---
title: Solutions for 3D Modelling
---

æ Stage Turntable

+ [Stage Turntable](stageturntable.pdf) (PDF)

# The Crane

+ [Step 1 and Figure](crane.pdf)

## Step 2

The co-ordinate system of the joint has not been completely specified
in the problem text, and some choices are free.  
It is probably simplest if we align the bases when the angles are zero, 
i.e. that the boom moves in the $xz$-plane of the basis of the joint.
This also calls for the boom being parallel to the $z$-axis of the hand system.

With these assumptions the position of $A$ in the joint basis is
$$ R_y(\beta)\cdot(A^T+(0,0,b))$$
where $R_y(\beta)$ is a rotation by $\beta$ around the $y$-axis, i.e.
$$
R_y(\beta) =
\begin{bmatrix}
  \cos\beta & 0 & \sin\beta \\
  0 & 1 & 0 \\
  -\sin\beta & 0 & \cos\beta 
\end{bmatrix}
$$

## Step 3

Similarly to above, a point $C$ in the joint basis becomes $C'$ in
the base basis by the transformation
$$ R_z(\alpha)\cdot(C^T+(0,0,a))$$
where $R_z(\alpha)$ is a rotation by $\alpha$ around the $z$-axis, i.e.
$$
R_z(\alpha) =
\begin{bmatrix}
  \cos\alpha  & \sin\alpha & 0\\
  -\sin\alpha  & \cos\alpha & 0 \\
  0 & 0 & 1
\end{bmatrix}
$$

To find the global co-ordinates of $A$, we combine the two transformations
and write
$$ A = R_z(\alpha)\cdot( R_y(\beta)\cdot(A^T+(0,0,b)) + (0,0,a)) )$$

## Exercise 2.1 (Ma 2004)

**Definition** of a linear function

$$f(C\cdot X) = C\cdot f(X)$$

$$f(X+Y) = f(X) + f(Y)$$

If we take a function $f(X) = X\cdot A\cdot X$ we can test if this is true, e.g.

$$f(X+Y) = (X+Y)A(X+Y) = (XA+YA)(X+Y) = XAX + XAY + YAX + YAY = f(X) + XAY + YAX + f(Y)$$

In this case we get two terms which we should not have had if the function were linear, and they are not zero in the general case.  Hence this function is not linear.  Similar calculations can be made for the function in the exercise.