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I need to find equation of the curve as shown below, for which, I need to find equation for upper part. lower part is half circle. upper part is a constant distance from circle with line passing through a point which is eccentric to the circle center as shown in the image. updated picture with information

edit 1: I have updated picture with more informative drawings, in current picture, I need to figure out equation of the Green curve.

new picture, with names

The line $L$ has constant length, and passes through a fixed point $E$. It's lower end (point $P$) moves along the lower semi-circle. It's upper end (the point $Q$) traces out some curve. We need to find the equation of this curve.

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  • $\begingroup$ There's honestly infinitely many different curves that could be given that would match your curve, if you supply a set of points on the perimeter i'll attempt to give the most basic example, do you have an idea of some additional properties you want? $\endgroup$ Aug 13, 2014 at 6:45
  • $\begingroup$ property is, all the points ( i.e. curve ) above that point( from which all the lines are passing through )should be equidistance from the circle point on the other end. $\endgroup$
    – iamgopal
    Aug 13, 2014 at 7:30
  • $\begingroup$ How about giving names to some things, for clarity. Let's call the "eccentric" point $E$ (this is the point through which all your lines pass). Suppose I draw some line $L$ through $E$. It meets the lower semi-circle at some point $P$. The upper end of the line $L$ is a point I'll call $Q$. As the line $L$ rotates, the point $Q$ traces out the curve you're interested in. The question is: how is $Q$ defined? Something is equidistant from two other things, or some things have equal lengths, or something ??? Please clarify. $\endgroup$
    – bubba
    Aug 13, 2014 at 11:14
  • $\begingroup$ Are you saying that, as it rotates, the line $L$ stays constant length? $\endgroup$
    – bubba
    Aug 13, 2014 at 11:30
  • $\begingroup$ @bubba yes, L ( PQ ) stays constant and with one end butting to the circle, other end trajects the curve, of which we need to find the equation, I have updated more drawings for more clarification $\endgroup$
    – iamgopal
    Aug 14, 2014 at 8:58

2 Answers 2

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Let's work with a unit circle, centered at the origin, and suppose the "eccentric" point $E$ is at $(0,a)$, where $a<1$. Let $c$ be the length of the line $L = PQ$, where $c > 1+a$.

A little vector reasoning shows that $$ Q = P + \frac{c}{ \| E - P \| }(E - P) $$ If the point $P$ is on the circle, then it can be described by coordinates $P = (\cos\theta, \sin\theta)$. If we substitute $P = (\cos\theta, \sin\theta)$ and $E = (0,a)$ into the equation above, and simplify, the $xy$ coordinates of $Q$ are given by $$ x = \cos\theta - \frac{c \cos\theta}{\sqrt{1 -2a\sin\theta + a^2}} $$ $$ y = \sin\theta + \frac{c(a - \sin\theta)}{\sqrt{1 -2a\sin\theta + a^2}} $$ Here is a plot for the case $a=1/3$, $c=3/2$, for $\tfrac34\pi \le \theta \le \tfrac94\pi$: enter image description here

Another one, this time showing the lines. Again with $a=1/3$, $c=3/2$, but now just the portion $\pi \le \theta \le 2\pi$:

enter image description here

The interesting thing is that the curve actually looks nothing like the pictures that you and I drew.

The situation is fairly simple, so this sort of mechanism (and the curve it traces out) might be well-known in the kinematics field, but I couldn't find any references.

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  • $\begingroup$ absolutely fantastic. :D, thanks for the answer, i think eccentricity at 1/3 , we get the curve like we drawn. $\endgroup$
    – iamgopal
    Aug 16, 2014 at 7:56
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A very simple equation in polar is coordinates is obtained if the center of the system of axis is not located at the center of circle, but at the "exentric" point :

enter image description here

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