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When designing a clock, I moved the rotation point of the text into the center of the clock, then copy/pasted the text and rotated by 30° for each number, like so:

enter image description here

Note the rotation point marked with red.

Now, the bottom numbers (like 6) are printed upside down. I want to correct that by rotating the text 30° back. Because the numbers need to stay in their places, I need the rotation center be back at the center of the text, not the center of the clock.

How do I reliably (not estimated by moving the mouse) set the rotation center back to the center of the text?

I have already tried snapping to the center of an object, but it does not seem to apply for the rotation point.

In the end I want something like this, just more accurate:

expected result

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    As you seem to speak German, here's a link to a guide I made for this purpose: vektorrascheln.de/posts/2016/Jun/… (in short: use Polar Arrangement dialog instead of manually rotating) – Moini Jan 4 at 2:49
  • @Moini: oh, I didn't know that. Nice feature – Thomas Weller Jan 4 at 13:00
4

You can do this using the XML editor Shift+Ctrl+X

Select the object, and in the XML editor, set the inkscape:transform-center for both x and y to 0.

Example

enter image description here

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I figured out that grouping provides at least an acceptable workaround:

  1. Select only one of the text objects
  2. Press Ctrl+G to group the object with itself, thus giving a new rotation point for the group at the center of the single object
  3. Rotate 30° back
  4. Press Ctrl+U to ungroup and thus restore the original rotation point
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As I understand it, your initial rotation is used to place the numbers at the right location on the clock. Instead of rotating the numbers in the first place, a better approach might have been to place 12 small circles as helping objects at the right positions by rotation, and then put the numbers on these helper objects with adjusting them vertically and horizontally centered. In the end, make the helping circles transparent or remove them. With a unique color selecting all of would be trivial.

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