Path Simulation Using Computational Methods

About this project

This is an effort in modeling the path of a trajectory using multiple methods including

• Epicycles and deferents (complex Fourier series) (In this code repository)
• Forward kinematics transformation matrix (matrix manipulation, commonly used in robot manipulators) (refer to this repository)
• Closed loop 4 bar linkage system (for future attempts)

Modeling path with Fourier series

Fouries series in continuous time domain

Base form $$f(t)=a_0+\sum_{n=1}^{\infty}a_ncos(\frac{2\pi nt}{T})+\sum_{n=1}^{\infty}b_nsin(\frac{2\pi nt}{T})$$

where:

• $$a_0=\frac{1}{T}\int_{0}^{T}f(t)dt$$
• $$a_n=\frac{2}{T}\int_{0}^{T}f(t)cos(\frac{2\pi nt}{T})dt$$
• $$b_n=\frac{2}{T}\int_{0}^{T}f(t)sin(\frac{2\pi nt}{T})dt$$

Compact form $$f(t)=c_0+\sum_{n=1}^{\infty}c_ncos(\frac{2\pi nt}{T}+\phi)$$

where:

• $$c_0=a_0=\frac{1}{T}\int_{0}^{T}f(t)dt$$
• $$c_n=\sqrt{a_n^2+b_n^2}$$

Exponential form $$f(t)=\sum_{n=-\infty}{\infty}e_0e^{-in\omega_0t}$$

where:

$$c_n=\begin{cases} \frac{a_n-ib_n}{2}, n < 0 \\ a_0, n = 0 \\ \frac{a_n+ib_n}{2}, n > 0 \\ \end{cases}$$

Discrete Fourier transform in time domain

$$X[k] = \sum_{n=0}^{N-1} x[n] \cdot e^{-j \frac{2\pi}{N}kn}$$

Amplitude: $A_k = \frac{2}{N} |X[k]|$

Phase: $\phi_k = \arg(X[k])$

Angular frequency: $\omega_k = \frac{2\pi k}{T}$

Modeling path in close loop with 4 bar linkage system

Assuming a 4-side geometry that consist of 4 bar of different dimensions and they form 4 angles so that the total inner angles is 360 degree.

4 bar linkage problems often come in 2 types:

• close loop
• open loop

In a close loop system, there are a unique constraint of the sum of 4 inner angles must add up to 360 degrees

Formula used to compute the location of this system (also applied to open system):

• $\frac{BD}{sinBCD}=\frac{BC}{sinBDC}=\frac{CD}{sinCBD}$
• $\frac{BA}{sinBDA}=\frac{BD}{sinBAD}=\frac{AD}{sinABD}$
• $\frac{AB}{sinACB}=\frac{BC}{sinBAC}=\frac{AC}{sinABC}$
• $\frac{AD}{sinACD}=\frac{AC}{sinADC}=\frac{CD}{sinCAD}$
• $BD=\sqrt{AB^2+AD^2-2\times AB\times ADcosBAD}=\sqrt{BC^2+CD^2-2\times BC\times CDcosBCD}$
• $AC=\sqrt{AB^2+BC^2-2\times AB\times BCcosABC}=\sqrt{AD^2+CD^2-2\times AD\times CDcosADC}$

The known parameters are the previous positions of each linkage and their angles with one known angle change that drive the rest of the parameters.