Does rotation by exact angle become low-precise in blender?

Question

I faced a strange behavior of Blender while trying to make something like this:

in one of the projections. It's clear that angle B (see picture below) should be 60 degrees for trivial trigonometrical reasons: just assume that the side of the square has the length 1 and calculate sin(A). Also note that D-D1 is divided into two equal parts by the inclined segment.

I try to reproduce that in Blender in orthographic projection of 3d view. Steps:

1. Create a new file
2. Delete cube
3. Press numpad 1 to view XZ plane
4. Press numpad 5 to switch to orthographic projection
5. Shift+s -> Cursor to center to ensure that cursor is centered
6. Shift+a -> Mesh -> Plane, so in selected projection we see the new plane as a segment.
7. Press N and enter "-60" to "Rotation: Y:" in the panel appeared.
8. Enter "-0.5" to "Location: X:" in the same panel.
9. Scale to ensure that we didn't get the required picture.

After this steps I get the following (open image in new tab or window):

If we fulfill this image to squares like above (see fulfilled image below) and count smaller squares, then we get 10 squares in horizontal line and less than 18 squares in vertical line, so sine of A on the picture from blender is not equal to 1/2 (it's greater) => angle A itself is greater than 30 degrees, but angle B is equal to 60 degrees by our definition! So the sum of degree measures in triangles drawn in blender gets greater than 180 degrees.

Looks like a weird logic in blender. Or are there any mistakes in my argumentation and I don't understand something about Blender?

Note 1.

It's not a computational precision issue. We use plane of the default size and there is no bouncing if we zoom the scene either in or out.

Note 2.

Scaling the plane (i.e. lengthening it's projection) will homogeneously scale all construction and will not affect angles and trigonometric functions of them.

Explanation of the "paradox":

I confused lengths of what I have on the "paint" picture with what I want to get in blender. Of course, half-length of the inclined segment is not equal to 1 => sin(a) != 1/2.

Answer 1

I believe the issue comes more with the maths than with Blender.

If you remember SOHCAHTOA, we know the adjacent side to angle B (0.5) and the length of the side opposite angle B (1). Adjacent and Opposite sides work with tan as:

tanθ = Opposite/Adjacent

Substitute in the known lengths with arctan as

arctan(1/0.5) = 63.43...`

This is why Blender looked to be slightly imprecise, where you thought it was 60 degrees, it's actually 63.4 degrees, a small difference, but it explains what seemed to be an error on Blender's part.

As a little tip, you can use shift while scale(s), rotating(r) and translating(g) objects to do it more precisely and this could help you test things a little bit by judging by eye.

Answer 2

It seems you have made a slight mathematical error.

The plane has a length of 2 (by default), so if you look at the diagram, the hypotenuse of one of the smaller right triangles is half the length of the plane, so it is $\frac{2}{2} = 1$ unit long. Now, $\cos(60) = \frac{1}{2}$ so the shorter leg will have length $(1)\cos(60) = (1)\cdot(\frac{1}{2}) = \frac{1}{2}$. Here is where you made a mistake: $\sin(60) = \frac{\sqrt{3}}{2}$, so the longer leg will have length $(1)\cdot\sin(60) = \frac{\sqrt{3}}{2}$, not $2$.

Since I stink at making mathematical diagrams on paint and that is all I have on my tablet, here is a diagram I drew:

Answer 3

your math is wrong...

angle 1: 90° angle 2: 63.435° angle 3: 26.565°

side 1: 1 side 2: 0.5 side 3: 1.118

correct this may be in: triangle calculator

Answer 4

To create a rectangle like that, where angle B is 60°, I'd do like this.

Create a circle with 3 vertices. This will create an equiangular triangle.

Tab into Edit mode, select one edge and subdivide it once.

Create an edge. Select the middle vertex of the edge you subdivided and the vertex to which it's not connected and hit F.

Delete half of it. Select one of the corner vertices of the edge you subdivided and hit X > V.

Select the other corner vertex of the edge you subdivided and snap cursor to selected (ShiftS > U). Set pivot point to 3D Cursor (see red marking in the image). Select all (A twice), duplicate but cancel the translation (ShiftD > Esc) and rotate 180° around the Z¹ axis (RZ180).

Select all (A twice). Duplicate but cancel the translation (ShiftD > Esc) and rotate 180° around the X¹ axis (RX180).

Select all (A twice). Remove doubles (CtrlV > D). Select the diagonal you don't want (two edges, three vertices) and delete it (X > E).

Select two adjacent, unconnected vertices and make an edge between them (F), then do the same with the other two adjacent, unconnected vertices.

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¹_{I'm in top view. If you're in another view, you will need to rotate around other axes.}

Answer 5

If what you're after is that Z shape, here's how I'd go about it.

Add a circle with 3 vertices.

Tab into Edit mode. Select the edge to the right and delete it (X > E).

Select the top vertex and snap cursor to selected (ShirtS > U). Set pivot point to 3D cursor (see red marking in the image).

Select all (A twice). Duplicate but cancel the translation (ShiftD > Esc). Rotate 180° around the Z¹ axis. Select all (A twice). Remove doubles (CtrlV > D). Optionally, select the middle vertex and dissolve it (X > D).

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¹_{I'm in top view. If you're in another view, you will need to rotate around another axis.}