wrog

(why I should not be allowed to write textbooks, part 342)

This is mainly because I wanted to have all of the formulas in one place. Also curious to see how much I can compress the derivation and still have it vaguely make sense. Also I wanted to learn more MathML.

The 2 body problem

We have two bodies with mass. Gravity attracts them to each other. How do they move?

The first order of business is to arrange our own seating so that their center of mass is stationary from our point of view. Once we do that, what we see is

a big mass $M$ at a small distance $r$ away from the center, moving at small velocity $v$ and
a small mass $m$ at a big distance $R$ away moving at big velocity $V$ ;

M r = m R

;

M v = m V

; and they're always opposite each other, mirroring each other's motions. Thank you, conservation of momentum.

We also notice that, at any given moment, you can always draw a line through the origin and both bodies. The velocities will usually not be parallel to the line, but there's always a plane that contains the line and one of the velocities, and since the other velocity is always the exact opposite direction from the first, it, too will be in the plane. And then there's nothing that's ever trying to pull either of the bodies out of that plane, we realize that plane doesn't move, they stay in it forever and so we can just rotate our coordinates so that it's the x-y plane, forget about z, and pretend we're in a 2-dimensional universe from now on.

And now for The Trick.

Whether you're following Hamilton's or Langrange's way of doing things (skipping how that works for now), the energies, as functions of positions and velocities, give you everything you need to know to solve for how things are going to move. So let's just write them down. Total kinetic energy will (because I hate dividing by 2) be half of

m V^{2} + M v^{2} = (m {(\frac{M}{M + m})}^{2} + M {(\frac{m}{M + m})}^{2}) {(V + v)}^{2} = \frac{M m}{M + m} {(V + v)}^{2}

while the potential energy, which will be all due to gravity here, is

- (\frac{G M m = G (M + m) (\frac{M m}{M + m})}{R + r})

What's interesting/weird here is how we have managed to recast everything in terms of just the relative distance $R + r$ and relative velocity $V + v$ and that these are the same formulas we'd get if we were trying to figure out what happens with a single object of "reduced" mass $μ = M m / (M + m)$ bouncing around a gravity well generated by something at the origin with mass $M + m$ (that has been magically nailed in place so that it doesn't move). Any solution we get for the latter problem will readily translate back.

One possibly-slight-surprise here is that, despite my talk of "small" and "big", we are, at no point ever actually depending on, say, $m ≪ M$ , or even $m < M$ . m could even be bigger and this will all still work; you just have to be sure to remember to do that last step (i.e., translate back to a 2-body problem and not be surprised when it turns out the "big" object is moving around, too).

So, scratch one body. Yay.

Also, we will not actually be mentioning the reduced mass μ ever again because from now on, we can just do everything in terms of energy/whatever per unit mass (of the one object).

Alsoalso, we can introduce a new constant $k ≔ G (M + m)$ so as not to have to look at that ever again, either. As it happens, in real life, k tends to be way more accurately known than any of G, M, or m, individually, so good riddance.

One body inverse-square force problem

Now that we're down to one body, in a $- k / r$ gravity well generated by (SHUT UP), what happens?

If the body's position in polar coordinates is $⟨ r, θ ⟩$ , then we have radial and angular velocities being $\overset{•}{r}$ and $r \overset{•}{θ}$ , respectively, where the • is time derivative, i.e., $\overset{•}{r} = d r / d t$ , and so on. (Yes, we're doing the Dot Thing.)

All forces come out of the center, so angular momentum per unit mass

L ≔ r^{2} \overset{•}{θ}

will be constant. Which, first of all, means $θ$ will be always increasing or always decreasing as time goes on (good to know!).

Also, if you can imagine lots of really thin triangles with base $r dθ$ and height $r$ (and therefore area $= ½ r^{2} dθ$ ), we see that $L$ is is twice the area being swept out per unit time, and therefore if we eventually get an orbit that's periodic with period $T$ (spoilers!), then the total area enclosed will be $½ L T$ (save for future reference).

For any $L$ , there will be a corresponding circular orbit radius $r_{0} = L^{2} / k$ , which I am introducing now solely for my own convenience and may or may not have anything to do with where the object is actually spending its time.

