---
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.