VoIP over NBN requires EWAN

Earlier this year I opted to use VOIP for phone calls, which seemed cheaper than using the phone port on my NBN box. I bought a new wireless router with VOIP, but it failed because it lacked a feature called EWAN. Then I bought an EWAN router, and it failed too because I forgot to check if it had VOIP, too. Then I bought another router which worked. Now I'm back on ADSL2, so my original router would have done the job. Then I made a loss on selling the routers, because of how much I paid for postage.

Earlier this year I opted to use VoIP for phone calls, which seemed cheaper than using the phone port on my NBN box. The home phone plugs straight into the router. I had NBN internet fine on my old router. Then I bought a new router with VoIP (Netgear DVG1000, $40), but it failed because it lacked a feature called EWAN (full ethernet-internet functionality). I called technical support at iiNet, and they didn’t even know the name of the feature I was missing. After solving the mystery, I bought an EWAN router (TP-Link TD-W8960N, $30) and it failed too because I forgot to check if it had VoIP. Then I bought another router which worked. Now I’m back on ADSL2+, so my original router would have been almost as good. Then I barely broke even selling the routers, because of how much I paid for postage.

Inverse square law for concentric cylinders

Consider two very long concentric cylinders with radii R_1 & R_2 (m) and scalar angular velocities \omega_1 & \omega_2 (rads/s). I assume there is a fluid in between the two whose velocity is irrotational and free of divergence, and who is not compressed along streamlines. Then the fluid’s velocity throughout, due to the first cylinder, should be \frac{\omega_1 R_1^2}{r}\hat\theta, where (r,\theta,z) are cylindrical coordinates and \hat\theta = \frac 1 r\frac{\partial\vec r}{\partial\theta} is a circulating unit vector. The fluid’s velocity should match that of the second cylinder at its surface if there’s no drag despite there being a fluid coupling, which I assume. The outer cylinder has velocity \omega_2 R_2\hat\theta. Therefore I set

\begin{array}{rcl} \frac{\omega_1 R_1^2}{R_2} \hat \theta &=& \omega_2 R_2 \hat \theta \\ \omega_1 R_1^2 &=& \omega_2 R_2^2 \end{array}

Repeating this argument for any other concentric cylinders reveals that \omega r^2 for all of them is the same constant k (\text m^2/\text s), assuming the intermediate ones are all solid but very thin. All cylinders should rotate according to

\omega = \omega(r) = \frac{k}{r^2}

The constant k corresponds to \alpha in my post regarding the flow around an asymmetric cylinder. The constant does not depend on any of the fluid’s properties. The fluids between the different cylinders may or may not be the same. If two concentric cylinders are very close together, with radii differing by \Delta r, with dense fluid trapped in between, then their angular velocities should differ by

\Delta \omega \approx -2\omega\frac{\Delta r}{r}

If the cylinders are audible, then the difference \Delta\omega might be heard as a beat. I don’t know how long the cylinders must be for the inverse square law to accurately apply. The intermediate cylinders can all have different lengths, which are not longer than the outer ones. Rings are allowed within the fluid coupling.

Update (3 Oct): I’ve added the “not compressed” assumption, because otherwise the speed could vary along a streamline with varying density, and I’ve added the fluid coupling assumption. The centripetal force experienced per unit volume (\text N/\text m^3) is equal but opposite to the pressure gradient, assuming uniform density and Bernoulli’s principle, and that provides another derivation of the same outcome. The pressure (\text N/\text m^2), according to Bernoulli’s equation with density \rho=\rho(r) (\text{kg}/\text m^3), assuming negligible viscosity, should be

P=P_1+\frac 1 2 (\omega_1R_1)^2\left(\rho_1-\rho\left(\frac{R_1}{r}\right)^2\right)

where P_1 & \rho_1 are the fluid’s pressure and density at the first cylinder. I’ve checked in Matlab that the outcome is exactly the same if 1 is replaced with 2. I’m not sure what happens when the innermost cylinder is shrunk, i.e. as r\rightarrow 0. I haven’t been able to apply the same idea to spinning concentric spheres.