The second constant of motion will be the energy per unit mass, $E$ , expressed thusly (again multiplying by 2, just because I can):

2 E = {\overset{•}{r}}^{2} + (r^{2} {\overset{•}{θ}}^{2} = \frac{(L^{2} = k r_{0})}{r^{2}}) - \frac{2 k}{r}

Now if what we want is to get the shape of the orbit/trajectory, then we don't care quite so much about $\overset{•}{r}$ or $\overset{•}{θ}$ as we do about their ratio $dr / dθ$ , which inspires the following truly awesome and sneaky maneuver:

\overset{•}{r} = \frac{dr}{dt} = (\frac{dθ}{dt} = \overset{•}{θ} = \frac{L}{r^{2}}) \frac{dr}{dθ} = L \frac{d}{dθ} (\frac{−1}{r})

which we can square and substitute into the previous equation. Then, dividing both sides by $L^{2} = k r_{0}$ we get

\frac{2 E}{k r_{0}} = {(\frac{d}{dθ} (\frac{−1}{r}))}^{2} + \frac{1}{r^{2}} - \frac{2}{r r_{0}}

where we now see two terms on the right as a square crying out to be completed, so we add $1 / {r_{0}}^{2}$ to both sides, while at the same time noticing that we can also add arbitrary constant crap to the thing we're differentiating in the middle, hence

\frac{1}{r_{0}^{2}} (ε^{2} ≔ 1 + \frac{2 E r_{0}}{k}) = {(\frac{d}{dθ} (\frac{1}{r_{0}} - \frac{1}{r}))}^{2} + {(\frac{1}{r_{0}} - \frac{1}{r})}^{2}

where on the left we have also consolidated yet more constant crap into a new constant, the eccentricity (but introducing it as $ε^{2}$ because the stuff on the right can't be negative).

And now we stop to ask the class, "So, does anyone know of some $f (θ)$ that does $f^{2} + {(df / dθ)}^{2} = {f_{0}}^{2}$ (a constant)?"

Inevitably, some nerd in the front who did the assigned reading or maybe they really are able to figure out on the fly that you can rearrange this into $\pm df / \sqrt{{f_{0}}^{2} - f^{2}} = dθ$ and then of course you want $f = f_{0} sin-or-cos (something)$ which both the df and the square root expression change to cos-or-sin()s which cancel leaving you with $d (something) = \pm dθ$ and then the something has to be ±θ plus a constant will be all, "Oo, pick me!" and then we get our answer (this is actually how all differential equations get solved, in case you didn't know):

\frac{1}{r_{0}} - \frac{1}{r} = - \frac{ε}{r_{0}} \cos (θ - θ_{0})

where the particular choice to have it be −cos() rather than +sin() makes no difference — because of that extra $θ_{0}$ which covers all of the bases — but happens to be easy to rearrange to get the version more people are familiar with:

r_{0} = r (1 + ε \cos θ)

And yes, I quietly rotated the coordinate system to make the $θ_{0}$ go away again because I was pretty sure I would get tired of looking at it. Good catch.

The first thing to notice is that if $ε = 0$ then this is a circular orbit and now, hopefully, the mysterious name I gave for $r_{0}$ should start making sense.

The second thing to notice is that if $ε > 1$ there's a range of angles where the stuff in parentheses can go negative, i.e., everything around 180° between the two magic angles where $\cos θ = −1 / ε$ is actually forbidden. Which would then mean we have something coming in from infinity at one of the magic angles, swooping around to do a close flyby at $θ = 0°$ , and then fucking off back to infinity at the other magic angle … what you might call a hyperbolic trajectory.

If instead $ε < 1$ , then θ goes all of the way around without any problems, … must, in fact, do so and keep doing so (thank you, angular momentum), which then means we are indeed stuck in something periodic, with r varying between a closest approach $r_{0} / (1 + ε)$ , at $θ = 0°$ , to a maximum distance $r_{0} / (1 - ε)$ , at $θ = 180°$ . Adding these and dividing by 2 gives us the semi-major axis

a ≔ r_{0} / (1 - ε^{2}) = - k / 2 E

the distance to the actual geometric center of this thing, which usually will be some distance away ( $a ε$ ) from where all of the gravity is coming from. (Also, don't freak about the minus sign; $ε < 1$ means $E$ has to be negative.)

