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Insulation Math

All that follows has admitted simplification, for easy hand-calculation. Rigor can be added, to a point of paralysis. Anyone, please, treat this as a wiki. I will edit in attributed contributions of better analysis. And know, wherever my simplification comes with numbers specific to my climate and heating appliance, your numbers can readily be imagined, and substituted.

Costs Associated With Heat Loss By Convection and Conduction:

In the absence of heat loss by draft airflow, the relationship between total (system) R-value of a wall, ceiling, or floor, and the dollar cost of the heat lost through that surface, is:

Cost = Fuel Cost * HDD * 24 / Eff * (Area/R)

Fuel Cost is $/Btu, and HDD is Heating Degree Days, Eff = Furnace Fuel Efficiency

Fuel Cost (natural gas) = $.00002/BTU ($2 per added therm. This is a reasonably high number consistent with enforcement of conservation. I see no reason to use a current number, such as my most recent bill, at $1.27 per added therm. With a trend to replace oil or electric heat, apply math only for natural gas heat, at this mildly inflated number. I wish we could call it, and pay it, at $4 per therm with a carbon tax, promoting conservation.

Assume a modern furnace with Eff = 0.88.

HDD = 4400, for Portland Oregon.

Find HDD values for other locations in the United States, at degreedays.net, linked from discussion at Wikipedia.

Area =  Area of heat transfer, measured in sq ft

R = System R-value of airtight boundary and insulation = 3 + Ri

Ri = Insulation or component inherent R-value

The number 3 of the formula R = 3+ Ri is a conservative allowance for heat transfer resistance at surfaces of contained insulation and framing, mainly in heat convection to air at all borders. Convection and the conduction resistances of cladding add to a total ranging from more than 2, to about 3, and a highest number gives wished under-estimate of savings with added insulation. Inherent R-value with parallel paths is computed with both rigor and simplification. Ri is strongly affected by poor insulation placement, and it is best and easiest to discount all insulation value where there are large loose-fill voids or lack of convecting surface contact with batts. Wanting incentive to do right in avoiding bare areas, voids and lack of contact at little cost done-right at installation, accept only high estimate of installation error.

1/(Reff + 3) = fjoists/(Rjoists +3) + fbatts in contact/(Rbatts +3) + fbare/3

where fjoists + fbatts in contact + fbare= 1

Annual Cost of Heat =  (.00002 * 4400 * 24  / .88) * (Area/R)  = 2.4* (Area/R)

Moderation in an Attic or Crawl Space:
The above is for heating conditions only, and does not account for moderating by sky and ground effects. The temperature above an attic floor has daytime rise above ambient, and may fall below ambient at night. For an attic, I stick with the formula. For a moderately-ventilated crawl space, ground heat keeps temperature above ambient. I assume a 25% reduction of HDD. Annual Cost of Heat = 1.8 * (Area/R), for Portland, Oregon.  I invite critique of all postings here; this on crawl spaces, especially, is just a blind step in the right direction. 

Heated Surfaces (Hydronic, Radiant Heat):
Now, what if the surface losing heat to ambient is HEATED, is the source of building heat? That is the case with ceiling hydronic or electric radiant heat. I think the acknowledgement may be in increase of the HDD number. How about by 20%?
With this adjustment, with surface radiant heat, Annual Cost of Heat, Portland Oregon numbers, is 2.88 * (Area/R)

With justice, we disdain electric resistance heating. There is 100% efficiency in conversion, but dollars per therm are high. $2.93 per therm at ten cents per KWH. $2.20 at a corruptive 7.5 cents per KWH in Oregon, near enough the $2 per therm, but NOT an inflated number. At a common 15 cents per KWH,  motivate insulation over hydronic heat, with a number much larger than the 2.88 factor above. 

How about $5 * (Area/R) anywhere in the USA?

Carbon Footprint:
The cost of replacement heat as  carbon footprint, is 1.2 * (Area/R), in therms of natural gas.  Where a therm of natural gas is 100,000 BTU, and is approximately the energy equivalent of burning 100 cubic feet Ccf, of natural gas. A therm corresponds to emission of about 12 pounds of CO2. As pounds of CO2 emitted for this locale and furnace efficiency, the round number is 14.4 * (Area/R).  

