# Geometry: move of “x” centimeter on a curve

Definition: I have two connected line segments (in black on picture) defined by 3 points: P1 P2 and P2 P3.I subdivise first 5 centimeters of line segment (in grey on picture). I connect corresponding subdivisions (in blue on picture). If I subdivide again the first 5 centimeters at infinite, I will get a perfect curve (in red on picture) formed by connection lines.

Problem: I'm on point P1 and I want to move of "x" centimeter toward P3 passing by line segment P1 P2, then on red curve and finally on line segment P2 P3. How compute the exact new point coordinate of this move ?

Solution: I only have an approximate solution where I compute the blue lines intersection points (in orange on picture). Then, I compute length between orange points. Finally, I move from P1 to 4nd subdivision of line segment P1 P2, then to first orange point, then on second orange point, etc. until I reach the "x" centimeter.

Hints:

There are two subproblems: finding the equation of the curve and rectifying the curve (i.e. expressing the curvilinear abscissa).

Assume that $P_2P_3$ forms an angle $\alpha$ with the horizontal. For a segment of unit length, if one endpoint is at $(0,t)$, the other is at some $(u\cos\alpha,u\sin\alpha)$ and the distance is

$$u^2\cos^2\alpha+(u\sin\alpha-t)^2=u^2-2ut\sin\alpha+t^2=1.$$

From this we draw $u$:

$$u=t\sin\alpha\pm\sqrt{1-t^2\cos^2\alpha}.$$

Now the equation of the supporting line of the segment is

$$(u\sin\alpha-t)(x-0)-(u\cos\alpha-0)(y-t)=0.$$

The searched curve is the envelope of these lines, and to find it we need to eliminate $t$ (and $u$) between the equation and its derivative wrt $t$.

$$\begin{cases}(u\sin\alpha-t)x-u\cos\alpha(y-t)=0 \\(u'\sin\alpha-1)x-u'\cos\alpha\,(y-t)+u\cos\alpha=0\end{cases}$$

where $u'=\sin\alpha\mp\dfrac{t\cos^2\alpha}{\sqrt{1-t^2\cos^2\alpha}}$.

In the special case $\alpha=0$, the solution is known to be the astroid curve, of parametric equations

$$\begin{cases}x=\cos^3\theta,\\y=\sin^3\theta\end{cases}.$$

You will find several Web entries under "sliding ladder envelope" that address it.

The curvilinear abscissa can be found by integrating the element of arc,

$$ds=3\sqrt{\cos^4\theta\sin^2\theta+\sin^4\theta\cos^2\theta}\,d\theta=3\cos\theta\sin\theta\,d\theta$$

so that

$$s=\frac32\cos2\theta$$

and

$$\theta=\frac12\arccos\frac{2s}3.$$

I don't know if the general case $\alpha\ne0$ is tractable.