If you dig a pit, you will need drainage. Without drainage, meltwater will increase the rate of ice loss.
Let's put some science behind the icehouse.
Heat is the enemy. It moves through 3 primary methods: Conduction, Convection, and Radiation.
In an icehouse, the solid mass of ice in an stagnant envelope of air or insulating material has little convection to worry about. If you open the door in the summer, some heat will move in and flow upward, but it is expected that you will shut the door shortly.
Radiation is not much of a concern because you will have insulated the structure and/or the ice.
The primary issue with which you must contend is the conduction of heat from the outside to the inside.
This is where Fourier's Law comes into play: q = k A dT / s
where
q = the quantity of
energy conducted through a wall. You want this to be really low.
A = heat transfer area (m2, ft2) This is the size of your ice house or your ice heap
k = thermal conductivity of the material (W/m.K or W/m oC, Btu/(hr oF ft2/ft)) The lower the value, the better the insulating property of the material
dT = temperature difference across the material (K or oC, oF) If it's 90 degrees outside, 20 degrees inside, the dT is 70
s = material thickness (m, ft) Thicker insulation will make your ice last longer
A list of k values for selected materials. 'k' is the Coefficient of Thermal Conductivity. Low k, low heat transfer. In the list, sawdust has a k value of .08, so heat moves through sawdust slowly. Dry earth is on the list with a value of 1.5. You would do well to add material between the earth and the ice.
Now let's look at a practical example. I'll use my old neighbors icehouse, and convert everything to metric.
A: 4 walls 12x10, roof and ceiling 12x12, A=768 sqft =71.3 square meters m^2
k: polyurethane foam, k=.03 W/(m.oC)
dT: 80F summer days, 50F summer nights, 65F degree average outside temp, 20F inside, dT=45F = 7.2 degrees Celcius oC
s: The insulation was sprayed on to a thickness of a foot and a half, s=1.5 ft = .45 meters m
q = k A dT / s
q= (.03 x 71.3 x 7.2)/.45 =W/(m x oC) x m^2 x oC / m
q=34.2 Watts
1 watt = 3.41214163 BTU / hour
q=116.7 BTU per hour
1 BTU is the energy required to raise 1 pound of water by 1 degree F
A pound of ice at 20 degrees, raised to 32 degrees requires 12 BTU. There is about another degree required to change the state from solid to liquid.
In laymans terms, his ice melts at a rate of about 3.5 pounds per hour in the summer, about 10 gallons per day.
To keep ice all year, 275 days of melting, around 2750 gallons of ice will be lost to melting, 370 cubic feet, the top 2.5 feet
If he used sawdust (k=.08 ) instead of polyurethane foam(k=.03), his rate of ice loss would be 27 gallons per day, 1000 cuft, 7 feet over the season from a 10 foot pile of ice. Would not leave him much to work with, but not every day is the heat of summer.
Edited for 2 calculation errors