# Corner Cube Area

Some pictures of the corner of my outbuilding shop that could become a (sort of) cCube.

This shows the corner the cube would be in (SW). The arrow lines show the area to be used.
The west wall of the cube would include the first window.
The south wall would include an 8 ft width of the collector (two 4 ft bays).
The vertical white glazing supports are 2 ft apart -- every other one is the end of a collector bay, and is a solid bulkhead.

The active area of each collector bay is 46.5 inches wide by 93 inches high.

Looking at the south wall from the inside.
The cube would end just to the left of the generator.
The width inside the cube would be 114 inches.
The bat insulation is R19 FG.

The west wall.
The window is 36 inches wide by 60 inches tall -- double glazed, low e.
The north cube wall would come in just where the picture edge is -- 12 inches to the north of the window.
West cube wall width would be 91 inches.

The ceiling is 14 inch deep engineered joists.  The spacing is uneven in this area due to the stair up to the top level.
Above the ceiling is a full height loft level -- the roof line of the loft is insulated with R19 FG bats.  The loft has windows that could be opened to keep the temp somewhere near ambient.

Ceiling height (slab to bottom of joists) is 114 inches.

The floor is a 4 inch (more or less) thick slab with thickened edges -- no insulation.  So, the slab in the cube area would probably require insulation on top of it to prevent a large heat loss.

South wall area (incl collectors) =  (114*114)/144 = 90 sqft
North wall area = 90 sqft
West wall area (including window) = (114*91)/144 = 72 sqft
East wall area = 72 sqft

Floor area = (114*91)/144 = 72 sqft
Ceiling area = 72 sqft.

Total surface area = 468 sqft

Tentative construction thought:
(idea is to keep the 4X8 panels as intact as possible for reuse)

- Ceiling: attach polyiso to bottom of ceiling joists using washer screws of some kind -- two layers with crossed joints.

- West wall (already has R19 FG): Maybe only 1 layer of 2 inch polyiso screwed with washer screws on top of the studs.  Cut insulation around the window so that it remains a window.

- South wall -- maybe fill in stud cavities with R19 FG, then add 1 layer of 2 inch polyiso inside the studs?

- East wall (added):  Screw 1 horizontal 2X4 to the bottom of the ceiling joists.  Temporarily anchor a 2nd 2X4 to the floor.  Add a couple vertical studs between the top and bottom horizontal studs.
Place the rigid polyiso panels against the outside of the 2X4's -- first layer with long way running North-South, 2nd layer with long way running vertical.  Since the ceiling is 9.5 ft high, part panels will be needed above the 8ft.
Maybe tape seams?

- North wall (added):  Do it the same way as east all, but add a door -- maybe just a removable 4X8 panel?

- Floor:  Just lay 1 or two layers of 2inch rigid on the floor, and lay some half inch CDX over the areas that get walked on.

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Is this enough insulation/collector area to maintain 70F?

(468 sqft)(1F)(1/R28 BTU/hr-sqft-F) = 16 BTU/hr-F
For an average 25F outside and 70F inside, (70F - 25F)(16 BTU/hr-F) = 720 BTU/hr, 17K BTU/day.
Not counting infiltration or greater loss in window area.

Sunny day gain for 60 sqft collector might be 48K BTU ish.

Seems like it might be in the general ballpark?

Gary  January 30, 2010