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Relationship between truncated tetrahedral and its enclosed tetrahedral...

  1. Oct 4, 2015 #1
    What is the mathematical formula to calculate "a" for a given "b" in the picture below of a tetrahedron with sides "b" inside of a truncated tetrahedron with sides "a". I think that this is a more challenging problem that what it appears to be.
     
  2. jcsd
  3. Oct 4, 2015 #2
  4. Oct 4, 2015 #3
    Put 4 tetrahedra with edge=a back on the truncated tetrahedron. This results in a big tetrahedron with edge=3a. Find the proportion between the 3a edge and the center-to-vertex length. Subtract the height of the top tetrahedron with edge=a from the center-to-vertex length. This results in the center-to-vertex length of the inner tetrahedron. Then b is found because the proportion between the edge and the center-to-vertex length is known.
     
  5. Oct 5, 2015 #4
    OK. Now I have a "new big tetrahedral with side "3a" that encloses my "original truncated tetrahedral with side "a".
    Question1:
    How do I find the proportion between the 3a edge of this new big tetrahedron and the center to vertex length.
    Question2:
    Which vertex do you mean? The vertex if the "new big tetrahedral", or the vertex of the small original tetrahedral enclosed by the truncated tetrahedral?
    Question3:
    Is the "center to vertex length" of the small original tetrahedral with side "b" called the "exsphere radius" of this tetrahedral?
    Question4:
    What are the "exsphere radius" and the "midsphere radius" of a truncated tetrahedron as refereed to in the truncated tetrahedral calculator found at: http://rechneronline.de/pi/truncated-tetrahedron.php
    Question5:
    What are the "insphere radius", "midsphere radius" and the "exsphere radius" of tetrahedrals as refered to in the tetrahedral calculator found at: http://rechneronline.de/pi/tetrahedron.php
    Sorry that I have so many questions.
    kind regards
    andrew
    PS
    In my problem, I know the value of "b" and need to solve for the value of "a".
     
  6. Oct 5, 2015 #5
    The problem is solved by comparing the big un-truncated tetrahedron to the inner tetrahedron. I suppose the center-to-vertex length is called the exsphere radius. We must find the proportions;
    3a / b = outer exsphere radius / inner exsphere radius
    I see that the inner exsphere radius is equal to the outer exsphere radius minus the height of the little tetrahedron at the top. It has edge=a.
    All there is needed to be known is the height and exsphere radius for any tetrahedron. You could look this up on wiki but it's not hard. The triangles of a tetrahedron subdivide into six 30-60-90 triangles. From these segments I can see how to calculate the height with pythagorean's law. The triangle that finds the height has a similar triangle inside that finds the exsphere radius.
     
  7. Oct 5, 2015 #6
    Thanks for the great help Helios.
    I used the radius of the circumsphere for the tetrahedrons instead of their exsphere radius to calculate the proportions which was the key to solving this problem.
    The radius of the circumsphere for a tetrahedral (a circumscribed sphere of a polyhedron is a sphere that contains the polyhedron and touches each of the polyhedron's vertices).https://en.wikipedia.org/wiki/Circumscribed_sphere. For a tetrahedron with sides "b", the circumsphere =(square root of 6 divided by 4) times b. The exsphere for the tetrahedral https://en.wikipedia.org/wiki/Exsphere_(polyhedra) (i am not really sure what it really is)=b/(square root of 6). https://en.wikipedia.org/wiki/Tetrahedron.
    The hight of a tetrahedron is (root of 6 divided by 3) times the length of its side.
    When I did all the calculations, I got
    a=0.6b.
    I made a paper model and it seemed to be right.
    Thanks again
    andrew
     
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