What are the angles formed at the center of a tetrahedron if you draw lines to the vertices?
I'm trying to make these:
I need to know what angles to bend the metal.
What are the angles formed at the center of a tetrahedron if you draw lines to the vertices?
I'm trying to make these:
I need to know what angles to bend the metal.
The required tetrahedral angle is $\arccos\left(-\frac13\right)\approx109.5^\circ$. You can use the law of cosines to show this... or more transparently, you can exploit the fact that a tetrahedron is easily embedded inside a cube:
I suppose now's as good a time as any to post the synthetic proof.
One can use the Pythagorean theorem to show that a square with unit edge length has a diagonal of length $\sqrt 2$. The Pythagorean theorem can be used again to show that a right triangle with leg lengths $1$ and $\sqrt 2$ will have a hypotenuse of length $\sqrt 3$ (corresponding to the triangle formed by an edge, a face diagonal, and a cube diagonal). We know that the diagonals of a rectangle bisect each other; this can be used to show that the diagonals of a cube bisect each other. From this, we find that the side lengths of the (isosceles!) triangle formed by two half-diagonals of the cube (corresponding to two of the arms of your caltrops) and a face diagonal are $\frac{\sqrt 3}{2}$, $\frac{\sqrt 3}{2}$, and $\sqrt{2}$. From the law of cosines, we have
$2=\frac34+\frac34-2\frac34\cos\theta$
where $\theta$ is the obtuse angle whose measure we are seeking. Algebraic manipulation yields $\cos\,\theta=-\frac13$.
One way is to write the vertices as vectors $a,b,c,d$ with norm $\|\cdot\|=1$. Then $a+b+c+d=0$. But
$ 0=\|a+b+c+d\|^2=4+2{4 \choose 2}\cos\theta,$
so $\theta = \arccos(-1/3)$.
(Assuming the tetrahedron is supposed to be regular) Take the tetrahedron with vertices $(1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1)$, which has centre at the origin, and use the dot product formula:
$a\cdot b = |a| |b|\cos\theta$
which gives $\cos\theta=-\frac13$
If the tetrahedron is regular, we can use statical equilibrium of four equal isotropic forces with angle $ \theta$ between any two of them. Referencing with reference to any one force,
$ F \cos \theta + .. + .. + F = 0$
$ \cos \theta = -\frac13 $
To generalize for all $i$ direction forces on a particular Z direction the vector dot product sum can be used:
$ \Sigma F_i . Z =0 $
Using the illustration of the tetrahedron embedded into the cube, you can lay out some dimensions. Assign a length of 2 to the sides of the cube, and then according to an old guy name Pythagoras, the hypotenuse of each side will be d 2*√2. Since the center of the tetrahedron is also the center of the cube, then you can draw an isosceles triangle using the hypotenuse and the center that will be 2*√2 at the base and 1 unit high. Split that in half to make two right triangles, each with a base of √2 and a height of 1. Now comes the real trig. Tan = Opposite / Adjacent, or √2/1, or just √2. Grab your calculator and you'll find that the angle you're looking for a the top of the right triangle is 54.735°. Put those two triangles back together, and you get an angle of 109.47°. For fabrication, I think you'd be fine with an angle of 110°.