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I've read several similar questions but they keep leading back to eventually becoming raster images, in which case the aliasing makes sense. However I haven't quite figured out why an image made entirely in Illustrator with vector objects would display with what appears to me like an aliased line on the bezier curve like my little example shows below.

enter image description here

So as in the example, if that's a vector ellipse with a stroke, can someone please enlighten me as to why that aliasing appears so dominantly on the Top/Bottom/Left/Right edges (EXAMPLE: RED ARROW) of that vector ellipse to make it appear all jagged on the four sides....EXCEPT while around the rest of the ellipse it appears as a clean & smooth line (EXAMPLE: GREEN ARROW)?

Also, and more importantly, how do I fix it so that the entire ellipses stroke appears as the same uniformed clean & smooth line all the way around the ellipse on both the artboard, and any exported RGB vector graphic types?

Thanks!

marked as duplicate by Scott adobe-illustrator Jun 9 '15 at 21:21

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Your monitor uses pixels to display anything. There's no such thing as a digital display where pixels are not present, regardless of how an image is constructed.

Some construction method do not output pixel data, but all displays use pixels.

  • If that were the cause, with resolution aside, why would the aliasing be so much more dominant at the four points of the ellipse and not elsewhere? – Chris W. Jun 9 '15 at 21:22
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    Because those are the sharpest angles of the curve and the most difficult part to visually represent with a grid of square pixels. – Scott Jun 9 '15 at 21:23
  • Hmm, I didn't consider that I suppose but it makes sense. I'm just trying to understand the math of it and trying to understand if there's maybe a technique to kind of blend the transition better to the human eye so it's not as noticeable. Cheers for giving me something to think about at least! Will accept the answer once it's past the 5minute thingy. – Chris W. Jun 9 '15 at 21:25
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    @ChrisW.: Since the ("ideal") line's slope is very small, there should not be individual pixels in black at all, but gray instead. The smaller the area covered by the "ideal" line, the lighter gray should be. But using that method, you'd get a fat gray line (2 pixels thick) for the horizontals, that transitions into a (mostly) black line towards the larger slope. So this is a 'less bad' solution for 1-pixel thin lines. (Thicker lines have the same problem but then it's less noticeable, because it will have valid entirely-black pixels.) – usr2564301 Jun 10 '15 at 8:01

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