If you're having trouble figuring out what this shape is, it may be more recognizable in Cartesian coordinates. However, life will be easier if we also shift the origin, so we set $⟨ r \cos θ, r \sin θ ⟩ = ⟨ x - a ε, y ⟩$ , at which point the orbit becomes

(r_{0} = a (1 - ε^{2})) = r + ε (x - a ε)

or, canceling the $a ε^{2}$ s, getting $r$ by itself on one side, and squaring everything:

{(a - ε x)}^{2} = r^{2} = {(x - a ε)}^{2} + y^{2}

or, eventually

1 = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{(a^{2} (1 - ε^{2}) = a r_{0})}

which, as long as $ε < 1$ , is taking a unit circle and stretching it, — in the $x$ direction by $a$ and in the $y$ direction by $b ≔ a \sqrt{1 - ε^{2}} = \sqrt{a r_{0}}$ — which is what an ellipse is.

Since the unit circle has area $π$ , our ellipse has to have area $π a b = ½ L T$ , per that thing we saved for future reference earlier, which, after we square it and cancel an $r_{0}$ from both sides, rearranges into

{(\frac{2 π}{T})}^{2} a^{3} = k

which is that square-cube law that's awfully useful for getting orbital periods from orbit sizes and vice versa.

If $ε > 1$ , then $a$ is negative, which is weird but not a showstopper, and we end up with the $y^{2}$ term being negative, which means we have a hyperbola instead of an ellipse (surprise!). And then the stuff after that (defining $b$ as the square root of a negative number, etc) fails and you don't get a period out of this (again, surprise! it's not periodic anymore).

We will leave the $ε = 1$ case — where the energy $E$ is zero and the object is always travelling at exactly the escape velocity, no more, no less — as an exercise.

Well okay, I guess there is a way to get Trilorne to hover.

If we're willing to toss the usual definition of "North" (rotation axis points that way), we can put the sun over the equator, and then we can have it be tide-locked. Meaning this planet is basically Mercury but farther away so that only the stuff directly underneath the sun is getting fried to shit.

And maybe having an actual atmosphere will help, too, in various ways, though I can't imagine there not being freaky weather patterns, e.g., some kind of permanent cyclone storm around the solar-maximum point wherever it happens to be at the moment, but that won't damage the story too much because nobody ever goes to the Fire Lands anyway.

The Wall then runs around ( surprise )

Crossposts: https://wrog.livejournal.com/70683.html

So there's this Arthur C. Clarke short story, "The Wall of Darkness" (1949). I read it as a kid and found it really haunting. Clarke does Haunting really well.

If you haven't read it already and want to go read it before I completely and totally ruin it, feel free.

(I found the whole thing by searching for "Trilorne" in google, which then gave me a google books hit; we'll see how much longer The Algorithm lets people do that).

But you've already had 70 years, so… onward…

( let the ruining begin... )

Crossposts: https://wrog.livejournal.com/70528.html

Continued from Part 3, what happens when there is no "parallel", the rules for circles aren't what you thought they were, and so on.

Napier's Rules

So how does trigonometry work in this world?

See, I belatedly realized that spewing walls of equations like this is not actually going to be much use when you're stuck in a rowboat in the middle of the North Atlantic having to navigate by the stars with no cell phone and no GPS. Because, chances are, you also have No Internet, and then my blog entries with their handy tables go to waste.

It would be far better if I can teach you how to derive these relationships instead, i.e., in a way that you might actually be able to vaguely remember while sitting in a boat in the middle of the North Atlantic.

But first I'm going to introduce a bit of gratuitous extra notation. Write

ᶜᵒθ

— pronounce it "co-theta" if you want — to mean (90° − θ). I do this because:

I can,
it's less typing,
it's way less degree vs. radian waffling, which I already do too much of,
but also,
you get all of the following useful and amusing equivalences (no, really; read them aloud):

sin ᶜᵒθ = cos θ
cos ᶜᵒθ = sin θ
tan ᶜᵒθ = cot θ	(= 1/tan θ, in case you've forgotten)
cot ᶜᵒθ = tan θ
csc ᶜᵒθ = sec θ	(= 1/cos θ, and no, I don't know why reciprocals get these special names)
sec ᶜᵒθ = csc θ	(= 1/sin θ, because, seriously, WTFF?)

You'd almost think they planned it this way.

Hopefully, it goes without saying that ᶜᵒ(ᶜᵒθ) = θ, except I had to go and say it, didn't I? (Damn.)