Usage of Electricity:
The work product of Bonneville Power Administration/ Regional Technical Forum, as it presumes to be a director of weatherization, is an Excel spreadsheet, last noted March 2011, as RESWXSF FY10v2.3, stating electricity savings for insulation measures, and more. We need a lot less hydroelectric or nuclear power in heating homes. Still, the numbers for Utility management purposes, are there for comparison with other predictions. Savings are expressed in KWH/sq ft/ year. One therm is 29.3 KWH. With different matters of efficiency, it is messy to relate hydro power and gas-furnace heat costs. I apply KWH numbers converted to therms, and to dollars at $2 per therm. The review is published at this site, here.

Insulation With Parallel Paths:

Use inverse summation to calculate effective Ri with parallel paths, over areas of unequal resistance. 

For lumber of framing and for applied insulation, find unbiased statement of materials R-value at Wikipedia .

For 2x joists 16” on-center:

1/Reff  = (1.5/16)/ Rjoists + (14.5/16)/ Rinsulation

where Rjoists is taken as 0.94 per inch (douglas fir), 2.4 for 2x3, 3.3 for 2x4, 5.2 for 2x6, 6.8 for 2x8, 8.7 for 2x10.

Above joists, add crossing insulation Ri for all area.

For ordinary fiberglass insulation batts, apply "thickness" as claimed R-value, divided by R3.5 per inch. For high density fiberglass insulation batts, apply "thickness" as claimed R-value, divided by R4.2 per inch. Find then that R19 batts have thickness 5.4" and are not sufficient to fill 2x6 framing. R15 HD batts have thickness 3.6" and do fill 2x4 framing, just barely. Prefer R19 batts in 2x4 walls. Yes, use R19 in 2x4 walls, but not in 2x6 walls. R19 compressed to 3.5" is then R15. The applied cost of R19 in 2x4 walls is about 25% below that of R15. Likewise prefer R25 batts in a 2x6 wall, compressed to R21, for less applied cost than R21.

Here do math in an attic floor, where batts compress in tight packing and in bearing covering batts. Compression to "thickness" is without penalty vs. free thickness. For all batts on an attic floor, apply bag-claimed R-value.

Example 1: 2x6 attic floor, joists 16” oc, with R19 fit between joists (partially filling to depth perhaps 4 1/2").

1/Reff = 1.5/16/5.2 + 14.5/16/19. 

Reff  = 15.2

Example 2: 2x4 attic floor, joists 16” oc, with R15 fit between joists, and R25 crossing batts.

1/Reff = 1.5/16/3.3 + 14.5/16/15. 

Reff  = 11.3.

Ri = 11.3 + 25 = 36.3.

This is fairly close to the intended R40.

Example 3: 2x8 attic floor joists at 16” oc, with R38 batts fit between joists, no covering insulation on joists. Batt insulation would slightly over-fill the 7.25" joists. Such framing is rare in older homes, and in newer homes would likely be found with loose-fill covering joists. 

1/Reff = 1.5/16/6.8 + 14.5/16/38.

Reff  = Ri =  26.6

Here the joist thermal shorting is painful.

To have >R38 in this example, starting with R26.6, I would add R19 crossing over joists, for total R45. I would not add thinner batts, as R19 is especially affordable, and comes in convenient sizes. I mainly use Johns Manville AU333 here, nice postage stamps for the puzzle, 24" by 48", and AU395, 16" by 96", often cut in half. It is important to not fight the roof-joist spacing. It is important to not have long strips to pick up and stow, for service access.

With 2x8 framing, learn to accept total insulation beyond an R38 target, when using batt insulation. 

Example 3a: 2x6 attic floor joists at 16” oc, with R38 batts fit between joists, no covering insulation on joists. Batt insulation would slightly over-fill the 7.25" joists. Such framing is rare in older homes, and in newer homes would likely be found with loose-fill covering joists. 

1/Reff = 1.5/16/5.2 + 14.5/16/38.

Reff  = Ri =  23.9

Example 3b: 2x10 attic floor joists at 24” oc, with R38 high-density batts fit between joists, no covering insulation on joists. Batt insulation just fills the 9.25" joists. This is uncommon framing, but is another case in the topic of uncovered framing thermal shorts. 

1/Reff = 1.5/24/8.7 + 22.5/24/38.

Reff  = Ri =  31.4

The wider joist spacing, and greater joist thickness, help. Getting to true R38 requires some crossing insulation. 

Example 3c: 2x10 attic floor joists at 24” oc. Batt insulation just fills the 9.25" joists. Add 2x4 decking 24" oc, with contained R19 unfaced batts. Call the base layer R38 between joists. Call the top layer R15 with over-compression, though covering plywood adds compensating points.

