I have a svg path as below:
'M12,2 C 8.13,2 5,5.13 5,9 c 0,5.25 7,13 7,13s7,-7.75 7,-13C19,5.13 15.87,2 12,2zM7,9c0,-2.76 2.24,-5 5,-5s5,2.24 5,5c0,2.88 -2.88,7.19 -5,9.88C9.92,16.21 7,11.85 7,9z M12,9m-2.5,0a2.5,2.5 0,1 1,5 0a2.5,2.5 0,1 1,-5 0'
Now the tip point at (12,22), I would like to keep shape the same but the tip point at (0,0).
Which tool can do such position shift?
One possible solution would be using this tool to convert the path to to all-relative path commands.
For example convert:
M12,2 C 8.13,2 5,5.13 5,9 c 0,5.25 7,13 7,13s7,-7.75 7,-13C19,5.13 15.87,2 12,2zM7,9c0,-2.76 2.24,-5 5,-5s5,2.24 5,5c0,2.88 -2.88,7.19 -5,9.88C9.92,16.21 7,11.85 7,9z M12,9m-2.5,0a2.5,2.5 0,1 1,5 0a2.5,2.5 0,1 1,-5 0
to:
M12,2c-3.87,0,-7,3.13,-7,7c0,5.25,7,13,7,13s7,-7.75,7,-13c0,-3.87,-3.13,-7,-7,-7zm-5,7c0,-2.76,2.24,-5,5,-5s5,2.24,5,5c0,2.88,-2.88,7.19,-5,9.88c-2.08,-2.67,-5,-7.03,-5,-9.88zm5,0m-2.5,0a2.5,2.5,0,1,1,5,0a2.5,2.5,0,1,1,-5,0
Now in order to move it in the 0,0 just change the first two values after the initial M command from 12,2 to 0,-18.
The first value is 0 and represents the x value of both tue top and the tip. The second value is -18 and represents the y value of the starting point of the path located almost at the top. In order to get this value I did: 2 (actual y coordinate of the point) - 20 (height of the path).
In order to get the height of the path I'm using getBBox()
console.log(g.getBBox())
svg{width:30vh;border:solid}
<svg viewBox="-7 -18 14 20">
<path id="g" d="M0,-18c-3.87,0,-7,3.13,-7,7c0,5.25,7,13,7,13s7,-7.75,7,-13c0,-3.87,-3.13,-7,-7,-7zm-5,7c0,-2.76,2.24,-5,5,-5s5,2.24,5,5c0,2.88,-2.88,7.19,-5,9.88c-2.08,-2.67,-5,-7.03,-5,-9.88zm5,0m-2.5,0a2.5,2.5,0,1,1,5,0a2.5,2.5,0,1,1,-5,0"/>
</svg>
Having dealt with converting the Bezier Patches into triangles, I need to do a Binary Space Partition in order to draw the projected triangles using the Painter's Algorithm.
I've implemented the algorithm from Wikipedia with much help with the math.
But it's making a Charlie Brown tree! That is most of the nodes have one branch completely empty. The whole strategy is all wrong. Since the teapot is essentially spherical, the entire shape is only on one "side" of any particular component triangle.
So I'm thinking I need partitioning planes arranged more like an apple-corer: all passing through the line of the y-axis. But I'm kind of going off book, you know? What's the best way to partition the teapot?
Here's my bsp-tree generator. It uses other functions posted in the linked question.
Edit: Extra juggling to avoid dictstackoverflow. Complete program available here (requires mat.ps and teapot). The numerical output shows the depth of the tree node under construction.
% helper functions to insert and remove triangles in lists
/unshift { % [ e1 .. eN ] e0 . [ e0 e1 .. eN ]
exch aload length 1 add array astore
} def
/shift { % [ e0 e1 .. eN ] . [ e1 .. eN ] e0
aload length 1 sub array astore exch
} def
/makebsp { % [ triangles ] . bsptree
count =
%5 dict %This is the tree node data structure
<</P[]/PM[]/plane[]/front[]/behind[]/F<<>>/B<<>>>>
begin
dup length 1 le{ % If 0 or 1 triangles
dup length 0 eq { % If 0 triangles
pop % discard
}{ % If 1 triangle
aload pop /P exch def % put triangle in tree node
}ifelse
}{ % length>1
shift /P exch def % P: Partitioning Polygon (triangle)
P transpose aload pop
[1 1 1] 4 array astore % make column vectors of homogeneous coords
/PM exch def
[ % Compute equation of the plane defined by P
PM 0 3 getinterval det
[ PM 0 get PM 2 get PM 3 get ] det
[ PM 0 get PM 1 get PM 3 get ] det
PM 1 3 getinterval det 3 mul
] /plane exch def
% iterate through remaining triangles, testing against plane, adding to lists
/front [] def
/behind [] def
{ %forall [P4 P5 P6] = [[x4 y4 z4][x5 y5 z5][x6 y6 z6]]
/T exch def
T transpose % [[x4 x5 x6][y4 y5 y6][z4 z5 z6]]
{aload pop add add} forall % (x4+x5+x6) (y4+y5+y6) (z4+z5+z6)
plane 2 get mul 3 1 roll % z|C| (x) (y)
plane 1 get mul 3 1 roll % y|B| z|C| (x)
plane 0 get mul % y|B| z|C| x|A|
plane 3 get add add add % Ax+By+Cz+D
0 le { /front front
}{ /behind behind
} ifelse
T unshift def
} forall
%front == ()= behind == flush (%lineedit)(r)file pop
% recursively build F and B nodes from front and behind lists
%/F front makebsp def
front currentdict end exch
makebsp
exch begin /F exch def
%/B behind makebsp def
behind currentdict end exch
makebsp
exch begin /B exch def
/front [] def
/behind [] def
} ifelse
currentdict end
} def
Output:
BSP was invented for geometries like levels in Quake-like games and it may be hard to use for some specific geometry sets. BSP uses one of the existing triangles to split your level, so just imagine how it would behave when you want to use it on a sphere...
