Subsections

• Formulation
• Performance

4 Shape Optimization of a Cam

Maximize the area of the valve opening for one rotation of a convex cam with constraints on the curvature and on the radius of the cam.

Formulation

The formulation of this problem is due to Anitescu and Serban [1]. We assume that the shape of the cam is circular over an angle of $\frac{6}{5}\pi$ of its circumference, with radius $r_{\min}$. The design variables $r_i$, $i = 1,\ldots, n$ , represent the radius of the cam at equally spaced angles distributed over an angle of $\frac{2}{5}\pi$. We maximize the area of the valve opening by maximizing

$$\ f(r) = \pi r_v^2 \left ( \frac {1}{n} \sum_{i=1}^{n}r_i \right )$$

subject to the constraints on $r$. The design parameter $r_v$ is related to the geometry of the valve. We also require that $r_{\min} \le r_i \le r_{\max}$. The requirement that the cam be convex is expressed by requiring that

$$\mbox{area} ( r_{i-1}, r_{i+1} ) \le \mbox{area} ( r_{i-1} , r_i ) + \mbox{area} ( r_i, r_{i+1} ) ,$$

where $\mbox{area} (r_i , r_j )$ is the area of the triangle defined by the origin and the points $r_i$ and $r_j$ on the cam surface. This convexity constraint can also be expressed as

$$2 r_{i-1}r_{i+1}\cos(\theta) \leq r_{i} ( r_{i-1} + r_{i+1} ) , \qquad i=0,\ldots,n+1 ,$$

where $r_{-1}=r_{0}= r_{\min}$, $r_{n+1}=r_{\max}$, $r_{n+2}= r_n$ and $\theta = 2\pi / 5(n+1)$. The curvature requirement is expressed by

$$- \alpha \le \left(\frac{r_{i+1} - r_i}{\theta}\right) \leq \alpha , \qquad i = 0,\ldots, n.$$

This is a departure from [1], where the curvature constraint was expressed in terms of $( r_{i+1} - r_i )^2$. Data for this problem appears in Table 4.1.

Table 4.1: Optimal design of a cam problem data

Variables	$n$
Constraints	$2n+2$
Bounds	$n$
Linear equality constraints	0
Linear inequality constraints	n + 1
Nonlinear equality constraints	0
Nonlinear inequality constraints	$n + 1$
Nonzeros in $\nabla ^2 f(x)$	0
Nonzeros in $c'(x)$	$5n$

We follow [1] and use $r_{\min}= 1.0$ and $r_{\max} = 2.0$ for the bounds on $r$, $r_v = 1.0$ in the area of the valve, and $\alpha= 1.5$ in the curvature constraint. Since the optimal cam shape is symmetric, we consider only half of the design angle. The problem was originally [1] formulated for the full angle of $\frac{4}{5}\pi$.

Performance

Results for the AMPL implementation are summarized in Table 4.2. We use a starting guess of $r_i \equiv ( r_{\min} + r_{\max} )/2$. The cam shape for $\alpha = 1.5$ appears in Figure 4.1. We note that the number of active constraints increases with $\alpha$ up to a threshold of $\alpha_1\approx 3.0$, after which increasing $\alpha$ does not change the optimal solution.

Table 4.2: Performance on optimal cam shape problem

Solver	$n=100$	$n=200$	$n=400$	$n=800$
LANCELOT	44.02 s	188.24 s	939.51 s	1877.47 s
$f$	4.30178e+00$\dagger$	4.35538e+00$\dagger$	4.45009e+00$\dagger$	4.85693e+00$\dagger$
$c$ violation	4.50980e-06$\dagger$	5.14160e-06$\dagger$	3.22620e-06$\dagger$	4.50740e-06$\dagger$
iterations	330	482	765	797
LOQO	0.8 s	2.37 s	3.82 s	15.49 s
$f$	4.28414e+00	4.27850e+00	4.27568e+00	4.27427e+00
$c$ violation	2.0e-12	2.7e-13	1.5e-12	4.8e-13
iterations	78	129	91	161
MINOS	0.83 s	1.3 s	4.46 s	21.41 s
$f$	4.28414e+00	4.27850e+00	4.27567e+00$\dagger$	4.27426e+00
$c$ violation	4.4e-16	1.4e-14	9.3e-14$\dagger$	3.1e-13
iterations	9	7	13	16
SNOPT	0.58 s	1.8 s	6.49 s	25.12 s
$f$	4.28414e+00	4.27340e+00	4.27022e+00	4.23739e+00
$c$ violation	1.3e-15	1.8e-07	2.3e-06	6.2e-07
iterations	6	4	4	5
$\dagger$ Errors or warnings. $\ddagger$ Timed out.

LANCELOT stops prematurely with the message step got too small for $n = 100,\ 200,\ 400$; and its solution for $n = 800$, while showing the best value, violates the problem constraints to an extent obvious in a graph of the solution. MINOS quits for $n = 400$ because the current point cannot be improved.

Figure 4.1: Cam shape for $\alpha=1.5$.