Finneas the Undulator

Hugo Guckert · Tay Han · Henry Heathwood · Will Sedo

Introduction and Project Background

Cam-mechanism undulating fin robot diagram.
Source: mdpi.com/2077-1312/10/9/1327

Requirements and Specifications

Qualitative

Quantitative

Design Features of Main Subsystems

Significant Design Decisions

Evaluation of Design

Analysis

CAM MATLAB

An eccentric cam was chosen as the cam shape to produce a sinusoidal follower motion. The total amplitude of the follower is given by two times the shaft hole offset. The sinusoidal motion of the followers is translated directly to the motion of the fin rods which can be seen in the sinusoidal fin motion. The cams are press fit onto a ⅜″ D-shaft to restrict both axial motion and rotational motion.

MATLAB plot of follower position profile.
Cam profile dimensioned drawing.

Fastener

Fastener free-body diagram.
Fastener shear-failure calculations.

CAM Torque analysis with motor

(equations only for now)

Inputs
SymbolValue
$R$1.25 in
$e$0.40 in
$s = 2e$0.80 in
$n$60 RPM
$\omega = 2\pi n/60$6.28 rad/s
$N_\text{cam}$8
$\Delta\phi$90°
$m$0.084 lbm (38 g)
$k$20 lbf/in (3.5 N/mm)
$x_0$0.10 in
$\mu$0.25
$g$386.09 in/s²
Kinematics
$$x(\theta) = e(1-\cos\theta)$$
$$v(\theta) = \dot{x} = e\,\omega\sin\theta$$
$$a(\theta) = \ddot{x} = e\,\omega^{2}\cos\theta$$
$$x_{\max} = 2e = 0.80\ \text{in}$$
$$v_{\max} = e\omega = 2.51\ \text{in/s}$$
$$a_{\max} = e\omega^{2} = 15.79\ \text{in/s}^{2}$$
$$\tan\phi(\theta) = \frac{e\sin\theta}{\sqrt{R^{2}-e^{2}\sin^{2}\theta}}$$
$$\tan\phi_{\max} = \frac{e}{\sqrt{R^{2}-e^{2}}} = \frac{0.40}{1.184} = 0.338$$
$$\phi_{\max} = 18.7^{\circ}$$
Force analysis
$$N\cos\phi + mg - F_{s} = m\,a, \qquad F_{s} = k(x_{0}+x)$$
$$N\cos\phi(\theta) = m\,e\,\omega^{2}\cos\theta + k\bigl(x_{0}+e(1-\cos\theta)\bigr) - mg$$
$$N\cos\phi(0) = m\,e\,\omega^{2} + k\,x_{0} - mg \;\ge\; 0$$
$$k\,x_{0} \;\ge\; mg - m\,e\,\omega^{2}$$
$$x_{0,\min} = \frac{mg - m\,e\,\omega^{2}}{k} = \frac{0.084 - 0.0034}{20} = 0.0040\ \text{in}$$
$$N_{\max} = k(x_{0}+2e) - mg - m\,e\,\omega^{2} = 20(0.10+0.80) - 0.084 - 0.003 = 17.9\ \text{lbf}\;(79.6\ \text{N})$$
Drive torque
$$T\,\omega = (N\cos\phi)\,\dot{x}$$
$$T_\text{cam}(\theta) = \bigl[m\,e\,\omega^{2}\cos\theta + k(x_{0}+e(1-\cos\theta)) - mg\bigr]\,e\sin\theta$$
$$T_\text{fric}(\theta) = \mu\,N(\theta)\,(R-e\cos\theta)$$
$$T_\text{cam,peak} \approx 4.8\ \text{lbf}\cdot\text{in}\quad (\theta \approx 120^{\circ})$$
$$\bar{N} \approx \frac{N_\text{min}+N_\text{max}}{2} = \frac{1.9 + 17.9}{2} \approx 10\ \text{lbf}$$
$$\bar{T}_\text{fric} \approx \mu\,\bar{N}\,R = 0.25 \times 10 \times 1.25 \approx 3.1\ \text{lbf}\cdot\text{in}$$
$$T_\text{shaft} \approx 8\,\bar{T}_\text{fric} \approx 25\ \text{lbf}\cdot\text{in}\;(2.8\ \text{N}\cdot\text{m})$$
$$P = T_\text{shaft}\,\omega \approx 18\ \text{W}$$

Future Work

Conclusions