Are so-called fossil fuels at least partly primordial?

For example, perhaps coal existed within the Earth, as it formed. Or maybe it can form geologically from some mineral. Maybe there’s a percentage of each, and also a percentage which is biological in origin. It’s an extraordinary claim that most coal is organic in origin, and I’d have to see the proof to believe it. The carbon to make all the plants had to come from somewhere.

Flow passed an asymmetric circular cylinder

According to Wikipedia, there is at least one solution to Laplace’s equation which could model fluid moving passed a cylinder. I’ve found an asymmetric flow (below) satisfying the necessary boundary conditions, with continuity. A spinning cylinder is asymmetric because it’s going one way, and not the other.

Suppose an infinitely long cylinder, with axis coinciding with z-axis, is in the flow of an inviscid, incompressible fluid, which would otherwise have velocity U\hat i in the x-direction. I assume, like Wikipedia, that the velocity field is irrotational and free of divergence. Then there should exist a scalar velocity potential, \phi(r,\theta), whose gradient equals the velocity, and who satisfies Laplace’s equation in 2D. My velocity potential (scalar, units \text m^2/s) is given by

\phi (r,\theta) = U\left(r+\frac{R^2}{r} \right) \cos\theta + \alpha \theta

where \alpha is any real number (\text m^2/\text s), (r,\theta) are 2D polar coordinates and R (metres) is the cylinder’s radius. The far-away velocity is handled by the well-known “d’Alembert” solution, and I’ve added an asymmetry. I assume a potential may be discontinuous, since it’s not unique, but the unique real velocity must be continuous. The gradient

\nabla\phi = \vec v (r,\theta) = U\left(1-\left(\frac R r\right)^2\right)\cos\theta\hat r+\left(\frac{\alpha}{r}-U\left(1+\left(\frac R r\right)^2\right)\sin\theta\right)\hat\theta

is the velocity field. The formula shows that at the cylinder’s radius, there is no radial component to velocity. Furthermore, the limit as r\to \infty is U\hat i. Therefore I’ve verified all boundary conditions are satisfied, and by inspection the velocity field is well-defined and continuous everywhere outside the cylinder. Furthermore, my asymmetry term in the scalar potential is the imaginary part of \alpha\log\frac{x+iy}{R} over the complex plane, so it must satisfy Laplace’s equation, except at the negative x-axis, where the velocity is still divergence and rotation-free, outside the cylinder. That is, my velocity field is correct.


Steady flow of U=1 m/s (3.6 km/h) current around cylinder of radius R=1 m, where I’ve assumed α (=a) is 1 m²/s. The asymmetry should be due to the cylinder spinning anticlockwise with frequency 1 radian/s ≈ 0.16 Hz ≈ 10 rpm, assuming coupling to the fluid. The fluid is dragged fast under the bottom. I’ve been trying to model inviscid drag. The image is from pplane8.

I have uncountably many partial solutions to Laplace’s equation, who satisfy all necessary boundary conditions, may model inviscid drag, but do result in discontinuity in velocity along the negative x axis. I can’t solve d’Alembert’s paradox, yet, which I encountered after pondering my reversible venturi experiment.

Update (2 Oct): The value of \alpha should be given by

\alpha = \omega R^2

where \omega is the cylinder’s anticlockwise (scalar) angular velocity (radians/s). I base this on the case where U=0 – the fluid should move at the same rate as the cylinder, at its surface, when there is no external current. In the cylinder’s inertial frame of reference, when the current and cylinder have opposite non-zero velocities, the current should always win.

Update (3 Oct): Suppose there’s uniform density in a fluid (\text{kg}/\text m^3) with no external current. Then the fluid’s total angular momentum per unit length along its axis (Ns) diverges with \left(\frac r R\right)^2. This is plausible for an eternal rotating cylinder with fluid coupling. The kinetic energy per unit length (J/m) also diverges, except in proportion to \log\frac r R. The fluid’s velocity is just \frac{\omega R^2}{r} \hat \theta. To obtain the cylinder’s rpm, multiply \omega by \frac{30}{\pi}.