12 Tetrahelix-Tetrahedron Intersection Points

We calculate the (x, y, z) coordinates for the 3x4=12 points (3 per Tetrahedron triangular face) for the triangle face coordinates (7, 3). This will allows us to then calculate other properties of the Tetrahelix.

Recall that the Tetrahelix axis passes through the (7, 3) points of the Tetrahedron's triangular face.

We position the Tetrahedron so that its 4 vertices are at

V1 = (1, 1, 1)
V2 = (1, 1, 1)
V3 = (1, 1, 1)
V4 = (1, 1, 1)

The Tetrahedron edge length is then

EL = dist(V2 - V1) = sqrt( (-1 - 1)^2 + (-1 - 1)^2 + (1 - 1)^2)
EL = sqrt( 4 + 4 + 0) = 2 sqrt(2) = 2.828427125...

For the V1-V2-V3 triangular face, we let

Va = (V2 V1) = (2, 2, 0)
Vb = (V3 V2) = (0 , 2, 2)
Vc = (V1 V3) = (2, 0, 2)

Then

P1 = V1 + (7/10)Va + (3/10)Vb = (1 14/10, 1 14/10 + 6/10, 1 6/10)
P1 = (4/10, 2/10, 4/10) = (0.4, 0.2, 0.4)

P2 = V2 + (7/10)Vb + (3/10)Vc = (1 + 6/10, 1 + 14/10, 1 14/10 + 6/10)
P2 = (4/10, 4/10, 2/10) = (0.4, 0.4, 0.2)

P3 = V3 + (7/10)Vc + (3/10)Va = (1 + 14/10 6/10, 1 6/10, 1 + 14/10)
P3 = ( 2/10, 4/10, 4/10) = (0.2, 0.4, 0.4)


For the V1-V2-V4 triangular face, we let

Va = (V2 V1) = (2, 2, 0)
Vb = (V4 V2) = (2 , 0, 2)
Vc = (V1 V4) = (0, 2, 2)

Then

P1 = V1 + (7/10)Va + (3/10)Vb = (1 14/10 + 6/10, 1 14/10, 1 6/10)
P1 = ( 2/10, 4/10, 4/10) = (0.2, 0.4, 0.4)

P2 = V2 + (7/10)Vb + (3/10)Vc = (1 + 14/10, 1 + 6/10, 1 14/10 + 6/10)
P2 = ( 4/10, 4/10, 2/10) = (0.4, 0.4, 0.2)

P3 = V4 + (7/10)Vc + (3/10)Va = ( 1 6/10, 1 + 14/10 6/10, 1 + 14/10)
P3 = ( 4/10, 2/10, 4/10) = (0.4, 0.2, 0.4)


For the V1-V3-V4 triangular face, we let

Va = (V3 V1) = (2, 0, 2)
Vb = (V4 V3) = (2 , 2, 0)
Vc = (V1 V4) = (0, 2, 2)

P1 = V1 + (7/10)Va + (3/10)Vb = (1 14/10 + 6/10, 1 6/10, 1 14/10)
P1 = ( 2/10, 4/10, 4/10) = (0.2, 0.4, 0.4)

P2 = V3 + (7/10)Vb + (3/10)Vc = (1 + 14/10, 1 14/10 + 6/10, 1 + 6/10)
P2 = ( 4/10, 2/10, 4/10) = (0.4, 0.2, 0.4)

P3 = V4 + (7/10)Vc + (3/10)Va = ( 1 6/10, 1 + 14/10, 1 + 14/10 6/10)
P3 = ( 4/10, 4/10, 2/10) = (0.4, 0.4, 0.2)


For the V2-V3-V4 triangular face, we let

Va = (V3 V2) = (0, 2, 2)
Vb = (V4 V3) = (2 , 2, 0)
Vc = (V2 V4) = (2, 0, 2)

P1 = V2 + (7/10)Va + (3/10)Vb = (1 + 6/10, 1 + 14/10 6/10, 1 14/10)
P1 = (4/10, 2/10, 4/10) = (0.4, 0.2, 0.4)

P2 = V3 + (7/10)Vb + (3/10)Vc = (1 + 14/10 6/10, 1 14/10, 1 + 6/10)
P2 = (2/10, 4/10, 4/10) = (0.2, 0.4, 0.4)

P3 = V4 + (7/10)Vc + (3/10)Va = ( 1 14/10, 1 + 6/10, 1 + 14/10 6/10)
P3 = (4/10, 4/10, 2/10) = (0.4, 0.4, 0.2)


Now that we know the (7,3) points we can draw the small triangle red triangle on each of the 4 Tetrahedron's faces.

We know that the Tetrahelix axes make crossing lines through these (7, 3) triangle face points.

For eaxmple, if we look at just 2 of the red triangles then two of the Tetrahelix axes pass through the 4 points as follows.

"A" goes to "A" and "B" goes to "B".

The two "A" point coordinates are

A(123) = (0.2, 0.4, 0.4), A(134) = (0.2, 0.4, 0.4)

The length of the line segment A-to-A is given by

dist = sqrt( (0.2 + 0.2)^2 + (0.4 - 0.4)^2 + (-0.4 - 0.4)^2 )
dist = sqrt(0.16 + 0.0 + 0.64) = sqrt(0.8)
dist = sqrt(8/10) = 2 / sqrt(5) = 0.894427191...

