Day length and ice melt

[Tim the plumber]

FACT. Feedback between variables creates very difficult-to-predict chaos. You reject scientific discipline and realities, don't act surprised that no one gives your alchemy the same respect as actually informed, supported and educated analysis. And "supported" doesn't mean selectively disregarding established data, knowledge and science that doesn't fit with your premise.

How the hell does the wind end up creating a net change in angular momentum over a long period???

Where is it supposed to be gripping an external fulcrum to allow it to reject to need to obay Newton's laws???

It's simple stuff. Your appeal to authority will not wash here.

 

[DirtyRat]

How the hell does the wind end up creating a net change in angular momentum over a long period???

Where is it supposed to be gripping an external fulcrum to allow it to reject to need to obay Newton's laws???

It's simple stuff. Your appeal to authority will not wash here.

Again you take a small piece and exclude all other data. Wind pushes on land, and coupled with the low pressure it also creates storm surges -- dragging a significant mass of water with it. Over time, over an extended period, this adds up.

You dismiss contravening variables as "drivel" and don't consider them ... this is beyond junk, this is outright male bovine faeces you peddle. You are completely oblivious to what objective research actually. /Every/ variable must be addressed, /none/ can be ignored and /none/ can be dismissed merely because its inconvenient to the conclusion you asserted before you looked into it -- but that's exactly what you've done here.

You have less than zero clues on how actual objective research is conducted. The Scientific Method was not some religious fad, it took centuries for researchers to realize the fallacies of the age of astrology, but you go right back to it because playing with numbers, however you do it, must be "sciencey."

Wrong, wrong, wrong, wrong.

 

[Surface Detail]

Drivel.

The changes to weather will balance out. There may be some temporary noise in the system due to this but there will be no long term effect on day length.

The movement of land at the pole where the ice is supposed to be melting from will have very little effect upon spin rate due to it's location, it's vertical movement parallel to the axis of rotation and the fact that it is not moving far as opposed to the 6,000+km that the water is moving.

The point is that the excusses that are given for the lack of observed changes to the day/year length do not add up. The idea that we have been able to measure the mass changes/ice melt of Greenland by the deviation of a satellite's path going over it is just silly. There are, in that case, any number of factors which will give wrong results.

But the length of the day is changing. Look at this graph:



Source: What is Earth Orientation?

You can see that the length of a day shortened by about 2 ms between 1975 and 2005, but it now seems to be lengthening again. What do you conclude from this?

 

[Tim the plumber]

Again you take a small piece and exclude all other data. Wind pushes on land, and coupled with the low pressure it also creates storm surges -- dragging a significant mass of water with it. Over time, over an extended period, this adds up.

No. I understand that this must balance out. That any change that the weather produces today will be automatically negated by the counter effect next week. There is a slow influence of the tides producing friction of the world which is slowly slowing the world's rotation and speeding up the orbital velocity of the Moon. The strength of this is know.

You dismiss contravening variables as "drivel" and don't consider them ... this is beyond junk, this is outright male bovine faeces you peddle. You are completely oblivious to what objective research actually. /Every/ variable must be addressed, /none/ can be ignored and /none/ can be dismissed merely because its inconvenient to the conclusion you asserted before you looked into it -- but that's exactly what you've done here.

Bollocks. I have clearly addressed the factors you have fired in. I have expalained why they would be of no significant influence. If you wish to bring more detail into this do so. How much mass is involved and by how much is it position away from the axis of spin changed for how low? If you have such data I will gladly work out it's comparitive influence.

You have less than zero clues on how actual objective research is conducted. The Scientific Method was not some religious fad, it took centuries for researchers to realize the fallacies of the age of astrology, but you go right back to it because playing with numbers, however you do it, must be "sciencey."

Wrong, wrong, wrong, wrong.

I understand basic rotational dynamics. You don't.

My claim on this thread has been and will continue to be subject to the scrutiny of others here. On this sub forum there are lots of people who are decently scientifically aware. Collectively we are able to read any decent scientific paper, or at least most of them. Nobody has produced any reasonable resolution to this paradox which has was not fist thought of by me but is a know issue in the field either here of elsewhere other than the ice is not melting.

 

[Tim the plumber]

But the length of the day is changing. Look at this graph:

View attachment 67207586

Source: What is Earth Orientation?

You can see that the length of a day shortened by about 2 ms between 1975 and 2005, but it now seems to be lengthening again. What do you conclude from this?

Having a think.

 

[Threegoofs]

Having a think.

Be careful- don't hurt yourself!

 

[Poor Debater]

Which is very similar to the result I got because I worked out the value for a year. You get 3.7 thousanths of a second per year for a 1mm sea level rise.

Swamped dy to day but over a year it's clear and over a decade it's massively more than the other factors. Thanks.

Nope.

