This started as a Christmas discussion and when I went to look, I could not easily find anything on the physics of ice roads. So, I thought I’d try to work out some of the basics from first principles and a bit of guess work.

Let’s start with an assumed 10tonne vehicle. If we assume the density of ice is 9/10 of normal water, then if 100tonnes of ice is pushed down into the water, it will support the weight of a 10 tonne lorry. If we then assume the ice depresses by 10cm (which seems a lot), then the area that is pressed down is around 1000m² or 31 x 31m..

If I were to guess, I’d say that typically ice sheets break so that the length is around 10x the thickness (total guess!!). If true that suggests that the ice needs to be at least 3m thick to produce slabs over 30m x 30m.

However, lorries don’t sit on the ice, instead they drive over it. And indeed I’ve seen rules that say “don’t stop on the ice!!”

# Dynamic calculations

Let’s again assume a 10tonne vehicle and this time choose a slab size of 10m x 10m which corresponds more closely to the size of a typical lorry. Now the pressure under the ice is 10tonne x 9.8 /100m² or near enough 1pascal (*As shown below this calculation proved unnecessary*).

If we assume the layer of water that moves under the ice is 1m deep, then we have 100tonnes. So the acceleration of depression is given by:

mass x velocity = force x time

force/mass = velocity/time = acceleration

100,000 N / 100,000 kg = acceleration

So the acceleration based on these simple assumptions is 1m/s². So how long can we stay on each slab before it depresses by 0.1m?

S = ut + 0.5 a t²

0.1 = 0 + 0.5 x 1 x t²

t = (0.1/0.5)^0.5 = 0.45s

Speed is given by:

v = 10/0.45 = 22m/s or 50mph

This figure is clearly far too high because recommended travelling speeds are under 25 km/h (16 mph) or between 40–70 km/h (25–43 mph).

Where have I gone wrong? The biggest unknowns (for me) are the size of ice-slab and the amount of water that is moving. But as both effect the amount of mass moving, we can estimate how much these must change as follows: We can also see that if the mass being moved increases by 9 times, that the speed reduces by root 9 to become 17mph. As vehicles do move more slowly, if we assume the only error is the mass, then the mass being moved must be perhaps 100x larger than my estimate, allowing speeds down to 5mph. But we can also see, that if we assume that instead of 0.1m that the ice depresses by 1m, then the mass of water being moved must be 10x greater or 1000 tonnes at 5mph.

# The wave effect

We now see that in effect the lorry is riding “on water” in the sense that what is keeping the lorry up is not the floatation of the ice, but the force needed to start the water moving. However, once the water starts moving, it keeps moving, in what we normally see as a wave. But these waves can be problematic on ice. Because at a speed between 16 to 25mph there is the possibility of a resonance in the ice which can cause it to break.

This is not a wave travelling through the ice like a sound wave, but instead it is an upward-downward wave like typical wave on the beach.

The speed of a wave in water is approximately

v=√((gλ/(2π)) tanh(2πd/λ)),

where

v is velocity (m/s)

g is the acceleration due to gravity (m/s²)

d is the depth of the water (m)

λ is the wavelength of the wave (m)

For shallow water (d≪λ) this simplifies to

v=√(gd)

d = v² / g

In the range of 25 to 40 km/h for water, this corresponds to depths between 5 and 13 meters deep, which seems not unreasonable for depths of many frozen rivers and lakes.

# Aquaplaning

Another way that we might think about this is that the lorry is effectively “aquaplaning” across the water with the ice spreading the load so as to create a very low ground pressure. As aquaplaning speed varies as the square root of tyre pressure, this might tell us how fast we need to travel to move across ice,

Typically cars with tyre pressure of 30psi aquaplane at 50mph, with the speed proportional to the square root of the pressure. If we assume 100m² and 10 tonnes, the pressure (as above) is 1 pascal or 0.00015 psi. That suggests the aquaplaning speed reduces from 50mph to 0.1mph. For this situation, that figure is clearly nuts. And thinking about it, the reason is obvious: aquaplaning occurs on thin sheets of water a few mm thick. Yes, if the water under the ice were a few mm thick, then this might tell us something, but clearly the deeper water makes the situation too different to use this simple calculation.

# Discussion

These calculations were simply to get an idea of the factors involved and to see whether I could get values that were not too unreasonable based on the assumed effects. I found that I had to increase the depth of water assumed to take place in the effect from 1m to around 10m as well as assume depression of about 1m of ice. Looking at the wave equations and the recommended driving speed, the 10m depth of moving water does not look out of place. But 1m movement seems far too high. In reality, the total amount of water moving is not just that under the immediate ice, but also that around about. This would dramatically reduce the necessary ice-depression. If we assume a Noughts and crosses grid, with the lorry in the centre, then the total ice “next to” the centre is 9x the central block or almost an order of magnitude larger, suggesting that the ice-depression would be closer to 0.1m rather than 1m.

Thus the general principle is tending toward the right kind of numbers.

However, one huge area of uncertainty, is the average size of slab. Indeed, whether we can even talk about individual slabs, because the joints would undoubtedly be irregular making it hard for one slab to move without effecting the neighbours. That makes the situation very complex. Also, I’m not sure how I can work out the “shear strength” or how much force is required to break apart a slab of a certain thickness. My guess is that these figures are very complex in real ice which is subject to freeze and thaw and to wave movement and so I cannot readily estimate them.

Also, I have throughout assumed a 10 tonne vehicle, because I’m basing my estimates on “ice-road truckers”. But in reality, I think I’ve seen much heavier vehicles. However, 10tonne is an easy number to use. Indeed, many of the ice-roads in Europe have a limit of a couple of tonnes. So 10tonne seems to be on the high side.

One final thought – I’ve calculated the depression when the lorry is present and suggested that is the total depression. But in reality, once the ice starts moving, it will continue to move. And if the depression is a small fraction of what would occur if the lorry sat on the ice, then the upward force is much smaller than the downward force of the lorry, so it will take much longer to decelerate. So perhaps a 0.1m depression as the lorry crosses, will continue for another perhaps 0.5m after it leaves that section of ice.

It all sounds very hairy!!

*NB. The actual density of ice is 0.9167 –0.9168 kg/litre at at 0 °C which means slightly less buoyant than I suggested at the start.*