# Diminishing Returns Of Insulation Question

This is diminishing returns of insulation performance per amount of insulation thickness as I understand it. It is based on doubling the amount of insulation and cutting the heat loss in half. But it seems like you get different results depending on the amount of insulation you start the progression with. What am I missing?

Below are two tables, each with two columns.

The left column is the inches of insulation. The right column is amount of heat loss as a percentage of the total heat loss occurring at the smallest thickness of insulation, which is expressed as 100%.

Table #1 starting with 1-inch of insulation:

1” 100

2” 50

4” 25

8” 12.5

16” 6.25

32” 3.125

64” 1.562

128” .781

Table #2 starting with 1/16-inch of insulation:

.062” 100

.125” 50

.250” 25

.500” 12.5

1” 6.25

2” 3.125

4” 1.562

8” .781

Both tables stop at an arbitrary cutoff point of .781, which is a 99.188% reduction of total heat loss.

Table #1 arrives at that reduction with 128” of insulation.

Table #2 arrives at that reduction with 8” of insulation.

Moreover, the total reduction of heat loss is greater in table #2 than it is in table #1 because table #1 starts with less insulation as a base point.

## Replies

Beats me

But 8" of insulation # 2 beats 10' of #1.

I go with #2, 10' thick walls went out of fashion in the Middle Ages.

Joe H

Yes, but here is the point:

Which table is true?

You would not want to use a false assumption and shiver all winter just to be fashionable.

Neither

table is "true". Both are abitrary and mean little if anything other than demonstrating the ability to work a calculator.

The whole concept of diminishing returns has to be based on a cost analysis.

There is a point at which spending more money for more insulation would never result in an acceptable payback. What is an acceptable payback? That depends upon who is spending the money and what their goals are.

In other posts, I reverted to using a formula provided by another poster that plugs in the actual heat quantity. But still, as you point out, this decision is hard to make in a practical sense. A builder could easily come to conclusion that is far different from what an owner/builder might decide for their own house. There is so much that goes into the decision besides just the cost of the insulation.

There are even subtle things like the fact that any given R-value of wall is likely to have some defects below that R-value. So if you add another increment of insulation, it brings the whole wall up to the intended nominal R-value.

As a owner/builder building my own house, I would error on the side of excessive insulation. Generally, I have decided on insulation values of approx. R-50 for walls, and R-90-100 for the ceiling.

I too

would err on the side of greater insulation as the owner/occupant of a building.

IF I were to set criteria for my own home, I would consider any payback less than 20 years (at today's prices, simple returns, nothing complex like a commercial lifecycle cost anaylsis) to be worthy of further investigation. Anything less than 10 years when a change/replacement was necessary would be a given.

You would want to model walls as composites, including the glass. R-50 walls with 20% glass at U= 0.1 would probably not be worth the money.

Should you ever have the opportunity, it is really nice to use a full featured heat loss/gain energy anaysis software, like Trane Trace 700 or even the simpler System Analyzer. I'm sure there are a multitude of excellent choices but I am most familar with those. Analyzing the entire energy use of a complex collection of systems is quite eye opening. Doesn't apply much to the single family home, though.

I too

would err on the side of greater insulation as the owner/occupant of a building.

IF I were to set criteria for my own home, I would consider any payback less than 20 years (at today's prices, simple returns, nothing complex like a commercial lifecycle cost anaylsis) to be worthy of further investigation. Anything less than 10 years when a change/replacement was necessary would be a given.

You would want to model walls as composites, including the glass. R-50 walls with 20% glass at U= 0.1 would probably not be worth the money.

Should you ever have the opportunity, it is really nice to use a full featured heat loss/gain energy anaysis software, like Trane Trace 700 or even the simpler System Analyzer. I'm sure there are a multitude of excellent choices but I am most familar with those. Analyzing the entire energy use of a complex collection of systems is quite eye opening. Doesn't apply much to the single family home, though.

