Is this shape definition an ellipse?

Question

I want to define an ellipse-like shape with two radii. The radius at 0 degrees is 3. The radius at 90 degrees is 2. But in order to close the loop, we have to specify some smooth transition between them.

If the transition is linear (radius at 45 degrees is 2.5), does that make it an ellipse?

EDIT: "Linear" means the radius decreases linearly from 3 to 2 as the angle increases linearly from 0 to 90 degrees. So the equation is r = 2 + t/90 for quadrant 1.

"The angle" means the angle made by the positive x axis and whatever radius we're looking at, centered at the origin. The loop I'm describing is also centered at the origin.

Answer 1

A common way to define an ellipse centered at origin is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where $a$ is the horizontal semi-minor axis, and $b$ is the vertical semi-minor axis.

You often see this written parametrically as $$\begin{cases} x(t) = a \cos(t) \\ y(t) = b \sin(t) \end{cases}$$ with $0 \le t \le 2 \pi$, but it is important to notice that $t$ is not an angle here. If we use $P = (x(t), y(t))$, $O = (0, 0)$, and $A = (1,0)$, the angle $\angle AOP$ is actually $\theta$, $$\theta = \arctan\left(\frac{y(t)}{x(t)}\right) = \arctan\left(\frac{b \sin(t)}{a \cos(t)}\right) = \arctan\left(\frac{b}{a}\tan(t)\right)$$ i.e. $\theta$ is the angle from positive $x$-axis counterclockwise to the point. Conversely, $$t = \arctan\left(\frac{a}{b}\tan(\theta)\right)$$

In the circle case, $a = b = r$, and we have $t = \theta$, so for circles, $$\begin{cases} x(\theta) = r \cos(\theta) \\ y(\theta) = r \sin(\theta) \end{cases}$$ is true.

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If we assume the angle is measured counterclockwise from the positive $x$ axis as usual, then you could use $$\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1 \tag{1}\label{1}$$

In general, we can also describe the same ellipse as a function of $x$, by solving equation $\eqref{1}$ for $y$: $$y(x) = \pm \frac{b}{a}\sqrt{a^2 - x^2} = \frac{2}{3}\sqrt{3^2 - x^2}$$

At 45° from origin (center of the ellipse) we have $y = x$, and the distance to the center is $$r(45°) = \sqrt{x^2 + (y(x))^2} = \sqrt{\frac{2 a^2 b^2}{a^2 + b^2}} = \sqrt{\frac{72}{13}} \approx 2.3534$$

So, if your shape has $r(45°) = 2.5$, it is not an ellipse.

Answer 2

Try the usual parametrization of an ellipse with radius $\;3,2\;$ :

$$r(t)=\left(\,3\cos t,\,2\sin t\,\right)\;,\;\;0\le t\le 2\pi$$

and we get

$$r\left(\frac\pi4\right)=\left(\frac3{\sqrt2}\,,\,\,\frac2{\sqrt2}\right)\;,\;\;\text{and}\;\;\left\|\,r\left(\frac\pi4\right)\,\right\|=\sqrt{\frac92+2}=\sqrt\frac{13}2\approx2.55$$

which is pretty close to what you want...though I'm not sure what you call "the transition" to, and even less what does it mean that is linear.

Answer 3

The principal radii of curvature are

$$ \frac{a^2}{b}, \frac{b^2}{a} $$

By looking at the ellipse evolute center and radii of curvature positions close to that of evolute can be chosen for a smooth transition tangential junction find two profiled ellipse approximations of any eccentricity.

Sketches of EvoluteBasedEllipses