# Which curve has the smoothest curvature : Clothoid or Hypocycloid? Benefits of Hypocycloid in general for making smooth path between two points.

I am investigating hypocycloids for making smooth paths. I would be thankful if you could share some insights on the following two questions:

Ques 1) Whose curvature is smoother: Clothoid or Hypocycloid (say we limit the discussion only to astroid and deltoid, or in general if possible).

Ques 2) What could be the potential advantages of using Hypocycloid over Clothoid for making smooth paths?

Smoothness [source: wiki]: Smoothness of a function is a property measured by the number of derivatives it has which are continuous. A smooth function is a function that has derivatives of all orders everywhere in its domain. In terms of robot, I am particularly interested in 1st derivative (velocity), and 2nd derivate (acceleration) only. In this regard, a straight line is not smooth as robot has to very abruptly turn at points A and B. I am interested in which curve has more curvature, which curve keeps more clearance from the obstacle P in the [edited] figure attached.

Regarding 1) I found the parametric equations for curvatures of hypocyloid and clothoid from Hypocloid curvature and for clothoid too. But I am unable to interpret the general case. Regarding 2) I found that parametric equations of hypocycloid are easier to compute than clothoids which involves Fresnel Integrals.

Concretely, I am trying to fix a hypocycloid or a clothoid between two points A and B in the figure attached. OA = OB. I would like to know mathematically about whether a hypocycloid (say astroid) or a clothoid curve would be smoother?

Curvature Equations Curvature of a general hypocycloid is given by [from Wolfram Hypocycloid] where $a$ and $b$ are the radius of larger circle and rolling smaller circle: $c(\phi) = \dfrac{2b - a}{4b(a-b)} cosec \biggl( \dfrac{a\phi}{2b} \biggr)$

For a specific astroid case, curvature is: $c(\phi) = \dfrac{-2}{3} cosec (2t)$

Curvature of a clothoid = $-t$ (equal to the parameter)

Hence, I cannot generalize which is greater. For astroid, at $t=0$, the curvature is $\infty$ (which is equal to that of a straight line). Then it gradually decreases. Hence the robot should experience no (or less) jerk. For clothoid, since its curvature is equal to the parameter $t$, it would degenerate quikcly. In these terms, I want to compare hypocycloid and clothoid. Any suggestions?

Fitting Hypocycloid(astroid) or clothoid between two points for robot

Thank you in advance.

• So, in addition to the curve between A and B being smooth, you want the curve to join smoothly with the lines at A and B. That makes a big difference. Commented Mar 5, 2016 at 11:18
• Yes, that is also a requirement. I have updated the question with curvature equations. Please have a look. Thank you. Commented Mar 7, 2016 at 10:03
• I believe that what you want is the clothoid (Cornu or Euler spiral). This was originally developed to provide a curve with constant acceleration, i.e., no jerk, for railroad track turns. It is still used today for that as well as highway design. See the Wikipedia page "Euler spiral:" en.wikipedia.org/wiki/Euler_spiral. Commented Apr 7, 2017 at 20:05
• @CyeWaldman Thanks. I have already solved the problem. You are right. Euler spiral is useful in my case. If you submit your comment as a proper answer, I will mark it as accepted. Commented Apr 9, 2017 at 3:02

The Euler spiral (also known as the Cornu spiral or clothoid) is the natural choice for a smooth transition curve, where by smooth we mean one that will not cause an abrupt change in acceleration (i.e., a jerk).

The Euler spiral is usually expressed in terms of the Fresnel integral, which can be expressed in closed form as follows,

$$z(s)=\int_0^s e^{i \pi s^2/2} ds=\frac{1+i}{2} erf\left(\frac{1-i}{2} \sqrt{\pi} \cdot s\right)$$

where $s$ is the arc length. Now it's also known that a spiral can be expressed in terms of it curvature, $\kappa$, namely,

$$z=\int e^{i\int \kappa(s)ds}ds$$

It follows that the natural equation of this spiral is $\kappa \propto s$, or in term of the radius of curvature, $\rho$

$$\rho s=\text{constant}$$

It is this property that we are interested in. When traversing a bend we seek to assure that the centrifugal force changes continuously in order to avoid a jerk. If this easement is not applied there would be an abrupt change in the acceleration at the point of transition from a straight path to a curved one.

You have to define what you mean by "smooth".

If you just want to join $A$ and $B$, you could use either a straight line or a quarter circle. Both of these are extremely smooth by any reasonable definition.

The quarter circle is a clothoid, though a very simple one.

Suppose you want the curve to join smoothly with the straight lines at $A$ and $B$, and be suitable for a robot path. One very simple solution is the cubic Bezier curve whose 4 control points (in order) are $A$, $O$, $O$, $B$. This curve matches the two lines in both tangent direction and curvature, so there should be no sudden "jerk" force as the robot passes through the points $A$ and $B$.

• Thanks Bubba. Sorry for not being specific and unable to ask question in mathematical jargon. I have edited the question and the figure. I am discussing in terms of robot traversing a path. In case of straight line, it has to abruptly turn at pts A and B. But not so in case of curve fitted between A and B. Also curve should avoid collision of robot from obstacle P in figure. Any idea in these terms is welcome. Please tell me if you need more specifications. Thanks in advance. Commented Mar 4, 2016 at 1:42
• Thanks Bubba. The robot should experience no jerk and Bezier curves can achieve it. But I am studying hypocycloids to achieve this task. I have edited the question, please have a look. From the curvature equations it seems that initially the curvature of astroid is infinite i.e. same as straight line, then gradually decreases, so robot should not experience a jerk or there should be less jerk (within bounds). I am wondering what benefits hypocycloid has over clothoid in this case. Any suggestion is highly welcome. Thank you in advance. Commented Mar 7, 2016 at 1:50