You don't keep halving heat loss with every doubling. Because you atart with more in your first example you reach the point of diminishing returns quickly, with foam it's after about 4 inches, maybe 5. With your other example you're starting with such a small amount it takes more doublings to reach the same point. I thought Mongo's explanation in the other thread was very clear in explaining this.

Clarification

florida wrote:

Because you atart with more in your first example you reach the point of diminishing returns quickly,...

But it is the opposite case. Starting with more insulation in the first example reaches the point of diminshing returns much slower than the second example.

You're right.. I totally messed that up. What I meant to say is that you don't have savings with every double.

But, I think what you think you see in your two examples is more of an artifact of math than information about insulation.

I didn't notice this until just now but the OP in the firsr thread included a document at the bottom of his post that actually had all the answers all along. Then I posted a link to some graphs and data that said exactly the same thing. Then Mongo came along and did all the math in the thread so we could all see it and his calculations proved exactly the same thing again. So, the tables are there, the graphs are there, the formulas are there and the answers are there. I'm getting confused as to why you're still unwilling to accept the answers?

Forget the math for a moment, we all understand that 1/16" of insulation is not the same as 1" right? And thtat 1" is not the same as 4 inches?

Explanations

I know Mongo gave a formula, and I will review it to see if I can get a handle on this doubling insulation cutting the remaining heat loss in half. There is something about that that just seems like the sound of one hand clapping. One thing I have noticed is that when people show a chart, graph, or some mathematical explanation, there is often something missing, or a wrong conclusion conveyed. People just take the explanation for granted even if the numbers don’t add up.

Another thing to consider is that spay foam cost much more per unit of R-value than other insulations such as fiberglass or cellulous. So the spray foam industry has a financial interest in trying to make the case that R-value is not as important as it once was believed to be. So they come up with statements that place the economic threshold of diminishing returns very low for foam. And they also tell you that fiberglass is totally defective and offers no insulating value because it is air permeable. So there is plenty of motivation to stretch the truth.

You mention the insulation thickness pdf linked by the original poster in the other thread along with other explanations that all say the same thing. Do they? They all talk about diminishing returns, but I am not convinced that they agree. I cannot confirm what the pdf on insulation thickness is saying. It would like us to believe that 2” of foam is all you need, but you will have to add a little more to meet codes. Do you believe that there is no economic payback for foam exceeding 2 inches? The report assures us that it is all proven in a very complex mathematical spreadsheet.

The engineered calculations in the original document gave these values:

Inches of

SPFR-Value Efficiency

0 0.64 0.0%

0.5 3.99 84.0%

1 7.34 91.3%

1.5 10.69 94.0%

2 14.04 95.4%

2.5 17.39 96.3%

3 20.74 96.9%

3.5 24.09 97.3%

4 27.44 97.7%

4.5 30.79 97.9%

5 34.14 98.1%

5.5 37.49 98.3%

6 40.84 98.4%

The calculations in the engineered study I linked to gave values of :95% of potential savings after 4 inches of insulation, agreeing with the first calculations.

Mongo did the calculations on the thread and came up with 96% savings at the fourth inch of insulation.

So you have 3 engineered studies showing exactly the same results.

Building Science Corporation recommends 5.5 inches of foam.which tkes you somewhere right at or above 96% of potential savings.

So, reviewing all the information it appears to me that they DO say exactly the same thing. I believe that there is no economic payback for foam after 5.5 inches and so little after 4 inches that I would stick with 4 inches in my own house.

Your mistake is starting from 100 in both cases. You're not counting the reduction in heat loss from the first increment.

How To Correct the Progression???

I wondered about that, but how would you change it to make it correct? The two charts have to start at different basis points in order to demonstrate that results differ. If one is right and the other is wrong, which is which? You can’t start with zero.

But the larger point that I see is that the chart that arrives at the diminishing return the earliest is also the one that has the most return to diminish. It starts with the most heat loss and yet gets the job done with only 8” of insulation. Whereas the other chart starting with less heat loss requires 16 times more insulation to achieve the same result.

