Building a Cone
Date: 01/28/2002 at 02:43:43 From: Surendra Kumar Chordia Subject: Building a larger radius cone My question is about cones. I am trying to draw a larger radius size cone (frustum). The details are as follows: Example: x = diameter of small end of cone x is 6.5 meter y = diameter of large end of cone y is 7.0 meter z = height of cone z is 5.0 meter For the above cones the the inner and outer radius comes out to: R = larger radius of developed cone R is 70.087 meter r = smaller radius of developed cone r is 65.081 meter the above radii are difficult to draw to 1:1 scale, so please gives some alternative. Any help or suggestions would be greatly appreciated. Thank you. S.K.Chordia
Date: 01/28/2002 at 12:58:34 From: Doctor Peterson Subject: Re: Building a larger radius cone Hi, Surendra. It appears that you are forming a frustum by cutting out a portion of an annulus and rolling it into the lateral surface of the frustum. You have not told me how you found the radii; I presume you have a formula such as those explained here: Pattern for Lampshade http://mathforum.org/dr.math/problems/loraine.02.17.01.html Flattening the Frustum of a Cone http://mathforum.org/dr.math/problems/scholl.10.15.00.html I think what you are saying is that you want a convenient way to draw an arc with a 70 meter radius without having to find a center 70 meters away and use a 70-meter string. The simplest solution I can think of is to plot a set of points on the arc, through which you can draw a smooth curve. One way to do this uses the properties of a segment of a circle: Circle Radius from Chord Length and Depth http://mathforum.org/dr.math/problems/duranleau.6.16.99.html The right triangle in the picture on that page shows that r-h --- = cos(theta/2) r Assuming you know the angle theta of your arcs (as given in the earlier references), you can solve this for h: h = r(1 - cos(theta/2)) The same formula is found here: Segments of Circles - Dr. Math FAQ http://mathforum.org/dr.math/faq/faq.circle.segment.html#13 So draw the chord of the appropriate length, find its center, and measure h units perpendicular to the chord to locate the midpoint of the arc. Now repeat the process using the two new chords you've found, repeatedly bisecting the arc until you have points close enough to form an accurate arc. There are probably some quick methods you can develop for doing these calculations more efficiently; I seem to recall a compass and straightedge construction that can accomplish this, but the calculations are probably good enough, and more accurate in practice. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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