I wasn't clear about what I did.

I used the formula:

D = sqrt( (pi * d) ** 2 + pitch ** 2) / pi

For the inside diameter that's:

sqrt ( (pi * 6) ** 2 + 6 ** 2 ) / pi = 6.30 inches

for the outside diameter it gives 12.15 inches

The problem seems to be in the amount of distortion required
to stretch the inside into a helix of the appropriate length.

I wonder if I could start a ribbon 12.15 inches wide and twist
it. Then I'd have to cut out space for the vortex finder.

Bob S

On 10 Oct 2003 04:39:37 GMT, (Bob Summers) wrote:

I need help fabricating an air ramp for a cyclone - Think of
an auger flight that is on a six inch shaft (the vortex finder)
inside of a 12 inch pipe (the barrel). The air ramp doesn't
turn - it's fixed solid to both the barrel and the vortex finder.

The pitch is 6 inches per turn (it only needs to have one turn).
The ramp needs to be a snug fit to both the vortex finder
and the barrel. It also needs to be smooth to prevent
unnecessary air turbulence.

So far, I've calculated the inside and outside diameter of the
ring required to make the flight (if that is the right word)
and cut it out of galvanized iron (26 gage). The diameters seem
pretty close to correct.

I'm having two problems: getting it to stand perpendicular to the
vortex finder and barrel and getting it to fit snugly, even though
the ring is cut to within a thirty second of the line I scribed.

When I stretch out the ring, it distorts and the ramp isn't perpendicular
to the axis; maybe the inside of the ring needs to stretch or shrink?
It also seems to very sensitive to small errors in curvature and it has
good sized (quarter inch) gaps between the barrel and the air ramp. I
haven't checked carefully against the vortex finder yet.

TIA

Bob S