Combining prisms

Step 1: Transform \(P_1\) to match the rotated base of \(P_2\)

Let

\[\phi=\frac{1+\sqrt5}{2}.\]

Define the first prism as

\[P_1=\left(\frac{15}{2\phi},\ \frac{8}{\sqrt3\,\phi},\ \sqrt3\,\phi\right), \qquad P_2=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ \frac{2}{\sqrt3}\right).\]

Here \(P_2\) is the prism currently produced by the geometry pipeline, measured from the actual assembled oblate-prism vertices, up to axis order.

Write

\[a=\frac{15}{2\phi},\qquad b=\frac{8}{\sqrt3\,\phi},\qquad c=\sqrt3\,\phi.\]

So initially,

\[P_1=[0,a]\times[0,b]\times[0,c].\]

For Step 2, rotate \(P_2\) so that its base is

\[\frac{5\sqrt3}{\phi^2}\times 2\]

and its stacking height is

\[\frac{2}{\sqrt3}.\]

So we want to convert \(P_1\) into

\[P_1'=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ h\right).\]

The height \(h\) is determined by volume preservation:

\[\operatorname{vol}(P_1)=\frac{15}{2\phi}\cdot\frac{8}{\sqrt3\,\phi}\cdot\sqrt3\phi=\frac{60}{\phi},\]

while the new base area is

\[\frac{5\sqrt3}{\phi^2}\cdot 2=\frac{10\sqrt3}{\phi^2}.\]

Hence

\[h=\frac{60/\phi}{10\sqrt3/\phi^2}=\frac{6\phi}{\sqrt3}=2\sqrt3\,\phi.\]

So the Step 1 target becomes

\[P_1'=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ 2\sqrt3\,\phi\right).\]

Step 1A: Reshape the \(xy\)-face so the first dimension becomes \(\frac{5\sqrt3}{\phi^2}\)

Keep the \(z\)-depth fixed at \(c\), and replace the base rectangle

\[a\times b\]

with an equal-area rectangle

\[\frac{5\sqrt3}{\phi^2}\times d,\]

where

\[d=\frac{ab}{5\sqrt3/\phi^2}.\]

Compute \(ab\):

\[ab=\frac{15}{2\phi}\cdot\frac{8}{\sqrt3\phi} =\frac{60}{\sqrt3\,\phi^2} =\frac{20\sqrt3}{\phi^2}.\]

So

\[d=\frac{20\sqrt3/\phi^2}{5\sqrt3/\phi^2}=4.\]

After Step 1A, the target block is

\[B_1=\left(\frac{5\sqrt3}{\phi^2},\ 4,\ \sqrt3\phi\right).\]

Geometrically:

  1. Take the face rectangle
    \[R_{xy}=[0,a]\times[0,b].\]
  2. Dissect it into finitely many polygons.
  3. Rearrange those polygons into
    \[R'_{xy}= \left[0,\frac{5\sqrt3}{\phi^2}\right]\times[0,4].\]
  4. Extrude each polygon through the full interval \(z\in[0,c]\).

This gives a 3D dissection of \(P_1\) into prisms that reassemble into \(B_1\).

Step 1B: Reshape the \(yz\)-face to get the final desired dimensions

Now start from

\[B_1=\left(\frac{5\sqrt3}{\phi^2},\ 4,\ \sqrt3\phi\right).\]

Keep the \(x\)-length fixed at \(\frac{5\sqrt3}{\phi^2}\), and reshape the \(yz\)-face.

Current \(yz\)-face:

\[4\times \sqrt3\phi.\]

Target \(yz\)-face:

\[2\times 2\sqrt3\phi.\]

These have the same area, because

\[4\cdot \sqrt3\phi = 4\sqrt3\phi,\]

and

\[2\cdot 2\sqrt3\phi = 4\sqrt3\phi.\]

Because this is the special case

\[4\times h \longrightarrow 2\times 2h \qquad\text{with}\qquad h=\sqrt3\phi,\]

Step 1B can be done with a single straight cut.

After this step, we obtain

\[P_1'=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ 2\sqrt3\phi\right).\]

Geometrically:

  1. Take the \(yz\)-rectangle
    \[R_{yz}= [0,4] \times [0,\sqrt3\phi].\]
  2. Cut it by the single line
    \[y=2.\]
    This splits the face into two congruent rectangles
    \[[0,2]\times[0,\sqrt3\phi] \qquad\text{and}\qquad [2,4]\times[0,\sqrt3\phi].\]
  3. Keep the lower rectangle fixed.
  4. Translate the upper rectangle by
    \[(0,-2,\sqrt3\phi)\]
    in \((y,z)\)-coordinates, so it lands on
    \[[0,2]\times[\sqrt3\phi,2\sqrt3\phi].\]
  5. Extrude both rectangles through the full \(x\)-interval
    \[\left[0,\frac{5\sqrt3}{\phi^2}\right].\]

At the end of Step 1, the first prism has been converted into

\[P_1'=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ 2\sqrt3\phi\right).\]

Step 2: Stack \(P_1'\) with \(P_2\)

Now we have

\[P_1'=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ 2\sqrt3\phi\right)\]

and

\[P_2=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ \frac{2}{\sqrt3}\right).\]

These already share the same first two dimensions, so Step 2 is just stacking along the third axis.

Place \(P_1'\) at

\[\left[0,\frac{5\sqrt3}{\phi^2}\right]\times[0,2]\times[0,2\sqrt3\phi]\]

and place \(P_2\) at

\[\left[0,\frac{5\sqrt3}{\phi^2}\right]\times[0,2]\times \left[2\sqrt3\phi,\ 2\sqrt3\phi+\frac{2}{\sqrt3}\right].\]

Together they form the rectangular block

\[B=\left(\frac{5\sqrt3}{\phi^2},\ 2,\ 2\sqrt3\phi+\frac{2}{\sqrt3}\right).\]

No additional cuts are needed in Step 2; it is just a placement of the two rectangular prisms along the third axis.