Find angles of two intersected planes.

I'm really bad at math so I'll try to explain as best as I can. Here's a visual representation of what I need to do. Basically it's a pop-up book.

There is a plane which can be folded on the blue line, so it makes two planes. I have dimensions of planes, angles of planes on Z axis and angles of the pages on which these planes are positioned.

I need to get the X axis angles for each of two planes to have them appear seamless as a folded one while I turn pages.

• I don't understand this: "Folding on the blue line." – zoli Jul 19 '15 at 12:58
• The pop-up part IRL is actually a single piece folded in half. Here's a picture. – RCKT Jul 19 '15 at 13:36
• Cute! Cute! Cute! – zoli Jul 19 '15 at 15:04
• What do the red lines represent? – Carter Pape Jul 20 '15 at 1:02
• The red lines represent bottom sides of the planes. Here's the picture. – RCKT Jul 20 '15 at 8:43

I am going to refer to the different planes and vectors seen on the image as fallows.

The red right page (with the script "N'T STAND IT!") will be called $r$. The red left page (with the script "BIG FROG CA") will be called $l$. The green left hand side of the frog will be called $gl$ the green right hand side of the frog will be called $gr$. When the pop up book is open then $l$ and $r$ coincide, $gl$ and $gr$ are perpendicular to $l$ and $r$, respectively. $gl$ and $lr$ intersect in a line perpendicular to the open book when $l$ and $r$ coincide.

$a$ $b$, and $c'$ are unit vectors on the common lines of $r$ and $gr$; $l$ and $gl$; and $gl$ and $gr$, respectively. $c'$ is perpendicular to $a$ and $b$ when the book is open, i.e. when the two red pages $l$ and $r$ coincide.

The next figure simplifies the two red and the two green pages. The upper figure depicts the fully open book.

Let $\alpha$ characterize the two vectors $a$ and $b$ when the book is flat open. Let $\beta$ characterize an in between situation when the two red pages do not coincide. At the beginning $\beta=0$.

Let's fix our coordinate system as shown above. Now,

$$a=\begin{bmatrix} \cos(\alpha)\cos(\beta)\\ \sin(\alpha)\\ \cos(\alpha)\sin(\beta). \end{bmatrix}$$

and

$$b=\begin{bmatrix} -\cos(\alpha)\\ \ \sin(\alpha)\\ 0 \end{bmatrix}.$$ Vector $c$ is perpendicular to the flat red plane when the book is flat open; it is perpendicular to $a$ and $b$. And, $c$ remains perpendicular to the plane determined by $a$ and $b$ while the book is closing or, in other words while $\beta$ goes from $0$ to $\pi$. This is because $c$ and $a$, and $c$ and be are rigidly connected to each other. See the model

Let $c$ be a vector pointing in the direction of $c'$. For $0\le \beta \le \pi$

$$c=a\times b= \begin{vmatrix} i&j&k\\ \ \cos(\alpha)\cos(\beta)&\sin(\alpha)&\cos(\alpha)\sin(\beta)\\ -\cos(\alpha)&\sin(\alpha)&0 \end{vmatrix}=$$ $$= \begin{bmatrix} - \sin(\alpha)\cos(\alpha)\sin(\beta)\\ \cos^2(\alpha)\sin(\beta)\\ \sin(\alpha)\cos(\alpha)(1+\cos(\beta)) \end{bmatrix}.$$

My solution will be rather sketchy from this point on.

Watch out: $c$ is not a unit vector and $\lim_{\beta \rightarrow \pi}=0$ which result is misleading. So, let's normalize $c$ and, obviously $c'$, mentioned above, is its normalized version.

Then calculate $d=c'\times a$ and $e=c'\times b$ and let the corresponding unit vectors be denoted by $d'$ and $e'$, respectively.

Now, $c'$ is the unit normal vector of the plane $r$ and $d'$ is the unit normal vector of the plane $gr$; $c'\cdot d'$ is the cosine of the angle between $r$ and $gr$. So the angle of these two planes can be calculated. Similarly, $e'\cdot c'$ equals the cosine of $l$ and $gl$. Also, $e'\cdot k$ is the cosine of the angle between $gl$ and $r$. The angle between $l$ and $r$ is $\pi-\beta$.

• Thank you! I think I'm almost there. Your explanation cleared up a couple of things for me. I'll post the result here later. =) – RCKT Jul 21 '15 at 8:30
• Here's a tech demo =) – RCKT Jul 25 '15 at 11:16
• @rockit: I don't dare to believe that you did this beautiful thing based on my advice... – zoli Jul 25 '15 at 11:53
• To be honest my solution was based not only on your advice, but it really helped to shed light on several unclear parts. Again, thank you very much. – RCKT Jul 25 '15 at 12:32