Wow. Cold too, without much sun... NREL says Fairbanks, AK has:
Jan Feb Mar... July... Oct Nov Dec
daily avg -10.1 -3.6 11.0 62.5 25.1 2.7 -6.5 F
daily max -1.6 7.2 23.8 72.3 32.0 10.9 1.8 F
horiz sun 38 240 720 1630 300 83 5 Btu/ft^2-day
south sun 240 740 1280 1180 590 350 94 Btu/ft^2-day
A small structure (or one with lots of internal walls and posts) could be
stout. A steep one could use less materials, with more solar heating and
wind load. But the upper part might be wasted space. Shedding snow needs
somewhere to go. If the sides are sloped or encumbered, snow might build
up to the roof before someone removes it. Vertical walls near the ground
could help, but that makes the structure even taller...
Our 24'x96' PA quonset greenhouse with 1.6" aluminum pipe bows on 4'
centers collapsed under 2-3' of snow. Not pretty, but we rebuilt it.
A new one with hinged portable wooden Korean surplus quonset bows on
8' centers hasn't seen snow yet, altho it tends to puddle rain on top.
The houses with steel pipe bows on 4' centers and 10/12 gable roofs had
no problem with snow. The pipes buckle before the plastic film rips...
I've built sunspaces with doubled curved 1x3s on 4' centers with 1x3
spacer blocks every 2'. Screw blocks to a 12' 1x3 bent to an 8' radius
(about 30% of them snap, when bent dry), then bend and screw another 12'
1x3 to the blocks. But this leaves a flat top and a wide structure, and
more height allows more wintertime solar heat, and the polyethylene film
only needs a slight curve (maybe 8" in 8') to avoid wind fatigue. So
maybe it's better to just bend the bows down between 2 concrete blocks
on the ground, and hinge them together at the top to make something more
like an equilateral A-frame sitting on top of stemwalls, for snow space.
It might be 32'x48', with a 32' warm equilateral triangle on 9' walls with
3 stacked 55 gallon drums for columns, about 37' from ground to ridge. If
each square foot of double-glazed south wall and roof collects 0.8x94 = 85
Btu/day of sun-warmed air and loses 1h(70-1.8)/R2 = 34, that's 51 net, or
32(9+28)51 = 60.4K Btu/day. The triangular top has 887 ft^2 of endwalls
and 3072 ft^2 of roof, totaling 3959 ft^2, so we need 24h(60-(-6.5))3959/Rv
= 60.4K, ie Rv = 105, eg 21" of fiberglass insulation. Too much. Looks like
it needs compost or fire in December. Or maybe the 1536 ft^2 base of the
triangle is heated, with unheated 887 ft^2 2nd and 441 ft^2 3rd floors.
But January has 185K = 24h(60-(-10.1))3959/Rv, so Rv = 36, eg 8" of
fiberglass or some sort of mulch, which is more reasonable. The rest
of the months look easier.
With 32(9+28)20 = 23680 pounds of 90 mph wind load and 438K ft-lb of
overturning moment, it needs 438K/32' = 13690 pounds on each side, eg
30 55 gallon water drums, or maybe 7 columns of 3 on 8' centers, with
deadmen purlins supporting the sides and at least 8480 pounds of water
under the triangle floor to store solar heat for cloudy days. If P
pounds of water cool from 70 to 60 F and (70-60)P = 5x185K, P = 92K,
eg a 92K/(32'x48'x62.33) = 1' pond under the triangle. With a pump,
most of that water could be on the ground.
Need an answer or a method for figuring which window, insulation, heating
system, etc to use?
Join PE Drew Gillett and PhD Rich Komp and me for a workshop on Solar House
Heating and Natural Cooling Strategies at the first Pennsylvania Renewable
Energy Festival on Saturday September 24, 2005 near Allentown. See