Base layer:  1/Reff = 1.5/24/8.7 + 22.5/24/38. Reff  = Ri =  31.4.

Top layer:  1/Reff = 1.5/24/3.3 + 22.5/24/15. Reff  = Ri =  7.3.

Total insulation R38.7. Add R3 as system R value.

And do math for wall insulation:

Example 4: 2x4 wall joists at 16” oc, with R15 batts fit between joists. R15 is any insulation completely filling the wall volume, stilling air circulation in bypass of insulation. 

1/Reff = 1.5/16/3.3 + 14.5/16/15.

Reff =  11.3

Example 4a: 2x4 wall joists at 16” oc, with batts not completely filling the wall volume. The insulation may have no value. The least penalty may be with R13 nearly filling the air space assuming blocked air circulation within the wall.

1/Reff = 1.5/16/3.3 + 14.5/16/13.

Reff =  10.2. This drop by 1.1 in effective R-value is with very large opportunity cost. The drop could be avoided with negligible effort.

Example 5: 2x4 knee wall joists at 16” oc, with R15 batts fit between joists, with crossing 2x3 supports and a second layer of R15 outboard (on attic side).

1/Reff, inner = 1.5/16/3.3 + 14.5/16/15.

Reff, inner =  11.3

1/Reff, outer = 1.5/16/2.4 + 14.5/16/15.

Reff, outer =  10.0

Ri for the doubled R15, is 21.3, with simplified math.

Think how this compares to a result if placement were not crossing, joists R = 2.4 + 3.3 + 5.7. R30 insulation.

1/Reff = 1.5/16/5.7 + 14.5/16/30.

Reff  =  21.4

I think the crossing should give a large improvement, and know the simplified math is not rigorous. 

Annual Heating Cost Calculations for These Examples:

My definition, System R-value of airtight boundary and insulation = 3 + Ri, is intended to not over-estimate found heating cost, and thus be high in prediction of savings. The "three" is system convection and R-value of the ceiling, elsewhere taken as as little as 1.2. With "three", I, too, avoid the impossible area-averaging of resistance of floor joists, with the "zero" between. The choice matters especially for found conditions with no insulation. Here consider starting conditions with no insulation. Annual heating cost for 1000 sf  is 2.4*1000/3 = $800. 

The heating cost for Example 1, is 2.4*1000/18.2 = $132. Savings vs. no insulation are $668.

The heating cost for Example 2, is 2.4*1000/39.3 = $61. Savings vs. no insulation are $739.

The heating cost for Example 3, is 2.4*1000/29.6 = $81. Savings vs. no insulation are $719.

The heating cost for Example 4, is 2.4*1000/14.3 = $168. Savings vs. no insulation are $632.

The heating cost for Example 4a, is 2.4*1000/13.2 = $182. Savings vs. no insulation are $618.

The heating cost for Example 5, is 2.4*1000/24.4 = $98. Savings vs. no insulation are $702.

Investment Payback Calculations for These Examples:

For Example 1, I would place about 910 sf of R19. I would use only unfaced insulation, and would ask about $560. Simple payback is in less than one year, with simple return on investment of more than 100%.

At January, 2012, with addition of Example 6 above, and a link to critique of the math of Bonneville Power Administration, study "Line 12" information that corresponds to Example 1. The critique, and numbers here, are consistent. Payback in less than a year. 

For Example 2, I would place about 910 sf of R15, and 1000 sf of R25. I would use only unfaced insulation, and would ask about $1760. Simple payback is in less than three years, with simple return on investment of about 42%.

For Example 3, I might place two layers of R19 within framing, 1820 sf. I would use only unfaced insulation, and would ask about $1120. Simple payback is in less than two years, with simple return on investment of about 64%.

For Example 4, I would place one layer of R15 within framing, 910 sf. I would use only unfaced insulation, and would ask about $910. Simple payback is in less than two years, with simple return on investment of about 70%.

For Example 5, For insulation only,  I would ask about $1820. Simple payback is in less than three years, with simple return on investment of about 39%.

Best return on investment, that might be achieved with less added insulation, is not the goal.  Choose maximum insulation that gives a return of more than, say, 7%.

Investment Payback for Attic With Starting Condition of R18:

Consider a 1000 sf attic, with joists covered. Add R25 crossing batts. The found Annual heating cost is 2.4*1000/21 = $114/ yr. The improved heating cost at R43 is 2.4*1000/46 = $52/ yr. Savings $62/ yr. I would ask about $1000. Simple payback is in about sixteen years, with simple return on investment of about 6%. This is near the boundary at which rule-makers authorize incentives.