So if the difference in starting insulation thickness is skewing the results, it does not seem to be doing so as one might expect.

You already have your numbers. If 1/16" is 100 and 1" is 6.25, that's where you'd start for the list starting with 1".

Hmmm...

But that would mean that the second table is correct and the first one is not. Why should it be an accurate progression only if it starts with 1/16” of insulation?

If you start with 1 mil (the thickness of a hair), you stop over 99% of the heat loss at about 1/8” of insulation.

They're both accurate progressions, just to different scales. Your 100 starting point is arbitrary -- it could be 27.6 or 3582.

???

But, how can both tables be accurate when both arrive at the same diminishing return, except one arrives after 8" of insulation and the other arrives after 128"?

I know what I am trying to convey, but I am not sure where to start... Your example, assuming you are using the same insulation material, is flawed because you are starting with 100% each time... 100% of what.... Both tables are correct, because you are using percentages, and the two tables would have different starting points as far as heat loss, and are therefore in different scales... If I offer to double two peoples salary, would they then be making the same amount of money? It would depend on how much money they were making to start with!

Wood varies in R value from about .7 to 1.4 per inch depending on species, density, etc. But for this discusion, let's just say that wood has R-1 per inch. Now lets build a building that has no windows or doors, and we put 1 inch of wood on it (perfectly sealed of course...) If that building loses 1000 units of heat (we'll call them units) in this state, we have a place to start.

But if we build an identical building, except we only put 1/16 inch of wood on it (perfectly sealed, of course) it would be losing a LOT more than 1000 units of heat. It would theoretically be losing 16,000 units of heat. (assuming that the temp differentials and all aspects besides thickness of wood are the same...)

Here are some tables that might help you understand...

Table #1 starting with 1-inch of insulation (r-1 per inch wood):

Inches Percent Units of heat loss

1” 100 1000

2” 50 500

4” 25 250

8” 12.5 125

16” 6.25 62.5

32” 3.125 31.25

64” 1.562 15.62

128” .781 7.81

Table #2 starting with 1/16-inch of insulation (r-1 per inch wood):

.062” 100 16000

.125” 50 8000

.250” 25 4000

.500” 12.5 2000

1” 6.25 1000

2” 3.125 500

4” 1.562 250

8” .781 125

Now, to determine where to stop insulating using these diminishing returns on preventing units of heat loss, you have to know two more pieces of information... The cost to add the next inch of insulation and the long term cost to produce enough heat to offset not adding that next inch, including the cost and maintenence of the heating system and the fuel to run it......

They're both accurately predicting the reduction in heat loss RELATIVE TO THE STARTING POINT. Different starting points, different mother-in-laws.

Another Example

Rather than compare two tables, let’s just look at one table and compare it to common sense expectations. Let’s set aside the issue of R-values, heat loss units, and look only at insulation thickness at the start, and what happens when doubling that insulation halves the remaining heat loss.

So let’s start with insulation that is 1 mil (1/1000th inch) thick. Whatever common insulation that is, 1 mil is not going to stop much heat loss. So whatever heat loss is occurring, it is going to be almost all that would occur if no insulation exists. So just focus on the total amount of heat loss that is occurring with 1 mil of insulation, and regardless of the quantity of that amount, just call it 100%. It is 100% of whatever heat loss is occurring with 1 mil of insulation.

When you double the 1 mil of insulation, you are adding 1 mil, which is the thickness of a hair. That tiny amount of added insulation will cut the heat loss in half. Does that seem plausible? It doesn’t to me.

If you keep doubling the insulation, by the time you get to 1/8th inch of insulation, you have cut the heat loss by over 99%. Now granted, we are not talking about the maximum amount of heat loss that would exist with no insulation at all, but with a start of only 1 mil, surely the heat loss we are considering is nearly all of the potential heat loss with no insulation.

