replying to willshak, Don Forcash, Mechanical Engineer wrote:
willshak wrote:

Don't really know precisely. But do you know that the material used in
insulation has no effect on the R value?
It is the amount of stagnent air trapped in the material that has any
effect on the true R value.
If you crush 4" thick insulation down to 2", you can expect the R value
to decrease. Perhaps even by half.

The R-Value per inch goes up within a band of approx 25% compression, but
the R-value of the purchased thickness goes down. Once you pass this
critical compression point, the R-value per inch begins to go down again
and the R-Value of the purchased thickness continues to go down. The
science has to do with the trade from compression of the increased
resistance to airflow (convection heat transfer) versus the increased heat
conductivity. A practical way to visualize this is a limit analysis at
the extreme end of the total compression vs. heat flow curve. At the
extreme end of the curve, fiberglass is compressed to the point of being
glass, with essentially no ability to stop heat flow (maybe R=1). To
save money, fiberglass insulation manufacturers go a little light on the
compression - Just shy of the optimal density. Compressing to approx 75%
of purchased thickness achieves the optimum R per inch, which will yield
approx 6% is R-value per inch improvement. So putting a 6" batt in a 4"
stud cavity can get you to R-15. Interestingly, some manufactures sell a
high-density more expensive 3.5" fiberglass insulation that will actually
give you an R15 at purchased thickness. If you were to further compress
this insulation you would see a reduced R per inch and a reduced total
R-value because they have already compressed it to the optimal point.

This is real science - Not opinion. If one wants to spend the time, there
is empirical and analytical data on the web that explains this in both
mathematical and measurement terms. Hope this helps.


--

"Don Forcash"; "Mechanical Engineer"
wrote in message
roups.com...
replying to willshak, Don Forcash, Mechanical Engineer wrote:
willshak wrote:

Don't really know precisely. But do you know that the material used in
insulation has no effect on the R value?
It is the amount of stagnent air trapped in the material that has any
effect on the true R value.
If you crush 4" thick insulation down to 2", you can expect the R value
to decrease. Perhaps even by half.

The R-Value per inch goes up within a band of approx 25% compression, but
the R-value of the purchased thickness goes down. Once you pass this
critical compression point, the R-value per inch begins to go down again
and the R-Value of the purchased thickness continues to go down. The
science has to do with the trade from compression of the increased
resistance to airflow (convection heat transfer) versus the increased heat
conductivity. A practical way to visualize this is a limit analysis at
the extreme end of the total compression vs. heat flow curve. At the
extreme end of the curve, fiberglass is compressed to the point of being
glass, with essentially no ability to stop heat flow (maybe R=1). To
save money, fiberglass insulation manufacturers go a little light on the
compression - Just shy of the optimal density. Compressing to approx 75%
of purchased thickness achieves the optimum R per inch, which will yield
approx 6% is R-value per inch improvement. So putting a 6" batt in a 4"
stud cavity can get you to R-15. Interestingly, some manufactures sell a
high-density more expensive 3.5" fiberglass insulation that will actually
give you an R15 at purchased thickness. If you were to further compress
this insulation you would see a reduced R per inch and a reduced total
R-value because they have already compressed it to the optimal point.

This is real science - Not opinion. If one wants to spend the time, there
is empirical and analytical data on the web that explains this in both
mathematical and measurement terms. Hope this helps.

Eeegads! There's alotta big words here ^@^