new approach => pyramidal instead conical Fermenter - Page 2 - Home Brew Forums

 Home Brew Forums > new approach => pyramidal instead conical Fermenter

06-02-2009, 12:28 AM   #11
HomebrewJeff
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Jan 2009
Lincoln Park, MI
Posts: 299
Liked 1 Times on 1 Posts

Quote:
 Originally Posted by Bavarian_ambassador I also had found a nice tutorial on a conical fermenter out there in the web ( How to Build a Conical Fermentor Cheap ) but I still worry to bend the metal into the circles, if the metal is stressed while welding
Eeks. A copper fermenter? I would be much more worried about copper poisoning!

I really need to get myself a welding setup. It's posts like this really make we want to take the plunge.

02-01-2011, 08:53 PM   #12
CamT
Recipes

Feb 2011
Posts: 26

I don't know much about any of this, but, from what i understand about conical fermenters, they use the cylindrical shape of the cone/fermenter to create a convection current in the fluid in order to get to the yeast and turb that would accumulate in the bottom most part of the cone. Would a square cone allow the same process?

02-01-2011, 08:57 PM   #13
Chipman
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Feb 2010
Jackson, WY
Posts: 166
Liked 4 Times on 4 Posts

Quote:
 Originally Posted by hammer one I LIKE IT ! It's problem solving like this that sets us apart from the apes.
It's having problems that sets us apart too
__________________
Remember, 'cold' is a flavor

Cooking with beer? Been doing it for years.
Add it to the food? I'll have to try it sometime.

He who has tasted sailing will walk the earth with his head turned into the wind.

02-03-2011, 03:04 AM   #14
DeafSmith

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Jan 2009
Richardson, TX
Posts: 1,448
Liked 34 Times on 32 Posts

Reading this thread got me interested in the mathematics of making a cone from a flat piece of metal, so I amused myself by working out some equations which may be of use to those who need to make a cone. In the link above (How to Build a Conical Fermentor Cheap) the cone was made from a semicircular template. This gives a fixed angle between the sides of the cone and the centerline of 30 degrees, for the end dimensions specified. I don't know what the standard is for conical fermentors, but it is easy enough to make a template to give whatever angle is desired; i.e., either a taller, more slender cone, or a shorter, fatter cone.

All cones can be made from a sector of an annulus - the area contained between two concentric circles, and between two radii extending out from the common center, cutting through both circles. Call the radius of the smaller circle r1 and the radius of the larger circle r2, and let theta be the angle between the two radii extending out from the center. The problem is, given the cone dimensions, to determine the values of r1, r2, and theta so you can cut the flat piece of metal for the cone.

Definitions:
D1 = the diameter of the small end of the cone
D2 = the diameter of the large end of the cone
H = the height of the cone, along its centerline
L = the length of the cone measured along the cone from end to end
phi = the angle which the side of the cone makes with the center line of the cone
NOTE: all angles in the following equations are measured in radians - multiply by 57.3 to convert to degrees, or divide by 57.3 to convert from degrees to radians.

Typically you will know D1, D2, and either H or phi. If you know H, then:
L = H^2 + ((D2-D1)/2)^2 and phi = arccos(H/L)

If you know D1, D2, and phi, then H = (D2-D1)/(2*tan(phi)), and L can then be determind as above, or also by L = H/cos(phi).

Now from D1, D2, and L, we can easily find r1, r2, and theta:

theta = (PI*(D2-D1))/L
r1 = (PI*D1)/theta
r2 = (PI*D2)/theta

Another useful calculation is to determine the minimum dimensions of the rectangular sheet from which the cone can be cut. These numbers can easily be obtained from r1, r2, and theta. There are four cases, depending on the value of theta:

Call the length of the two sides of the rectangle S1 and S2.

Note - angles here are in degrees

* theta < 90 degrees
S1 = r2 - r1*cos(theta)
S2 = r2*sin(theta)

* 90 <= theta <= 180
S1 = r2*(1+cos(180-theta))
S2 = r2

* 180 <= theta <= 270
S1 = 2*r2
S2 = r2*(1+sin(theta-180))

* 270 <= theta < 360
S1 = S2 = 2*r2

Disclaimer - I believe these equations are right, but the calculation was kind of "quick and dirty", so before you use them to cut an expensive piece of metal, remember - "Measure twice, cut once."

02-03-2011, 09:18 AM   #15
Pinworthy
Recipes

Jan 2011
Ft. Wayne, IN
Posts: 5

Interesting idea but, how bout this. Make it out of any metal you want, either round or square and instead of welding the joints use lead free silver solder. You could get your smooth inner joints without grinding them smooth. You could also leave a lip on the outside of the cone/cylinder/pyramid/square that could be used to clamp the joints together while soldering.

Heck be the first to have a transparent fermenter, you can do it with plastic and one of those plastic solvent glues, silicone or food grade epoxy. They have been building aquariums that way for years. It would be like caulking a bath tub.

The hardest part is going to be getting an air tight seal on what ever you use as a lid. How much pressure builds up inside a fermenter anyway? 5 to 10 psi?

Just my 2 cents, Pinworthy.