Flat layout of frustums and cones

First, please note that, apparently, the correct spelling is frustum, not frustrum. Hmmm….  You learn something new everyday!

Second, you might ask why am I talking about this?  Isn’t this subject more appropriate to a blog on sheet metal work?  Well, think about it.  What if you’re designing a chair with a back that is laid out as a portion of a frustum?  What if you want to make a brass lamp shade?  How about a coopered pail?  How about a dunce cap for your friends?  I mean, the potential for this method is unlimited.  Right?  Okay.  Maybe unlimited is a bit of a stretch.  But understanding this layout may help visualize measurement of these shapes (or portions of them) in a number of scenarios.  Remember diameter x 3.1417 = circumference.

If you should ever find yourself in the highly unlikely situation of having to layout a truncated cone, here’s an excellent tutorial site:  http://leonjane.hubpages.com/hub/How-to-develop-a-Truncated-Cone#

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4 Comments on “Flat layout of frustums and cones”

  1. Tico Vogt Says:

    Excellent. I would have no idea how to do that and actually have wanted to make some funnels for various purposes. thanks!

  2. Sylvain Says:

    The link shows a nice classical exercise of descriptive geometry. Although it also could have shown how to construct the real shape of the top.
    Your example is useful for those wanting to build a cyclone.
    A slightly more difficult exercise is the intersection of the inlet tube with the cylindrical part on top of the cone. Altough the method is is similar to the one showed in the link.

    For those who like graphical construction of compound joinery, visit
    http://thecarpentryway.blogspot.be/
    look for “treteau” and “Mazerolle”.
    Sylvain

  3. Sylvain Says:

    It seems you just eyballed the 0.1415…D you added after reporting 3X the diameter D.
    If B is the radius of the big circle on your drawing, then the angle Alpha of the yellow sector must be such that
    Pi X D = B X Alpha ; all angles being expressed in radians.
    Knowing that 2 X Pi radians = 360°
    the angle in degree is = Alpha X 360 / 2Pi
    After simplification we obtain : angle in degree = (D/B) x 180
    On your drawing, you easely measure B without having to calculate it.
    (The diameter B of the big circle is the square root of (D/2)²+H² where H is the height of the full cone.)
    Sylvain


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