I've just made a regular hexagon in MS Powerpoint. The 1/sqr(12) comes from the fact that the 0.25 has to be adjusted for the 2/sqr(3) ratio, so (1/4) * (2/sqr(3)) = 1/sqr(12). This corrects the hexagon to a regular hexagon which can be rotated to any multiple of 60 degrees angle and fit together with hexagons with perfect match. Select your hexagon, press Alt+F11 (opens vb-editor), press Ctrl+G (opens Immediate window). Regardless if you draw the hexagon with or without holding shift, your hexagon needs to be adjusted at its corner setting. So you have to adjust the Adjustments(1) by this offset.Īs suggested by jmac I also recommend using vb-editor, not changing widths or heights since it won't correct the false position of the corner point, which you will se when you rotate the hexagon and join with other similar hexagons. Microsoft has the point at h/w*p, meaning that the proportion will only be valid if the w/h-ratio is 1, which it isn't, it's 2/sqr(3). So one thinks the Adjustments(1), which is the point where Microsoft determines where to put the corner point, for the hexagon should be 0.25, but no. The first corner, the top left one, is at 25% of the width.
28706, and lo and behold, that was the magic number.Ī regular hexagon has a width-to-height ratio of 2/sqr(3).
28002, and my many attempts to get it right with none of them working, I tried taking half of. Given the starting value of the close-but-not-quite hexagon of. Then I tried setting it as far right (turning the hexagon into a diamond) and got. So I set it to the furthest left it could go (turning the hexagon into a square) and the value was 0. 28002, so I started fiddling around and trying to do math assuming that this value was somehow based on angles. The value for the close-but-not-quite hexagon was. I used some VBA to check what the position of the handle was (with the hexagon selected): ?(1).Adjustments(1) To get a regular hexagon in PowerPoint, create a hexagon using shift, then run the following VBA command with that hexagon selected: (1).Adjustments(1) = 0.28706Īfter I created the approximate shape using the overlay on the close-but-not-quite equilateral triangle, I decided to get programmatic.
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