# PSA: new UoA print queues

If you have been trying to get the new print service working wirelessly (i.e. without having to click through five or six pages on the web interface), here’s the info I used to get it working through CUPS:

• Make and model: Ricoh Aficio MP C7500 PS

Hope this saves someone else some time…

Also, don’t use the “print all” button on the printer for the time being, it was charging me the full amount but not actually printing. Hopefully it will be fixed soon.

# “Trackless trams”

According to the dompost,

Trackless trams could become Wellington’s favoured mass transit system.

Now correct me if I’m wrong, but isn’t that just… a trolley bus?

(I have no idea how much of that article is just the dompost talking out of their Mt Vic tunnel, but it sounds like someone is trying to push through either a budget solution or a high-tech futuristic project that will cost a boatload of money and eventually fail. If light rail had been funded when it was needed, last decade, it would have been built by now…)

# Latest Metlink report to GWRC

As I like to do from time to time, here are the highlights of last week’s meeting of the GWRC sustainable transport committee.

(I’m skipping the sections on reliability and on-time performance, I think the Dom Post has written about them, and you can go read them yourself, but I’m more interested in infrastructure and future plans on this blog.)

It is anticipated double-decker buses will commence operation on routes 31x and 36 during week of 25 March 2019, subject to deployment of driver guidance aids and driver training in Hataitai Bus Tunnel.

Double-decker buses will be operational on the Eastbourne routes once these routes have been cleared for double-decker buses (estimated to be ready in April 2019).

So we have confirmation now that the double deckers (a) can go through the Hataitai tunnel (!) and (b) will be on the routes 83 and 81. Is the Hataitai tunnel larger in bore size than Seatoun? (I thought it was smaller.) Since the Karori tunnel is larger, is there any chance of double deckers on the route 2?

Our current focus is on the Wellington Bus Interchange at the Railway Station as there are a number of pressing operational and safety issues that need to be addressed here. We are working to agree design principles with WCC.
Does the “number of pressing… issues” include the hundreds of people walking across the bus parks and bus lanes to get from one side to the other, dodging double-deckers, in order to avoid using the pedestrian crossings?
Demolition of the depot buildings on the Waterloo site is also due to commence in April.
RIP Valley Flyer, formerly Cityline Hutt Valley, and even before that NZR Road Services Hutt Valley. Does this mean that Airport Flyer buses are now running out of a City depot?
The work in the Wellington yard to allow safe storage of a 9-car long Wairarapa train is now underway. The work remains on target for completion in the week of 18 March – however, this does rely on timely delivery of a new track turnout.
Will there be changes to Waterloo to allow a nine-car train to berth at the platform there? I remember the last carriage being locked at Waterloo due to the shorter platform.
The second procurement component of the national ticketing solution to procure a range of financial services is currently in the market as a Request for Tender. The tender phase closes on 15 March 2019 after which evaluation for the three services sought will commence. Evaluation is anticipated to take approximately two months.
I think integrated ticketing is supposed to be on trains (we’ve heard) by 2021 according to the RLTP. Is this timeframe reasonable? I doubt it.
In advance of a public announcement of the recommended programme and finalisation of the Programme Business Case, the programme team has been developing an Early Improvements programme, focussed on actions that can be
delivered in the first few years. Preparatory work is also ongoing for a delivery
structure for the programme.
…….

Edited 19/03/2019: I would like also to share the Prime Minister’s statement made in Parliament. “He sought many things from his act of terror, but one was notoriety. And that is why you will never hear me mention his name. He is a terrorist. He is a criminal. He is an extremist. But he will, when I speak, be nameless.”

I have nothing to say.

# Physics Diary (#012)

I promised back in Introspection Week, when I talked about why I study mathematics, that I would eventually write a post about why I decided to study physics. This is not that post, but hopefully it appeals a little to my readers (both of them!) who study physics as well.

Indeed, this semester I am taking two physics papers: mechanics and thermodynamics, and electronics and imaging.

I think I have spoken at length about why I have spent two years(!) basically treating physics as the poor cousin in my university career; for this reason I won’t repeat myself here. On the other hand, I think it is worth saying that I am finally getting back to the state I was in back in 2016 when I actually enjoyed physics. I feel like it is proper to quote Feynman here, so have a link to a random webpage mirroring the semi-relevant section of Surely You’re Joking.

