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the first wofati greenhouse design

 
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here are the drawings that will be the basis for Kyle's new 3D model.
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Josiah Kobernik wrote:Tonight's design meeting is at 6 p.m. mountain time



I've gotten emails with links to the zoom meetings that contain just the link. Also, it seems that the links are invalid after the end of the meeting, so I've got no clue when it even was? I'm lost.

Would it be possible to add the planned DATE and TIME of the meeting into the SUBJECT of the email (from permies special events), to be able to see at a glance if I'm able to join in? or if I've missed it? (or lead off the BODY of message with date and time, followed by the link?)



 
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Wow, amazing discussion!

I'm wondering about the existing designs (that have a fan, but also 10' boreholes)--could those people be involved in running some tests to get some data to inform the current project?

If it takes 3 years to charge but then works for 50, that's a valid investment.   If you have to cheat with photovoltaics for those first 3 years but then can remove them and pass them on to a neighbor building their new greenhouse, then there's minimal loss.  

It makes it harder to judge the effectiveness of the experiment and is less kickstarter-attention-span compatible, but it is permaculture. Permanence.

However, let's say the power needed in year 1 or 2 from photovoltaic is less than any competing greenhouse design in the same latitude, that is a good result to show for the experiment.

It would be great to know what the 10' borehole design needs for power in year 1, 2, 3, even without deliberate thermal mass involvement--useful information and also a point of comparison to know what number to beat to show progress.
--
I see almost no downside to making extra boreholes and capping them as options for the future.  
 
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Josiah Kobernik wrote:here are the drawings that will be the basis for Kyle's new 3D model.



Looking forward to seeing that model, what the engineering numbers indicate the log sizes need to be, etc. When is it due? When are you all trying to start construction?
 
Joshua Myrvaagnes
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Joshua Myrvaagnes wrote:Wow, amazing discussion!

I'm wondering about the existing designs (that have a fan, but also 10' boreholes)--could those people be involved in running some tests to get some data to inform the current project?

If it takes 3 years to charge but then works for 50, that's a valid investment.   If you have to cheat with photovoltaics for those first 3 years but then can remove them and pass them on to a neighbor building their new greenhouse, then there's minimal loss.  

It makes it harder to judge the effectiveness of the experiment and is less kickstarter-attention-span compatible, but it is permaculture. Permanence.

However, let's say the power needed in year 1 or 2 from photovoltaic is less than any competing greenhouse design in the same latitude, that is a good result to show for the experiment.

It would be great to know what the 10' borehole design needs for power in year 1, 2, 3, even without deliberate thermal mass involvement--useful information and also a point of comparison to know what number to beat to show progress.
--
I see almost no downside to making extra boreholes and capping them as options for the future.  



Also if you test the new design with no inputs, someone can still build it and use an input in years 1-3 if they need to.  It seems best to keep the testing of the new design 100% passive for clarity and best information gathering
 
Joshua Myrvaagnes
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Josiah Kobernik wrote:In critique of the strategy of capping things off for later testing, I will relay two things that Paul has told me.

thing one, instead of testing each innovation independently with controls, paul likes to heap ten or more innovations into one experiment and then if the experiment is successful, you can successively divide the innovations in half to sort for relative influence.

thing two, the annualized thermal inertia aspect of wofati structures takes years to test. It may take 2 or more years for the mass to be fully charged and operating in semi-stable seasonal temperature fluctuations. So capping and uncapping earth tubes within the first 5 years muddies the results of the thermal inertia. That being said, If it takes several years for the greenhouse to start working, then it's not very attractive as a design solution.



All the more reason to put in all the extra tubes, and leave them capped (or leave them uncapped) for the first three years.  Then you can fiddle around with variables in year 4 without having to restart from the beginning.
 
Joshua Myrvaagnes
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EDIT--I completely screwed up my math, mixed up units by 2 orders of magnitude.  Oops!



OK, I finally thought of another piece of math that I believe should at some point be considered:  what energy does the 1"-black pipe actually carry? what's the limiting factor at least (hard to know what it will actually carry but the best-case scenario is at least a point of reference.). I figure the engineering folks have already worked through this but I didn't see it in the thread, so this is more for us laypeople.

The 1" black pipe will get about 2500 watts maximum per meter.  

1" = 2.5 centimeters = .025 meters
x 1 meter = .025 square meters meeting sunlight
sun energy is a _MAXIMUM_ of 1,000 watts per square meter (no cloud or other obstruction)

So .025.5x 1000 = 25 watts if 1 meter of pipe is exposed to sunlight.  (if it's closer to 2 meters then 50 W).

2 2m 1" pipes would be 100w; 10 would be 500w.

Comparing to fans that operate in a greenhouse in the non-passive designs, a cursory look shows wattages of between 120 and 450 watts.  I don't know how many fans they're using in these greenhouses (these are exhaust fans, not stratification fans).  But how it plays out in real life is still a question.  

