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title: Solutions for 3D Modelling
---
# Stage Turntable
+ Position in the local co-ordinate system $\mathbf{x}=(x,y,1)$
+ Translation from the local to the global system $T=(0,10,0)$.
+ Rotation
$$R_{-\alpha} = \begin{bmatrix}
\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha
\end{bmatrix}$$
+ Homogeneous transformation
$$H_{-\alpha} = \begin{bmatrix}
\cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 10 \\
0 & 0 & 1
\end{bmatrix}
$$
+ Global position of the actress
$$
H_{-\alpha}\mathbf{x} =
\begin{bmatrix}
x\cos\alpha + y\sin\alpha \\ -x\sin\alpha + y\cos\alpha + 10 \\ 1
\end{bmatrix}$$
+ Global position in heterogeneous co-ordinates
$$(x\cos\alpha + y\sin\alpha, -x\sin\alpha + y\cos\alpha + 10)^T$$
# The Crane
## Step 2
In homogeneous co-ordinates we had the transformation
$$ A^T\mapsto R_y(\beta)\cdot(A^T+(0,0,b)^T)$$
Because the translation is made before the rotation, we have
to rewrite them as homogeneous co-ordinates separately.
The translation is
$$
H_T =
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & b \\
0 & 0 & 0 & 1
\end{bmatrix}
$$
and the rotation is
$$
H_y(\beta) =
\begin{bmatrix}
\cos\beta & 0 & \sin\beta & 0 \\
0 & 1 & 0 & 0 \\
-\sin\beta & 0 & \cos\beta & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
$$
The actual homogeneous transformation is
$H(\beta)=H_z(\beta)H_T$, or
$$ \tilde A^T\mapsto H_y(\beta)\cdot H_T\cdot \tilde A^T$$
By completing the multiplication, we get
$$
H(\beta) =
\begin{bmatrix}
\cos\beta & 0 & \sin\beta & b\sin\beta \\
0 & 1 & 0 & 0 \\
-\sin\beta & 0 & \cos\beta & b\cos\beta \\
0 & 0 & 0 & 1
\end{bmatrix}
$$
## Step 2 Validation
Consider an arbitrary homogeneous point $(x,y,z,1)$ and
let's check if the heterogeneous and the homogeneous
transformation give the same result.
**Heterogeneous**
$$(x,y,z)\mapsto R_y(\beta)\cdot(x,y,z+b)^T
= \begin{bmatrix}
x\cos\beta + (z+b)\sin\beta \\
y \\
-x\sin\beta + (z+b)\cos\beta
\end{bmatrix}$$
**Homogeneous**
$$(x,y,z)\mapsto H_y(\beta)\cdot(x,y,z,1)^T
= \begin{bmatrix}
x\cos\beta + z\sin\beta + b\sin\beta\\
y \\
-x\sin\beta + z\cos\beta + b\cos\beta \\
1
\end{bmatrix}$$
## Step 3
Similarly to above, the transformation from the joint basis
to the base basis is
$$
H(\alpha) =
H_z(\alpha)H_{T'} =
\begin{bmatrix}
\cos\alpha & \sin\alpha & 0 & 0\\
-\sin\alpha & \cos\alpha & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & a \\
0 & 0 & 0 & 1
\end{bmatrix}
=
\begin{bmatrix}
\cos\alpha & \sin\alpha & 0 & 0 \\
-\sin\alpha & \cos\alpha &0 & 0 \\
0 & 0 & 1 & a \\
0 & 0 & 0 & 1
\end{bmatrix}
$$
Step 3 can be validated in the same way as Step 2.