Or is it? Say we start with 1/10th mil of insulation. Double that, and you cut the whole remaining heat loss in half. By they time you double the insulation 7 times, you cut the heat loss by over 99%. And you do that with only 13 mils of insulation, or about the thickness of a few sheets of paper. Does that seem plausible?

The problem is, when you have a layer only 1 mil thick "skin effects" provide more insulation than the layer itself. Depending on the circumstances you can have from maybe 1/4" to a couple of inches of air insulating, with an R value from roughly 0.15 to about 1.0.

Okay, then just deduct the skin effect plus the R-value of the 1 mil of insulation from the total heat loss for zero insulation, and use that result as the basis for the progression. It should not change what I am trying to illustrate.

Then you basically have to start with an assembly that has an estimated R of 1 or so (from 0.30 to 3.0 for the double air skin, depending on wind and other factors) and begin doubling that. The first doubling will not be doubling of thickness because the original assembly includes the skin effect. 1/4" or so of polystyrene would be about R1, so could be used for the first doubling. (And, in fact, 1/4" of polystyrene by itself could comprise both the infinitely thin membrane and the first doubling, for about R2 total. The next doubling would add 1/2" of poly.)

http://ufgscriteria.tpub.com/3_440_03/3_440_030073.htm

BTU mulligan

KDESIGN wrote:

Table #1 starting with 1-inch of insulation:

1” 100

2” 50

4” 25

8” 12.5

16” 6.25

32” 3.125

64” 1.562

128” .781

Table #2 starting with 1/16-inch of insulation:

.062” 100

.125” 50

.250” 25

.500” 12.5

1” 6.25

2” 3.125

4” 1.562

8” .781

Both tables stop at an arbitrary cutoff point of .781, which is a 99.188% reduction of total heat loss.

Table #1 arrives at that reduction with 128” of insulation.

Table #2 arrives at that reduction with 8” of insulation.

Moreover, the total reduction of heat loss is greater in table #2 than it is in table #1 because table #1 starts with less insulation as a base point.

Here are a couple of points that I'll try to reinforce to you:

1) In your two tables you're simply focusing on the fact that doubling the insulation thickness seven times will sequencially halve the heat loss seven times, from a starting value of "100% heat loss" to a final value of "0.781% heat loss". It's a mathematical cerrtainty that halving ANYTHING it seven times will always lead you to the final figure being .781% of the initial figure. But by using arbitrary starting points, it doesn't relate to real heat loss in the real world.

2) In your two tables you're assigning a value of 100% heat loss to your first values of 1/16th" and 1" of insulaton. That may be fine for arbitrary discussions, but not for heat loss discussions, because it's not real.

Example: If I start my own Table 3 with Value 1 being 10" of insulation and I consider it to have 100% heat loss. After seven doublings I'm going to end up with 1280" of insulation and a final value of .781%. Is that a valid comparison? It is in terms of knowing that seven doublings gets me to .781% of my initial value. But my initial value is flawed in the heat loss world because we know you don't have 100% heat loss with 10% of insulation. Nor is is economically feasible to have 1280" of insulation.

Or I make up Table 4, where I assign Value 1 a thickness of 1/250th of an inch of insulation and consider it to have 100% heat loss. After seven doublings, I'll have a little over a half-inch of insulation, But I'll still have a final value of 0.781% compared to the starting value of 1/250th of an inch. I still achieved the .781 number. But it's not economically feasible to have just a half-inch (R-3) of insulation in a structure.

None of the above are valid comparisons of anything, other than I'm mathematically proving that if I double ANY number seven times (in this case the thickness of the insulation), the initial number will be .781% of the final number. And vice-versa; if I halve a value seven times (the heat loss), the final value will be 0.781% of the initial value.

This is about heat loss, and the end game is to reduce heat loss to a sustainable level. Using the two extremes of Table 3 and Table 4 as examples, both give me .781% as a final value. But as a final value of what? Table 3 has me with a final thickness of 1280" of insulation, Table 4 with a final thickness of a little over a half-inch, or 0.512", of insulation. Which would retain more heat?