And now let's start with a right triangle, with vertices/angles and sides/lengths labeled a,b,c,A,B, the way you usually see it in trigonometry class, ( and then derive stuff about it )

Crossposts: https://wrog.livejournal.com/70126.html

Continued from Part 2, exploring the benighted universe where "parallel" is Not a Thing.

How circles work

So, to review the weird things we've seen so far:

When we have a circle radius of 90°, otherwise known as a straight line, and we're traversing the circumference, i.e., measuring the total length along it as we sweep out 360° from the pole in the middle, we get 360° worth of path (phrasing it this way so that if this turns out we're on a projective plane rather than a sphere and what we're really doing is traversing the same 180° path twice, I won't have been lying to you), which, being 4 times the radius, is slightly less than one might have expected (2π being roughly 6.28).
If we attempt a circle of radius of 180°, we stay firmly nailed to the antipode of the center, our circumference traversal goes nowhere and thus we get a circumference of zero.

Meaning if we have to explain to the residents what "π" is, we're going to lose horribly. Best we can do is, "So: Circumference to radius? That's a ratio. It's literally all over the map. But as radius approaches zero, once you're under 90°, you'll notice the ratio is always getting bigger. If you work at it, you can prove that it's bounded and it converges to this weird transcendental number like e. And, no, don't ask us how we came up with this…"

We need to understand better how curved paths work. ( and so ye shall... )

Crossposts: https://wrog.livejournal.com/69769.html

Continued from Part 1, in which we discover at least one consequence to doing away with the concept of "parallel lines".

Let's talk about Area

Having noticed that isoceles right triangles give us a natural way to define/measure distances, we see that we can do area this way as well. That is, the area of ΔAPX is clearly the angle at P times some constant, which we may as well just take to be 1 if we haven't defined a unit of area yet. ( so let's do that ... )

Crossposts: https://wrog.livejournal.com/69460.html

So, as part of my possibly-continuing "Geometry on Drugs" series, here is a prequel to my post on spherical geometry, which was more of a "hey, this is useful" post in which much there's a whole lot you're expected to take on faith. It was really more intended for the hardcore engineering type who needs to see that use case up front.

This version is going back to first principles, where we do the axiom wanking and you (hopefully) get a sense of why things turn out the way they do.

Also, this is the practice run before I launch into the Essence of Hyperbolic Geometry, so, … Onward …

The Geometry Axiom Everybody Hates

Start with this diagram and the inevitable question that comes up:

Start with a line ℓ and a point A not on it. How do you put a line through A that doesn't intersect ℓ?

(In other news, I am now convinced that the Unicode committee contained at least one disgruntled geometry teacher. How else to explain why there's this isolated script ℓ code point?)

We can drop a perpendicular from A meeting ℓ at some point X, and then it's obvious that the line you want (dotted) is the one perpendicular to XA. If you tilt it even slightly away from 90°, then it simply must intersect ℓ somewhere.

Proof by diagram. We're allowed to do that, right? ( Read more... )

Crossposts: https://wrog.livejournal.com/68946.html

So here's a puzzler

Brief review of the physics of wind instruments

Your standard wind instrument is a pipe attached to some kind of sound source (lip or reed). Being of a certain length it resonates at particular frequencies and suppresses the others, ulimately being capable of producing tones that are multiples of a certain fundamental frequency f, 2f being an octave higher than f, 3f being a fifth above that and so on.

The main things to keep in mind about the multiples of f (the "harmonic series"):

going up by a particular interval means multiplying the frequency (or dividing the wavelength) by a particular ratio (2 ↔ octave, 3/2 ↔ perfect fifth, 5/4 ↔ major third, etc.…), and
intervalwise, the notes are getting closer together the higher you go.

Eventually, the notes get so close to each other that you can do scales of a sort -- this is how the natural (i.e., valveless) French horn works; the tube is so damned long that you're always playing in the 8f-16f range and then you just fudge the few notes that are out of tune with lip and hand-in-the-bell tricks. ... and likewise for why baroque trumpet parts are so insanely high-pitched.

