Oval() point placement

ryan

Hi there,

I'm curious as to why the output from oval() has on-curves on angled extrema rather than traditional 0/90 extreme points? Sometimes I build patterns, export to PDF, and then work with them in Illustrator—but this point structure makes it more difficult. Screenshot in Illustrator below.

Thanks,
Ryan

gferreira

hello @ryan,

I dont’t know why the extremes are angled…

here’s a way to get the desired result by modifying oval:

def twist(func, angle):
    def wrapper(x, y, w, h):
        with savedState():
            translate(x + w/2, y + h/2)
            rotate(angle)
            func(-w/2, -h/2, w, h)
    return wrapper

oval = twist(oval, 45)

when working with BezierPath, it’s possible to rotate the shape:

B = BezierPath()
B.oval(x, y, w, h)
B.rotate(45, (x + w/2, y + h/2))

hope this helps!

frederik

I don't know why but my wild guess is it has todo with drawing ovals at more extreme w, h ratios, when the points are on the extremes you will need a pushing point in the middle to nice curve back.

@gferreira you method only works for circles...

the drawBot RoboFont extension has points on the extremes, cause the context is type... see https://github.com/typemytype/drawBotRoboFontExtension/blob/master/DrawBot.roboFontExt/lib/glyphContext.py#L23

see the big difference when the oval is getting thinner...

def straightOval(x, y, w, h):
        c = 0.55
        hx = w * c * .5
        hy = h * c * .5
        path = BezierPath()
        path.moveTo((x + w * .5, y))
        path.curveTo((x + w * .5 + hx, y), (x + w, y + h * .5 - hy), (x + w, y + h * .5))
        path.curveTo((x + w, y + h * .5 + hy), (x + w * .5 + hx, y + h), (x + w * .5, y + h))
        path.curveTo((x + w * .5 - hx, y + h), (x, y + h * .5 + hy), (x, y + h * .5))
        path.curveTo((x, y + h * .5 - hy), (x + w * .5 - hx, y), (x + w * .5, y))
        path.closePath()
        drawPath(path)
        

x, y, w, h = 10, 10, 8.0, 146.0


oval(x, y, w, h)
straightOval(x + w + 10, y, w, h)