Where Does River Water Go? - Numberphile
video description
Questions:
1) Diffusion? The relatively fresh river water must take up salt from the ocean and thus become more dense with time. Thus, the g' term must grow smaller. Also, chemical and micro-pollutants must disperse. Eventually, an identifiable current must fade away.
2) Temperature? Diffusion rates will have a temperature dependency. Colder water will be more dense; warmer water will be less dense. Again, g' is affected. Again, pollutant dispersion is affected. Inevitable heat transfer will normalize the temperatures. Thus, again, g' will change over distance.
3) What role does momentum transfer serve? The velocity of the river will inevitably be transferred to the more quiescent sea. The velocity difference will promote slowing, mixing, and broadening of the initial -jet- of incoming river water. Not only the volume of incoming water, but its velocity will play a role. Is the apparent absence of momentum transfer consideration another simplifying assumption?
4) What useful purpose do your calculations serve? Repeatedly, direct measurements are cited. Being you have the real data, why perform the assumption-laden approximate calculations? I can understand academic interest (I like to play with numbers too. Is making a mathematical model the sole purpose for your PhD?
Observation:
The sea in the area studied is not very deep. Doggerland is not far below the surface. A quick Google search (not an elaborate study) said that the depth over Doggerland is 15 to 36 meters. The depth surrounding that area is approximately 20 meters more. The infinite depth assumption, while mathematically attractive, is far from correct.
Topic Suggestion:
The Navier-Stokes equation has a lot of -traction- in the academic fluids community. However, it is notoriously difficult to apply and to solve (reference your own statements in this video. So why is this equation so revered, so perpetuated? It seems that its acceptance lies in the necessity to apply numerous assumptions before arriving at a -solution. - Thus, it seems that it is this magical thing that you whack with a hammer until it looks like what you want. Presto. PROOF! And a proof with the backing of the mystical, magical name. Navier-Stokes (bowing at this point is appropriate. Defend the equation. Defend use of the equation.
Date: 2022-04-09
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Comments and reviews: 9
Aurelia
I think it's amazing how far the water actually moves so far and clings to the coast of Denmark like that. I would have thought that once it got to the top of Jutland it would just -let go- and then crash into Norways coast or something like that. How does it still cling to the coast after going around Jutland? I would have thought that when it does a 180 degree turn like that, the coriolis effect would push the water away from the coast. So how accurate is this model? and is there no water that just flows directly out into the sea and mixes with that instead of either getting trapped in that whirlpool or clings to the coast? Oh and why does that whirpool form? It seems like something that should happen often and/or be visible in all situations where you have water flowing into a large -basin- but this is the first time I've ever heard about this.
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I think it's amazing how far the water actually moves so far and clings to the coast of Denmark like that. I would have thought that once it got to the top of Jutland it would just -let go- and then crash into Norways coast or something like that. How does it still cling to the coast after going around Jutland? I would have thought that when it does a 180 degree turn like that, the coriolis effect would push the water away from the coast. So how accurate is this model? and is there no water that just flows directly out into the sea and mixes with that instead of either getting trapped in that whirlpool or clings to the coast? Oh and why does that whirpool form? It seems like something that should happen often and/or be visible in all situations where you have water flowing into a large -basin- but this is the first time I've ever heard about this.
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Doodelay
One thing that's interesting to me about mathematics is that it's very easy to work out most of the factors or parameters that contribute to something like river flow.
Math and physics would be similar to the quantitative difficulty of political science, writing, or art if that were a decent chunk of the work.
But it seems so much more difficult to work out the precise interactions between some set of parameters than it is to qualitatively list them out. The difference is just enormous, and I think that's really the thing that makes math and physics so much more difficult than other fields. It's just extraordinarily hard sometimes to nail down the precise influence things have on one other.
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One thing that's interesting to me about mathematics is that it's very easy to work out most of the factors or parameters that contribute to something like river flow.
Math and physics would be similar to the quantitative difficulty of political science, writing, or art if that were a decent chunk of the work.
But it seems so much more difficult to work out the precise interactions between some set of parameters than it is to qualitatively list them out. The difference is just enormous, and I think that's really the thing that makes math and physics so much more difficult than other fields. It's just extraordinarily hard sometimes to nail down the precise influence things have on one other.
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Aron
Here's a question/comment/concern: based on the numbers/estimates he gives, the volume flux doesn't quite add up. Typically for a fluid flow, the volume flux is equal to the product of the flow's width, height, and flow speed. But here are his numbers for the Rhine:
width: 10-20 km
height: 10-20 m
speed: -1 m/s
volume flux: -2000 m-3/sec
BUT, if you multiply the flow's width, height, and flow speed, you get:
10, 000m - 10m - 1m/s = 100, 000 m-3/second on the low end, and 400, 000 m-3/sec on the high end. What's the reason for this discrepancy?