Earth gains 10.53 μs / day from a 1mm sea level rise, regardless of how long it takes to add the 1mm. It could be a day, a week, a month, a year, a century, or a millenium.

 

[Poor Debater]

But the length of the day is changing. Look at this graph:

View attachment 67207586

Source: What is Earth Orientation?

You can see that the length of a day shortened by about 2 ms between 1975 and 2005, but it now seems to be lengthening again. What do you conclude from this?

Note that each tick on the y-axis is 500 μs, fifty times larger than the 1mm sea level rise value. Like I said, lost in the noise.

 

[Tpaine]

Not even close to being correct.

1. Compute Earth's moment of inertia.

For a sphere of uniform density, I = .4 MR²
But since Earth's density is non-uniform, the correct moment of inertia factor is .3307, hence for Earth, I = .3307 MR²
Mass = 5.97e24 kg
Radius = 6371 km
therefore I = 8.02e31 kg km²

2. Compute moment of inertia for a thin shell of water 1 mm thick

Volume of 1 km x 1 km x 1 mm of water = 1000 x 1000 x .001 = 1000 m³.
Mass of 1000 m³ water = 1e6 kg

Area of Earth's oceans: 361,060,000 km²
Mass of 1mm water added to earth's oceans: 1e6 x 3.6106e8 = 3.6106e14 kg

For a thin spherical shell, I = 2/3 MR²
Here we assume that the oceans are evenly distributed across the surface, which isn't true, but it's not far wrong.
Therefore moment of inertia of 1mm water shell = 9.77e21 kg km²

Notice already: TEN orders of magnitude difference!

3. Angular momentum is conserved. Compute angular momentum L = Iω where ω is angular velocity in radians per second. For earth, ω = 2 π / 86400.
Therefore angular momentum L = 8.02e31 2 π / 86400 = 5.83e27 kg km² rad/sec.

Assuming for convenience that polar ice is perfectly axial, its angular momentum is zero. Melting the ice adds the moment of inertia of the shell to the moment of inertia to the Earth as a whole. So post-melt moment of inertia is: 8.02e31 + 9.77e21 = 8.02e31 (plus a lot of decimals) kg km². Since angular momentum is conserved, the new tiny-bit larger moment of inertia must be balanced by a new tiny-bit smaller angular velocity.
So:
Initial angular velocity: 7.242205216643040e-5 rad/sec
Final angular velocity: 7.272205215756699e-5 rad/sec
with a difference of 8.8634e-15 rad/sec

which amounts to a longer day by a whopping 10.53 microseconds (that's millionths of a second).

Is this detectable? It's just barely within measureability limits, but it is swamped by daily effects of weather (~1 order of magnitude larger), seasonal effects (~2 orders of magnitude larger) and long-term lunar effects (~3 orders of magnitude larger). So it's lost in the noise.

I don’t think you can use the I value of a shell to calculate the total I value of a sphere with an additional thickness. I used the radius of the earth and calculated the momentum then increased the radius by 1 mm and calculated a new IV value. From that I calculated a new Velocity. By subtracting the two I got an increased in the velocity which came to 27 ms/day. For a 180 mm rise that would be 4,860 ms or 0.005 seconds.

I didn’t understand where you got the difference in the angular velocity. The difference in those two velocities don’t seem to equal to the last number.

 

[Poor Debater]

I don’t think you can use the I value of a shell to calculate the total I value of a sphere with an additional thickness. I used the radius of the earth and calculated the momentum then increased the radius by 1 mm and calculated a new IV value.

But the whole earth doesn't get 1mm larger, just the ocean does. And the density of that ocean is far, far less than the density of the Earth as a whole.

I didn’t understand where you got the difference in the angular velocity. The difference in those two velocities don’t seem to equal to the last number.

By convention, angular velocity is computed in radians per second. There are 86400 seconds in a day, and 2π radians in a full circle (i.e., 1 rotation of the earth). To convert rad/sec to seconds/day, the conversion factor is therefore (2π * 86400)/(1/86400) = 1.188e9. So 8.8634e-15 rad/sec * 1.188e9 = 1.053e-5 sec/day.

 

[Lord of Planar]

Note that each tick on the y-axis is 500 μs, fifty times larger than the 1mm sea level rise value. Like I said, lost in the noise.

Lost in noise...

Funny how that concept is not accepted by you on your pet matters...

 

[Tim the plumber]

Nope.

Earth gains 10.53 μs / day from a 1mm sea level rise, regardless of how long it takes to add the 1mm. It could be a day, a week, a month, a year, a century, or a millenium.

Yes. But the effect on year length is, according to my maths, 1.9 thousanths of a second per year and 3.7 thousanths according to yours. You are using a more detailed model of the density of the earth with the mass more centrally located.