Example 6: 2x8 crawl space joists at 16” oc, with R30 batts fit between joists.

1/Reff = 1.5/16/6.8 + 14.5/16/30.

Reff =  22.7

If batts don't fill joists, the joist thermal shorting is greater, and batts claimed 10" thick at R30, might be of lesser value. I find framing revealed at least 1" on average, where batts are pushed tightly to the floor sheathing. At maximum, think of containment in joists R5.8.

1/Reff = 1.5/16/5.8 + 14.5/16/30.

Reff =  21.6

I will apply  Reff  = 21 in critique of the math of Bonneville Power Administration,  at "line 10." This form of presentation is more interesting, and  is consistent with payback study numbers above, for Examples 1 through 5. 

Example 7: One baffled can light in every 100 sf of an "R38" attic floor. Say, cellulose twelve to fifteen inches depth. Real R38 insulation except for holes. Baffle 12" dia., area 0.8 sf uninsulated or worse, with driven leakage.

100/Reff = 99.2/38 + 0.8/3. Reff = 34.8. As if insulation were less deep by an inch or two. Why would one "waste" all that insulation, to avoid replacing the ugly can light? The baffle, with a cover in a deep insulation bed, must cost as much as the light replacement, say $60 for a T91 LED plate light in a junction box? Correct actions are achieved by prohibiting rebates if wounds and dangers in a house are not fixed.

With math for my home, the uninsulated area at each can light loses $2.4*0.8/3 = $0.64 per year. Bulb local heat is all lost in winter. Countable pennies are lost in heat-driven excess of house fresh air exchange through any can gaps.

As Applied for Attic Ladders and Portal Doors (Attic Hatches):

At December 2015, please know that wooden ladders I work with, Fakro and Calvert, are rapidly moving toward offering of very high insulation value, at higher cost that may hurt sales. The Fakro LWT has a 3 1/8" door thickness, called R12.5, and costs $495. The LWP with 1.25" door thickness costs $340.

This page has had math now in italics, as the only offered form. 

Best ladders are of European design: the high-end Rainbow steel ladder (door with 2" styrofoam, claimed R15 though that is questionable, and affordable wood ladders, Fakro (made in Poland, door with 1 1/4" polystyrene, claimed R6.2),  MidMade (made in Sweden, door with 2" polystyrene, claimed R10.5 although the manufacturer claims a door European U-value of 0.78 (R7.3)). We must cope with inconsistency in the various R-value claims. Let the claim for MidMade  be a template, where "system" R-value is door R-value plus three. I guard against over-estimating insulation benefits, by using this perhaps-large +3 increment from door panel, to system R-value, despite a commonly-published increment of only 1.2 for convection of energy approaching and departing an airtight boundary. With this template and a Wikipedia value of R4 per inch for high density expanded polystyrene,  work out annual heat costs, for a door with area 8.4 sq ft, 22.5" by 54". Allow for thermal bridging in a strengthened actual door with internal wood framing. Make this allowance assuming one fifth of door area is structural wood at R0.94 per inch; expanded polystyrene foam, EPS, is R4 per inch. 

 1/Reff = 0.2/Rwood + 0.8/Rfoam

Per inch of door thickness less facings, using EPS, this is evaluated as: Reff = 1/(0.2/0.94 + 0.8/4) = 2.423 per inch.

Approximate system R  is three plus R-value of door facings, plus Reff of the space between facings.The R-value of combined 1/4" wood facings is about 0.2.

Reff  =  0.2 + 2.423*(Thickness - 0.25).

System R = Reff + 3

Annual Heat Cost = $2.4*8.4/System R 

Set these formulas in an Excel file now, to populate the following table:

Thickness    Reff      System R    U = 1/ System R    Annual Heat Cost

Bare Ceiling                      3                   0.33                        $6.72

   1"               2.02            5.02               0.199                       $4.02

1.25"             2.62            5.62               0.178                       $3.59

  1.5"             3.23            6.23               0.161                       $3.24  

   2"               4.44            7.44               0.134                        $2.71 

   3"                6.2             9.86                0.101                      $2.04 

   4"               9.29          12.29                0.081                      $1.64

   5"              11.71         14.71               0.068                       $1.37

R38 Ceiling                       41                 0.024                       $0.49

R49 Ceiling                       52                  0.019                      $0.39

An R5 door is really easy to achieve. 

A door with U = 0.17 is about 1 1/2" thick and without much added cost. 

