# To find that shape and length

A circle of radius 'r' is centered at origin, consider an imaginary line y=r (tangent to the circle at 0,r say A)

Now what kind of shape should an arc (not necessarily circular) have that if initially it's one end is at A and being tangential to circle at A, then on rolling the arc over the circle (just like tyre rolls without sliding over road) the 'A' end of the arc always traces/'have locus on' y=r

• Are you referring to a cycloid? – mrnovice Feb 24 '17 at 16:40
• How can it be said, let's visualise it. If a smaller circle is inside and is touching a larger curve (like bigger circle for e.g.only) and the latter rolls over it such that the smaller circle is unmoved while the point of latter which was usually touching the smaller circle WILL TRACE some curve. If that curve is along a line then what shape is that bigger curve. Cycloid should come in picture in relative motion in circle and line, how is this case arising here!! – Anshuman Dwivedi Feb 24 '17 at 17:47
• I am not sure what you mean by the "end" of the arc. Clearly as you roll down the circle the arc gets longer and longer. I am thinking some sort of spiral could remain always tangent to y = r. – victoria Feb 24 '17 at 17:48
• @Victoria. Refer en.m.wikipedia.org/wiki/File:EpitrochoidOn3-generation.gif here if outer rolling circle would have been some other (could be non closed also) arc/shape but if on rolling, it traces a 'red' line like x=3 in that diagram. Then what is that curve? I wish I am understood! – Anshuman Dwivedi Feb 24 '17 at 17:58

Your question is not very clear at all.

In the differential geometry of curves, a roulette is a kind of curve that is described by a point (called the generator or pole) attached to a given curve and that curve rolls without slipping, along a second given curve that is fixed.

We have :

As you are working with "a circle of radius 'r' is centered at origin", we have to deal with epicycloids, hypocycloids or involutes.

Because of your english, I suspect you are referring to the involute.

• Refer s7.postimg.org/4zpmr02az/Curve_Sliding_over_circle.jpg here the shape to be found is the red arc which although shown as a line segment but should be some different curve/'shape', but the condition to be satisfied is that one end of the arc (here 'A') should remain on y=r while the curve rolls over circle, like here A, A', A'' all remain on y=r – Anshuman Dwivedi Feb 25 '17 at 2:20
• Please, correct me if I am wrong, in order to understand your problem. So we have a segment AB on line $y=r$. This segment rolls and translates in such a way that extreme A is always on line $y=r$, obtaining A', A'', ..... You want the curve made by B, B", B"", .... Is this correct? – cgiovanardi Feb 25 '17 at 2:38
• Not the curve made by B but the AB curve itself such that this condition fulfills. – Anshuman Dwivedi Feb 25 '17 at 2:54
• OK, I understand now. I will try to make something during the weekend. By the way, I think there are some meanings to make clear: en.wikipedia.org/wiki/Line_segment and en.wikipedia.org/wiki/Curve. Besides, with the motion described, point B traces a curve and segment AB defines an area. – cgiovanardi Feb 25 '17 at 4:12

After the comments mantained with Anshuman Dwivedi, I am posting a new answer.

**

I am solving the following problem:

A circle of radius $r$ is centered at origin.

Consider an imaginary line $y=r$, tangent to the circle at A = (0,r), and a segment AB of length $l$ on this line.

What shape will this segment describe if it moves rolling around the circumference and maintaining its point A on line $y=r$.

For any position of the segment around the circumference, we have:

We advance point A and rotate the segment till it is tangent to the circumference.

Coordinates of tangential point are calculated as follow

$Alfa= 2*atan((x-coordinate-of-point-A) / r)$

$Tx = r * cos( \pi/2- Alfa)$

$Ty = r * sin( \pi/2- Alfa)$

Now, new position of point B will be

$Bx = ( x-coordinate-of-point-A ) + l * cos(Alfa)$

$By = r - l*sin(Alfa)$

Following is the curve obtained

http://imgur.com/a/uf4sW