From: Rik Cabanier <cabanier@gmail.com>

Date: Mon, 23 Jul 2012 10:22:58 -0700

Message-ID: <CAGN7qDC7Wa=C6rsoangoEgdbtkXsxQ_V8wT+uLCyP4PaPzxFow@mail.gmail.com>

To: Alan Stearns <stearns@adobe.com>

Cc: "www-svg@w3.org" <www-svg@w3.org>

Date: Mon, 23 Jul 2012 10:22:58 -0700

Message-ID: <CAGN7qDC7Wa=C6rsoangoEgdbtkXsxQ_V8wT+uLCyP4PaPzxFow@mail.gmail.com>

To: Alan Stearns <stearns@adobe.com>

Cc: "www-svg@w3.org" <www-svg@w3.org>

I think the formula comes from the average of a ellipse: http://math.wikia.com/wiki/Ellipsoidal_quadratic_mean_radius Maybe you create a hypothetical ellipse with the SVG's width and height and use the length at 45 degrees. Rik On Fri, Jul 20, 2012 at 11:51 AM, Alan Stearns <stearns@adobe.com> wrote: > Hey all, > > There's a typo in section 7.10 in the formula for percentage lengths that > are not heights or widths. It's just an extra parenthesis before the > slash. So it should read: > > sqrt((actual-width)**2 + (actual-height)**2)/sqrt(2) > > > But I'm also wondering what motivates this particular formula. I'm looking > at how a percentage value for the radius of a circle works. In a square > 10x10 viewport it's a percentage of 10, which makes sense. You're getting > the length of the diagonal divided by the square root of 2, which gets you > back to height or width. > > In a 20x10 rectangle, you get a percentage of 15.81. I don't understand > how that's useful in this case. You have the length of the diagonal on top > again, but how is the square root of 2 relevant for non-square viewports? > Is there a non-radius use case for percentage lengths in rectangles where > this result is useful, or was this formula chosen mainly to make a square > viewport work as expected? > > Thanks, > > Alan > > >Received on Monday, 23 July 2012 17:23:27 GMT

*
This archive was generated by hypermail 2.3.1
: Friday, 8 March 2013 15:54:51 GMT
*