I sought for a way to plot pendulum direction fields and found this superb post:
It's been said that this direction field of the Differential equation: $\ddot{\varphi} = \frac{g}{R}\,\sin(\varphi)$ can be interpreted 2 ways:
one via $V = [\omega,-\frac{g}{R}\,\sin(\varphi)]$ $\quad$ ($\dot{\varphi} = \omega$)
the other by $\dfrac{\mathrm{d \omega}}{\mathrm{d\varphi}} = -\frac{g}{R}\,\dfrac{\sin(\varphi)}{\omega}$. I took that to make it $V = [1, -\frac{g}{R}\,\dfrac{\sin(\varphi)}{\omega}]$ because I've been taught the direction field of a first order ODE $y'(x) = f(x,y)$ can be given by $V = [1, f(x,y)]$
However the plots are not matching:
Why is that? The last one seems not right.