Interim lengths of an oval (bathtub)

Question

I have a real life problem to solve. I have two pillars in a bathroom that are 52" apart at their center. I want to slide a 59" oval bathtub (lengthwise) in between as far as it can go. Use this bathtub (59" long by 33.5" wide) as an example--with the following specs. How much of the width be tucked into the pillars and consequently how far into the room would the bathtub come out from the center of the pillars? Is there a way to calculate the interim lengths from two equal points from the center width point? And then calculate the width of that interim line to the furthest width point? To re-state, measuring from the center point between the two pillars, how do I calculate what would be the width into the room and what would the width be going behind the pillars?

Answer 1

Bathtubs are usually not mathematical ellipses. We like squarer corners on them so they come closer to superellipses, which have equation $\left(\frac xa\right)^n+\left(\frac yb\right)^n=1$ for some $n$ larger than $2$. The larger $n$ is the more the corners poke out. In the limit $n \to \infty$ you get a rectangle. We can't even count on that as they could just be drawn to look good. I would take the top drawing on your spec sheet and measure with a ruler. Even better, go down to the store with a tape measure.

To answer the question as asked, the equation of the ellipse would be $\left(\frac x{29.5}\right)^2+\left(\frac y{16.75}\right)^2=1$. Plugging in $x=26$ gives $y\approx 7.91$, so only $8$ inches of the width fit between the pillars and $25.5$ sticks in front of the pillars.