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How to form an air ramp?

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

Reply to
Bob Summers
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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:

Reply to
Richard Ferguson

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

Reply to
Howard R Garner

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

Reply to
Bob Summers

Yes, close.

Yep.

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

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Reply to
Doug Goncz

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