[Math] Rotation Matrix with (Cos(theta) = 0,Sin(theta) = 1) as Identity

linear programmingmatricesorientationrotations

In my program, all rotations are handled with unit-vector orientations:

$$[x,y] = [\cos{\theta}, \sin{\theta}]$$

In the game engine I'm using for visualization (Unity3D), the $Y$ axis is forwards – so the identity orientation direction is $(0,1)$. In other words $(0,1)$ relative to orientation $A$ is equal to orientation $A$.

To make my life easier, I'd like to use $I[0,1]$ as the identity for rotations as well.

What is the rotation matrix/algorithm I can use to rotate points about the origin where $[0,1]$ is the identity instead of $[1,0]$?

With $[0,1]$ as the identity, $0$ degrees starts at the top of the unit circle and increases clockwise.

In summary, I'm trying to find a matrix of the form $[[a,b],[c,d]]$, where each element is $\sin$ or $cos$ of a single scalar value theta and rotating by an angle where $\cos{\theta} = 0$ and $\sin{\theta}= 1$, the point stays the same.

Best Answer

I'm not sure how to summarize the chat linked in the comments, but here are the key takeaways:

The sine and cosine functions are defined in such a way that $\sin(0)=0$ and $\cos(0)=1$. This is important -- more important than the usual association of $\cos$ with $x$ and $\sin$ with $y$.

To rotate a column vector $(u,v)$ or $(u\;v)^T$ clockwise by $\theta$, multiply by $\pmatrix{\cos\theta&\sin\theta\\-\sin\theta&\cos\theta}$. This is the same as rotating counterclockwise by $-\theta$, since $\sin(-\theta)=-\sin(\theta)$. This works for any vector, including $(0,1)$ and $(1,0)$.

If a rotation has been stored (for performance reasons) as the effect of such a matrix upon the vector $(0,1)$, then it is equal to $(\sin\theta\; \cos\theta)^T$. This is similar to the usual convention in pure math, except that the order of the sine and cosine are reversed. The reversal arises because when we flip "counterclockwise from $+x$" across the $x=y$ line, we get "clockwise from $+y$".

Given the above, from $(\sin\theta\; \cos\theta)^T$ we can reconstitute the full matrix $\pmatrix{\cos\theta&\sin\theta\\-\sin\theta&\cos\theta}$ on-demand in order to rotate other vectors by the same amount.

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[Math] A 4×4 homogeneous matrix for a 90 degree rotation about Y axis

To express ordinary $\mathbb{R}^n \to \mathbb{R}^n$ linear transformation into homogeneous coordinates just add another row and column where every term is equal to $0$ but the diagonal, which should be $1$.

For example if $A$ is your transformation matrix, then the new matrix would be $$A_H = \left[\begin{matrix}A& 0 \\\ 0& 1 \end{matrix}\right].$$ In your example $$A = \left[\begin{matrix}1& 0 & 0 \\\ 0& \cos\theta &-\sin\theta\\\ 0& \sin\theta &\cos\theta \end{matrix}\right],$$ so $$A_H = \left[\begin{matrix}1& 0 & 0 & 0 \\\ 0& \cos\theta &-\sin\theta & 0 \\\ 0& \sin\theta &\cos\theta & 0 \\\ 0 & 0 & 0 & 1 \end{matrix}\right].$$

Hope that helps ;-)

[Math] Java Tetris – Using rotation matrix math to rotate piece

The operations you are wanting to perform can be easily implemented without the need of rotation matrices since the transforms you desire are limited to right angle rotations and mirroring.

Assuming $B$ is a two dimensional array representing a block and $B[i,j]$ is the $i^{\text{th}}$ row and $j^{\text{th}}$ column of the block taking on values of 1 for a single block segment and 0 for an empty space.

The first operation you need is a transpose which is effectively a 90 degree clockwise rotation:

B T(B b)
    B b' <- new B[b.columns, b.rows]

    for i = 0 to b.rows - 1
        for j = 0 to b.columns -1
            b'[j,i] <- b[i, j]

    return b'

The second operation is mirroring a block about the y-axis:

B Y(B b)
    B b' <- new B[b.rows, B.columns]

    for i = 0 to b.rows - 1
        for j = 0 to b.columns - 1
            b'[i,j] <- b[i, b.columns - 1 - j]

    return b'

And finally, mirroring a block about the x-axis:

B X(B b)
    B b' <- new B[b.rows, B.columns]

    for i = 0 to b.rows - 1
        for j = 0 to b.columns - 1
            b'[i,j] <- b[b.rows - 1 - i, j]

    return b'

Given a block $B$ in a neutral position, a 90 degree clockwise rotation is $T(B)$, 180 is $T(X(T(B)))$, 270 is $Y(T(B))$ and 360 is simply $B$. counter clockwise rotations are just 360 minus the complementing clockwise rotations.

Also, as @StevenStadnicki pointed out in the comments, you can hard code much of this and you will find that if you are replicating the original Tetris, you'll encounter more that requires a hard coded solution than not.

Best Answer

Related Solutions

[Math] A 4×4 homogeneous matrix for a 90 degree rotation about Y axis

[Math] Java Tetris – Using rotation matrix math to rotate piece

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