There seems to be the thought that Table 4 is "better" because it gets you to the .781 value with only 1/2" of insulation, where the other three tables get you there with greater thicknesses.

Remember, these comparisons are all fluff when it comes to heat loss. They aren't real apples-to-apples heat loss comparisons. They aren't heat loss comparisons at all. Because we're not comparing heat loss. All we are comparing is the mathematical fact that when you halve something seven times, you eventually end up with .781% of your starting value. That's is.To add heat loss into this comparison you need to use the "Q=" heat loss equation I posted in the other thread. And when you use that equation you have to honor R-value, since "R" is part of the equation. And that's where your Tables 1 and 2 and my Tables 3 and 4 fall apart when it comes to comparing those heat loss tables. We're assigning 100% heat loss to the initial insulation value in each table.The premise of your two tables is that 1/16th" of polyiso and 1" of polyiso have 100% heat loss and the premise of my two tables is that 1/250th" of polyiso and 10" of polyiso have 100% heat loss. We know that's not true, so our tables can't be used to compare heat loss.

All they are doing is comparing doubled numbers. That if you double a number seven times, you get the same fractional result regardless of what initial number it is that you are doubling.

3) The actual idea of diminishing returns. You are correct, in that when you start with R-0, any insulation will be a marked imrovement over R-0. Each initial thickness will save you BTUs because from the start you are losing all your BTUs. If my initial insulation thickness is 1/32nd of an inch, I'm saving some BTUs compared to an uninsulted wall, but I'm still losing a huge amount of heat. If I double that to 1/16", even though I've doubled my insulation, I only added another 1/32nd of an inch. I've doubled my savings, but I'm still losing a huge numbers of BTUs.

Then I double again, to 1/8th", then 1/4", then 1/2, 1, 2, and 4, each of those doublings saves me a signficant number of BTUs per doubling, again, because I was still losing so much heat prior to the doubling. My savings in the earlier doublings is more significant than the savings in the later doublings, because the initial doublings used less insulation per doubling and the heat loss was greater earlier, so there were more BTUs available to be saved.

Again, for the theoretical part, after seven doublings of the R provided by the just the insulation, from 1/32nd" to 4", I'll have reduced my heat loss to about 0.781% of what it was initially. Financially it was worth doing all those doublings because my initial insulation was so thin that I was losing huge BTUs. At an end thickness of 4" thick, I finally got to the point where based on my design load of a deltaT of 40F, I was retaining enough BTUs for the structure to remain comfortable.

To go to the other extreme, if I START with an uninsulated wall and then go to 4" of insulation with my initial thickness, that inital thickness will cut my heat loss significantly. We just showed it'll reduce it to .781% of the initial loss. Even though I've only put on one layer of insulation, I have 4" of insulation, the same thickness I had after the seven doublings in the first example. I could double it again to 8", but since I saved so much with the first 4" layer, there's less to be saved with the second 4" layer.

If I went from 4", to 8", then 16, 32, 64, 128, etc, etc...after seven doublings I'll have reduced my heat loss to 0.781% of what is was with 4" of insulation. But financially, was it worth doing that? Especially when the first 4" of insulation had already reduced the loss to .781% of the uninsulated wall number?

I think the point you are missing is that the 1/32nd" to 4" doublings, they are not independent of the 4" to 512" doublings. The 1/32" to 4" doublings are incorporated into the initial 4" layer of insulation in the second doubling example, from 4" to 512".

Diminishing returns.

When do we stop doubling? When you get to a sustainable heat loss. In Mexico City I might only have to design to a deltaT of 10F and with solar gain the R-vaue of the uninsulated walls alone might be adequate. In New England 40F might be more appropriate, and 4" of polyiso might be best with a decent oil burner. In Antarctica it might be a 90F deltaT and with a minimal heating source I might need 12" of polyiso. It's getting enough insulation for your design criteria. Minimizing the structure's heat loss just enough so that passive and/or active heat gain can keep the structure comfortable.

