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I'm trying to make a mesh like this, except where the loop cuts are exactly the same distance x from the edges (this is just a eyeballed approximation of what I'm trying to get):

want

I've tried looking around for an answer, but the only thing I come across people using beveling or insets, but I need overlapping cuts making the little squares in the corners. I don't want diagonal lines from the corners pointing inwards like this:

don't want

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  • $\begingroup$ I don't know if it's possible except moving the edges to the border then entering the unit value, if I had to do this maybe I would inset then cut with the knife $\endgroup$
    – moonboots
    Oct 21, 2019 at 7:46

4 Answers 4

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The way I would do that is:

  1. Decide the height of the object, but leave the square shape
  2. Apply two loop cuts
  3. Bevel them
  4. Grab one side
  5. Move it on an axis to get the shape you want

enter image description here

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Select all and Bevel Ctrl+B with Segments 2, Profile 1.

enter image description here

Select unwanted edge loops ... Delete X > Dissolve Edges.

enter image description here

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  • $\begingroup$ @vklidu Much improved, thanks! $\endgroup$
    – Robin Betts
    Jul 27, 2020 at 20:02
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In editing mode, do Ctrl-R for loop-cut then page-up for more loop-cuts equidistant from each other. Make the loop-cuts without changing the location of them (just press enter), then select the loop-cuts and press 's' and whichever axis to change location.

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  • $\begingroup$ it will works for 2 of the edge loops, but he wants all the 4 margins to be the same length $\endgroup$
    – moonboots
    Oct 21, 2019 at 7:42
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Using two separate single loop cuts, this can be done on a single axis with some equations using edge lengths and the loop factor. The steps can be applied again independently on the other axis for the OP problem.

If you have a specific x in mind, the factor required to generate it can be obtained with this: $$LoopFactor1 = 1 - \frac{2 * x }{InitialEdgeLength}$$

The second loop factor can be calculated from the first with this: $$LoopFactor2 = 3 - \frac{4}{LoopFactor1 + 1}$$

Use LoopFactor1 and LoopFactor2 as the magnitudes for single loop cuts in succession on a single axis.

Applying this as an example, with $$x = \frac{1}{16}$$ $$InitialEdgeLength = 2$$

$$LoopFactor1 = 1 - \frac{2}{2*16} = \frac{15}{16} = .9375$$ $$LoopFactor2 = 3 - \frac{4}{\frac{15}{16}+1} = \frac{29}{31} = .935484$$

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