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Here's a question/comment/concern: based on the numbers/estimates he gives, the volume flux doesn't quite add up. Typically for a fluid flow, the volume flux is equal to the product of the flow's width, height, and flow speed. But here are his numbers for the Rhine:
width: 10-20 km
height: 10-20 m
speed: -1 m/s
volume flux: -2000 m-3/sec
BUT, if you multiply the flow's width, height, and flow speed, you get:
10, 000m - 10m - 1m/s = 100, 000 m-3/second on the low end, and 400, 000 m-3/sec on the high end. What's the reason for this discrepancy?
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byFiscus
Just wrote a report on T. J. Crawford's thesis, gotta say I did not do it nearly the justice it deserves. If any viewers are watching this, I highly encourage you to give his work a read if you are in the field of mathematics or engineering or just find it neat. It was super cool but a 10 pager doesn't even come close to being long enough to discuss a 200+ page doctoral thesis. Very cool vid, and very inspiring work from a (hopefully) future mechanical engineer
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Just wrote a report on T. J. Crawford's thesis, gotta say I did not do it nearly the justice it deserves. If any viewers are watching this, I highly encourage you to give his work a read if you are in the field of mathematics or engineering or just find it neat. It was super cool but a 10 pager doesn't even come close to being long enough to discuss a 200+ page doctoral thesis. Very cool vid, and very inspiring work from a (hopefully) future mechanical engineer
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Sujith
At 6: 58 Tom assumes that the ocean is infinitely deep. I understand that this may hold true when we are looking at a point very far from the coast. However, I don't understand how the ocean can be so deep near the coastline. Doesn't the ocean depth decrease when we move further from the coastline (i. e. shallowest near the coast? So is the assumption that the ocean is infinitely deep (near the coast) correct?
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At 6: 58 Tom assumes that the ocean is infinitely deep. I understand that this may hold true when we are looking at a point very far from the coast. However, I don't understand how the ocean can be so deep near the coastline. Doesn't the ocean depth decrease when we move further from the coastline (i. e. shallowest near the coast? So is the assumption that the ocean is infinitely deep (near the coast) correct?
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Robert
I don; t understand why he keep making Navier-stokes sounds so far fetching and impossible. It is tough, sure. Most practically applications require a computer to solve Navier-stokes equations, but they really aren't that confusing or impossible after you spend a few weeks/months with them. They're still very applicable and useful and understood to a satisfactory degree.
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I don; t understand why he keep making Navier-stokes sounds so far fetching and impossible. It is tough, sure. Most practically applications require a computer to solve Navier-stokes equations, but they really aren't that confusing or impossible after you spend a few weeks/months with them. They're still very applicable and useful and understood to a satisfactory degree.
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Robert
I don; t understand why he keep making Navier-stokes sounds so far fetching and impossible. It is tough, sure. Most practically applications require a computer to solve Navier-stokes equations, but they really aren't that confusing or impossible after you spend a few weeks/months with them. They're still very applicable and useful and understood to a satisfactory degree.
reply
I don; t understand why he keep making Navier-stokes sounds so far fetching and impossible. It is tough, sure. Most practically applications require a computer to solve Navier-stokes equations, but they really aren't that confusing or impossible after you spend a few weeks/months with them. They're still very applicable and useful and understood to a satisfactory degree.
reply
David
As a physicist, my 1st check on these results is to make sure the dimensions work out, but there is a problem. For your formulas to give [h]=[w]=m and [u]=m/s, you need [Q]=m-3/s, [f]= 1/s, and [g'] = m/s-2. But you say g' is formed from the density difference (kg/m-3) and g (m/s-2. There needs to be some other factor with units of mass to make kg disappear. What is it?
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As a physicist, my 1st check on these results is to make sure the dimensions work out, but there is a problem. For your formulas to give [h]=[w]=m and [u]=m/s, you need [Q]=m-3/s, [f]= 1/s, and [g'] = m/s-2. But you say g' is formed from the density difference (kg/m-3) and g (m/s-2. There needs to be some other factor with units of mass to make kg disappear. What is it?
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Alex
When he says that in a hemisphere flows to the right and in the other to the left. Is that correct? I mean, what if 2 rivers reach the ocean in the same latitude of the continent but one does it in the east coast and the other one in the West coast? If they are influenced by coriolis effect shouldn't they both flow north or south? Why is left and right?
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When he says that in a hemisphere flows to the right and in the other to the left. Is that correct? I mean, what if 2 rivers reach the ocean in the same latitude of the continent but one does it in the east coast and the other one in the West coast? If they are influenced by coriolis effect shouldn't they both flow north or south? Why is left and right?
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