Ice thickness

Safe ice thickness for various activities requires varying levels of thickness based on the weight of what is venturing out onto the ice.
The guidelines vary based on a number of factors, a chart in the Farmer’s Almanac at http://www.almanac.com/outdoors/safeice.php indicates that ice would need to be roughly 4 inches thick to be considered safe. 

Ice thickness should always be tested before heading out onto any frozen body of water.  There are many variables and factors that influence thickness besides air temperature.  However, an estimate is useful for deciding if it might even be worth bothering with making a trip to your favorite ice skating, ice fishing, or snowmobiling river or lake.  Using a mathematical model is only a start--if the weather conditions seem supportive and the model gives a positive indication, it then becomes worthwhile to spend the time and effort to obtain firsthand observations to verify the true ice thickness. 

The following is a formula from the US Army Corps of Engineers Cold Regions Research & Engineering Laboratory in Hanover, NH:
(See also www.crrel.usace.army.mil/library/technicalnotes/TN04-3.pdf)

Ti = C √ AFDD

where:
Ti = ice thickness in inches
AFDD = Accumulated Freezing Degree Days (in °F days)
C = Coefficient for use in modified Stefan equation with ice thickness in inches, AFDD in °F (see table below)

Condition                            Typical Value of C
Windy lake with no snow        0.8
Average lake with snow          0.5 to 0.7
Average river with snow          0.12 to 0.15
Sheltered small river              0.21 to 0.41

The AFDD is derived from FDD, or
FDD = (32 – Ta), where Ta = Average Daily Temperature

For AFDD, data can be obtained from the NOAA data or other and is sort of a net FDD. 

An example, it appears for December 8, for the Hennepin canal area, the AFDD was in the 100 to 200 range.  However, just to the south, Peoria was in the 0 to 100 range.  If we assume AFDD = 100 and treat the Hennepin as a sheltered small river, using the upper end of the range for C—as a canal, the water isn’t moving in most spots and is closer to a lake—we then can assume C = 0.4.  

Thus, with AFDD = 100 and C = 0.4, an approximate ice thickness of 4 inches is obtained, as Ti = (0.4)*(√ 100) = 4.0.  I had reports that people were out ice fishing in the area near the Hennepin Visitor Center—but this was on a lagoon, which more closely resembles a lake, thus one could assume a higher coefficient, say 0.5, there.  The actual canal was mostly frozen but I also heard that the ice may not have been as thick or solid there as in the lagoon.

This formula uses broad assumptions.  While the metrological data is going to be fairly accurate, the choice of the coefficient C involves much guesswork.  While at places the canal is somewhat like a windy lake, in others, moving water at the locks or due to a spring may make it more like a river, with a much lower value for C.  And the ice strength is also influenced by the quality and nature of the ice—white, bubble-filled ice counts for only half the strength of clear ice.   

Sure, one could blithely say that the canal is like a lake and assign a high value to the parameter C, but making assumptions about parameters in mathematical formulas (in the finance world, assumptions about parameters such as default rates, potential real estate value decline percentages, liquidity, etc.) is what those investment bankers and financial whizzes at Lehman Brothers, Bear Stearns, A.I.G., et al did when constructing their models about CDOs, CDSs, funding, and other complex financial instruments.  So, bottom line is, don’t ever be too confident in any formula, even if the formula is right it won’t matter if you are making the wrong assumptions about parameters.  TEST THE ICE ALWAYS FIRSTHAND.  We’ve already spent something like a trillion dollars because people made the wrong assumptions about parameters in formulas.