IsoAxis

Animated IsoAxis
Field: Geometry
Image Created By: J.A Gutierrez

Animated IsoAxis

A dynamical system that shows the ideal motion of the model is not possible without a slight deformation.

Basic Description

Systematic animation of a mechanical system with multiple ties between the elements that compose it creates a variety of degrees of freedom that gives place to movement. The IsoAxis was discovered by Wallace Walker in 1958 (patent USA nº 3302321). The discovery took place in a project that was trying to find configurations for paper. In view of the IsoAxis discovery a new type of figures called Kaleidocycle was developed.

System's Description

The animated IsoAxis is formed by a two-dimensional folded paper that creates a grid of sixty isosceles triangles. These triangles can form a ring that has an outside in rotation forming beautiful shaped faces.

Figure 1 is the starting grid, there are a total of sixty isosceles triangles which are divided into six sectors of ten triangles. To further explain the construction, we will isolate one of these sectors.

In order to obtain the systematic explanation of the IsoAxis, the degrees of freedom have to be found. In figure 2 we take a given sector mimicking what the figure looks as the actual IsoAxis, from the individual sector we consider a vertex labeled 0. The opposition of this vertex has space that is appointed by three coordinates. If we take two additional angular coordinates we can determine the position of point 5 which is adjacent to point 0. If the points are defined, the union of this forms a straight line and with the help of another coordinate we can find the coordinate that closes the triangle, in this case it is point 1. If we take the same consideration for periods 1 to 6, these ones can be found, and 7,8 and 9 can be found by symmetry. In light of this information we need 10 points to find the answer to the system.

In order to make the ring structure, we have to add the necessary restrictions to the figure. Points 0, 1 and 2 have to be situated in the plane Y= $\sqrt{3}$ X; as we see in figure 3, the points 3 and 6 have to be situated on top of the plane X=0 and from this information we can obtain 7 equations lowering down the degrees of freedom to 3. Figure 3 has points 4 and 5 as the center of the plane above the Y-Axis, if we give a latitude of the sector in the Z-Axis we can lower down the degrees of freedom to two. Figure 3 names the angles of the triangles alpha (α), beta (β), gamma (γ) and delta (δ) for further explanation.

Delta and alpha can be taken as degrees of freedom, associating delta as the direct variable for movement of the IsoAxis.

A More Mathematical Explanation

[Click to view A More Mathematical Explanation]

The coordinates of the points 0 to 6 can be found to be

Point 0= ($\sin \alpha$ [...]

[Click to hide A More Mathematical Explanation]

The coordinates of the points 0 to 6 can be found to be

Point 0= ($\sin \alpha$ , $\sqrt{3}$ $\sin \alpha$, $-\cos \alpha$ $\sin \delta$ )

Point 1= ($\frac{Y1}{\sqrt{3}}$ , $Y_1$, $Z_1$ )

Point 2 = ( $\frac{Y2}{\sqrt{3}}$ , $Y_2$, $Z_2$ )

Point 3= ( $0,\sqrt{3}$ $\sin \alpha + \cos \delta \cos \alpha- \sin \delta + \sin \beta, \cos \delta - \cos \beta$)

Point 4 = ( $0 , \sqrt{3}$ $\sin \alpha + \cos \delta \cos \alpha - \sin \delta, \cos \delta$ )

Point 5 = ( $0,\sqrt{3}$ $\sin \alpha + \cos \delta \cos \alpha + \sin \delta, - \cos \delta$)

Point 6 = ( $0,\sqrt{3}$ $\sin \alpha+ \cos \delta \cos \alpha + \sin \delta + \sin \gamma, - \cos \delta + \cos \gamma$)

We can find six unknown coordinates: $Y_1$, $Z_1$, $Y_2$ , $Z_2$, $\beta , \gamma$ This coordinate can be found analytically using equations that can be derive in the conditions of perpendicularity in between segments and the distance in between points.

System's Solution

One uses the coordinates [n, m] in order to appoint the segment that connects the vertex between n and m; $||[n, m]||$ designates the distance between n and m.

The first step is to find point 1. This would be found by taking the following equations:

i.$[0,5]\cdot [1,5] =0$

ii.$||[0, 5]|| = \sqrt{2}$

By using the dot product we can expand to the following (if the vector is perpendicular then the dot product is equal to 0)

($X_0 - X_5) (X_1 - X_5) + (Y_0 - Y_5) (Y_1 - Y_5) + (Z_0 - Z_5) (Z_1 - Z_5) = 0$

Replacing

$X_5$ by $0$

$X_1$ by $\frac{Y1}{\sqrt{3}}$.