For the teapot you could get better results using OCTree, which doesn't need to split your geometry along existing triangles. But I'm not sure how well does it work with the Painter Algorithm.
If you really need to use BSP tree here, then you should pick your triangles carefully. I don't understand all of your code but I don't see this part here. Just iterate over all triangles in your current tree branch and for each of them compute the number of triangles in front of it and at the back. The one that has similar number of front-triangles and back-triangles is usually the best one to be used as a split plane.
I didn't quite do an octree, but I modified the bsp-tree builder to use an explicit list of planes which I filled with axis-aligned planes slicing the space -4 < x,y,z < 4.
/planelist [
0 .2 4 { /x exch def
[ 1 0 0 x ]
[ 1 0 0 x neg ]
[ 0 1 0 x ]
[ 0 1 0 x neg ]
[ 0 0 1 x ]
[ 0 0 1 x neg ]
} for
] def
Postscript program available here (requires mat.ps).
The lighter green artifact is the result of a "preview" shown during construction of the bsp. Once built, subsequent pages (images) are drawn quickly and with no artifact as the camera revolves around the teapot.
The join of the body with the spout and handles (not shown from this angle) still needs work.
With the bsp better behaved, backface culling isn't strictly necessary. But it makes the preview nicer.
Another way to improve the BSP for this image is to use a hierarchical decomposition. The teapot isn't just a bunch of bezier surfaces, it has some surfaces that describe the body, some others that describe the handle, the spout, the lid (, the bottom? ).
So the first few levels of the tree ought to be the top-level pieces. Is the handle in front of or behind the body? Is the spout in front of the body? Answers to these questions would be a useful guide for the painter's algorithm.
I was wondering how can I rotate a graphic, say a rectangular by certain angle in post script.
Or at least is there any way to draw a very bold ! like, with an angle !?
I have list of sentence around a circle, so each or in 1 direction, and now, I would like to put each in a rectangular and make hyperlink for them.
The power of Postscript is its ruthless pursuit of the ideal of "delayed binding". The implementation of rotations is no exception. It works by making use of a more general tool, the Affine Transformation Matrix.
You can rotate both text and graphics (because text IS graphics) because all user specified coordinates are first multiplied through this matrix to produce device coordinates.
To perform all the necessary tricks (scaling, rotation, shear, translation), we first have to extend the 2d points to 3d points on the plane z=1 (don't ask me why; read Bill Casselman's Mathematical Illustrations or the Adobe Blue Book for more).
[ x [ a b 0
y * c d 0 = [ x' y' 1 ] = [ ax+cy+e bx+dy+f 1 ]
1 ] e f 1 ]
Since the 3rd column of the matrix is always [ 0 0 1 ] it is omitted from the external representation, and the matrix is described in postscript as:
[ a b c d e f ]
So when you use a coordinate pair for, say, a moveto operator, moveto first transforms it to device coordinates, x' = ax+by+e, y' = cx+dy+f, before adding a <</move [x' y']>> element to the current path.
Change the matrix: change the "meaning" of user coordinates.