As far as heat loss curves and "seeing" the diminishing returns, seeing "where" they occur:

Each of those doubling runs I wrote about earlier; from 1/32" to 4" and from 4" to 512" of insulation, are part of the

samecomparative heat loss curve because we are talking about the same type of insulation. Combine all those "insulation thickness" versus "BTUs lost" values into a single heat loss curve and they represent 15 points plotted on the curve. If you take any two points on the curve that are eight points (seven doublings) apart, one of the points will always be .781% of the other point. That's a mathematical certainty.The R-value for 1/32nd" of polyiso is R-0.1875. The R-value of 4" of polyiso is R-24. And 0.1875 is .781% of 24. So

1/32nd" is .781% of 4"

1/4" is .781% of 32"

1" is .781% of 128"

4" is .781% of 512"

That's the math. But those thicknesses alone relate to nothing in terms of BTUs, or money, or R-value.

Were you to assign those thicknesses an R-value per inch and then plug them into the Q heat loss equation designed around a certain deltaT, now you have something you can apply apart from the basic math, you can relate insulation thickness to heat loss.

Remember, in Table 1 through 4 we each assigned the starting values of the first thicknesses insulation as having 100% heat loss. That's horribly flawed.

When we plot "insulation thickness R-value" versus "BTU loss" for those 15 thicknesses above using the Q equation, we get a "BTU savings curve", or a heat loss curve. The steepest part of the "BTU savings curve" will be with your initial insulation doublings. Going from no insulation to 1/32nd, then 1/16th, then 1/8th, etc. You have no insulation at first, so at first you're losing all your BTUs. Any insulation you do add will give you significant savings. The savings curve will be steep. When you get into the thicker doublings later on, you've already saved significant BTUs, so the savings curve flattens out because not only is there less to save, but with each doubling you're adding a huge (and comparatively expensve) thickness of insulation, so with fewer BTUs to save and with thicker layers being added, you're saving fewer BTUs per inch added.

So you get a big bang for your insulation buck with your first thicknesses. You get lesser bang for you buck with later doublings.

Diminishing returns.

Mongo,

Thanks for that comprehensive explanation. Yesterday, I went back to your earlier thread, reviewed the formula, and plugged in values to see how it works. It seems to work fine and does not raise any questions in my mind. I understand the concept of diminishing returns, but I could not reconcile what I was seeing with the way I was looking at simply doubling insulation. My use of 100% was intended to define the heat loss remaining after the first increment of insulation used as a starting point.

I was reasoning that when you start with a very small increment of insulation, the heat loss remaining would be nearly all of the heat loss possible. Looking at the whole thing that way required starting with a small increment because I could not start with zero.

But now I see that the total heat loss with no insulation is infinite. That is why you can’t start with zero the way I was looking at it. So even when starting with a small increment, the remaining heat loss is so enormous that even a 99% reduction after 7 insulation doublings still leaves a lot of heat loss. That is the part I was not seeing.

But still, it is a little hard to see how the total heat loss with no insulation can be infinitely large when it is stipulated to be within a given area with a given temperature differential. I know it works with the formula, but I want to be able to picture these things. Although, I suppose that, in a vacuum, the speed of thermal transfer could be infinite. Hmmm…

Mongo,

Thanks for that comprehensive explanation. Yesterday, I went back to your earlier thread, reviewed the formula, and plugged in values to see how it works. It seems to work fine and does not raise any questions in my mind. I understand the concept of diminishing returns, but I could not reconcile what I was seeing with the way I was looking at simply doubling insulation. My use of 100% was intended to define the heat loss remaining after the first increment of insulation used as a starting point.

I was reasoning that when you start with a very small increment of insulation, the heat loss remaining would be nearly all of the heat loss possible. Looking at the whole thing that way required starting with a small increment because I could not start with zero.

