The values of the data that I took was as follows:
results with no cover in bright sunlight
{ "tslLUX":"65536.00", "uvVIS":"1677.00", "uvIR":"13354.00", "uvUVRaw":"761.00", "uvUV":"7.61" }
reading from site was 12.0 for UV
Kingstowne says 7.4 UV, 835 watts/m^2
results with plastic cover in bright sunlight
{ "tslLUX":"65536.00", "uvVIS":"1594.00", "uvIR":"12876.00", "uvUVRaw":"716.00", "uvUV":"7.16" }
results with polycarbonate in bright sunlight
{ "tslLUX":"65536.00", "uvVIS":"1600.00", "uvIR":"13228.00", "uvUVRaw":"720.00", "uvUV":"7.20" }
Kingstowne says 6.9 UV, 796 watts/m^2
results with soda glass in bright sunlight
{ "tslLUX":"65536.00", "uvVIS":"1627.00", "uvIR":"12333.00", "uvUVRaw":"734.00", "uvUV":"7.34" }
Kingstowne says 7.2 UV, 833 watts/m^2
results in the shade with no cover
{ "tslLUX":"2152.00", "uvVIS":"312.00", "uvIR":"962.00", "uvUVRaw":"30.00", "uvUV":"0.30" }
{ "tslLUX":"2168.00", "uvVIS":"315.00", "uvIR":"976.00", "uvUVRaw":"31.00", "uvUV":"0.31" }
Kingstowne says 6.7 UV, 817 watts/m^2
According to this data, I should not have a problem if I used the lid to my sandwich container (plastic cover in the test). I just have to compensate for the loss; there doesn't appear to be a loss associated with the UV sensitivity as far as I can tell. So I should multiply my UV readings by 1.0628 to get the same as the Kingstowne values? I should multiply my uvVIS values by 0.4979 to get the watts/m^2 equivalent to the values from the site where I am comparing my values.