Anyway, we’ll see how we go. I always preferred mechanics to electromagnetism or optics anyway, so hopefully it will go well. We have spent some time writing the various differential equations that relate the fundamental Newtonian quantities in a fixed reference frame (linear and angular momentum, work, force, position, and so on) and we are now moving towards studying motion in a changing reference frame via elementary differential geometry. (Interestingly, I’ve never been particularly excited by differential geometry. Is that just because do Carmo makes it feel like glorified calculus? I suppose I should pick up Spivak’s book at some point and try to work out why it’s interesting.)

Question. In the Bohr model of the hydrogen atom, the electron orbits the proton in the nucleus due to the electrostatic force between them. The “standard” problem with this model is that it violates energy conservation: the electron is accelerating, thus work is being done, and so (if no energy is added to the system) eventually the electron will spiral in and collide with the nucleus. Problem: the electrostatic force is always orthogonal to the path of the electron, so who’s doing this work? [This problem is also a standard one posed when one considers magnetism: the magnetic force always acts on a moving particle orthogonally to the direction of travel, so it can’t do any work either. Of course, in this case the solution is that the energy source driving the magnetic field is doing work, and the field itself is merely “directing” that work into the proper direction. See Griffiths, sec. 8.3.]

Just as an aside, I am finding electrodynamics much more interesting now that every little piece of work I do isn’t being assessed.

## Complex analysis

Now on to some mathematics. I am only doing one paper this semester, on complex analysis. We have only just (three lectures in) proved that there is a difference between the notions of differentiability in C compared to R by deriving the Cauchy-Riemann equations; however, we did so in a much more beautiful way than the way I saw last semester (briefly, the approach I had seen before is to approach the limit defining the derivative in the purely real and purely imaginary directions; the approach seen in this course was to make explicit the natural bijection R<->C, differentiate f in C, and then calculate the skeleton of the differential of f when passing through the bijection).

Philosophically, I am leaning more and more towards the ‘linear approximation’ formulation of calculus and analysis: the role of the limit, in my view, is to act as an “easy to define but hard to use” tool that we use precisely once in order to define ‘little-oh’ functions (functions which tend to zero faster than their argument: $f \in o(h)$ iff $f(h)/h \to 0$ as $h \to 0$) and then use these functions to formulate things like the derivative via ‘infinitesimal approximations’ (a tool that is hard to define without limits, but is easy and intuitive to use once it has been defined and its properties derived).

Also, this video is the greatest thing ever.

# New bus replacement pattern on HVL?

I just noticed an announcement that Metlink has made about a new trial pattern of services for when the HVL is bus replaced in evenings and on weekends; instead of buses leaving from P9 at WELL, with express and stopping buses, trains will run as far as MELL and then separate all-stops buses via LH and UH will leave from the bus stop there. A couple of thoughts:-

• AVA isn’t being serviced. Apparently the plan is for WEST to be the ‘replacement’ station for the area, but I’m surprised Metlink isn’t telling people to transfer onto a 110 at PETO instead – surely that would provide better service?
• Are the trains after 6:00 (when evening replacements start) usually more than one unit when not bus-replaced? Will we see 4-car or 6-car trains on the Melling branch, then? (Do 4-car trains usually run into MELL on peak? If not, maybe they’ll be the longest trains into there for some time. I think WEST still has an unnaturally long platform left over from when it was the mainline, so maybe it’ll be put to slightly better use than it is usually.)
• If someone is travelling from (say) HERE to TAIT, they’ll need to transfer buses at either SILV (long wait) or MELL (longer journey than it should be) – going the other way (from south to north) the buses look like they take drop-off requests, but is there a better way to organise that? I wouldn’t expect ridership to be too high, and the 110 is probably the ‘expected’ alternative (it’s slower than the train would be, but faster than going via MELL).

# Cubes

In two dimensions, consider the following standard problem: Show that the area of a rectangle of given perimeter $\mathcal{P}$ is maximised when the sides all have length $\mathcal{P}/4$, i.e. we have a square.

This is easy to solve, we simply write $\mathcal{P} = 2(x + y)$ and substitute into $\mathcal{A} = xy$, which simplifies to a quadratic function; then we simply complete the square to find the vertex of the parabolic graph.

In three dimensions, something a little more interesting happens. I wish to frame the problem as follows:

Find $x$, $y$, and $z$, such that a rectangular prism with perimeter (sum of all sixteen sides) $\mathcal{P}$ and surface area $\mathcal{S}$ has maximal volume.