But now I see that the total heat loss with no insulation is infinite. That is why you can’t start with zero the way I was looking at it. So even when starting with a small increment, the remaining heat loss is so enormous that even a 99% reduction after 7 insulation doublings still leaves a lot of heat loss. That is the part I was not seeing.

But still, it is a little hard to see how the total heat loss with no insulation can be infinitely large when it is stipulated to be within a given area with a given temperature differential. I know it works with the formula, but I want to be able to picture these things. Although, I suppose that, in a vacuum, the speed of thermal transfer could be infinite. Hmmm…

To Infinity...And Beyond!!

KDESIGN wrote:

But still, it is a little hard to see how the total heat loss with no insulation can be infinitely large when it is stipulated to be within a given area with a given temperature differential. I know it works with the formula, but I want to be able to picture these things. Although, I suppose that, in a vacuum, the speed of thermal transfer could be infinite. Hmmm…

That's why much of this discussion is theoretical and very roughly shaped and not 100% practical in nature. When I wrote "R-0", you are correct, you never really have a true R-0, because the earths atmosphere itself, air films, etc...even when there is "nothing there" in terms of foam insulation there is still "something there" providing

someresistance to heat transfer.It's the "open window concept" in a house. If you open a window, does all the "heat", or do all of the BTUs, escape instantaneously due to their being "R-0" in the window space? Of course not. A true R-0 in the Q formula would, as you wrote, indicate "infinity heat loss per hour". It doesn't work that way in reality, because we both know in our typical environment we don't see R-0. But sometimes we have to talk simplistically like that to relate certain concepts, or compare certain ideas.

But as you wrote, the larger issue to grasp for diminishing returns is the idea that it's a function of how many BTUs you have saved versus how many more you still have to save. When you're spending a little to save a lot and there's still a lot more to be saved, it might be prudent to spend more to save more. When you're spending a lot to save a little, you've gotten to a good place.

There is an analogy with quantum mechanics: With virtually all phenomena, when you keep going smaller and smaller you usually get to a point where the physics change character. It's not turtles all the way down.

Reply

Thanks Mongo. My reply seems to be stuck in the system somewhere.

Edit:

Well I just tried it again, but it will not post because it says it is being held for moderator approval. I don't get how that works with this forum.

Of course, as I pointed out earlier there's no such thing as R-0. Microscopically thin aluminum foil still has an R of 0.3 to 3, due to the skin effect. R-1 is probably a fair general-purpose estimate.

being succinct isn't one of your strong suites is it? :)

nope

this time it is

Point of no return.

Experiance shows that after 8 inches of polyurethen foam is installed in a ceiling, the result becomes immeasurable.

Buying and fitting the wireless remote of a Weather Station in the roof, enables the 24/7 monitoring of the humidity and temperature in the roof on your laptop.

You're doing it wrong. You need to take the insulation thickness and the total U-value of the construction with each layer of insulation. Both graphs will be the same.

The difference is that in one case you start out at one point on the graph and move to another point (by adding insulation). In the second example, you start out at a different point on the graph and move to another point (again by adding insulation).

Dimiminshing returns: If I start w/ e.g. 1 or 2 inches of insulation, my savings are much smaller than if I start out w/ no insulation (assuming I add the same amount of insulation). If I add more insulation, I can save more, but relatively not as much.

Adding insulation depends also on: do I have the space to add insulation (e.g. 2x4 vs. 2x6 wall or unlimited attic space and how much more does the extra inch(es) of insulation cost given that the insulator is already on site and mobilized.

Diminishing returns have to be crossed w/ relative cost. If you have extra money in your pocket, put in the extra. But remember, once you've added a few inches, you've eliminated a large percent of your heat loss. Also remember, though that to add insulation LATER will cost you proportionally a LOT more and you'll get a LOT less benefit out of it.

So the moral of the story .... do all you can afford, NOW.

Insulation

I don't think this table takes into consideration thermal drift.