Why does the Durango curve problem puzzle mathematicians?

[rick1231]

This is one I haven't worked out a solution for yet. It's probably not too hard. Hopefully someone can provide a clear explanation. http://rickmckeon.com/mathfun/puzzles/The Durango Curve Problem.pdf

rick123

 

[skeeter]

Could be mistaken, but I believe the shadow curve is part of the ellipse ...

$\dfrac{(x-h)^2}{(r+s)^2} + \dfrac{(y-k)^2}{r^2} = 1$

... where $s$ represents the shadow length at the start of the curve.

https://www.physicsforums.com/attachments/6635

 

[rick1231]

I believe so too, but how do we prove it? It must have something to do with the projection of the circular barrier onto the freeway. A special case would be when the sun is directly overhead. Then it would be the quadrant of a circle (special case of an ellipse).

Other thoughts are:
1. It is a smooth curve
2. It's slope is zero at the south end and infinite on the west end.
Are there any other curves that meet those parameters?

Rick

 

[MarkFL]

Let's use a simple model where the sun rises due east and sets due west. Let $\theta$ be the angle of the sun above the horizon, and only consider \(\displaystyle 0\le\theta\le\frac{\pi}{2}\) so that the shadow is on the outside of the described curved barrier. We then find the length of the shadow of an object of height $h$ to be:

\(\displaystyle s=h\cot(\theta)\)

Now suppose $\beta$ is the measure of the angle we have traveled through on the curve, where $\beta=0$ corresponds to first entering the curve and $\beta=\dfrac{\pi}{2}$ corresponds to first exiting the curve. Then the length $\ell$ of the shadow cast by the barrier as measured along the radius of curvature would be:

\(\displaystyle \ell=s\cos(\beta)=h\cot(\theta)\cos(\beta)\)

 

[rick1231]

So, what's the equation of the ellipse and where are the foci? The barrier is 3 feet high.

 

[MarkFL]

So, what's the equation of the ellipse and where are the foci? The barrier is 3 feet high.

We would have an ellipse with horizontal semi-major axis $r+s$ and vertical semi-minor axis $r$, exactly as described by skeeter.

If we orient our coordinate axes such that the center of the circle on which the curved road lies, then we would have:

\(\displaystyle \frac{x^2}{(r+s)^2}+\frac{y^2}{r^2}=1\)

And the foci would be located at:

\(\displaystyle \left(\pm\sqrt{(r+s)^2-r^2},0\right)=\left(\pm\sqrt{s(2r+s)},0\right)\)

 

[I like Serena]

Wouldn't it be a circle with the same radius as the turn? (Wondering)

With the sun due east and 45 degrees high, every point on the turn casts a shadow of 3 feet to the west.
This corresponds to a translation.
If the sun is at any other altitude, then that only affects the magnitude of the translation.
And if the sun is at a different azimuth, then that only affects the angle of the translation.
In all cases the curve of the shadow throughout the turn is a quarter circle with radius $r$.

 

[skeeter]

Wouldn't it be a circle with the same radius as the turn? (Wondering)

With the sun due east and 45 degrees high, every point on the turn casts a shadow of 3 feet to the west.
This corresponds to a translation.
If the sun is at any other altitude, then that only affects the magnitude of the translation.
And if the sun is at a different azimuth, then that only affects the angle of the translation.
In all cases the curve of the shadow throughout the turn is a quarter circle with radius $r$.

Good question ... makes the sketch in the problem statement seem deceiving.

I tried to get a shadow picture of a cylindrical lid illuminated by my skylight, unfortunately it's overcast & raining today. I want to repeat this observation outside on a sunny day using my bride's circular cake pan.

 

[skeeter]

Sun came out and took this pic on the back porch ... note the interior and exterior shadows form a circle. Interesting.

 

[rick1231]

Interesting! I wonder if the pan being flared out (instead of being cylindrical) effects the shadow? I've driven around the Durango curve many times, but my observations were always pretty casual.

Rick

 

[skeeter]

Interesting! I wonder if the pan being flared out (instead of being cylindrical) effects the shadow? I've driven around the Durango curve many times, but my observations were always pretty casual.

Rick

I wouldn't think so ... a vertical obstruction and a slanted one with equal vertical heights should cast the same shadow.