 

[Tim the plumber]

I don’t think you can use the I value of a shell to calculate the total I value of a sphere with an additional thickness. I used the radius of the earth and calculated the momentum then increased the radius by 1 mm and calculated a new IV value. From that I calculated a new Velocity. By subtracting the two I got an increased in the velocity which came to 27 ms/day. For a 180 mm rise that would be 4,860 ms or 0.005 seconds.

I didn’t understand where you got the difference in the angular velocity. The difference in those two velocities don’t seem to equal to the last number.

The big difference is that the extra 1mm of water is effectively new water from a position where it has no effect on angular momentun being over the pole. A simplification but...

 

[Tpaine]

But the whole earth doesn't get 1mm larger, just the ocean does. And the density of that ocean is far, far less than the density of the Earth as a whole.

My argument would be that when you calculate the original L value you combine the oceans and the earth. Then you add an additional mass when there is no mass added to the system. The mass is the same for both cases it’s just redistributed to a new R. I understand the redistribution is just in the water and the water is not the same density as the earth but that wasn’t a distinction made in the original I value. I think we agree the velocity varies by the inverse of the angular momentum. So to get the new velocity you need to find the new inertia which is where I disagree with the method you and Tim are using. This is the way I see the equations not being equal:

Method 1:

I =0.4 M(R + ΔR) = 0.4 M(R^2 + 2RΔR + ΔR^2) = 0.4 MR^2 + 0.8 MRΔR + 0.4MΔR^2

The last term will pretty much go to zero.

Method 2:

I = 0.4 MR2 + 0.66 ΔMR^2

Maybe the density of the earth being about 5 times the density of water is making the difference.

By convention, angular velocity is computed in radians per second. There are 86400 seconds in a day, and 2π radians in a full circle (i.e., 1 rotation of the earth). To convert rad/sec to seconds/day, the conversion factor is therefore (2π * 86400)/(1/86400) = 1.188e9. So 8.8634e-15 rad/sec * 1.188e9 = 1.053e-5 sec/day.

It doesn’t seem your units are working out. Shouldn’t it be:

(2 pi rad/day) / (rad/sec) = 2 pi sec/day?

Your units appear to be:

(2 pi rad/day) * sec/day * sec/day = rad-sec^2/day^2

Am I missing something?

Also 2 pi * 86400 / (1/86400) = 4.69e10

 

[Tpaine]

The big difference is that the extra 1mm of water is effectively new water from a position where it has no effect on angular momentun being over the pole. A simplification but...

Yes but you are adding mass to the system when there is no mass added to the system. When the skater comes out of a spin she doesn't add something to her shoulders she expends her arms.

 

[Tim the plumber]

My argument would be that when you calculate the original L value you combine the oceans and the earth. Then you add an additional mass when there is no mass added to the system. The mass is the same for both cases it’s just redistributed to a new R. I understand the redistribution is just in the water and the water is not the same density as the earth but that wasn’t a distinction made in the original I value. I think we agree the velocity varies by the inverse of the angular momentum. So to get the new velocity you need to find the new inertia which is where I disagree with the method you and Tim are using. This is the way I see the equations not being equal:

That's where we differ I think. The inertia change due to thermal expansion would be as you describe. That is no significant change. But for a redistribution from the pole to the general surface of the ocean this is wrong.

 

[Tim the plumber]

Yes but you are adding mass to the system when there is no mass added to the system. When the skater comes out of a spin she doesn't add something to her shoulders she expends her arms.

By considering that the ice is sat directly over the pole in a hugh colum, which is wrong I agree but close enough, the mass of the melt water is new to having any effect on the spin of the earth. Again, close enough.

If you wish to calculate the inerrtia for the mass when it is lying on Greenland and take this away from it's effect on the whole earth be my guest but I think it would be negligable due to the closeness to the axis of spin.

 

[Tim the plumber]

But the length of the day is changing. Look at this graph:

View attachment 67207586

Source: What is Earth Orientation?

You can see that the length of a day shortened by about 2 ms between 1975 and 2005, but it now seems to be lengthening again. What do you conclude from this?

OK, so, given that the graph is of day length and clearly there are general trends with noise over it the world is, or at least has, speeded up in it's rotation.

I can't think of any mechanism which would do this to anything like the degree that moving mass to/away from the poles in the form of glacial ice does.

So perhaps we are seeing the result of a build up of ice in Antarctica and possibly central Greenland. I know there are papers which say the opposite but given that WWII planes lost on the ice of Greenland are now many meters below the surface and too difficult to get out even though they are worth millions.

In 1988, two explorers sponsored by the Greenland Expedition Society finally found a lead. Patrick Epps and Richard Taylor led an expedition to the the ice cap which used steam to bore a hole and locate airplane parts buried under the Greenland ice. The two men found that in the forty-six years since the planes had crash-landed, an astonishing 268 feet of ice had accumulated over them, and they had been carried three miles by the drifting glacier.

https://www.damninteresting.com/exhuming-the-glacier-girl/

That would do it.