Then we obtain

($X_0$ $\frac{Y1}{\sqrt{3}}$) + ($Y_0 - Y_5$) ($Y_1 - Y_5$) + ($Z_0 - Z_5$) ($Z_1 - Z_5) = 0$

Next

$\frac{X0 Y1}{\sqrt{3}} + Y_1 (Y_0 - Y_5) + Z_1 (Z_0 - Z_5) = Y_5 ( Y_0 - Y_5) + Z_5 ( Z_0 - Z_5$)

Giving

$Y_1$ ( $\frac{X_0}{\sqrt{3}}$ + $Y_0 -Y_5$) + $Z_1$ ($Z_0 -Z_5$) = $Y_5$ ( $Y_0 - Y_5$) $Z_5$ ( $Z_0 - Z_5$)

Now

$||[1,5]|| = \sqrt{2}$

Means

$(X_1- X_5)^2 + (Y_1-Y_5)^2 + (Z_1 - Z_5)^2 = 2$

Replacing

$X_1$ by $\frac{Y_1}{\sqrt{3}}$

and

$X_5$ by $0$

Gives

$(\frac{Y_1}{\sqrt{3}})^2 + (Y_1 - Y_5)^2 + (Z_1 - Z_5)^2 = 2$

$(\frac{Y_1}{\sqrt{3}})^2$ is equal to $(X_1)^2$

If $Z_0 =Z_5$ Then $Y_1(\frac{X_0}{\sqrt{3}} + Y_0 - Y_5) = Y_5 (Y_0-Y_5)$

So $Y_1= \frac{Y_5(Y_0 - Y_5)}{\frac{X_0}{\sqrt{3}} + Y_0 - Y_5}$

We already know $X_1= \frac{Y_1}{\sqrt{3}}$

We also have

$Z_1= Z_5 + \sqrt{2 - X_1^2 - (Y_1 - Y_5)^2}$

If $Z_0\neq Z_5$ then divide both sides by $(Z_0 - Z_5)$

To get

$Y_1 (\frac{\frac{X_0}{\sqrt{3}} + Y_0 - Y_5}{Z_0 - Z_5}) + Z_1 = \frac{Y_5 (Y_0 - Y_5)}{Z_0 - Z_5} + Z_5$

Setting

$Z_1 =B- A Y_1$

Where

$A = \frac{(\frac{ X_0 }{\sqrt{3}} + Y_0 - Y_5 )}{( Z_0 - Z_5 )}$

$B = \frac{Z_5 + ( Y_0 - Y_5 )}{ ( Z_0 - Z_5 )}$

$U = Y_5$

$V = Z_5$

In which is possible to obtain $Y_1$ in function of $(A,B,U,V)$ Giving

$Z_1 = B - A Y_1$

$X_1 = \frac{Y_1 }{\sqrt{3}}$

We see that

$(\frac{Y_1}{\sqrt{3}})^2 + (Y_1 - U)^2 + (A - A Y_1 -V)^2 =2$

This expands to

$\frac{Y_1^2}{3} + Y_1^2 - 2U Y_1 + U^2 + B ^2 - 2 AB Y_1 + A ^2 Y_1 ^2 - 2 (B - A Y_1) + V ^2 =2$

Which simplify

$Y_1^2 (\frac{1}{3} + 1 + A ^2) + Y_1 (-2U - 2A B + 2A V)^2 -2 = 0$

Giving

$Y_1 = \frac{-2(A V - U - A B ) + \sqrt{4(A V - U- AB)^2 - 4(\frac{4}{3} + A^2)(U ^2 + (B - V)^2 -2)}}{2(\frac{4}{3} + A ^2)}$

And then

$Y_1 =\frac{ U +A (B - V) + \sqrt{(U + A (B - V))^2 - (\frac{4}{3} + A ^2)( U ^2 + (B - V)^2 - 2 )}}{\frac{4}{3} + A^2}$

The solution for point 2 is analogous to point 1. This one can be found by substituting point 4 for point 5.

