1
$\begingroup$

I am trying to find the equation of a parabola given the following conditions.

Given:

  • circle with radius $R_1$ centered at $(0, R_1 + R_2)$
  • $R_2$ is given
    • $\frac{R_1}{R_2} = 0.382$
  • $R_3 = R_2 * \sqrt{16}$
  • $L$ is a given constant which is the length of the conic section from the vertex to points $D$ and $E$
  • parabola does not have a vertex at the origin
  • parabola is tangent to circle at some point to the right of the x-axis, '$A$'
  • parabola has points: $B(x_A, y_A)$, $C(x_A, -y_A)$, $D(x_D, R_3)$, $E(x_, -R_3)$
  • points B and C are symmetric wrt. $x$ axis
  • points D and E are symmetric wrt. $x$ axis

My approach:

Use the general equation of a horizontal conic section: $x = ay^2 + by + c$

By substituting in points B and C or D and E, I find that: $b = 0$. Therefore, $$ x = a y^2 + c $$

I want $\frac{dy}{dx}$, so through implicit differentiation I get: $$dx = 2a*ydy$$ $$\frac{dy}{dx} = \frac{1}{2ay}$$

Next, I attempted to express $x_n$ and $y_n$ in terms of a common unknown. $$x_n = R_1 * \sin{\theta_N}$$ $$y_n = R_2 + (R_1 - R_1 * \cos{\theta_N})$$

Substituting $y_n$ into the equation for $\frac{dy}{dx}$, $$\frac{dy}{dx} = \tan{\theta_N} = \frac{1}{2a*(R_2 + (R_1 - R_1\cos{\theta_N}))}$$

Substituting $y_n$ and $x_n$ into the equation of a parabola yields, $$R_1 * \sin{\theta_N} = a*(R_2 + (R_1 - R_1 * \cos{\theta_N}))^2 + c$$

Substituting $y_n$ into the equation for $\frac{dy}{dx}$, $$\frac{dy}{dx} = \tan{\theta_N} = \frac{1}{2a*(R_2 + (R_1 - R_1\cos{\theta_N}))}$$

Since $\frac{R_2}{R_1} = 0.382$, $R_2 = 2.6178R_1$ $$\frac{dy}{dx} = \tan{\theta_N} = \frac{1}{2a*(3.6178R_1 - R_1\cos{\theta_N})}$$ $$R_1 * \sin{\theta_N} = a*(3.6178R_1 - R_1 * \cos{\theta_N})^2 + c$$

At this point, I am not sure how to proceed.

Context:

This problem is used in finding the diverging section of a bell nozzle for rocket engines. I am attempting to find the equations of the plots on page 80 of Rocket Propulsion Elements 7th edition by Sutton and Biblarz.

A more general form of this problem would have left the $16$ in the equation $R_3 = R_2 * \sqrt{16}$ in variable form. This would have allowed the plots in RPE to be generated. $16$ in this context represents the area ratio between the exit plane and the throat.

  • 1
    Who is author of book? – 2017-01-04
  • 0
    Thanks for pointing that out. It has been corrected. – 2017-01-04
  • 0
    You're almost there. As you not, at the point $x_n$ and $y_n$, the slopes of the circle and parabola must be the same. But there's another equation you've missed: the circle and parabola must both contain the point $(x_n, y_n)$. That is to say $R_1 \sin \theta_N = a (R_1 - R_1 \cos \theta_N)^2$. This, together with the equation youv'e derived, should let you determine $\theta$. – 2017-01-04
  • 0
    Does it have symmetry about $x$ axis? – 2017-01-04
  • 0
    Also: when you write $dy/dx = \theta_N$, surely you meant to have $\tan$ or $\cot$ in there somewhere. – 2017-01-04
  • 0
    Center of circle is on $y$ axis but parabola is symmetric wrt. $x$ axis? Sounds like a mistake. Your $y_n=R_1-R_1\cos\theta_N$ sounds even more confusing. Shouldn't you add $R_2$ to that? Personally I dislike trigonometric functions and prefer using [the tangent half-angle formulas](https://en.wikipedia.org/wiki/Tangent_half-angle_formula) instead so I could stick with polynomials. – 2017-01-05
  • 0
    @MvG The circle is indeed on the $y$ axis with the parabola symmetric wrt. $x$ axis. You are correct. I should have added $R_2$ to $y_N = R_1-R_1\cos{\theta_N}$ – 2017-01-05
  • 0
    @JohnHughes I think the same mistake addressed above caught you.. The equation you point out, $R_1\sin{\theta_N} = a(R_1 - R_1\cos{\theta_N})^2$ should have a $+c$ term on the right side since we do not know if the vertex is located at the origin. – 2017-01-05
  • 0
    Ah....good point. I was assuming that the initial stuff, which was a little too garbled for me to make sense of, was actually correct. Sigh. – 2017-01-05

0 Answers 0