Greenland's ice sheet is 1.71 million Km2 and if it's 28m thicker since WWII then 215.8 thousand billion tonnes has been deposited there. Obviously there could also be more ice in Antarctica.

Again, this disagres with the numbers coming out of the climate science community which are sonehow generated by the path of satelites going over Greenland. Which do you think are better, less fudgable, numbers?

 

[Tpaine]

By considering that the ice is sat directly over the pole in a hugh colum, which is wrong I agree but close enough, the mass of the melt water is new to having any effect on the spin of the earth. Again, close enough.

If you wish to calculate the inerrtia for the mass when it is lying on Greenland and take this away from it's effect on the whole earth be my guest but I think it would be negligable due to the closeness to the axis of spin.

I showed the formulas were different. Can you show where I was wrong?

 

[Tim the plumber]

I showed the formulas were different. Can you show where I was wrong?

I don’t think you can use the I value of a shell to calculate the total I value of a sphere with an additional thickness. I used the radius of the earth and calculated the momentum then increased the radius by 1 mm and calculated a new IV value. From that I calculated a new Velocity. By subtracting the two I got an increased in the velocity which came to 27 ms/day. For a 180 mm rise that would be 4,860 ms or 0.005 seconds.

I didn’t understand where you got the difference in the angular velocity. The difference in those two velocities don’t seem to equal to the last number.

Before we get to the formulae to use we have to work out how we should be modeling this.

My method is to imagine that the ice which melts starts with no significant angular velocity or distance from the axis of spin. I don't think you take that into account.

By just looking at the rise in sea level you miss the point of the mass moving from pole to general sea surface.

 

[Tpaine]

Before we get to the formulae to use we have to work out how we should be modeling this.

My method is to imagine that the ice which melts starts with no significant angular velocity or distance from the axis of spin. I don't think you take that into account.

By just looking at the rise in sea level you miss the point of the mass moving from pole to general sea surface.

Why do you say that? I calculate the rotation without the 1 mm water then again with the 1 mm water. The first assumes no angular velocity due to the added water and the second assumes there is added water. How is that different from what you are saying?

 

[Tim the plumber]

Why do you say that? I calculate the rotation without the 1 mm water then again with the 1 mm water. The first assumes no angular velocity due to the added water and the second assumes there is added water. How is that different from what you are saying?

Because whilst I agree that the mass of the earth does not change it is better to model it as though it has water added.

As I say if you feel like it you could model the ineretia for the ice as it is in place then work out the shell and use the difference to see what fraction of the total of the earth's inertia that was.

I don't like the going into m/s or anything else because it will add needlessly to the complexity of the maths.

 

[Tpaine]

Because whilst I agree that the mass of the earth does not change it is better to model it as though it has water added.

So you think using a model that doesn’t match reality is better. That makes perfect sense.

I don't like the going into m/s or anything else because it will add needlessly to the complexity of the maths.

What does that mean?

 

[Tim the plumber]

So you think using a model that doesn’t match reality is better. That makes perfect sense.

How does your model reflect the previous position of the ice? How would it differ if the water had come from glaciers melting in the Himalyas?

If there is no difference you have not got the idea of the model.

 

[American]

When the ice melts at the poles it spreads the water all around the world. This moves mass from the poles to the equator and everywhere else.

Here is my maths to show what this would do to the day length;

Take the moment of initeria of a hollow sphere of negligable thickness 2m[SUB]1mm[/SUB]r[SUP]2[/SUP]/3 and divide by the total moment of inertia for the entire earth m[SUB]e[/SUB]r[SUP]2[/SUP]/2.

The r[SUP]2[/SUP] cancells as it's the same. The m[SUB]1mm[/SUB] is the mass of a 1mm layer of water over the whole earth. The m[SUB]e[/SUB] is the mass of the whole earth.

So, m[SUB]1mm[/SUB] x4
________________ = (360 x 10[SUP]12[/SUP]kg / 6 x 10[SUP]24[/SUP]) x 4/3 = 6 x 10[SUP]-11[/SUP]

m[SUB]e[/SUB] x3


This is the fraction that the world's spin is slowed by.

Multiply this by the number of seconds in a year 31.5 x 10[SUP]6[/SUP]

So that is 1.9 x 10[SUP]-3[/SUP] or 1.9 thousanths of a second per mm of sea level rise.

This is not noticable in human terms but it is easily measurable with atomic clocks. It has not happened. We are supposed to have had at least 180mm of this ea level rise since 1900. It has not happened. There is no explaination for why this has not happened other than the measurement of the ice melt from Greenland etc is wrong.

My maths is very rusty, like 30 years since I did any of this, so please correct me if I've droped one.

Here's an illustration of what will happen if you are correct about the movement of water to different parts of the Earth.