The units conversion is a little confusing. Let's try it this way, with known values to test:
R = t/k
R: Thermal resistance
t: thickness
k: thermal conductivity coefficient
This lists foamed plastic insulation with k=0.03
we are looking for R-value per inch, so t in meters = 1/39.37 = 0.0254
R = (0.0254 m)/(0.03 W/m*K) = 0.847 m^2*K/W
to convert this SI R-value to US R-value, we multiply by 5.682
0.847 x 5.682 = US R-value of 4.81/inch of foam plastic.
This appears correct. Now we just need some k-values of soil to calculate with.
The same page lists dry & wet soil with a range from 0.15 to 4. This is a huge range, dependent on soil moisture. Let's first check these values, in units of US R-value per
foot.
thickness, t, of one foot in meters: t = 12/39.37 = 0.3048
R = t/k = (0.3048 m)/(0.15 W/mK) = 2.03 x 5.682 = 11.5 US R-value/ft, best case
R = t/k = (0.3048 m)/(4 W/mK) = 0.076 x 5.682 = 0.43 US R-value/ft, worst case
Can we narrow our range of likely k-values?
IEEE 442-1981, Guide for Soil Thermal Resistivity Measurements
See Figure 3 on page 9. The Y-axis is in units of degreeC*cm/W. We can convert to k-value = 100/y.
I'm going to select two values, 200 and 80.
k = 100/200 = 0.5
R = (0.3048 m)/(0.5 W/mK) = 0.61 x 5.682 = 3.47 US R-value per foot
k = 100/80 = 1.25
R = (0.3048 m)/(1.25 W/mK) = 0.24 x 5.682 = 1.36 US R-value per foot
Now if we look at data listed
here, units of thermal conductivity in Figures 5 & 6 are in Btu/(hr ft F), the reciprocal will give us R-value per foot..
1/0.52 = 1.92 US-R/ft for loam
to 1/0.96 = 1.04 US-R/ft for saturated silt/clay
to 1/1.44 = 0.7 US-R/ft for saturated sand
Soil type & moisture will make or break it..