To B4man-as promised

[meika123]

Here ya go Barbara. Just follow this and you will have a 24" high-19" diameter cone. PM or email me if you have any questions.

[McRabbet]

David,

Is the radius of the lower (cone base) curve at 10"? Based on the vertical dimension, that's what I'd guess. Obviously, when the metal is formed, that opening will reduce to 6", right? Where are your plans from? Did you use Bill Pentz's spreadsheet? Curious minds want to know....

[walnutjerry]

Question-------------your layout shows a 24" dimension on the slope side of the cone, when that cover is wrapped, will the vertical height not be less than 24" ?

Jerry

[meika123]

Originally Posted by walnutjerry

Question-------------your layout shows a 24" dimension on the slope side of the cone, when that cover is wrapped, will the vertical height not be less than 24" ?

Jerry

A wee slip of the pencil Jerry. Good catch. The actual OA is indeed 24"

Rob, the calculations came from my "Fertile Brain" . I have been on Bill Pentz' site many times, but he is so far over my head I'll never catch up. I did use some of his logic though in setting up my own system.

[Jim Murphy]

Originally Posted by McRabbet

David,

Is the radius of the lower (cone base) curve at 10"? Based on the vertical dimension, that's what I'd guess. Obviously, when the metal is formed, that opening will reduce to 6", right? ....

Sorry about this. After 35 years of doing this sort of math for a living, it's hard to let an opportunity to actually use the stuff in something besides measuring land come into play.

Accept the lower radius at 10". Working through some curve formulae, I get the sweep angle (actually, the delta or central angle) to be 102.41°. That delta angle sweeping through a 10" radius yields an arc length of 17.879", which will create a radius of 5.689", not 6".

This is based on the Long Chord at the far (34") radius being 53".

Long Chord = 2 X R X sin(delta/2) ==> 102.41°

Arc = pi X delta X R / 180 ==> 17.879"

Circum=Pi X Diam so Diam = 1.789 / ~3.14159 or 5.689"

I also agree that the slope distance is 24" (i.e., the hypoteneuse of a right triangle), so the actual vertical height of the cyclone can be calculated by using the above formulae to compute the radius of the upper arc, and with your friend Pythagorus you can detemine height. If you care, I can do that, but I don't think it matters how high the thing is. Get a sawzall and make it fit.

[meika123]

Originally Posted by fernhollowman

Sorry about this. After 35 years of doing this sort of math for a living, it's hard to let an opportunity to actually use the stuff in something besides measuring land come into play.

Accept the lower radius at 10". Working through some curve formulae, I get the sweep angle (actually, the delta or central angle) to be 102.41°. That delta angle sweeping through a 10" radius yields an arc length of 17.879", which will create a radius of 5.689", not 6".

This is based on the Long Chord at the far (34") radius being 53".

Long Chord = 2 X R X sin(delta/2) ==> 102.41°

Arc = pi X delta X R / 180 ==> 17.879"

Circum=Pi X Diam so Diam = 1.789 / ~3.14159 or 5.689"

I also agree that the slope distance is 24" (i.e., the hypoteneuse of a right triangle), so the actual vertical height of the cyclone can be calculated by using the above formulae to compute the radius of the upper arc, and with your friend Pythagorus you can detemine height. If you care, I can do that, but I don't think it matters how high the thing is. Get a sawzall and make it fit.

To be perfectly honest Jim, I made several patterns from heavy cardboard to get the exact angles I needed. Not being as mathematically proficient as you, that is how I had to do it. All I know is-it works and will soon be sucking it's all-fired guts out through all that new-old 6" PVC I just acquired. Thanks my Brother-I very much appreciate it.
This is without a doubt the best Forum I have ever been associated with.

[walnutjerry]

Originally Posted by meika123

A wee slip of the pencil Jerry. Good catch. The actual OA is indeed 24"

.

The only reason that caught my eye was the fact I had to build some forms, years back, that involved cones and cylinders. Some of the cones intersected with the cylinder, off center but parallel with the center line of the cylinder.

I struggled with that job big time. Not being exceptional in math was a handicap to say the least. I had to lay it out actual size on my workbench top to get dimensions for drawing the arcs. I was lucky enough to find a formula in an old drafting book to calculate how much of a complete circle it would take to cover said cone/frustrum.

Having to struggle to figure it out made it stick with me. It is definitely not the quickest way to figure it but it is a way I can get it done. Give me the vertical height and diameters at each end and I can lay it out, as long as it is not bigger than my shop. After that, you have to call Jim

I am thankful for the members that excel where I come up short.

Jerry

[Jim Murphy]

Originally Posted by walnutjerry

I was lucky enough to find a formula in an old drafting book to calculate how much of a complete circle it would take to cover said cone/frustrum.

frus·tum (frŭs'təm)
n. Mathematics., pl. -tums or -ta (-tə).
The part of a solid, such as a cone or pyramid, between two parallel planes cutting the solid, especially the section between the base and a plane parallel to the base.

I love it. I simply love it. A woodworking site with FRUSTRUM as part of the discussion.

Last week it was opprobrium. It's those 'um' enders that crank me up.

[walnutjerry]

Originally Posted by fernhollowman

frus·tum (frŭs'təm)
n. Mathematics., pl. -tums or -ta (-tə).
The part of a solid, such as a cone or pyramid, between two parallel planes cutting the solid, especially the section between the base and a plane parallel to the base.

I love it. I simply love it. A woodworking site with FRUSTRUM as part of the discussion.

Last week it was opprobrium. It's those 'um' enders that crank me up.

OK----I stand corrected. Was not a genius at spelling or typing either.

Jerry

[Alan in Little Washington]

Frustum- sounds like one of those unmentionable parts of the human anatomy.

There are a couple of ways to get the dimensions for a cyclone. I used Bill Pentz's spreadsheet for my non-standard, 3:1 (cone length is 3 times the diameter of the cylinder) cyclone.

For most cyclones based on Bill's design and using a 2 - 5 hp blower, you will be looking at an 18" diameter (and approx 18" high) upper cylinder and a cone length of 1.64 X the diameter of the cylinder or approx. 29". You can get the more dimensions and see a sketch on the ClearVue site.

A spiral inlet ramp and slanted inlet are also recommended. The ramp directs the air down and around the central tube so it doesn't strike the side of the inlet chute (neutral vane). The inlet should also be angled to match the slope of the spiral ramp.

FYI, three years ago it cost me about about $40 in 26 gauge galvanized sheet steel and some scrap 3/4" ply to make my cyclone.