|


Finding a Plane Shape for a Truncated ConeDate: 05/10/2000 at 23:30:54 From: Jessica Hartman Subject: The plane shape of a truncated cone I need to cut a truncated cone shape out of some sheet metal, and I am having some problems with the math. Essentially, the base height of the cone must be 4.75 cm (the vertical line going through the center of the cone) and the diameter of the top circle must be 15.5 cm. The base (smaller circle on the bottom) must be 5.7 cm. What is the best way to cut a flat two-dimensional shape (the sheet metal) into a three-dimensional truncated cone? Thanks! Date: 05/11/2000 at 08:19:27 From: Doctor Jerry Subject: Re: The plane shape of a truncated cone Hi Jessica, I'll talk and draw as if the frustum has its smaller base on top. I "completed" the frustum by sketching in the cone of which it is a part. Next to the cone I drew a sector of a circle, with center of the sector at the top. Let the center of the sector be point A and the arc of the sector BC, where B is on the left. Next, draw an arc of circle with center at A and radius about one-half of BC. Let the arc be DE, where D is on the left and lies on AB, and E is on the right and lies on AC. The shape DBCE becomes the frustum when rolled up. All we have to do is to figure out its dimensions. Because arc BC comes from the bottom of the frustum, we see that its length is 2*pi*15.5; because DE comes from the top of the frustum, DE = 2*pi*5.7. Using the formula s = a*t for the length of the arc of a sector of a circle of radius a and central angle t, you can relate AB and AD to the angle DAE and the known arcs DE and BC. Also, using similar triangles in a vertical cross-section of the cone, you can figure out the "slant heights" of the cone in terms of the heights of the frustum and cone. Putting this all together will give you the angle DAE and the length DB = EC. I hope that I've said enough to get you started. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]


Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/