On 9 Jul 2008, at 4:02 pm, Chris Suter wrote:
so you can get a worst case bounds by ignoring the mitre limit.
No you can't. If the stroke width is 'w' then you can outset the
bounds by w/2 to enclose the path for straight edges and angles >= 90
degrees. When there is a more acute angle
Cool I got it right - finally! Its on page 154 of the spec (174 in PDF
viewer) section 4.3.
Me in another post "I believe the fully correct answer to calculate
the extension is: (strokeWidth/2)/ sin(angle/2);"
Peter
On 09/07/2008, at 4:02 PM, Chris Suter wrote:
"The miter limit help
Ok - one last time!!
I believe the fully correct answer to calculate the extension is:
(strokeWidth/2)/ sin(angle/2);
Take a stroke width of 20 half of which is 10.
At 90deg the extension is: 10/sin(45) = 14.2
At 45deg the extension is 10/sin(22.5) = 26.1
At the limit of approx. 11deg the
Hi Graham,
On 08/07/2008, at 10:16 PM, Graham Cox wrote:
A close second best would be a rect that was the worst case bounds
for a given stroke width (it wouldn't need to know the actual path,
just its basic bounds), which thus assumed that all angles were
acute enough to trigger the mitre
Actually Graham I think I got it a bit wrong. I was working by
induction and it just happened that the two data points more or less
worked. My picture at 45 deg wasn't actually 45 deg (I did it on
sight). However I believe the correct value uses the sin. eg
sin(11.5deg) = .199, sin(45) = .7
Graham,
Since no-one else has replied and I'm going to have to figure out
this myself, I thought I'd take a stab at it. I've uploaded some
screen shots http://www.fracturedsoftware.com/downloads/Stroke20.tiff
and http://www.fracturedsoftware.com/downloads/Stroke30.tiff of a
simple poly l
Hi,
have you checked CGContextReplacePathWithStrokedPath (CGContextRef
c) ? This together with CGRect CGContextGetPathBoundingBox
(CGContextRef c) might do the trick for you.
Best,
Kai
On 8.7.2008, at 14:16, Graham Cox wrote:
I need to know a rect within which all pixels will be painted f
Graham,
I know this issue has cropped up before (probably on the Quartz list).
Have your searched its archives?
A simple answer to this sort of question involves painting the path
into a bitmap and hunting for the boundary
Matt
On 8 Jul 2008, at 2:16pm, Graham Cox wrote:
I need to know
I need to know a rect within which all pixels will be painted for a
given path and stroke width. I'd prefer not to make this rect any
larger than it really needs to be. To compute this, I have to factor
in all sorts of bits and pieces such as the angles of the line joins,
miter limits and s
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