Standing On Ice

[seasalt]

Interesting.


Standing on ice - When is it possible?

By Nir Shaviv, Mon, 2006-08-14 00:38

Ice covering Grenadier Lake, High Park Toronto. The ice was 8 cm thick, enough to stand on. The small patch in the ice is a dead fish, frozen into the ice. How cold should it be and for how long to have ice thick enough to stand on?
Ever wondered whether it was sufficiently cold for sufficiently long to allow you to stand on ice, without falling in? I once did. Here is an estimate for the duration required to reach a given thickness. Actually, it is a lower limit, since we assume a few simplifying assumptions.

The most important assumption is that the water beneath the ice is still, such that it could have cooled down to a near freezing temperature (if it cannot, because of constant mixing for example, more heat will be available). In other words, the estimate is a lower limit. Don't try to stand on the ice just because you calculated it to be thick enough!

Having cleared that up, we can begin with the estimate. We assume that the time it takes the heat to diffuse across the ice is much shorter than the time it takes the ice to form. This implies that the temperature profile can reach its equilibrium which is a linear change from the bottom to the top. (We will later have to check the validity of this assumption). Thus, at every instant, we assume that the profile has the form:
   
where x is the width of the ice at the given instant. T0=-ΔT<0 is the external temperature (again, we assume it be 0 at the inner edge).

As time progresses, the thickness increases x→x+δx. This implies two things. First, freezing of a layer δx wide takes place. Second, the whole x wide layer cools. Both require energy. Per unit area of ice, it is:
   
The absolute sign is there because the layer cools, and we are looking at the total energy we need to take out of it. ε is the latent heat required to melt ice (per unit mass) while c is the heat capacity per unit volume.

We assume that this energy leaves the ice only from its top (i.e., there is no heating from the water beneath). The heat flux leaving is thus:
   
By combining the two equations, we obtain an equation for x:
   
which has the solution:
   
where tf is the final time and xf is the final width. We assumed of course that freezing commences at t=0. The time required to obtain an ice layer with a width x is:
   
We see that there are two terms. One depends on the latent heat and another on the heat capacity. The heat capacity term is going to be more important if
   
where we have plugged in the values of the latent heat and heat capacity of ice: ε=3.3 105 J Kg-1, c=2.1 103 J Kg-1 K-1. Clearly, the heat capacity is not important under normal terrestrial conditions. Note also that if the heat capacity is neglected, we can easily generalize the result to a time varying temperature. In such a case, the integral over dt is simple enough:
   
where ΔT is the average temperature. In such a case, we have
   
Using the actual numbers for the ice density and conductivity (ρ=920 Kg m-3, κ = 2.3 J m-1 s-1 K-1), we find that the time in days is given by
   
If for example the average temperature is 2 degrees below freezing, then it would require about 4 days to freeze 10 cm of ice, which is about the minimum necessary to safely travel on foot (and do some ice fishing, if it's your cup of tea). If you want to drive your car (e.g., to shorten the way back home to Yellowknife), you'll need 25 days at this average temperature. For 10 degrees below freezing, it would shorten to about 5 days. Again, this assumes optimal conditions, that the water was close to freezing before the ice began forming (and in particular, that there is no constant supply of warm water such as that from a nearby stream inlet).

One last point, we assumed in the calculation that the heat diffusion time through the ice is short (thereby allowing us to assume a linear temerature profile). Now we can verify that this assumption is valid. If the diffusion coefficient is λ, the diffusion time scale is tdiff = x2/λ. Since the heat diffusion coefficient of ice is λ=1.1 m2/s, tdiff ~ 2.5 hrs for 10 cm thick ice, indeed much shorter than a few days (in fact, since the thermal diffusion time is related to the thermal conductivity, one can show that this assumption is valid as long as the heat capacity, which we neglected above, is indeed not important).

[matfalk]

Good read but that made my brain hurt lol.

[seasalt]

Me too. That's a lot of info to take in.

[Jig_Head]

I took in a few sentences til i saw formulas lol

[chillinNgillin]

Geez, I think I'll just use my spud bar.

[Flypopper]

I've been trying to explain this to my fishing buddies for a long time!

[Jig_Head]

Can u explain it to me? Im that lazy

[seasalt]

Just read the second to last paragraph. 1 centimeter equals 25/64 of an inch. So 10 centimeters is just under 4 inches.

[rjflynn]

Who the heck needs four inches of ice to walk on?

[GIBBS]

 :oWOW!!!

[spring bobber]

This article is missing information. Must have gotten lost when it was copied and pasted. If you could provide the link, seasalt, I would be interested in making sense of this.

[Swedish__Pimple]

:oWOW!!!

           LMAO.......All we know is if we dont go down its safe huh AAron...lol

[musky8it]

When I got to this sentence, " where x is the width of the ice at the given instant. T0=-ΔT<0 is the external temperature (again, we assume it be 0 at the inner edge). " I stopped, went no further.

