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

What method did you use to calculate the dimensions? I assume that you
did not just cut a 12 inch OD circle with a 6 inch ID, and then try to
twist it to shape. If that is what you did, you need to start over and
rethink. Please explain what calculation you did.

Richard


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

Richard Ferguson wrote:
What method did you use to calculate the dimensions? I assume that you
did not just cut a 12 inch OD circle with a 6 inch ID, and then try to
twist it to shape. If that is what you did, you need to start over and
rethink. Please explain what calculation you did.

Richard


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.


You need to lay this out differently.

The inside edge is on a helix. Measure the distance from the
starting point around the shaft to the ending point. Then layout
a circle with that length as the circumfrence. Then layout the
outer line 6 inches further out.

This should give you the piece that you need.

Howard

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

From: (Bob Summers)

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.

Yes, close.

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?

Yep.

Think of
an auger flight that is on a six inch shaft (the vortex finder)
inside of a 12 inch pipe (the barrel).

The pitch is 6 inches per turn (it only needs to have one turn).

12 x 6 cylinder minus 6 x 6 cylinder. A 12 x 6 rectangular toroid in OD/ID
specification, or 12 x (6 x 3) in major/minor specification.

The circufrence of the six inch core is 6*pi. The length is 6. If you were to
lay out a rectangle 6*pi long and 6 high, its diagonal, L, would be the inner
circumfrence of the ramp disc. This is rather a bit longer than 6*pi.

The diameter that corresponds to

L = sqrt ( (6*pi)^2 + 6^2 ) is

Di = L/pi

Likewise, the wrapper is a 12*pi by 6 cylinder and the OD of the ramp blank is

Do = ( sqrt ( (12*pi)^2 + 6^2 ) ) / pi

of course there is some thinkness involved.

If the circumfrence of the outer shell is measured really accurately, to more
precision that just one thickness of steel, and is C, then the figure for 12*pi
above can be replaced by the ID of the shell:

C/pi - 2*(26 gage)

While measuring the true circumfrence of the inner core gives a good
computation for diameter without the thickness.

To arrange the ramp along the outer shell, lay out about 12 position lines
circumfrentially along the shell, and locate 1/2 inch increments along them.
Wait, that might not be right.

There will be two locations for the start and stop, along one line. There will
be eleven other positions. Therefore, the ramp needs to be divided at 13
locations including each end, or 12 sectors, same as the shell.

But along the six inch line, there are 13, not 12 positions, so they are laid
out closer than 1/2 inch, they are at 6/13 inch increments. Calculate and
measure, then drill little holes.

When you see the ramp at its associated hole, you can hold it there and plug
weld the ramp to the shell with a very thin arc welding electrode.



Yours,

Doug Goncz (at aol dot com)
Replikon Research, Seven Corners, VA

1200+ original posts at:
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