Welcome to the Onshape forum! Ask questions and join in the discussions about everything Onshape.

First time visiting? Here are some places to start:
  1. Looking for a certain topic? Check out the categories filter or use Search (upper right).
  2. Need support? Ask a question to our Community Support category.
  3. Please submit support tickets for bugs but you can request improvements in the Product Feedback category.
  4. Be respectful, on topic and if you see a problem, Flag it.

If you would like to contact our Community Manager personally, feel free to send a private message or an email.

Create tetrahedron?

alexander_potochkin
alexander_potochkin Member Posts: 45 ✭✭
edited October 2015 in Community Support
Hello

What is the right way to create a tetrahedron in onShape?
I can make an equilateral triangle and then loft it to a specific point,
but in this case I need to manually calculate the correct height of the tetrahedron.

Is there any better way to make it?

Thanks much!

Tagged:

Best Answers

Answers

  • andrew_troup
    andrew_troup Member, Mentor Posts: 1,585 ✭✭✭✭✭
    Once we have 3D sketches the construction will become trivially easy. In the meantime you'll have to be ingenious. 
  • brucebartlett
    brucebartlett Member, OS Professional, Mentor, User Group Leader Posts: 2,148 PRO
    Draw a triangle and blind extrude (with a arbitrary number larger than the desired height) add draft and your extrude should finish with a point giving 4 faces. You will have to do some math to work out the angle to extrude at. 
    Engineer ı Product Designer ı Onshape Consulting Partner
    Twitter: @onshapetricks  & @babart1977   
  • brucebartlett
    brucebartlett Member, OS Professional, Mentor, User Group Leader Posts: 2,148 PRO
    Could not help myself. I used 2 sketches, the second one works out the angle to extrude at, I have linked sizes so all faces come out equal. 

    I tried to link angle with a variable but that does not seem possible yet. 

    https://cad.onshape.com/documents/5f90e0c61c164cb5bf4141cf/w/b4350e5bf94349dc871ed0bb/e/de54a518e7af4756bf334182
    Engineer ı Product Designer ı Onshape Consulting Partner
    Twitter: @onshapetricks  & @babart1977   
  • alexander_potochkin
    alexander_potochkin Member Posts: 45 ✭✭
    Wow, Bruce thanks a lot!

    I need some time to comprehend your solution
    (not at my computer right now)

    Will it properly recalculate if I change the size of the base triangle?

    What do you mean by " link angle with a variable" ?

    thanks again
  • andrew_troup
    andrew_troup Member, Mentor Posts: 1,585 ✭✭✭✭✭
    @brucebartlett
    One slight snag with your construction: the side you have set equal to the side which is 200mm long measures only 173.205, which is to be expected because an equal relation is true in the plane of the sketch, not in 3D.
    So the equality is NOT with the length of the other line, but with the projection of that line onto the sketch plane

    This is discussed elsewhere on this forum, including here:
    https://forum.onshape.com/discussion/1576/equal-constraint-use-with-care-between-sketches

    I don't know offhand what the angle of a tetrahedron is, but if you check it I think you might find it's rather more than the 65.9 suggested by your model
  • brucebartlett
    brucebartlett Member, OS Professional, Mentor, User Group Leader Posts: 2,148 PRO
    andrew_troup said:
    @brucebartlett
    One slight snag with your construction: the side you have set equal to the side which is 200mm long measures only 173.205, which is to be expected because an equal relation is true in the plane of the sketch, not in 3D.
    So the equality is NOT with the length of the other line, but with the projection of that line onto the sketch plane

    This is discussed elsewhere on this forum, including here:
    https://forum.onshape.com/discussion/1576/equal-constraint-use-with-care-between-sketches

    I don't know offhand what the angle of a tetrahedron is, but if you check it I think you might find it's rather more than the 65.9 suggested by your model
    Good pick up, I have fixed so all sides are now equal (well almost 200 vs 200.002). When time permits I'll will have to read that post.
    Engineer ı Product Designer ı Onshape Consulting Partner
    Twitter: @onshapetricks  & @babart1977   
  • andrew_troup
    andrew_troup Member, Mentor Posts: 1,585 ✭✭✭✭✭
    edited October 2015
    @brucebartlett ;
    I  guess the obvious remedy would be to link the line length in sketch 2 to the line length in sketch 1 using a variable

     (I haven't had a chance to explore them thoroughly but I presume this would be straightforward, although your post suggests this does not yet work for angles)
  • brucebartlett
    brucebartlett Member, OS Professional, Mentor, User Group Leader Posts: 2,148 PRO
    edited October 2015
    I just put another line across the triangle on sketch 1 and linked to sketch 2 via a = constraint. 

    I just can not work out how to get the angle into the draft on the extrude.
    Engineer ı Product Designer ı Onshape Consulting Partner
    Twitter: @onshapetricks  & @babart1977   
  • andrew_troup
    andrew_troup Member, Mentor Posts: 1,585 ✭✭✭✭✭
    @brucebartlett
    The linked post is very wordy, because we were trying to find our way to the root of the issue.
    Here's the executive summary:

    # Constraints which operate between sketches ALWAYS project the remote sketch entity to the destination plane before solving the constraint.
    The only exception is the pierce constraint (where a projection would be inherently meaningless: it would not result in the specified result)

    Currently AFAIK the only other Onshape situation which does not ever rely on this sort of projection is constraint to a remote plane (rather than to a sketch entity).

