Rectangle Needed To Make ConeDate: 9/5/96 at 8:34:50 From: Dean Thompson Subject: Rectangle Needed To Make Cone This is practical problem. I want to make a lampshade (actually many of them). It is made by rolling a piece of paper into a non-closed cone shape. The base is wider than the top. Given: the length of a side, the top diameter, and the bottom diameter. I need a formula that will take those three variables and output the X and Y dimensions of the piece of paper which will just inscribe the pattern. I can't even figure out whether this is a geometry, trig, or calculus problem. Thanks. Date: 9/7/96 at 6:45:5 From: Doctor Pete Subject: Re: Rectangle Needed To Make Cone You probably don't need calculus for this. Here are a few ideas that might help: When you take the lateral surface of a cone (i.e., without the circular base), cut it down the side, and unroll it flat, you get a sector of a circle. Furthermore, the radius of the circle which is formed by the edge of the flattened paper is the length of the side. So picture your lampshade as part of a cone; the cone's apex is missing, and what we want to find out is the length of the cone's side in terms of the length of the lampshade's side and the two radii. In pictures, we have | | . /|\ / | \ m / | \ /___|___\ / | x \ / | \ n / | \ / | \ /________|________\ | y | So we want m+n in terms of n, x, and y. By similar triangles, we see that m m + n ----- = ------- x y and solving for m, we have m = nx/(y-x). So m+n = ny/(y-x). Now, the second quantity we need to find is the angle t which sweeps out the sector of the flattened lampshade. This is simple, since t(m+n) is the length of the arc of the sector, which corresponds to the circumference of the lampshade's bottom, 2*Pi*y. Thus t = (m+n)/(2*Pi*y) = n/(2*Pi*(y-x)) when t is in radians. Now, the hard part is figuring out the smallest rectangle that fits such a shape. If t is greater than Pi radians, it follows that the optimal rectangle is symmetrically arranged (this is really hard to describe, but the sector's axis of symmetry should coincide with one of the rectangle's axes of symmetry). If t is less than Pi radians, the optimal way to draw the rectangle will depend on the value of x, y, and n. In some cases, the optimal rectangle will be drawn such that one side will coincide with a radius of the sector. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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