Now repeat this math of parallel paths in a panel door in a manner that gives higher door R-value, where paths aren't fully separate, but have heat flow division also affected by common resistances in convection at surfaces and conduction through door skins. Reality is somewhere in-between, and is probably nearer the more-favorable result below. The newer and favored analysis is again not in italics.

 1/(Reff + 3) = 0.2/(Rwood + 3) + 0.8/(Rfoam + 3)

With 1" EPS, this is evaluated as: 1/(Reff + 3) = 0.2/3.94 + 0.8/7. Reff = 3.1 (up from 2.02). System R = 6.1. U = 0.17. Annual heat cost is $3.30, thought truer than the $4.02 value above.

However you do the math, save a bit more than $2 per year at system R12, vs. system R5. This $2 difference is with very large increase of ladder cost perhaps never repaid, but with strength and buyer satisfaction. The more-complicated ladder had better be more  durable. A compromise "high-end" ladder might be R7.5, a bit thicker than 2", and with not more than a dollar-per-year to gain or lose, with other options. Please see feeding of r5portals better-code-proposal , from math here.

Ladders at more than R12 are not on offer, with thickness impractical. Achievement of actual R12 in a 3" door should present structural compromise. Manufacturer-claimed R-values correspond to System R, 1/U, not panel Reff. Note that R values can not be measured or computed with great accuracy.

Wikipedia lists a higher R-value for foil-faced polyisocyanurate rigid panels, pentane expanded, aged five to ten years.  Manufacturers stick with expanded polystyrene. Expect that R12 claims for a ladder might not be defensible.

Heat Cost Due to Drafts:

Methods in Energy Center of Wisconsin Report 208-1,  (at Equation  (5), on Page 8), and in the Energy Conservatory Blower Door Operation Manual (at pdf page 45, labeled page 35), are applied to assess the virtue and cost-effectiveness of making a ladder installation air tight (fully gasketed and with sealing of the rough opening). 

The Equation 5 relationship is:

Qh = 1.06 · Ah · ( Ph)0.5 


Qh = rate of air flow through the hole (cfm)

Ah = hole area (square inches)

Ph = pressure drop across the hole (Pa)

With Ph = 50 Pa, Qh = CFM50 = 7.5 * Ah.

From Energy Conservatory:

Annual                   26 x HDD x Fuel Price x CFM50

Heating       =      ------------------------------------  x 0.6

Cost                           N x Seasonal Efficiency

- HDD is the annual base 65 F heating degree days for the building location (4400).

- The Fuel Price is the cost of fuel in dollars per Btu ($.00002).

- N is the Energy Climate Factor from the Climate Information Screen (adjusted for wind shielding and building height). (21 for Portland, Oregon).

- Seasonal Efficiency is the AFUE rating of the heating system, taken as 0.88 for a common value.

The formula is further explained by The Energy Conservatory, in an old edition of its Blower Door Manual:

The "26" factor is derived from multiplying the heat capacity of air (0.018 BTU/cubic ft/Degree Fahrenheit) by 60 min/hr and 24 hr/day. The "0.6" value is a correction factor which is used to adjust the theoretically calculated energy savings for the following interactions:

  • The theoretical model assumes that homes need heat at an outside air temperature of 65 degrees. In fact, many homes do not need heat until the outside air temperature is below 65 degrees F.
  • Some air that leaks into the house may not get heated to the inside temperature before it leaks out again.
  • Air leakage through small cracks reduces conduction losses by tempering surrounding surfaces. This process is referred to as the dynamic insulation effect.
The 0.6 multiplier has been described in error at this page, citing the same factors as for the "26" number, attributed to The Energy Conservatory. At March, 2013, the discussion  is deleted.

Annual Heating Cost = 0.074 * CFM50

(There is virtue in reducing math to a simple formula, without further statement and defense of details, independent of the economy in better heating systems.)

CFM50 is the air flow artificially outward through all leakage paths, under blower door test conditions, with a pressure differential of 50 Pascal (0.0073 psi, 0.2 inches water column). 

CFM50 is computed from leakage path area, under these conditions, as:

CFM50 = 7.5 * (Path Area, sq in), if the path is a common hole. Flow is higher where area is distributed in rectangular gaps.

Annual Heating Cost = $.555 * (Path Area, sq in)

The cost for a 1/4" annular gap about a 4" ceiling junction box is $0.555 * 3.14 /4 * (4.5² - 4²) = $1.85 per year. The area is 3.3 square inches.