We can calculate the crossing angle as follows:

Since the edge length of the Tetrahedron is EL = 2 sqrt(2) the edge length of the small red triangle is

ELs = EL / 10 = sqrt(2) / 5.

We let alpha = (1/2) crossing angle.

Then

sin(alpha) = (ELs/2) / ( (2/sqrt(5)) / 2)
sin(alpha) = ( sqrt(2) / 10 ) / ( 1 / sqrt(5) )
sin(alpha) = sqrt(5) / (5 sqrt(2) )
sin(alpha) = 1 / sqrt(10)

Then the crossing angle is

alpha = 2 arcsin(1/sqrt(10)) = 36.86989765... degrees.

We now wish to calculate the angle which a Tetrahelix makes with a table top when the a Tetrahedron face is placed flat on the table.

We first need a vector which is perpendicular to a Tetrahedron face. This can easily be found by summing the vectors to vectices V1, V2, and V3. (The red line in the next figure is perpendicular to the V1,V2,V3 face of the Tetrahedron.)

This is essentually taking the average of the 3 vectors and multiplying by 3. Recall that the average of the vectors is given by

Avg(123) = (1/3)(V1 + V2 + V3)

So, multiplying by 3 we get

Vn(123) = (V1 + V2 + V3) = (-1, 1, 1)

This is a vector pointing out through the Face Center of the V1, V2, V3 face.

This vector Vn(123) is normal (perpendicular) the the V1, V2, V3 face.

If we put the Tetrahedron of the table with its V1, V2, V3 face flat on the table, this vector will point down through the table. So, we reverse the direction of this vector so that it will point up from the table top.

Vn(123) = (1, -1, -1)

For calculation purposes, we make the length of this vector to be 1. In the Figures, the length is not 1 and is drawn as a red line.

Vn(123) = (1/sqrt(3))(1, -1, -1) = (1/sqrt(3), -1/sqrt(3), -1/sqrt(3))

We will use the vector dot product to calculate the angles.

A.B = |A| |B| cos(theta)
theta = arccos(A.B / (|A||B|))

The vector connecting the 2 "A" points above

A(123) = (0.2, 0.4, 0.4)
A(134) = (0.2, 0.4, 0.4)

is given by

VAA = A(134) - A(123) = (0.4, 0.0, -0.8) = (2/5, 0, -4/5)

(see the blue line in the figures.)

Its magnitude is

mag(VAA) = sqrt(0.4^2 + (-0.8)^2) = sqrt(0.16 + 0.64)
mag(VAA) = sqrt(0.8) = sqrt(8/10) = 2 / sqrt(5)

The unit vector in the VAA direction is then

UAA = (sqrt(5)/2)(2/5, 0, -4/5)
UAA = (1/sqrt(5), 0, -2/sqrt(5))

Note that vectors can be moved to any position as long as we don't change their direction nor their magnitude. So I move it (blue line) in the Figure so that it is at the face center. The red line (perpendicular line to the Tetrahedron face) also passes through the face center point.

The dot product is

(1/sqrt(3), -1/sqrt(3), -1/sqrt(3)).(1/sqrt(5), 0, -2/sqrt(5))
(1/sqrt(15) + 2/sqrt(15)) = 3/sqrt(15) = sqrt(3/5)

So theta is

theta = arccos(sqrt(3/5)) = 39.23152048... degrees

Which means the Tetrahelix makes an angle of

90 - arccos(sqrt(3/5)) = 50.76847952... degrees

with the table's surface.


There is another angle at which a Tetrahelix can make with the table when the Tetrahedron is placed flat on the table. See the green line in the above figure. This is another axis of symmetry for some Tetrahelix. It passes through the table top some distance away from the Tetrahedron as the following Figures show.

There are several line segments we could chose for the calculation. The following 2 Figures shows the line segement in green that I will use. Yes, its very hard to see the orientation.

The two points selected are:

P3 = (0.4, -0.2, 0.4) from face (1.2.4), and
P2 = (0.4, 0.2, -0.4) from face (1.3.4).

The vector along this line segment is then

V = (0.4, 0.2, -0.4) (0.4, 0.2, 0.4)
V = (0.0, 0.4, -0.8) = (0, 2/5, -4/5)

As before, the magnitude of this vector is

mag(V) = sqrt(0.4^2 + (-0.8)^2) = sqrt(0.16 + 0.64)
mag(V) = sqrt(0.8) = sqrt(8/10) = 2 / sqrt(5)

The unit vector in the direction of th egreen line segment is then

UV = (sqrt(5)/2)(0, 2/5, -4/5)
UV = (0, 1/sqrt(5), -2/sqrt(5))

We can place this vector anywhere we want. So we place it at the center of the 1.2.3 face.

The dot product with the face unit normal is given by

D = (1/sqrt(3), -1/sqrt(3), -1/sqrt(3)).(0, 1/sqrt(5), -2/sqrt(5))
D = 0 - 1/sqrt(15) + 2/sqrt(15) = 1/sqrt(15)

So, for this case, theta is

theta = arccos(sqrt(1/15)) = 75.0367825... degrees

Which means this Tetrahelix makes an angle of

90 - arccos(sqrt(1/15)) = 14.96321744... degrees

with the table.



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