1
$\begingroup$

I am trying to find the equation of a parabola that is tangent to an exterior circle (this is for designing a bell-shaped nozzle). I know the point (0.055, 0.9) lies on the parabola and also that the circle of radius 0.1625 is centered at (0.1875, 0.25). But I am having trouble getting the point of tangency and then finding the parameters for the parabola.

I tried to match derivates of the functions, but ran into some algebra problems. I started with a general parabola ($y=ax^2+bx+c$) and took the derivate ($y'=2ax+b$). Then I got the equation of a circle centered at (h,k) ( $(x-h)^2+(y-k)^2=r^2$ ) and took the derivate through Wolfram Alpha to get $y'=-\frac{x-h}{y-k}$ but then I got stuck trying to equate the two and find the point of intersection.

Please let me know what I could to try and solve this. Thank you!

enter image description here

$\endgroup$
  • $\begingroup$ Is the parabola symmetric about the $y$-axis? $\endgroup$ – Blue Feb 5 at 4:20
1
$\begingroup$

As noted in @Andrei's answer, the given information is insufficient. @JeanMarie has a detailed answer, so I may not need to post on my own, but I wanted to show that the curviness of the result won't be quite so dramatic as indicated in OP's sketch.


Below are situations where the parabola meets the circle at a point corresponding to an auxiliary angle relative to the circle's center.

At about $144^\circ$, the parabola is effectively a double-ray pointing straight up; at about $177^\circ$, the parabola degenerates to a straight line tangent to the circle. (Beyond $177^\circ$, the parabola opens downward.)

enter image description here enter image description here

At about $175.75^\circ$, the parabola is symmetric about the $y$-axis :

enter image description here

Here are a couple more cases:

enter image description hereenter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ My result $3.0684*180/\pi=175.81°$ coincide perfectly with yours ! $\endgroup$ – Jean Marie Feb 7 at 1:21
1
$\begingroup$

Making the hypothese that your parabola has equation :

$$y=ax^2+c\tag{1}$$

Taking into account that

$$0.9=a(0.055)^2+c\tag{2}$$

by difference between (1) and (2), we can eliminate $c$, and obtain the following equation for the parabola :

$$y-0.9=a(x^2-0.055^2)\tag{3}$$

Now consider a parametric representation of the current point of the circle :

$$M_{\alpha} = \binom{x(\alpha)=0.1875+0.1625 \cos \alpha}{y(\alpha)=0.25+0.1625 \sin \alpha}\tag{4}$$

The tangent vector at point $M_{\alpha}$ is obtained by differentiation of (4) :

$$T = \binom{x'(\alpha)=-0.1625 \sin \alpha}{y'(\alpha)=0.1625 \cos \alpha}\tag{5}$$

Let us consider that the contact point $M_0$ corresponds to the value $\alpha_0$ of the parameter, i.e.,

$$M_0=(x(\alpha_0),y(\alpha_0))\tag{6}$$

The tangent vector to the parabola at $M_0$ is

$$U = \binom{1}{2a x(\alpha_0)}\tag{7}$$

Let us express that the parabola passes through $M_0$ ; using (3), we have :

$$y(\alpha_0)-0.9=a(x(\alpha_0)^2-0.055^2)\tag{8}$$

from which we can extract

$$a=\dfrac{y(\alpha_0)-0.9}{x(\alpha_0)^2-0.055^2}\tag{9}$$

Besides, expressing that tangent vectors $T$ and $U$ are proportional is equivalent to express the equality of slopes :

$$-\dfrac{\cos \alpha_0}{\sin \alpha_0}=\dfrac{2 a x(\alpha_0)}{1}\tag{10}$$

which can be simplified into

$$a=\dfrac{\text{cotan} \ \alpha_0}{-2 x(\alpha_0)}\tag{11}$$

Relating the two expressions of $a$ in (9) and (11), using relationships (4) end up into the following trigonometric equation :

$$\text{cotan} \ \alpha_0=-2\dfrac{(0.25+0.1625 \sin \alpha_0)-0.9}{(0.1875+0.1625 \cos \alpha_0)^2-0.055^2} (0.1875+0.1625 \cos \alpha_0)\tag{12}$$

that can be solved in different ways.

I have in $(\pi/2,\pi)$ found the unique solution :

$\alpha_0=3.068462$ radians = 175.81 (decimal) degrees

giving in particular

$M_0 (0.0254, 0.2619)$

$a = 268.3$ which is incredibly straight... (almost a vertical line)

Conclusion : as my result has been checked by @Blue, I think that, by modifying the numerical values, but keeping the same final equation (12), you could obtain the nozzle shape with a satisfying curvature.

Remark : considering parabolas with equation $y=ax^2+bx+c$, we have too many parameters...

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Your solution of $3.068\ldots$ radians agrees with the approximation of $175.75^\circ$ for the symmetric-about-$y$-axis case in my illustrations. $\endgroup$ – Blue Feb 7 at 1:19
0
$\begingroup$

Unfortunately you do not have enough information. Let's simplify the problem, and suppose that the parabola is symmetric around the vertical axis. Then the formula for it is $$y=ax^2+c$$ You still have the following unknowns $a$, $c$, $x_i$, $y_i$. The last two are coordinates of the intersection point. But you have only two equations:

  • The two curves intersect
  • The two curves are tangent

The solution to your problem is unique only if you fix two of those variables. You can fix for example the intersection point. Or you can fix the shape of the parabola.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't agree with the fact that we haven't enough information (see my solution) $\endgroup$ – Jean Marie Feb 7 at 1:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.