Cut the knot: learn to enjoy mathematics
A math books store at a unique math study site. Shopping at the store helps maintain the site. Thank you.
Learning Math Online
Sites for teachers
Sites for parents
Terms of use
Awards
Interactive Activities

CTK Exchange
CTK Wiki Math
CTK Insights - a blog
Math Help

III Millennium Olympiad

Games & Puzzles
What Is What
Arithmetic
Algebra
Geometry
Probability
Outline Mathematics
Make an Identity
Book Reviews
Stories for Young
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Proofs
Math as Language
Things Impossible
Visual Illusions
My Logo
Math Poll
Cut The Knot!
MSET99 Talk
Other Math sites
Front Page
Movie shortcuts
Personal info
Privacy Policy

Guest book
News sites

Recommend this site

Sites for parents

Education & Parenting

Manifesto  |  Bookstore  |  Contents  |  Amazon store  |  Term index  |  What changed?  |  Contact  |  Recommend
RSS Feed: Recent changes at CTK

Problem

In my business we occasionally need to cut across large diameter plastic or steel pipe at a 45 degree angle. By wrapping a piece of thin carboard around a cut pipe I can make a template which, when layed flat, looks like a bell curve. Can you provide me with the general mathematical formula that defines this curve for any diameter pipe that we might have to cut?

Solution

Let's position the pipe in such a way as to have Z axis run along one of its generating straight lines, X axis pass through the center of its cross-section and Y axis tangent to the pipe (in the diagram on the left Y axis points away from the screen.) Assume also that the plane of the supposed cut is described by X+Z=2R where R is the radius of the pipe.

The top view is present in the diagram on the right. Let the central angle a be 0 at the bottom of the circle. The positive direction is taken counterclockwise, as usual. Then X=R-Rcos(a).

We want to express Z in terms of L - the arc length of the circumference. We can imagine unfolding of the latter to coincide with a segment of Y axis after being cut at the top point (2R, 0). Accordingly, let L change from 0 for a=-π to 2Rπ for a=π. This gives L(a)=Ra+Rπ. In other words a=L/R-π.

From here and X+Z=2R we get

Z=2R-X=2R-(R-Rcos(L/R-π))=R(1-cos(L/R))=2Rsin2(L/(2R))

 

 

Copyright © 1996-2010 Alexander Bogomolny

35697820Page copy protected against web site content infringement by Copyscape

Search:
Keywords:

Google
Web CTK