So I have wrapped my feeble mind around this for the last time……..

what am I doing wrong???????

I am calculating temp at the inside of my sheathing – but keep getting answers that don’t seem right…..
Zone 6A 8200 HDD – average mean temp over three months is 12.13F, but I am using 2F for my calculations. Inside temp is 70F, RH % 35, Dew Point is 41F.

Here is the wall…..from the outside to the inside

Hardieboard lap siding, 2×4 vent channel (does this have a R-value?), R18 (3″ ISO) or 2 layers of R18 (3″ ISO) for R36, then ice&water, then stripsheathing or plywood, then 3″ stud, then drywall, then latex paint.

Whether I use R 18 or R 36 ISO in my calculation, I get nearly the same temp on the inside of my sheathing……it can’t be right – right?????

Example 1:
delta T = C (comp R) / W (whole assembly R) * ( S (inside temp) – D (outside temp) )
delta T = 6 / 7.24 * (70 – 2)
delta T = .8287 * 68
delta T = 56.35

R6 (1″ ISO) + R1.24 (drywall, plywood, ice&water) = Total assembly R of 7.24.
70 (inside temp) – 2(outside temp) = 68
6(component R)/7.24 (total assembly R) = .828
.828 * 68 (temp change from outside to inside) = 56.35
does 56.35 get subtracted from the inside temp to let me know what the temp is on the inside of the sheathing? 13.65??? this makes more sense if there is only R6 of insulation in the whole wall…

Example 2:
delta T = C (comp R) / W (whole assembly R) * ( S (inside temp) – D (outside temp) )
delta T = 36 / 37.24 * (70 – 2)
delta T = .9667 * 68
delta T = 65.735

R36 (6″ ISO) + R1.24 (drywall, plywood, ice&water) = Total assembly R of 37.24.
70 (inside temp) – 2(outside temp) = 68
36(component R)/37.24 (total assembly R) = .9667
.9667 * 68 (temp change from outside to inside) = 65.735
NOW WHAT??? this is where the math seems right but the calculation seems wrong for the assembly