But if you want to do anything useful down in the lower register, you have no choice but to mess with the length of the tube somehow, the two most popular methods being

(the woodwind solution) poke holes in the tube (and cover them with removable keys)
(the brass solution) insert valves or slides in the tube to change its actual length

We'll focus on the second solution since I really have no freaking clue what's going on with woodwinds. Oddly enough, the brass solution is simpler but it didn't come along until fairly late in the game. Not until the 19th century did metalworking technology finally get good enough that Heinrich Stölzel was able to construct the first actual valved instrument in 1814. Once this happened it didn't take long to catch on -- hence the 19th century explosion in the use of brass instruments in orchestral works (cf. Wagner, Mahler...).

So how do valves actually get used?

Let's first get clear what a valve is: ( Yes, we are going deep into the weeds. You knew this was going to happen )

a.k.a., Space 11: How to do Interstellar Navigation

Various antecedents you may want to have peered at first:

the Spherical Trigonometry lesson
Space 10 on consequences of Relativity (why clocks slow down, lengths change, and how FTL breaks things), or
Space 9 if you need to brush up on spacetime diagrams and why simultaneity gets screwed up

Today's post is about Hyperbolic Geometry, wherein you learn what those "Warning, Evil, Don't Look" columns are about. It's now safe to look; well okay, no it isn't, but too late! AHAHAHAHAHAHAHA.

Hyperbolic geometry is basically Geometry On Drugs and we know that's never going lead anywhere good.

To be fair, Spherical Geometry is arguably also on drugs, but at least it's easier to explain in that, having had lots of experience with basketballs and whatnot, you already know what a sphere is. Having a concrete place for the "points" to live, I can then tell you

what "lines" are (great circles, or planes slicing the sphere through the origin / center of the sphere),
how to measure "distance" along a "line" segment (measure angle between endpoints from the center of the sphere),
how to measure "angles" between "lines" (the planes will intersect; there's an angle there; done), and
what "circles" are (they're um, circles, … or, if you like, planes that don't necessarily go through the origin, or cones coming out of the origin; whatever works for you),

and then you're basically good to go, ready to do all of the geometry/trigonometry you could ever want, once you've heeded my warnings that Certain Things Will Be Different (no such thing as "parallel", triangles add up to 180 plus area instead of just 180, do not feed them after midnight, etc…).

Unfortunately, the place where we're Doing Geometry today is this inside-out Hyperboloid Sheet Thing with a fucked up metric, … and if you've actually seen one of those in real life, I will be very surprised, especially since it's not something that can exist in ordinary 3D space. Oddly enough, it will end up relating to something you do have day-to-day experience with, namely (cue reverb and James Earl Jones voice)… Your Future,… but I'm not sure how much help that's going be in visualizing it.

( bring on the drugs... )

Crossposts: https://wrog.livejournal.com/68636.html

William has learned about Rock, Paper, Scissors. In honor of that, a puzzle:

Same game: Rock breaks Scissors, Paper covers Rock, Scissors cut Paper. So far so good. Now we add a couple twists:

I do research and find myself a Better Rock, a chunk of pure New England granite that rules, absolutely crushes all lesser Rocks. Even though it still loses to Paper I'm happy with it.
Meanwhile you've been doing your own research; being of a technological bent you know there are better ways to do Scissors; carbon steel with a diamond edge; not only cuts Paper but completely destroys other Scissors as well. Granted, even a diamond edge is no match for an any actual Rock, let alone mine, but you still win against everything else, so you're happy.

So... my Rock beats your Rock, your Scissors beat my Scissors, and Paper vs. Paper is the only draw possibility left.

What's my strategy? What's your strategy?

And how does this change if I go out and get really good, battle-ready Paper as well, e.g., some of that Tyvek stuff that they use to insulate houses; something that will entirely shred your Paper, even if your high-tech Scissors will still make short work of it. Meaning that not only is there no longer any possibility of a draw, but out of the 9 possible scenarios, I'm winning in five of them.

Unfair, right?

Discuss.

So, I remember this demo.

It was something that, back in grade school, the engineer father of one of my best friends really liked to put on -- did it on several occasions that I remember. It was one of these Physics is Cool demos.

Start with a room about 40-50 feet long, almost exactly like this one, in fact if we can trust Google's scale marker, that's pretty much what it was (yes, the room takes up the whole length of the building; yes, the Internet is really scary these days...). I remember the ceiling as being really high, but since I was smaller back then, that recollection is suspect. I also distinctly recall someone accidentally sticking the end of a flagpole into the ceiling (we had our Boy Scout troop meetings in that room), so it couldn't have been that high. From the photo, 12 feet seems likely.

