ma012 - Design of an Oscillating Sprinkler Bart Braden...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Design of an Oscillating Sprinkler Bart Braden Northern Kentucky University Highland Heights, KY 41076 Mathematics Magazine, January 1985, Volume 58, Number 1, pp. 29–38. T he common oscillating lawn sprinkler has a hollow curved sprinkler arm, with a row of holes on top, which rocks slowly back and forth around a horizontal axis. Water issues from the holes in a family of streams, forming a curtain of water that sweeps back and forth to cover an approximately rectangular region of lawn. Can such a sprinkler be designed to spread water uniformly on a level lawn? We break the analysis into three parts: 1. How should the sprinkler arm be curved so that streams issuing from evenly spaced holes along the curved arm will be evenly spaced when they strike the ground? 2. How should the rocking motion of the sprinkler arm be controlled so that each stream will deposit water uniformly along its path? 3. How can the power of the water passing through the sprinkler be used to drive the sprinkler arm in the desired motion? The first two questions provide interesting applications of elementary differential equations. The third, an excursion into mechanical engineering, leads to an interesting family of plane curves which we’ve called curves of constant diameter. A serendipitous bonus is the surprisingly simple classification of these curves. The following result, proved in most calculus textbooks, will play a fundamental role in our discussion. LEMMA. Ignoring air resistance, a projectile shot upward from the ground with speed v at an angle from the vertical, will come down at a distance (Here g is the acceleration due to gravity.) Note that gives the maximum projectile range, since then Textbooks usually express the projectile range in terms of the “angle of elevation’’, but since the range formula is unaffected when the zenith angle is used instead. The sprinkler arm curve In Figure 1, a (half) sprinkler arm is shown in a vertical plane, which we take to be the xy plane throughout this section. Let L be the length of the arc from the center of the sprinkler arm to the outermost hole, and let be parametric equations for the curve, using the arc length s , as parameter. Let denote the angle between the vertical and the outward normal to the arc at the point We’ll see that the functions and which define the curve are completely determined (once L , and have been chosen) by the requirement that streams passing through evenly-spaced holes on the sprinkler arm should be uniformly spaced y s 0 d a s L d y s s d x s s d s x s s d , y s s dd . a s s d 0 s L , y 5 y s s d x 5 x s s d , sin 2 s p y 2 2 u d 5 sin 2 u , p y 2 2 u ; sin 2 u 5 1. u 5 p y 4 s v 2 y g d sin 2 u . u
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
when they strike the ground. If there were a hole at the point on the sprinkler arm, the direction vector of the stream issuing from this hole would be and this stream would reach the ground at a distance The condition that evenly-spaced holes along the arm produce streams which are evenly spaced when they reach the ground is that be proportional to s :
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern