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Did Claude-Louis Navier and George Gabriel Stokes came up with their equations which accounted for viscocity by studying each other's "money shots?" I mean, there is very valuable data to be had there and I just wonder if they took it into consideration. Or would the difficulty in establishing a constant velocity render it moot?

2007-10-24 07:37:42 · 8 answers · asked by ZombieTrix 2012 6 in Science & Mathematics Engineering

8 answers

The equations account for velocity not being constant, which is apparent when it drips down the wall after some Navier-Stokes action, the acceleration of gravity causes it to speed up, but the viscous forces keep it from moving at free fall speed.

2007-10-24 07:46:32 · answer #1 · answered by Anonymous · 5 0

This one time I tried to prove the Navier-Stokes equations. Then I realized that I barely even understand the problem.

2007-10-24 07:42:37 · answer #2 · answered by Anonymous · 3 0

The constant velocity issue was resolved by capturing the fluids in a beaker that had a circulation pump and tubing inserted inside the beaker. They could then circulate the fluids at a constant velocity, although they were happy to replenish the supply frequently.

2007-10-24 08:14:55 · answer #3 · answered by Anonymous · 2 0

Are you sure you meant Claude-Louis Navier and not Peter North?

2007-10-24 07:44:06 · answer #4 · answered by Cat Stevens 6 · 3 1

I know one thing...misers have a very high viscosity for their money.....it does not move with the `gravity` of their needs.

2007-10-24 09:34:53 · answer #5 · answered by Gee Waman 6 · 1 0

This is a question I will ask the wife when she gets out of her "analysis of Structures" class. Promise.

2007-10-24 07:52:32 · answer #6 · answered by LabGrrl 7 · 4 1

If x and U are in one million to one million correspondence, you could of course use the chain rule to get d(dU/dx)/dU = d^2U/dx^2*dx/dU whether you need to to particular this in any different case. watching that for this reason dx/dU = one million/(dU/dx) or dU/dx*dx/dU =one million and differentiating wrt U you get d^2U/dx^2 (dx/dU)^2 +dU/dx*d^2x/dU^2 = 0 so yet another expression of the consequence is won via multipying the final equality via dU/dx to get: d^2U/dx^2dx/dU = -(dU/dx)^2*d^2x/dU^2

2016-10-13 22:35:13 · answer #7 · answered by ? 4 · 0 0

I liek puppies!

2007-10-24 07:43:24 · answer #8 · answered by Armless Joe, Bipedal Foe 6 · 3 1

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