The cost for a 1/4" annular gap about a 7" can light is $0.555 * 3.14 /4 * (7.5² - 7²) = $3.16 per year. The area is 5.7 square inches.

Blower Door Numbers For My Home:

I should wish to have 0.35 to 0.5 actual air changes in my home every hour, 0.35 to 0.5 ACHnat. This is accepted as necessary for good health. Corresponding blower door numbers are dependent on the air volume in my home. Say in round numbers for my modest 1955 ranch home, 1000 sq ft living space, eight foot ceilings, that volume is 8000 cubic feet. 0.35 changes per hour converts to CFM at natural conditions, as 0.35 Air Changes/hr*hr/60 min*8000 ft3/Air Change  = 46.7 CFMnat. Then multiply by twenty, to 933 CFM50. At 0.5 air changes per hour, numbers are 67 CFMnat and 1333 CFM50.

Each ACH50 for my home is 8000/60, 133 CFM50, and corresponds to makeup heat cost of $0.074*133 = $10 per year. The size of a hole that could carry this flow is 9.86/0.555 = 17.8 sq in, a circle 4.8" diameter.

Every ACH50 I may covet, could come through its 5" hole. 5" holes! No big deal. Driven by natural pressure differential without cost of electric power and with measurable increment of good health.

Think about this please. Math is for fuel cost bumped up to truer cost at $2 per therm. It's a 1000 sf ft home in a moderate climate formerly 4400 65° HDD and already down to 4000 HDD with global warming. If in silliness I have tightened my home to 3 ACH50, from 0.5 ACHnat, I am paying  less for cost of heating the air from ambient to 65°F, to the tune of $10 * (10-3) = $70. Now I need to pay for the driven fresh air. Say it is 25 watts, continuous, six months per year. I will be erratic in regulating the system while mindful of opening windows more in mild weather. 25 watts *24 hr * 365/2 * KW/1000 watts *$0.11/KWH = $12 per year. There are maintenance costs. I am not saving enough, for the bother. I admit though, I have an unmentionable reason to be on this path. I have a radon problem largely solved by letting ample fresh air in at bedside windows at night, and will be happy to relax that some if I achieve a plan to exhaust stale air through-roof, via a sealed, conditioned crawl space. With this solution, I don't visibly and audibly display the radon problem. I think I have answers and not many problems. Silly people with blower doors know none of this. They think they know better. They cause us so much trouble and expense. We are made to feel guilty about opening windows a bit, letting in the cold.

We should not allow careless leakage, as at can lights in the building envelope. In this, consider 2015 International Energy Conservation Code (Residential), Section R402.4.5:

R402.4.5 Recessed lighting. Recessed luminaires installed in the building thermal envelope shall be sealed to limit air leakage between conditioned and unconditioned spaces. All recessed luminaires shall be IC-rated and labeled as having an air leakage rate not more than 2.0 cfm (0.944 L/s) when tested in accordance with ASTM E 283 at a 1.57 psf (75 Pa) pressure differential. All recessed luminaires shall be sealed with a gasket or caulk between the housing and the interior wall or ceiling covering.

How much can light leakage area is allowed? Scale down first to 50 Pa differential, as 2.0 cfm * (50/75)1/2 = 1.6 CFM50. Area as a hole is CFM50/ 7.5 = 0.22 sq in. Area as a slot, more probable, is less. Say area is 0,2 sq in, for an annulus 7" OD. What is the annulus ID?

0.2 = 0.7854 *(72-ID2) ; ID2 = (49 - 0.2/0.7854)1/2 ; ID = 6.982". The average diametral gap is a tiny 0.018". And, this is ridiculous. An ordinary diametral gap is 1/8". Corresponding area is 1.36 sq in, and flow at 75 Pa is 1.36*7.5 * (75/50)1/2 = 12.5 cfm. Flow at 50 Pa is 10.2 cfm, and makeup heat with this leakage costs 0.555*1.36 = $0.75 per year. The amount of leakage and its cost might be higher with careless installation, easily by times four. Can lights should be sealed airtight. It does matter. But best savings with a can light, say beaming a 65 watt incandescent bulb, is in swapping the can contraption, for a ten watt LED disk giving more and better light. In that, save electricity, annually in running four hours per day: (65-10)watts*4*365 hr*KW/1000 watts*0.11/ KWH = $8.80 per year. All savings add up, and these are easy picking. Swapping a few can lights is pure profit, at that two or three locations, overwhelms all possible savings in so very difficult, even painful, squeezing of a few ACH50.