Work is shown

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$[0,4]\cdot [2,4]=0$

$[0,4]=<X_0 - X_4, Y_0-Y_4, Z_0-Z_4>$

$[2,4]=<X_2-X_4, Y_2-Y_4, Z_2-Z_4>$

$(X_0 - X_4)(X_2-X_4) + (Y_0-Y_4)(Y_2-Y_4)+(Z_0-Z_4)(Z_2-Z_4)=0$

where

$X_4= 0$ and $X_2 = \frac{Y_2}{ \sqrt{3} }$

replacing this numbers we obtain

$Y_2(\frac{X_0}{ \sqrt{3} } + Y_0-Y_4) +Z_2(Z_0-Z_4)=Y_4(Y_0-Y_4) +Z_4(Z_0-Z_4)$

Then

$||[2,4]||=\sqrt{2}$

$(X_2-X_4)^2 + (Y_2-Y_4)^2 + (Z_2-Z_4)^2=2$

where we know that

$X_2=\frac{Y_2}{ \sqrt{3} }$ and $X_4=0$

So

$(\frac{Y_2}{ \sqrt{3} })^2 + (Y_2-Y_4)^2 + (Z_2-Z_4)^2 = 2$

If $Z_0=Z_4$ then $Y_2 (\frac{X_0}{ \sqrt{3} } + Y_0 - Y_4) = Y_4 (Y_0-Y_4)$

So

$Y_2= \frac{Y_4(Y_0-Y_4)}{\frac{X_0}{\sqrt{3}} + Y_0 - Y_4}$

Again we know that $X_2= \frac{Y_2}{\sqrt{3}}$

Also

$Z_2=Z_4 + \sqrt{2-X_2^2(-Y_2-Y4)^2}$

If $Z_0\neq Z_4$ then divide by $(Z_0- Z_4)$ to get

$Y_2(\frac{\frac{X_0}{\sqrt{3}}+Y_0 - Y_4}{Z_0-Z_4}) + Z_2= \frac{Y_4(Y_0-Y_4)}{Z_0-Z_4} +Z_4$

Once one have found point 1, we are able to find point 6 using perpendicularity conditions.

$[1 ,6] \cdot [5, 6] = 0$

Where

$1 = ( Y_1 - Y_5 ) \sin \gamma - ( Z_5 - Z_1 ) \cos \gamma$

$Y_6 = Y_5 + \sin \gamma$

$Z_6 = Z_5 + \cos \gamma$

As shown in figure 4

The solution for point 3 is analogous to the solution for point 6, this is also shown in Figure 4.

$1 = ( Y_2 - Y_4 ) \sin \beta - ( Z_2 - Z_4 ) \cos \beta$

From figure 4 we see that

$(Y_3-Y_4) = \sin \beta$ and $(Z_3 - Z_4) = -\cos \beta$

Figure 5 shows how point 4 and 5 were found.

As Figure 5.2 shows,

$\alpha$ is cut by the plane $X=0$.

$X_0=\sin \alpha$

$Y_0=\sqrt{3}X_0$

which equals

$Y_0=\sqrt{3} \sin \alpha$

In figure 5, we can see that from point 4 to point 5 cut by c, this angles are transversal, giving as a result to have equal angles, this are represented as angle $\delta$

Angle $\omega=90^\circ - \delta$

$\cos \omega = \sin \delta$

$\sin \omega = \cos \delta$

An other important information is that

$0^\prime = (0,Y_0,Z_0)$

$0=(X_0,Y_0,Z_0)$

with this information we can asume that

$Z_0=-\cos \delta \sin \delta$

$Z_4=\cos \delta$

$Z_5=-\cos \delta$

We know:

$X_4=0$ $X_5=0$

this are place in plane $X=0$

Giving as a result:

$Y_4=Y_0 - K$ but $K= \sin \delta - \cos \alpha \cos \delta$

so

$Y_4=Y_0 + \cos \alpha \cos \delta - \sin \delta = \sqrt{3}\sin \alpha + \cos \alpha \cos \delta - \sin \delta$

Giving

$Y_5 = Y_0 + \cos \alpha \cos \delta + \sin \delta = \sqrt{3} \sin \alpha + \cos \alpha \cos \delta + \sin \delta$

Why It's Interesting

The Iso-Axis model is interesting mathematically because it shows the coordination between angles and the relationship between them, the Iso-Axis shows imply Geometry in all aspects of the model. The figure is a beautiful representation of a simple piece of paper, it shows how something so simple can be much more complicated but at the same time interesting. This model shows that mathematics can be taken in to an other libel, in other words, this is an example of mathematical art. An example of the use of this models is the making of paper lamps in modern decorating designs, simple lamps that are made out of folded paper. After the discovery of the Iso-Axis the origami and some other design techniques were developed.

Teaching Materials

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Related Links

Additional Resources

Degrees of freedom

Dot product

Kaleidocycles

References

J.A Gutierrez[1]

The kaleidohedron from the IsoAxis grid:[2]

Wikipedia:[3]

Sahara Gabrielle:[4]

Wolfram MathWorld:[5]

Encyclopedia Britannica:[6]

Matematische:[7]

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