    NOTE: This # behaviour is common to all similar MCAD modellers: it is not a bug. 
    In other words, this behaviour is     (justifiably, almost unavoidably) BY DESIGN.
  • philip_thomas
    philip_thomas Member, Moderator, Onshape Employees, Developers Posts: 1,381 image
    Ok, i am having flashbacks - why are we now doing a tetrahedron? :)
    Philip Thomas - Onshape
  • andrew_troup
    andrew_troup Member, Mentor Posts: 1,585 ✭✭✭✭✭
    edited October 2015
    Don't worry, @philip_thomas; only three more to go... (and two of them are trivial  ...)
  • clayton_ertley
    clayton_ertley Member Posts: 58 ✭✭
    edited October 2015
    Further more, you don't have to scale afterwards to get the size you need.

    You can see in my model that I used a cube with an edge length of 60mm. This method will always yield a tetrahedron with an edge length of 84.852. Meaning that the ratio of the
    Cube edge length to tetrahedron edge length is (1 : 1.3642). You can use this ratio to predetermine the edge length of the tetrahedron.
  • andrew_troup
    andrew_troup Member, Mentor Posts: 1,585 ✭✭✭✭✭
    @clayton_ertley
    Nice! Kudos to you
  • clayton_ertley
    clayton_ertley Member Posts: 58 ✭✭
    Thanks! I design rubik's cubes among plenty others. We were left to design our own platonic solids. I had to use actual cartesian coordinates for each and every vertex to get a dodecahedron. it was very tedious.
  • alexander_potochkin
    alexander_potochkin Member Posts: 45 ✭✭
    Thank you Clayton, Andrew and Viru!

    Now I definitely know what onShape tools I should have a look to.
    What a wonderful community here!

    Have a great day.
    Alex.
  • alexander_potochkin
    alexander_potochkin Member Posts: 45 ✭✭
    With big help from the community I made my very own version of tetrahedron.
    I made a cube and sliced a tetrahedron out of it.

    https://cad.onshape.com/documents/d21a14adc0ef4fa09095b29d/w/79bbbe32437a42df9980f7ea/e/30a56b8e579348c084f6f188

    Thanks again

  • patrick_moore
    patrick_moore Member Posts: 1 ✭
    I would use the fact that a cube contains all the points of a tetrahedron at it's corners. Not as quite simple to draw but requires no calculations and is driven by the one variable.

    Draw a square and extrude with side length as a variable.
    Create a plane that intersects three of the points of the cube so that they make an equilateral triangle.
    Sketch a circle on the new plane that also intersects all three points. (doesn't have to be a circle but I found it easiest)
    Extrude (by variable) and Subtract  the circle from the cube, lopping off the corner.
    Repeat the Plane/Sketch/Extrude for the remaining three corners.

    It might take a little more time to draw but it is much more satisfying as it's a perfect tetrahedron.
  • EvanReese
    EvanReese Member, Mentor Posts: 3,030 PRO
    since @patrick_moore decided to revive this thread I'll throw my solution in there too:
    1. sketch triangle
    2. sketch one of the other edges (using an equal constraint)
    3. loft to the point
    Here's a quick video. I like it because there's no math, no variables, and only 3 features.

    Of course, to do it in one feature you could use the Polyhedron feature too :D . Go here to see both examples.
    Evan Reese
    The Onsherpa | Reach peak Onshape productivity
    www.theonsherpa.com
  • bruce_williams
    bruce_williams Member, Developers Posts: 842 EDU
    nice @Evan_Reese !   I love the way equal constraint can be used across different sketches.  Instead using a construction line making the new line equal to one of the triangle sides works too.
    www.accuratepattern.com
  • EvanReese
    EvanReese Member, Mentor Posts: 3,030 PRO
    edited August 2020
    bruce_williams said:
    nice @Evan_Reese !   I love the way equal constraint can be used across different sketches.  Instead using a construction line making the new line equal to one of the triangle sides works too.
    Assuming I understand your suggestion, it would make it equal to the projection of the edge, so I don't think it would end up right. see what I mean?
    Evan Reese
    The Onsherpa | Reach peak Onshape productivity
    www.theonsherpa.com
  • alnis
    alnis Member, Developers Posts: 453 EDU
    Solution based on @brucebartlett's suggestion to draft an extrude, but without using the measure featurescript. The dihedral angle for a tetrahedron is arccos(1/3), so 90 deg - acos(1/3) gives the correct draft angle. This gives an exact result (60.00000 mm measured side length for all sides).
    https://cad.onshape.com/documents/93f2dfdea0555d930849c033/w/738de8482a5a2e6f9d440e97/e/a1b6010751a1e7e89ecc3d2a


    Get in touch: contact@alnis.dev | My personal site: https://alnis.dev
    @alnis is my personal account. @alnis_ptc is my official PTC account.

Categories