Modeling a tetrahedron in Shapr3D

Good morning,
about the video that was posted some time ago:

Something is unclear to me: why set an angle of 19,471??
Seems to me a workaround, if I set an angle of 20° degrees (which should be the correct one) then the sides would not be all equal to each other, nor the angles they make… not a correct Tetrahedron would come out. But if is so, how this workaround of 19,471° was determined? In short, I can’t understand if this is a problem in the Shapr3d Extrude tool, a specific quirk of a Tetrahedron shape drawn in Shapr or else…?

The exact number is (19,4712206345), you can input that as well (though we will only show 19,471 - in the background we store all the digits)

As for the calculation, this is how you can end up with this exact angle: 90 - 180*acos(1/3) / pi (courtesy of our in-house mathematician @Koko_Shapr3D )

You can read more about it here:

Thank you… this solid is much more complex than it appears to be… and for sure I’m not starting to argue with Koko :slight_smile:
The exact number with all the digits makes infact an exact 60° for all sides (and it is good to know that tracks are so deeply followed into the calculations).

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You can edit the corect body, with subract from a triangel based column.
I draw it a few month ago.
Use construction planes and project.


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Ineristing, JST, but I have difficulty to understand the underlying reasoning: you project each of the three external triangles planes on the inner planes with a 60° offset? Do I understand correctly? If so, how do you choose the dimensions of the external triangles in the first place to be able to come up with an exact tetrahedron in the end?

Its more simple…

Draw a triangle with egal edges.( a)
Extrude to column and hide
Draw the 3 heigts of the tringle (vertical in half of “a” to the peak. (h)
Draw an constuction axis “perpendicular to a face at point” in the crossings “h” scale to longer as column
Make an construction plane on the edge “h” with the tool “through edge at angle” set 90 degree
Ofset it outside the triangle
Project the construction axis and “h” to the plane
From the axis will the height of the body “H”
On the plane draw the “a” . Draw a line from the end of “h” to conection to the projected axis with the length of “a” will got the “H”
Draw a line from the peak to the another end of “h”

After it you can edit a triangle to extrude and subtract from the column
See the picture…


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JST, I thank you for your help… but I’m not sure I can follow all passages (I tried). While it is interesting as an alternative, applying a math formula to set the extruding slope angle, in the end is straightforward… the only problem is to know which formula to apply :roll_eyes:

Sorry for my English, I made a vid, i hope will help.
On the end i choose more simple way to subtract from three ways, rotated the body for subtract.
In the forum you can find a drawings from the octahedron, may be also help to change the thinking😎


It surely does help! Thanks for your time, it did work out perfectly this time :wink:

A question: are you proposing that this tethraedron build only geometrically in some subtle way is “more exact” than the one created throught a formula? If this would be the real world I could agree, but within software (with 10 decimals resolution) I tend to believe this two ways to solve it superimpose to the last decimal… they simply converge to the same intrinsic resolution built in the software.

Is there any way to check if the tetrahedrons are “truly” identical? For all practical purposes nothing appears to me would in any way change, this is just an academic question, but at the same time isn’t this a basic question any software engineer would ask himself in the process of building a (exact) 3d engine?

Out of any dispute, guys, just curious to understand better under the hood :slightly_smiling_face:

I tried to draw with 2000 mm basement with set the angle to seventh digit and got a few mm less edges on the sides (1980…)
In my opinion where the square root, and the “pi” or another infinite decimal fraction, are in the mathematical formula, the best way the geometric editing.
But im not software engineer. In the college we made the technical drawings with a compass and ruler😎.

Maybe we are not using the same Shapr3d version? I noticed when I reproduced your video, mine is 3.33.0. The idea to scale the solid is a great one to test it, but my findings are different…

With a triangle base of 5000mm and a 4 decimal Extrude draft angle (19,4712) I get a side with a 4μm error, see photo:

I tried to reach the limits of the software (which I found out is a 1km square) but since I couldn’t extrude more than 500mt, I resolved with a Tetrahedron of 500mt base. This, with a Extrude draft angle of 7 decimals (19,4712206) gives an error of 1μm, see photo:

Since we have still three decimals to go and we are at the boundaries of the software workspace (you see the earth is flat beyond :grin:) that appears to me to be truly enough to work with peace of mind… :slightly_smiling_face:

I fully agree that using (and rounding) infinite numbers adds errors not present in a pure geometric solution, but seems to me that within Shapr workspace (but THIS version, yours seems to give much worse results) they are beyond perception.

Very interesting conversation, started with a weird draft angle (for me) and progressed to a 500mt Tetrahedron… :beers:

As i see i typed wrong the 5th decimal character, left a”2”:face_with_monocle:. This is why i get worse result as you.