# Fold up Equilateral Triangle to make Tetrahedron with arc tool and guideline

Hi Friends: I’m working on folding up a triangle (Rotate) to the vertical height of central equilateral triangle. (I uploaded a PDF showing my steps)

I see that there is a problem: The arc tool is line segments. So the Tip does not actually touch in the correct vertical point.

Any ideas to make this technique work.

I’m wanting to make a “water-tight” model for 3-d printing.

thanks,

Jim

jmg-foldup01.pdf (238.8 KB)

How about if you use the Arc tool to draw the arc from the apex of the triangle to the vertical line? That will effectively put a vertex on the arc at the right point. place the center of the arc on the midpoint of the edge shared by the two triangles.

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Dave, I feel embarrassed asking this since I’m supposed to know what I’m doing, but how did you get the end of the arc to snap to the vertical line? When I try that, I can get an inference point on the vertical line, but nothing for the arc end and nothing when they intersect exactly. As a result, when I zoom in very close there is an evident gap between the arc end and the vertical line no matter how careful I thought I was.

You may or may not consider this cheating, but the dihedral angle of a tetrahedron is known to be 70.528779 degrees. Hence:

-Gully

Edit: You can also use the Trilateration plugin available from SketchUcation to plot the apex.

I guess it doesn’t show so well in the screen shot but I drew a vertical line up the blue axis. This gave the Arc tool something to snap to. The end indicator is a red X. I checked that the arc actually intersected the vertical line by checking to see it had been split.

@slbaumgartner, as Dave described, you should be able to get the ‘Intersect’ cross.
Bur be aware that the rotation uses geometry that is within the same (editing) context. Otherwise there is no true intersection, just like in versions prior to SketchUp 2015.

Gentlemen:
Here is more detail of the problem I’m finding with the technique of intersecting an arc and line to create vertex for tetrahedron tip

Jim

jmg-foldup02.pdf (305.0 KB)

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Why don’t you try what I showed? It works perfectly. And try drawing the initial triangle centered on the origin so you don’t have to locate the center. Saves a few steps.

a variation of dave’s method…

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Drawing from scratch a couple of times: Yes was able to find the ‘red X’ defining the top vertex!

So very helpful to know that this technique will work. I’m hoping to model all five of the platonic solids using this technique. Otherwise need to know inner rectangles and rotate them and so forth.

Have a good day all,

Jim

Thanks @DaveR and @Wo3Dan! I don’t know what was wrong before…I was not getting the red X but now I am. Maybe what @Wo3Dan suggested, as I had created a couple of groups, though I didn’t think the arc or the vertical line was in either of them.

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And if you want to go the next step, you can have a look at this blog post (uess by whom):

@Cotty - very clear video of finding the vertical angle where the gap in the flat pentagons meet in space.
Thanks for guiding me: will attempt this technique.
Have a good day

Say thank you to Dave…

@DaveR - thanks for helping me through my very first post on the forum!

Sincerely

Jim

Jim, you’re welcome. I’m glad the dodecahedron thing helped, too.

Couple of shapes on process

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Jim,

While the rotate-to-intersect-line topic is interesting, on its own merits, you should note that the fact that circles are composed of line segments causes inherent inaccuracy in the “intersection.” (The intersection is with a line segment, not the mathematical circle, itself.)

If you want precision, consider the perfect tetrahedron, circumscribed by a perfect cube:

Since a perfect cube is easy to draw, a perfect tetrahedron is subsequently easy, as well.

I join the discussion, because you are also going to have a problem with the icosahedron and dodecahedron. Use the initial steps of this model (employing the golden section,) for a perfect icosahedron:

Here’ a comparable model, for the dodecahedron:

-Taff

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Taff,

Here we approach one of the deeper questions I have about how SU works, beyond user technique.

Namely how does SU handle irrational numbers?

1. We use the rectangle tool to draw GM rectangles. We trust that these are inference points we can use, yet these are imaginary numbers.

2. Arcs and circles are also objects we draw. Yet these are also imaginary numbers, which we do not trust?

Are these contradictory statements to the logic of SU. Or is this still a lack of user technique?

Thanks for helping my define this question.

Jim
Note:
In mathematics, an irrational number is any real number that cannot be expressed as a ratio of integers. Irrational numbers cannot be represented as terminating or repeating decimals. As a consequence of Cantor’s proof that the real numbers are uncountable and the rationals countable, it follows that almost all real numbers are irrational.[1]

When the ratio of lengths of two line segments is irrational, the line segments are also described as being incommensurable, meaning they share no measure in common.

Numbers which are irrational include the ratio π of a circle’s circumference to its diameter, Euler’s number e, the golden ratio φ, and the square root of two;[2][3][4] in fact all square roots of natural numbers, other than of perfect squares, are irrational.

It handles them the same way it handles all floating point numbers … there is a finite limit to how accurately you can represent a decimal number (rational or otherwise) using binary numbers. The GM is equal to (1 + sqrt(5)) / 2, so its accuracy (or trustworthyness) is only limited to how accurately this calculation can be made. Note, for example, that the decimal number 0.1 has no exact representation in binary, so the precision issue is not just limited to transcendentals or irrationals.

Wikipedia has a nice explanation: Floating-point arithmetic - Wikipedia