I took the tuffest math classes in school(algebra, geometry, and trigonometry). But that was long ago,1970's, and I done forgot all the equations. So I didn't feel like reading it all.

Can someone sum it all up in a short short version, or will I have to read it.

[wax_worm]

4" will supposedly hold a horse.  I have seen some big ol' boys on the ice before, but none that weigh as much as a horse!

[blufloyd]

Ok is spud bar goes thru the dead fish do not drive on ice.  If spud bar goes thru else where do not walk out.  Same rule I always use really.

[Shawgill]

I'm pretty sure  I have not seen Nir Shaviv on the ice. Is he on " THE BIG BANG THEORY ?

[abishop]

Laymans english PLEASE. LMMFAO

[spring bobber]

When I got to this sentence, " where x is the width of the ice at the given instant. T0=-ΔT<0 is the external temperature (again, we assume it be 0 at the inner edge). " I stopped, went no further.

I took the tuffest math classes in school(algebra, geometry, and trigonometry). But that was long ago,1970's, and I done forgot all the equations. So I didn't feel like reading it all.

Can someone sum it all up in a short short version, or will I have to read it.

Like I said previously, there are several components to the original article missing for whatever reason, but the article is simply saying that by knowing the average temperature, one can formulate the amount of ice that will be made with respect to time. The article provides the example that if the average temperature is 2 degrees below freezing, then it will take 96 hours to make 10 cm of ice. However, there are several assumptions in these equations and reasonings that make the calculation more or less applicable only in ideal ice making conditions, which are never present.

That is what I got from it. Its nothing groundbreaking, just an icefisherman with a little knowledge of physics and an inquisitive mind, thats all.

[abishop]

My brother-in-law just left, he is a nuclear physicist. I am sure he would understand this equation perfectly.

[spring bobber]

The thing he should be talking about more is pressure. Pressure is force per area and that is the main reason why some of the smaller, more petite figures 'get' to go out on the ice first. Force is simply mass times acceleration. Acceleration, while standing, is the same for everyone: the acceleration of gravity (-9.81 meters/second squared).

Therefore, it comes down to mass, which is not weight! Since we all have the same acceleration (-9.81 m/s2) our mass is essentially our weight. So in regards to pressure, this force is exerted over a given unit of area. So if you have big shoes or are sitting on a bucket, the force is exerted over those areas. So this is the reason some of the bigger guys usually wait for more ice.

This is the same principal that applies to the spud bar. All of the force is centered on a 2 inch edge of the spud bar and the wielder of the spud bar is applying extra acceleration to the bar, making its force high. This high force divided by the small area of impact is what allows it to go through thin ice so easily.

Just a few things to think about. I could continue rambling, but its supposed to be my Christmas Break so I'll stop thinking for now.

[bret]

I knew that....

[musky8it]

Minn. Ice guidlines site
http://www.dnr.state.mn.us/safety/ice/thickness.html

2" or less -           STAY OFF
4" -                       Ice fishing or other activities on foot
5" -                       Snowmobile or ATV
8" - 12" -              Car or small pickup
12" - 15" -            Medium truck

US Army Corp of Engineers site:(I think they are incorrect)

http://www.mvp-wc.usace.army.mil/ice/ice_load.html

Required Minimum Ice           Description of
Thickness in inches               Safe Moving Load
 
1-3/4                                    One person on skies
2                                          One person on foot or skates
3                                           One snowmobile
3                                          A group of people walking single file
7                                           A single passenger automobile
8                                           A 2-1/2 ton truck
9                                           A 3-1/2 ton truck
10                                        A 7 to 8 ton truck

[walleyepac]

Pretty good theology, rite there

[Flat Bob]

Here is my equation. X+Y+Z= spud, step,spud, step....... etc.. etc...

[spring bobber]

Here are my equations:

  + =   

  - = 

Therefore: = 

[seasalt]

Here's the link to the article.


http://www.sciencebits.com/StandingOnIce

[shovelholdsdirt]

Lol...There are additional varibles left out of the equation.  This is useless information. Furthermore....really? 

[spring bobber]

Lol...There are additional varibles left out of the equation.  This is useless information. Furthermore....really? 

This isn't useless information, I was just saying that force and pressure should be considered as well. Also, if you look at the original post and the link, there are several equations that were missing.

For those of you who are still interested, the article goes through a series of mathematical simplifications to come to a conclusive equation which is:

Tf is time in days, x is thickness in cm, and DeltaT is average temperature below freezing in Celsius

You can insert a desired thickness, as well as the average temperature, and come up with the amount of days it will take to reach that thickness.

1cm is about .4" so 10 cm is about 4 inches. However, as the author clearly states, this is a LOW estimate and should only be considered as an estimate and nothing to be risking your life over.

[revice]

We need some good ice soon....

[spring bobber]

I would much rather be countin my gills on the ice than doing this math, but hey, math is math.