area on tapered extrusions

[lclew]

How do you figure out the surface area created by a tapered extruded arc? The arc in question has an extruded height of 6" and a taper of -30 degrees.

[Benson Shaw]

Select the object, then mouse to the menu and choose Modify/Volumetric Properties.

Some choices are presented along with the object's surface area and volume.

-B

[lclew]

Thanks for the tip! I didn't know that bit of info was there, but unfortunately, I don't think it really helps either. Let me try to rephrase:

The drawing the designer provided only shows the text rail that needs to be printed in plan view. They will NOT provide elevations. (don't ask...) But our sign fabricator needs the surface area flattened and in vectors to send to the router to cut. I can recreate the the rail in 3-d based on the measurements off the groundplan, but I just can't seem to figure out an easy way to lay it out flat since the shape of the curve will change when being flattened. I primarily use VW for Spotlight, but am the only one in my department who uses VW regularly, so of course, I get to solve this problem, although this is just not a problem I encounter often in light plots...

radius of the arc is 4'-7"

Chord of the arc 8'-5"

Height of rail 6"

Tilt Angle: -30 degrees.

Anyone have any tips????

Edited June 1, 2007 by lclew

[mike m oz]

TouchCAD can unfold. It would be a worthwhile complementary program to have in addition to VW. http://www.touchcad.com/

[MullinRJ]

For a simple conic section, which is what I believe you described, to describe the flattened shape you need to know the radii of 2 arcs that correspond to the top and bottom edges of your sign. The reference point you are looking for is the apex (or nadir in this case) of the cone your sign is wrapped around.

I drew the arc you described with a chord of 101" on a radius of 55", which comes out to a sweep of ~133.3234? and an arc length of 127.98129".

When you extrude this at a taper of 30? to a height of 6", then extend the sides to a vertex, you get a distance of 110.00" to the narrow edge (bottom edge, the one you specified) and ~116.93" to the wide edge (at the top, extruded edge).

To flatten this pattern, draw two arcs with a common center:

1) radius = 110.00" and Arc Length = 127.98" (This works out to a Sweep of 66.6617?),

2) radius = 117.30" and Arc Length = 136.47".

Connect the two arc ends with lines and Compose the 2 arcs and 2 lines to get the flattened pattern you were looking for.

HTH,

Raymond

Edited June 2, 2007 by MullinRJ

[lclew]

Brilliant! Thank you! The old geometry is rusty these days...