Now, this assumes that everything around the black pipe is completely reflective...i.e. perfectly white or mirror-shiny, bouncing all the energy away and not heating that air at all, or else is perfectly transparent, allowing all the light to pass through it without leaving any energy behind in the greenhouse.  If everything around the pipe is in fact closer to leafy green, then there's less contrast, but it's still considerable: the leafy greens will take up some of that energy in photosynthesis (a few percent), and then absorb and radiate heat somewhat, and reflect a bit back out of the greenhouse.  Some will pass by and strike the thermal mass behind the plants.


 
Joshua Myrvaagnes
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Even distribution of temperature laterally is another consideration.

“Often, these systems do not homogenize the climate across the entire growing area as well as they do vertically, near the fan,” Becker says. “Additionally, conflict or buffers between fan patterns can cause inconsistency across the crop.”

Horizontal fans, on the other hand, help humidity move evenly around the greenhouse facility, while also promoting transpiration.

(from a website on greenhouse fans https://www.greenhousegrower.com/technology/heating-cooling-ventilation/four-keys-to-optimal-air-flow-in-the-greenhouse/)
 
Joshua Myrvaagnes
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Another brainstorm--what about "cool pipes" for air down-flow?  In other words, pipes that are minimum energy absorbance (so, white colored, lighter than the plants or surrounding colors), or maybe shaded by a white-painted piece of wood? how much difference might that make?  would it make up for the lack of differential between black pipe and green plant color?  
 
Josiah Kobernik
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The build will start one week from today, Monday August 17th. We were planning to start this week but the Kickstarter funds have been delayed and so some crucial material has not been purchased yet.

Kyle's Model is coming along beautifully. I will wait to post pictures of it until he is done.

Paul and I decided to completely redesign the wing walls. One of the benefits of building a new structure right now is that several of us have been actively working on the older structures and are aware of their design flaws. We have the opportunity to refine our building techniques, while also testing something completely new.

We combined two techniques from framing the berm shed, the corner cell and the attic cell, into the new wing wall design. I'm still chewing on the specific details but here is the general design


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top down view of the wing wall
top down view of the wing wall
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Paul and I got out clay and sticks to play with 3D geometry
Paul and I got out clay and sticks to play with 3D geometry
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Josiah Kobernik
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I calculated that the total weight of the roof with three feet of soil soaking wet above the membrane is 12,660 lbs.
Each of the 8, 8 inch posts can support more than 22,000 lbs. So that's cool.
 
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Josiah Kobernik wrote:I calculated that the total weight of the roof with three feet of soil soaking wet above the membrane is 12,660 lbs.
Each of the 8, 8 inch posts can support more than 22,000 lbs. So that's cool.



13x safety factor!?! Hopefully it's not an unlucky number.

What is the purlin size needed to transfer loads laterally to the posts?
 
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Josiah Kobernik wrote:I calculated that the total weight of the roof with three feet of soil soaking wet above the membrane is 12,660 lbs.
Each of the 8, 8 inch posts can support more than 22,000 lbs. So that's cool.



Could we see your calculations, Josiah? The numbers I find say wet soil averages 3000 lbs per cubic yard. If I recall the greenhouse dimensions correctly, 10 foot by 9 foot, 90 square feet times 3 feet deep =270 cubic feet. 27- cubic feet divided by 27 cubic feet in a cubic yard = 10 cubic yards. 10 yards times 3000 equals 30,000 lbs. You still have a HUGE excess load capacity!

Edited for my atrocious spelling!
 
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Joshua Rimmer wrote:

Josiah Kobernik wrote:I calculated that the total weight of the roof with three feet of soil soaking wet above the membrane is 12,660 lbs.
Each of the 8, 8 inch posts can support more than 22,000 lbs. So that's cool.



Could we see your calculations, Josiah? The numbers I find say wet soil averages 3000 lbs per cubic yard. If I recall the greenhouse dimensions correctly, 10 foot by 9 foot, 90 square feet times 3 feet deep =270 cubic feet. 27- cubic feet divided by 27 cubic feet in a cubic yard = 10 cubic yards. 10 yards times 3000 equals 30,000 lbs. You still have a HUGE excess load capacity!

Edited for my atrocious spelling!



In a timber structure like this the vertical posts won't present the major problems because they are in compression. The real concern is with the sizing of the purlins since they are taking the load in the shearing direction. You need to figure the size of the purlin based on the span distance between posts and the load coming down from above.
 
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Alan Booker wrote:

Joshua Rimmer wrote:

Josiah Kobernik wrote:I calculated that the total weight of the roof with three feet of soil soaking wet above the membrane is 12,660 lbs.
Each of the 8, 8 inch posts can support more than 22,000 lbs. So that's cool.



Could we see your calculations, Josiah? The numbers I find say wet soil averages 3000 lbs per cubic yard. If I recall the greenhouse dimensions correctly, 10 foot by 9 foot, 90 square feet times 3 feet deep =270 cubic feet. 27- cubic feet divided by 27 cubic feet in a cubic yard = 10 cubic yards. 10 yards times 3000 equals 30,000 lbs. You still have a HUGE excess load capacity!

Edited for my atrocious spelling!



In a timber structure like this the vertical posts won't present the major problems because they are in compression. The real concern is with the sizing of the purlins since they are taking the load in the shearing direction. You need to figure the size of the purlin based on the span distance between posts and the load coming down from above.