There's a stepladder right at the back wall of the room.

Somewhere towards the middle of the room, there's a hook in the ceiling.

A piece of twine is tied to the hook, long enough to almost reach the floor; at the other end of the twine, a Rather Heavy Object of some sort, ... shotput or bowling ball or somesuch, I forget.

Up at the front of the room is a chair with a 1-2 foot diameter basket on it.

He pulls two volunteers out of the audience. Volunteer #1 is somebody's little sister. He positions her near the middle of the room, between the basket and the hook, and hands her this big-ass butcher's knife. "This is extremely sharp. I want you to hold this right here, just like that." Blade is pointed towards the back of the room, angled downward.

Volunteer #2 takes the bowling ball up the stepladder. "Yeah, that's about high enough. Just let go of it."

Ball swings forward, right down to the floor and up again, reaches the knife, the twine parts pretty much instantaneously, the ball sails through the air, and lands right in the middle of the basket.

And there was much rejoicing.

What's a bit frustrating now is that, since I was in 6th or 7th grade at the time, this was all before I'd had any Actual Physics. Meaning "Physics is Cool" was pretty much all of the content that I got out of this. It was mentioned that the ball needed to be heavy and the knife needed to be sharp, but beyond that I had no idea which other details of the setup were crucial and which didn't matter at all. Nor was there really much of an explanation of why it worked, what principle was being demonstrated -- or maybe there was one, and I just didn't have the background to appreciate it and so it all just went kind of went whoosh.

Actually, now that I think about it, there may indeed have done some kind of brief "inertial foo gravitational mass mumble mumble", but I'm sure he knew up front that when you've got something that works mainly because it just drops out of the math and you've got an audience that doesn't know a whole lot of math, you're just going to lose them if you try to explain too much, so you just skip the boring part and get on with the show.

To be sure, it's fairly cool to be able to set up a scenario, calculate out in advance where things are going to go, then pull the trigger and have them Actually Go There. Which perhaps describes pretty much every classroom physics demo ever. And this may indeed be enough of a point for 7th & 8th graders (i.e., "Study your math, kids. Keys to the universe!" or maybe just, "It works, bitches").

But still a bit frustrating, because, years later I learned some Actual Physics, and a few questions remain. Not that I'm particularly surprised that this worked, but,... what's the gimmick? Some particular sweet spot where to put the basket so that it doesn't matter quite so much where you put the knife or where you let the ball drop from? Or maybe it's a sweet spot in the knife placement or the original drop point? Was the scenario chosen the one that was easiest to calculate? Or the one that had the basket the furthest forward? Or the one had the longest flight time for the ball?

I suppose I could just call John up and ask, "Hey, that thing your dad did with the knife and the swinging shot put; what was the deal with that, anyway?" But this would be cheating. So I'm going to analyze this a bit. . .

( bet you didn't see *this* coming... )

So I've arbitrarily decided that more people need to know about spherical trigonometry. (e.g., just in case the GPS gets destroyed and we're stuck having to do our own navigation again.)

It's really the same solving of triangles that you learned to do in high school geometry/trig, i.e., gimme side-angle-side or angle-side-angle to nail down what the triangle actually is, then use Law of Sines or Law of Cosines or some combination thereof to work out the previously unknown sides/angles that you care about.

It's just that some of the rules for spherical trig are a Little Bit Different.

So, jumping right into the deep end, here's an application inspired by recent events:

You, an observer stationed on planet Earth at some particular latitude, want to know where the sun is going to be in the sky at some particular time of day, some particular day of the year.

And, cutting to the chase, here's a triangle to solve:

If you know b, A, c (side, angle, side), you can solve for a (opposite side) using the Law of Cosines

cos a = cos b cos c + sin b sin c cos A

and then B (next angle) using the Law of Sines

sinB/sinb = sinA/sina = sinC/sinc

( ...math wanking continues... )

You may not have been aware of this, but III.F.1 of the WSDCC's 2008 Delegate Selection Plan reads as follows:

Alternates shall be listed and seated in the order in which they were elected,
and must be of the same presidential preference, and, to the extent possible,
of the same gender and from the same electing jurisdiction as the delegate being
replaced.

This rule applies at all levels of caucus starting with the LD caucus, where the (bottom-level) precinct delegates go to elect the (2nd-level) delegates to the CD caucus and state convention.