I think that the blower door people know none of this math.

Study the example of a 22x54 attic ladder:

A gasketed ladder might replace an old ladder 22" by 54" with an average gap tabulated here. Path Area = Perimeter*Average Gap. Perimeter = 152". Assume distributed-gap air flow is increased by times 1.37, vs. a simple hole. 

CFM50 = 1.37*7.5*152*Average Gap. 

Divide CFM50 by twenty, to note leakage at  natural conditions, Qnat, cfm. 

Average Gap     Path Area, Sq. In.    CFM50     Qnat , cfm   Annual Heating Cost, Leakage   Annual Heating Cost, Convection   

    0.01"                       1.52                 15.6                    1                                $1                                                                 $6.60

    0.06"                       9.12                   94                     5                                $7                                                                 $6.60

    0.12"                       18.2                 187                     9                              $14                                                                 $6.60

    0.25"                        38                   390                    20                             $29                                                                 $6.60

This considers only the gap between the door and the ladder frame, probably less than 0.06" on average, in any not-decrepit ladder. In addition, consider gaps between the ladder frame and its rough opening. A 1/8" average gap is common here, where molding against a rough ceiling may only hide the problem from view.  Gaps about the ladder frame must be plastered or caulked.  I use no molding, ensuring a problem is not concealed.

Examine the Energy Star standard for allowed exterior door leakage (within the door frame):

Where Energy Star standards give an allowed maximum leakage for an exterior swinging door, Q <= 0.5 CFM/ft2 at  75 Pascals differential (ASTM E283), see if this is a suitable requirement for a gasketed 22x54 attic ladder. Door area is 8.4 sq ft. Maximum leakage, Q75 , is 4.2 cfm.

CFM50 = Q50 = (50/75)0.5Q75 = 0.82 Q75 = 3.4 cfm = 1.37*7.5*152*Average Gap.  Average Gap = 0.002" 

To achieve this, gasketing must be indefinitely resilient, and have large compression, say at least 1/8". Say more-reasonable leakage past gasketing is the allowed, times ten. What does that cost? 

CFM50 = 34 cfm. CFMnat = CFM50/20 = 1.7 cfm. Annual Heating Cost = 0.074 * CFM50 = $2.50 per year. 

Know though that a healthy home has 5 ACH50 fresh air exchange, with nearly unavoidable cost to heat incoming air. For my 1000 sf home with 8-ft ceilings, this is Qnat of (1/20) * 5 * 8000 cfh * hr/60 min = 33 cfm. 1.7 cfm is a small part of the needed ventilation, and I may afford its $2.50 per year heating cost. Compare to a total 660 CFM50 fresh air heating cost of 0.074 * 660 = $49 per year.

Require full gasketing of the door, and full sealing about the door frame:

Contending with attic ladder hype:

With a leaky, uninsulated attic ladder, the cost due to leakage may exceed the cost of convection and conduction through the door. Often, leakage paths around a ladder frame are larger than those between the door face and the frame. Yet, leakage at a most-awful attic ladder will not rival the 80 cfm Qnat of a common bath fan. We must contend against hype, that an attic ladder can lose an inordinate amount of energy. One such is this passed along by Habitat for Humanity, as written by southface.org: Don’t leave a hole in the ceiling : A ¼-inch gap around the perimeter of a standard pulldown staircase can potentially leak the same amount of air that is supplied by a typical bedroom heating duct (~100 CFM). Unsealed, the attic access in a home leaks energy dollars and causes the house to be less comfortable. During winter, conditioned room air may escape to the ventilated attic, while in the summer, hot attic air (which may contain airborne insulation fibers) can infiltrate into the home.

Leakage beyond the Energy Star standard must be prohibited, and that is entirely a matter of reforming installation methods. It is achievable at no cost, that there shall be no leakage between any exterior or service door, and its rough opening. 

As Applied for General Infiltration Heat Losses:

Imagine the cumulative effect of blocking a few "holes":

Item                                                        Path Area, Sq. In.    CFM50        Annual Heating Cost, Leakage**

Closet door not gasketed*                                  9                       68                                  $5  

Closet open to attic, if other doors open*         40                     300                                 $22

Hole in outside wall, draped by panel*              40                    300                                 $22

* - These are chosen for discussion from actual experiences. Numbers have consistent math, but have quite uncertain bases. The gasket consequence is not worth debating. We would like to fix it. The door impact is best inferred from cycling the door, while running a blower door test. How would one guess at the path area?  We can estimate the area of an obstructed hole, or infer it from before/ after blower door tests. We should offer modest rewards for air sealing, with action driven more by the rewards of comfort and freedom to not isolate rooms.

