Normal distance Recently, I saw a paper "G. H. Darwin, Periodic Orbits" and I don't undertand the concept of "normal displacement $\delta$p". 
"...
Now suppose that x, y are the coordinates of a point on an orbit,  and that x + $\delta$ x, y + $\delta$ y are the coordinates of a point on an adjacent orbit. Then if we put 
$\delta$p = $\delta$x cos $\phi$ + $\delta$y sin $\phi$
$\delta$s = -$\delta$x sin $\phi$ + $\delta$y cos $\phi$
$\delta$p, $\delta$s as are the distances, measured along the outward normal and along the arc of the unvaried orbit, from the original point x, y to the adjacent point x + $\delta$x, y +$\delta$y."
$\phi$ is a angle between outward normal and x-axis. Someone can help me to understand.
Appreciate. 
 A: I searched online to find a copy of the paper you're referring to at Periodic Orbits - 1485881980, with the particular section you're asking about being on page $36$. My understanding is that the "normal displacement $\delta p$" is basically the perpendicular change in distance between a periodic orbit $s$ and a

... path which varies by infinitely little from that of the periodic orbit.

as stated on page $35$ in that paper.
Although you didn't explicitly ask for clarification of the equations, you included a copy of them so I assume you may want to understand them. I'm also explaining them below so you can see how the various aspects interact. The following diagram illustrates what the author is talking about.

Here, the center is far off the lower-left corner of the diagram, the red line indicates the periodic orbit and the green line the infinitesimally perturbed orbit. $ABDE$ is a rectangle, with $EC$ being parallel to the $x$-axis. Also, $A(x,y)$ is a point on the periodic orbit and $D(x + \delta x, y + \delta y)$ is a point on the perturbed orbit. In addition, $AB$ and $ED$ is normal to the periodic orbit at $A$, with length $\delta p$, plus $AE$ and $BD$ are parallel to the periodic orbit arc at $A$, with length $-\delta s$. Note this assumes the orbit is counter-clockwise, so this is why the graph uses $-\delta s$ instead of $\delta s$.
Note that, as shown, $\angle DEC = \phi$, so $\angle CEA = \frac{\pi}{2} - \phi$. To make things somewhat simpler, horizontally and vertically translate $ABDE$ so $A$ is at the origin, i.e., $x = y = 0$. To convert from $(x,y)$ co-ordinates to $(p,-s)$ co-ordinates requires a counter-clockwise rotation of $\frac{\pi}{2} - \phi$, so $D(\delta x, \delta y)$ will then be mapped to $D'(-\delta s, \delta p)$. The formula for converting from $(x,y)$ by a counter-clockwise rotation of $\theta$ to $(x',y')$ is, as given in the Two Dimensions section of Wikipedia's article on Rotation:
\begin{align}
x'&=x\cos \theta -y\sin \theta \\
y'&=x\sin \theta +y\cos \theta \tag{1}\label{eq1}
\end{align}
For the case here, $\theta = \frac{\pi}{2} - \phi$, $x = \delta x$, $y = \delta y$, $x' = -\delta s$ and $y' = \delta p$. Thus, the first equation of \eqref{eq1} becomes
\begin{align}
-\delta s & = \delta x \cos\left(\frac{\pi}{2} - \phi\right) - \delta y \sin\left(\frac{\pi}{2} - \phi\right) \\
& = \delta x \sin\phi - \delta y \cos\phi \tag{2}\label{eq2}
\end{align}
Multiplying both sides by $-1$ gives
$$\delta s = - \delta x \sin\phi + \delta y \cos\phi \tag{3}\label{eq3}$$
The second equation of \eqref{eq1} becomes
\begin{align}
\delta p & = \delta x \sin\left(\frac{\pi}{2} - \phi\right) + \delta y \cos\left(\frac{\pi}{2} - \phi\right) \\
& = \delta x \cos\phi + \delta y \sin\phi \tag{4}\label{eq4}
\end{align}
As you can see, \eqref{eq3} and \eqref{eq4} match the $2$ equations you're asking about.