In other words, we want to maximise $V = xyz$ with the constraints $\mathcal{P} = 4x + 4y + 4z$ and $\mathcal{S} = 2xy + 2yz + 2zx$. Substituting:

$\displaystyle \begin{array}{ll} V = x(yz) = x\left(\frac{\mathcal{S} - 2xy - 2zx}{2}\right) &= \frac{1}{2}\mathcal{S}x - x^2 y - x^2 z\\ &= \frac{1}{2}\mathcal{S}x - x^2 (y + z) = \frac{1}{2}\mathcal{S}x - x^2 \left(\frac{\mathcal{P} - 4x}{4}\right) \end{array}$

hence $V$ is a cubic function of $x$. This is interesting, because it suggests that as $x \to \infty$, $V \to \infty$; a very different behaviour to the 2D case. However, as we vary $x$ to large values, we notice that $y$ and $z$ become non-real; so we’ve lost some information somewhere: namely, the largest value of  $x$ that `makes sense’ to form a cube with the given perimeter; we have $x = \frac{1}{4}\mathcal{P} - y - z$, so for $x > \frac{1}{4}\mathcal{P}$, at least one of $x$ and $y$ must be negative – which is absurd.

Taking the case $\mathcal{S} = 6$ and $\mathcal{P} = 12$, we find that the graph of $x$ versus $V$ is a cubic with a single point of inflection (the parabola graphed here is $dV/dx$):

The only critical point is $x = 1$, but the constraint we identified above tells us that the values of $x$ which might make sense are $x \leq \frac{12}{4} = 3$; for $x = 3$, we obtain $V = 9 - 27 + 81 = 63$, and thus we have the system of equations $63 = 3yz, \; 12 = 12 + 4y + 4z$  which again give us non-real values for $x$ and $y$; thus our constraint is not good enough to solve the problem!

By symmetry, I guess in this case that the optimal solution is a cube ($x = y = z = 1$); but how to prove this?

Well, if we solve for $y$ in terms of $x$ alone by substituting our original constraints together, we end up with a quadratic equation and the following:

$\displaystyle 4y = \frac{\mathcal{P}}{2} - 2x \pm \sqrt{2\mathcal{P}x + \frac{\mathcal{P}^2}{4} - 8\mathcal{S} - 12x^2}$.

Hence $y$ is real if and only if $2\mathcal{P}x + \frac{\mathcal{P}^2}{4} - 8\mathcal{S} - 12x^2 \geq 0$ (and the formula for the perimeter tells us that $z$ is real precisely when $y$ is).

In our concrete example, we need to find the maximal $x$ such that $-12 + 24x - 12x^2 \geq 0$; this $x$ will be the larger root, which is $1$, precisely as predicted. (In fact, $1$ is the only root; hence we can say that the only set of dimensions which produce the given perimeter and area are $x = y = z$, which is a much stronger statement than we can say in the 2D case!)

In fact, one might conjecture based on our calculations so far that the allowable values of $x$ are precisely those between the turning points of $y = V(x)$; this is almost true!

Indeed, $\frac{\mathrm{d}V}{\mathrm{d}x} = \frac{\mathcal{S}}{2} - \frac{\mathcal{P}}{2} x + 3x^2$; and $y$ and $z$ are defined between the roots of $0 = \frac{\mathcal{P}^2}{4} - 8\mathcal{S} + 2\mathcal{P}x - 12x^2$, which are the same as the roots of (after dividing through by $-4$) $2\mathcal{S} - \frac{\mathcal{P}^2}{16} - \frac{\mathcal{P}}{2}x + 3x^2$. These are so close to being the same polynomial; they simply differ by a constant term! We see that:

• in the case that $\mathcal{P}^2 = 24\mathcal{S}$, the permissible values for $x$ lie within the turning points of $V$;
• in the case that $\mathcal{P}^2 < 24\mathcal{S}$, the range of permissible values lies strictly between the turning points; and
• in the case that $\mathcal{P}^2 > 24\mathcal{S}$, the range of permissible values includes more than the range between the turning points.

The case that suggests another interesting line of enquiry is the final one: is it possible for there to be a permissible value of $x$ that gives a larger value of $V$ than the local maximum? (At first glance,  yes: simply take a really large value for the perimeter and a small value for the surface area. But does a large perimeter force us to pick a large surface area? I don’t think so, but I haven’t checked.)

Anyway, we have answered our original question: we simply find the values of $V$ at the endpoints of the range of permissible values of $x$, and the local maximum of the $V(x)$ cubic, and take the largest value of $V$ that we get in this way (and find the dimensions corresponding to it by solving the system of equations given by the constraints).