Now, the interesting thing about this is that, unlike at the higher levels, where male and female delegates/alternates are strictly segregated (i.e., you get so many delegates of each gender for a given jurisdiction and there are are likewise separate alternate lists for each gender), the delegation from the precinct caucus to the LD caucus is co-ed; and for each precinct for each candidate there's likewise just a single ordered list of alternates where the M-F pattern can be anything.

( ...they'll probably just cut me up and put me in a recycle bin under the Washington State Convention Center )

Current Mood: accomplished

So I guess that was too easy.

( if I turn into The Time Cube Guy, please shoot me... )

You have one (1) ordinary six-sided die (i.e., the traditionally marked, uniform-density, cubical kind that you could find in any Vegas casino within about 5 minutes).

You need to make a random choice between four (4) alternatives, same probability for each (25%, i.e., barring the occasional coin-landing-on-edge weirdness that supposedly can happen in real life, which we won't worry about).

You could, e.g., roll the die and if you don't get 1,2,3,or 4, just roll again and keep rolling until you do. That would be too easy, and also could conceivably take a while if you get unlucky with the 5s and 6s.

But the real problem is that, due to various contrivances entirely beyond your control, it just so happens that if you roll the die a second time, you will be plagued by frogs, and (trust me on this) you really do not want to be plagued by frogs.

So you absolutely have to do it in one (1) roll. Good luck.

( hint and potential spoilage... )

Tune her piano to Kirnberger III ( ** )

Of course, this being the 2nd time in my life that I've attempted to tune a piano, I imagine getting it to sound even remotely correct will be something of a triumph. (Last time around, people were saying "Gee, you ought to get that tuned" after I was done). On the other hand, last time around was 20 years ago and back then I didn't have clue #1 about temperament issues ( * ).

Suffice it to say, having a sense of absolute pitch --- and yes, I know people call it "perfect" pitch, but it just isn't, okay? --- is almost completely useless for this sort of job, just in case you were wondering.

On the other hand, given the state that her piano is in now (hasn't been tuned for maybe 20 years), almost anything will be an improvement, and since she's mostly tone-deaf anyway, I could probably tune it completely randomly and she wouldn't notice a thing (though

emmacrew and I almost certainly would...).

Also, I have this theory that Kirnberger III, with all of its perfect fifths, will be a lot easier to do than Equal Temperament. We'll see.

( more math/music wanking . . . )

Current Music: The Well-Tempered Clavier (what else?)

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Kepler Orbit Cheat Sheet
Getting Trilorne to Hover
Arthur C. Clarke and the Projective Plane
Spherical Geometry 4 — Mommy, Where Does Trigonometry Come From?
Spherical Geometry 3 -- A bit of Circular Reasoning
Spherical Geometry 2 -- Deficits Matter
The Essence of Spherical Geometry, Part 1
towards a truly chromatic Horn
Public Service Math Lesson: Hyperbolic Trigonometry
Today's game theory intro
physics is cool
public service math lesson: spherical trigonometry
Something I probably won't present to the state Rules Committee in 2012
more on S4 (really A5, but who cares)
Today's Puzzle a.k.a. Stupid S4 Tricks
Mean tricks to play on your mother-in-law #27

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Entries tagged with math

Kepler Orbit Cheat Sheet

(why I should not be allowed to write textbooks, part 342)

The 2 body problem

One body inverse-square force problem

Getting Trilorne to Hover

Arthur C. Clarke and the Projective Plane

Spherical Geometry 4 — Mommy, Where Does Trigonometry Come From?

Napier's Rules

Spherical Geometry 3 -- A bit of Circular Reasoning

How circles work

Spherical Geometry 2 -- Deficits Matter

Let's talk about Area

The Essence of Spherical Geometry, Part 1

The Geometry Axiom Everybody Hates

towards a truly chromatic Horn

Brief review of the physics of wind instruments

So how do valves actually get used?

Public Service Math Lesson: Hyperbolic Trigonometry

a.k.a., Space 11: How to do Interstellar Navigation

Today's game theory intro

physics is cool

public service math lesson: spherical trigonometry

Something I probably won't present to the state Rules Committee in 2012

more on S4 (really A5, but who cares)

Today's Puzzle a.k.a. Stupid S4 Tricks

Mean tricks to play on your mother-in-law #27