Creating an arc with a given thickness using PIL's Imagedraw

Question

I am trying to create a segmented arc using PIL and Imagedraw. The arc function allows me to draw an arc easily, but it is just a line. I need to be able to place an arc of given radius and thickness(ID to OD), but AI cannot find any type of thickness or width setting. Is there a way to do this? If not, is there some other way to do this using PIL?

Snippet:

import Image
import ImageDraw

conv = 0.1
ID = 15
OD = 20
image = Image.new('1',(int(ceil(OD/conv))+2,int(ceil(OD/conv))+1), 1)
draw = ImageDraw.Draw(image)
diam = OD-ID
box=(1, 1, int(ceil(diam/conv)), int(ceil(diam/conv))) #create bounding box
draw.arc(box, 0, 90, 0) #draw circle in black

Answer 1

I created the following arc replacement function based on Mark's suggestion:

https://gist.github.com/skion/9259926

Probably not pixel perfect (nor fast), but seems to come close for what I need it for. If you have a better version please comment in the Gist.

def arc(draw, bbox, start, end, fill, width=1, segments=100):
    """
    Hack that looks similar to PIL's draw.arc(), but can specify a line width.
    """
    # radians
    start *= math.pi / 180
    end *= math.pi / 180

    # angle step
    da = (end - start) / segments

    # shift end points with half a segment angle
    start -= da / 2
    end -= da / 2

    # ellips radii
    rx = (bbox[2] - bbox[0]) / 2
    ry = (bbox[3] - bbox[1]) / 2

    # box centre
    cx = bbox[0] + rx
    cy = bbox[1] + ry

    # segment length
    l = (rx+ry) * da / 2.0

    for i in range(segments):

        # angle centre
        a = start + (i+0.5) * da

        # x,y centre
        x = cx + math.cos(a) * rx
        y = cy + math.sin(a) * ry

        # derivatives
        dx = -math.sin(a) * rx / (rx+ry)
        dy = math.cos(a) * ry / (rx+ry)

        draw.line([(x-dx*l,y-dy*l), (x+dx*l, y+dy*l)], fill=fill, width=width)

Answer 2

PIL can't draw wide arcs, but Aggdraw can, and works well with PIL (same author).