The identity matrix is this:
[ 1 0 0 1 0 0 ] % x' = x, y' = y
To scale, replace the 1s with x and y scaling factors:
[ Sx 0 0 Sy 0 0 ] % x' = Sx*x, y' = Sy*y
To translate, replace e and f with the x and y translation offsets:
[ 1 0 0 1 Tx Ty ] % x' = x+Tx, y' = y+Ty
To rotate, replace a,b,c,d with sin and cos scaling and shearing factors:
[ cosW sinW -sinW cosW 0 0 ] % x' = x*cosW-y*sinW, y' = x*sinW+y*cosW, where W is angle(degrees) from x-axis
You "install" this matrix with concat which takes the Current Tranformation Matrix (CTM), multiplies it by your new matrix, and uses the product as the new CTM. So translate, rotate, and scale are just "convenience functions" which could be implemented like this:
/translate { [ 1 0 0 1 7 -2 roll ] concat } def
/scale { [ 3 1 roll 0 0 3 -1 roll 0 0 ] concat } def
/rotate { [ exch dup cos exch sin dup neg 2 index 0 0 ] concat } def
Since the CTM is part of the graphics state, you can use the graphics state stack to manipulate your transformations in a hierarchical manner:
/box { % x y w h %create a path in the shape of a box w*h with lower left corner at x,y
4 2 roll moveto
exch dup 3 1 roll
0 rlineto
0 exch rlineto
neg 0 rlineto
closepath
} def
/Courier 10 selectfont
100 100 100 100 box stroke % draw an oriented box
120 120 moveto (inside) show
gsave
150 150 translate % make the center of the box the new 0,0 point
45 rotate % rotate CCW 45 degrees
0 0 100 100 box stroke % a rotated, shifted box
20 20 moveto (inside) show
grestore
100 200 100 100 box stroke % another box, just north of the first, in the original coordinte system
120 220 moveto (inside) show
This produces the following image:
(source: googlecode.com)
I haven't used PostScript for a long time, but as I remember you could just use "rotate".
% do some steps
% ...
% ...
20 20 moveto % go to new position
30 /Times-Roman SetFont % select active font
45 rotate % set direction to diagonal
(Something)show % print text "Something"
showpage % show it all
cheers Kris
Postscript renders graphics in a given context - and it is this context that can be rotated (or scaled/translated) before drawing. Therefore any element on the image can be transformed as one wishes, all you have to do is to perform the necessary context transforms beforehand.
However, unfortunately, while I can give you an idea of it in this writing, i is a fundamental concept of Postscript, and you won't be able to do any real work in it without understanding that first. I suggest reading a brief tutorial such as the one in http://paulbourke.net/dataformats/postscript/ .
So, the "rotate" name is a function that does rotate the graphcis context - you use rotate, before drawing anything you want (rendering text also being "drawing" in this case).
%!PS
(Helvetica) findfont 12 scalefont setfont %select a font to use
300 300 translate % sets the orign at 300,300 points from the bottom left of page
/start 5 def % creates variable for keeping track of horizontal position of text
36 % pushes number of repeats on the stack
{
start 5 moveto % places cursor on the starting position
(postscript) show % renders the string in the starting position, within the current context
/start start 3 add def % increases the value on the variable
10 rotate % rotates the context 10 degrees clockwise (around the 300,300 new origin)
} repeat
showpage % renders whole page
You can obtain the width of a string in the current font with stringwidth and although this actually pushes offset coordinates on the stack, the y-value always seems to be useless. Is there a way to determine the exact height of a string, that may or may not include descenders?
stringwidth, as it says, doesn't return string's height. (In all cases I looked at, the second integer on the stack after executing stringwidth was 0 -- for strings that run in horizontal direction.) stringwidth gives the relative coordinates of the currentpoint after executing a (string) show.
The PLRM has this to say about stringwidth:
Note that the width returned by stringwidth is defined as movement of the current
point. It has nothing to do with the dimensions of the glyph outlines.
So what would work to take into account the string's height? The magic words to read up about in PRLM are charpath and pathbbox. Try this:
%!
/Helvetica findfont 60 scalefont setfont
200 700 4 0 360 arc fill
200 700 moveto (test test) dup
true charpath pathbbox
3 -1 roll sub 2 div neg 3 1 roll sub 2 div exch
1 0 0 setrgbcolor
200 700 moveto rmoveto show showpage
It calculates the string's (printed in red) height and uses that info to try and center a small filled circle (printed in black) into the center of its bounding box:
I have already answered this in How to determine string height in PostScript?, but it is useful here also.
Just adding to pipitas answer:
/textheight {
gsave % save graphic context
{
100 100 moveto % move to some point
(HÍpg) true charpath pathbbox % gets text path bounding box (LLx LLy URx URy)
exch pop 3 -1 roll pop % keeps LLy and URy
exch sub % URy - LLy
}
stopped % did the last block fail?
{
pop pop % get rid of "stopped" junk
currentfont /FontMatrix get 3 get % gets alternative text height
}
if
grestore % restore graphic context
} bind def
/jumpTextLine {
textheight 1.25 mul % gets textheight and adds 1/4
0 exch neg rmoveto % move down only in Y axis
} bind def
The method expects that some font is already set. It works over the selected font (setfont) and its size (scalefont).
I use (HÍpg) to get the biggest bounding box possible, using accentuated uppercase characters and "below line" characters. The result is good enough.
The alternative approach steals from dreamlax's answer -- some fonts do not support charpath operator.
Saving and restoring the graphic context keeps the current point in place, so it has no impact over the "flow" of your document.
Hope I've helped.
This seems to work most of the time:
/fontheight { currentfont /FontMatrix get 3 get } bind def
/lineheight { fontheight 1.2 mul } bind def
It won't work for all /FontTypes.