Agreed! I was mostly checking the info I found for wet soil weight. I've built several above-ground pole buildings, and never had an issue with the poles, but my early work had to be redone, because the door headers sagged badly.
 
Josiah Kobernik
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I've been looking unsuccessfully for trustworthy sources for information on beam load capacities to plug into my calculations. There is lot's of data on 2 inch wide lumber and glued laminate beams, but I'm having a hard time coming up with numbers for raw timber beams, anyone have a source for that they feel good about?

 
Josiah Kobernik
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here are my calculations for the weight of earth on top of the roof. The number I gave earlier included 1246 lbs for the 8 inch rafters and purlins as well as the roof pole sheathing.

Sample weights:
Soil 1 cubic ft. weighs 70.44 lb. when dry and 87.6 lb. when saturated to the point of runoff.

Area of triangle (A)
1.4 x 8.7 / 2 = 6.09 sq. ft.

Volume of (A)
6.09 X 11.5 ft. wide = 70.04 ft.^3

Weight of (A)
70.04 ft^3 x 87.6 lb (wet soil) = 6135.5 lb
Or
70.04 ft^3 x 70.44 lb (dry soil) = 4933.6 lb

Area of triangle (B)
1.5 x 8.7 / 2 = 6.52 sq. ft

Volume of (B)
6.52 sq. ft x 11.5 ft wide = 74.98.ft^3

Weight of (B)
74.98 ft^3 x 70.44 lb (dry soil) = 5281.6 lb

Weight of soil on roof (A) + (B) when (A) is saturated to running off.
11416.5 lb.

The next step is to determine how much of that weight will be on each roof pole, purlin, and rafter so that I can size them appropriately.
wofati-greenhouse-roof-load-diagram-side-view.png
[Thumbnail for wofati-greenhouse-roof-load-diagram-side-view.png]
 
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This only shows 4 species- spruce, hemlock, birch, cottonwood- but it will give you some helpful guidelines. Go to cespubs.uaf.edu and type HCM-00752 into the search box.

The two concerns are deflection and shear. The front of the greenhouse appears to bear hardly any weight at all, but the center and rear would need sturdy rafters/purlins. One other consideration is that in the spring, you could encounter a time when not only would you have 100% saturation of the soil, but a significant additional wet snow load.
 
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Josiah Kobernik wrote:here are my calculations for the weight of earth on top of the roof. The number I gave earlier included 1246 lbs for the 8 inch rafters and purlins as well as the roof pole sheathing.

Sample weights:
Soil 1 cubic ft. weighs 70.44 lb. when dry and 87.6 lb. when saturated to the point of runoff.

Area of triangle (A)
1.4 x 8.7 / 2 = 6.09 sq. ft.

Volume of (A)
6.09 X 11.5 ft. wide = 70.04 ft.^3

Weight of (A)
70.04 ft^3 x 87.6 lb (wet soil) = 6135.5 lb
Or
70.04 ft^3 x 70.44 lb (dry soil) = 4933.6 lb

Area of triangle (B)
1.5 x 8.7 / 2 = 6.52 sq. ft

Volume of (B)
6.52 sq. ft x 11.5 ft wide = 74.98.ft^3

Weight of (B)
74.98 ft^3 x 70.44 lb (dry soil) = 5281.6 lb

Weight of soil on roof (A) + (B) when (A) is saturated to running off.
11416.5 lb.

The next step is to determine how much of that weight will be on each roof pole, purlin, and rafter so that I can size them appropriately.



I see where the difference comes from,I clearly didn't have a good grasp of wofati construction!

I wish I could help with roundwood strength calculations, but I am nowhere near to being a civil engineer!
 
Josiah Kobernik
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I'm definitely not an engineer either. I need to study up on deflection force. I did a bit more math today and I think that learning to see the structure in terms of loads, compression, tension, etc. will make me a better builder.

Determining weight on roof poles

Break (A) and (B) up into smaller bits to determine the volume of soil above specific areas of the roof. In this case above the roof pole span (C). The weight of a2 + b2.

a1 + a2 = (A) and b1 + b2 = (B)

Area of a1
4.7 x 0.7 / 2 = 1.65 ft.^2

Volume of a1
1.65 x 11.5 = 18.92 ft^3

Volume of a2
70.04 ft.^3 - 18.92ft^3 = 51.12 ft^3

Weight of a2
51.12 ft^3 x 87.6 lb = 4478.11 lb

Area of b1
0.8 x 4.7 / 2 = 1.88 ft.^2

Volume of b1
1.88 x 11.5 = 21.62 ft.^3

Volume of b2
74.98.ft^3 - 21.62 ft^3 = 53.36ft.^3

Weight of b2
53.36ft.^3 x 70.44lb = 3758.68 lb

Weight of a2+b2 = 8236.79 lb

Roof poles used in the berm shed have a diameter of 3.5 inches spaced 3.5 inches on center giving an average of 34 poles per roof cell.

8236.79 lb / 34 roof poles = 242.25 lb. per pole
wofati-greenhouse-roof-load-diagram-side-view-2.png
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