** - I am distressed by these small numbers. I have wanted to believe there are larger penalties for such lack of care, and that rebates offered in Oregon for air sealing achievement, at $1 per cfm, are justified. Imagine a modest single-story house of 1000 sq ft, 8000 cubic ft volume. If it measures1333 CFM50 (at minus 50 pascal test conditions),  that is 1333 *60/ 8000 = 10.0 ACH50 (ten changes per hour of the house air volume). That house has more infiltration than is needed for a healthy home, say 7 ACH50. The ACH50 reduction from 10, to 7 is a blower door test reduction of 400 CFM50. The math above says the involved cost saving is $30 per year, and a $1 per cfm rebate is $400. Do these numbers make sense? If my math is in error, I ask: someone please contact me. I have cared deeply to understand this spectrum of conservation opportunity. 

Sealing measurable by a blower door tester sometimes includes furnace air duct leakage to an attic or crawl space. Leakage pressurized by a furnace blower has far greater significance, than other boundary leakage. I have experienced more than 200 CFM50 benefit from reattaching separated ducts, where the detachment had been hidden by crummy insulation wrap. Such repairs require only diligence, the conditions fixed are a result of negligence, and repairs cost almost nothing.  I am sure that savings far exceed $15 per year, yet don't see justification in much of a rebate incentive. I don't think measured air sealing achievement should ever be a basis for incentive rebates.

When air sealing achievement is rewarded, there is incentive to game the readings. Test results do wander. A large threshold for sealing-achievement rebates, 200 cfm, is a measure of test uncertainty. Conflict of interest problems are real, and should be avoided by a shift in concepts for "home performance testing." Plenty of this testing should be going on, but always with impartiality, and with inclusion of big-picture concepts including moisture control. Fewer providers of repair services should be in the business. What happens when they are not the best choice for a particular repair? The home owner becomes unfairly locked into employing the tester.

Attic Floor Pits:

The amplified heat loss under a big opening in the attic floor, to walls below, is typically larger than the loss through involved attic floor area, by an order of magnitude. I have blog posts on this. It is clear that big openings, like a pit to the ceiling of a stairwell, are - - pits. They are not "bypasses." They are almost entirely a convection opportunity. It is less clear what is going on at a far end of opening size to space below, in common wall header gaps, that might average 1/16" combined on both sides of a wall. In my house I expect that not more than 300 CFM50 of sealing to be achieved is in wall header gaps. The small expectation is consistent with a disrespect for such sealing in most weatherization, where change of blower door readings is so slight, and 400 CFM50 reduction targets are impossibly elusive. I can only allege that respect is important, for action as throttled floor pits, feeding convection loops.

Interior Wall Header Gaps As Attic Floor Pits:

In my own home, nearing completion of energy improvements in 2015, I find that the length of all wall headers is 107 ft, vs heated area now 1080 sf. The area of the gaps is 107*12*.062 = 80 sq in. That's a lot of area if it were fully engaged in infiltration. CFM50 = 7.5 * Area = 600, high by at least times two. The gaps are active and visible for upward flow by dirt deposition in insulation, but I believe there is a churning, flow both up and down, where cold air from the attic must fall, to feed the convection engine, tending to equalize in-wall and attic temperature. Gaps can be much greater than 1/16" both sides, raising the need to direct action for other than change of infiltration. I propose the math on-average , as part of the invention of "bare area equivalents:" The effect of sealing wall headers may also be estimated in equivalent bare area fully insulated. Say the area both sides of interior walls is the same as attic floor area, and stilling leakage/ floor pit bathing, is 10% of that which would occur if interior walls were directly open to the atticIn this, use further simplification, that interior wall area, two sides, is same as attic floor area. This is very unscientific, and none of the assumptions deserve sharpening for specific home details. Bare area equivalent as 10% of floor area gives an appropriate degree of importance. For my home with 107 ft of headers and eight-foot ceilings, the wall area involved with header throttled pits is 1700 sf. 170 sf not-insulated would be added to attic floor heat loss, if wall headers were not sealed. The cost is $2.4*170/3 = $136 per year, compared to infiltration cost for 300 cfm, of $0.074*300 = $22. I prefer the bigger number, despising an area 13 ft square left bare. Does it really seem less significant than that? The smaller number wrongly inspires inaction.