Monthly listing

January: PWEI (24th, Shepherd’s Bush)

February: Rammstein (5th, Brixton), The Killers (19th, Brixton), Dragonforce (21st, Mean Fiddler)

March: (no gigs)

April: Dead Can Dance (7th, Forum), Interpol (9th, Brixton), GWAR (28th, Mean Fiddler)

May: Danzig (15th, Forum)

June: Green Carnation (22nd, Underworld), Nuclear Assault (27th, Underworld)

July: Nine Inch Nails (4th, Brixton), Lake Trout+Kalev (28th, Barfly)

August: Skinny Puppy (10th, Astoria), Kalev+WIFE (17th, Turnmills), Zodiac Mindwarp (19th, Underworld), Nick Cave (25th, Alexandra Palace)

September: Puressence (8th, Islington), WIFE (20th, Betsey Trotwood), Dragonforce (22nd, Barfly), The Veils (26th, Barfly)

October: Killing Joke (14th, Astoria), WIFE (20th, Playroom), WIFE (25th, Water Rats), Breathless (29th, Vicenza)

November: Roger Chapman (10th, 100 Club), The Young Gods (25th, Barfly)

December: 3 Inches Of Blood (13th, Underworld)

Facts and Figures

• Total: 27 gigs

• Most: WIFE (4), Kalev (2), Dragonforce (2)

• Best: 3 Inches Of Blood, Breathless, The Killers, Rammstein

• Worst: Roger Chapman, The Young Gods, The Veils, Nick Cave

The quickest game is 11-0. If I serve first, I serve to 2-0, 6-0, 10-0 and thus 6 times. If I serve second, I serve to 4-0, 8-0, 11-0 for 5 serves.

The slowest game is 11-9—whether I serve first or not doesn’t matter, I always serve 10 times.

You’d expect that the average would be around 7.5. But what is it?

Games with even points mean you serve N/2 times whether you serve first or second.

Games with odd points mean you serve ceil(N/2) if you go first, or floor(N/2) if you go second. (This trivialises to N/2 when N is even, handily.)

Adding these up, we get: ((5+6)+(6+6)+(6+7)+(7+7)+(7+8)+(8+8)+(8+9)+(9+9)+(9+10)+(10+10))/20 Which is 7.75 or close to the rough = 7.5 original estimate.

A whole page of Zombie headlines

I love the 21st Century.

blech just posted the C90 project on #moo which, via last week’s mumbling about transistors , led me to wonder how much effective storage a tape holds (90 minutes of 128k MP3 is a charitable estimate giving 100MB) and how much you could fit into the same volume using modern storage.

Taking a typical tape popular amongst the #moo, the TDK D90, the dimensions are 4.25 × 2.7 × 0.50 inches or 107.95 × 68.5 × 12.7 mm.

Compact Flash is 42.8 × 36.4 × 3.3 mm meaning you can only fit 3 × 1 × 3 giving 9GB of storage (using cheap 1GB CF cards). That’s a factor of 90 increase.

xD (the smallest common form factor today) is even better: 25 × 20 × 1.7 mm giving 4 × 3 × 7 or 84GB—840 times more storage.

• They’re advertising a game called ‘N-TROPY’ where the point is to build something. That sounds dumb until you realise it’s an unstable structure which will eventually fall down. Props for the amusing nod towards The Second Law.
• Michael Johnson! Advertising Linux for IBM! The world has gone TOPSY-TURVY!
• Mega-party-poppers leave their little bits of tissue ammunition everywhere
• It’s Fred Dibnah Weekend on UKTV Documentary . I saw him fell the Alder Chimney in Leigh when I was a child (from a safe distance.)
• Trying to add a static light to an Ixus 30 is hard work. (The Ixus 30 won’t use the flash in Digital Macro mode so taking macro photos of insects at night is impossible. I’ve been using the camera light on my K700i so far but it’s faff and casts sharp, angled shadows. So I got an LED reading light from Foyles which is being reconstructed by my father.

Yesterday, Gryn wanted some 400MB SDK downloading (he’s on dialup) so I obliged but instead of writing a DVD, I stuck it on a 1GB Compact Flash card (I have four empty ones and it’s a lot quicker.)

Whilst walking to the station, it occurred to me that a CF card is much smaller than a CD and then I wondered what the smallest consumer form-factor for storage is (probably the xD cards.)

That lead to wondering about possible quad-stable circuits using fewer transistors or smaller area than two bi-stables (I’m guessing it’s not easy) which would lead to even larger amounts of storage for a given physical size.

Of course, that lead to calculating how many transistors I was carrying down the hill. Since you need at least two transistors to implement a single bit equivalent (using a bistable multivibrator), the calculation is simple: number of bytes × 8 bits in a byte × 2 transistors.

Flash RAM also uses two transistors per bit in a different configuration so we can use the same calculation.

Ignoring the associated electronics (almost certainly orders of magnitude fewer transistors) and (most of) my MP3 player (disk-based), we find:

• Starting with the 1GB CF in my left pocket, we have 1,000,000,000 × 8 × 2 transistors = 16,000,000,000 transistors. 16 billion transistors. Eesh.
• In my other pocket, I have my Ixus 30 with a 256MB SD card. 256,000,000 × 8 × 2 = 4,000,000,000 transistors. 4 billion. Also a lot.
• The MP3 player has a 14MB play buffer. 14,000,000 × 8 × 2 = 224,000,000 transistors. 224 million transistors just to give me gap-less playback. Inconceivable.
• Lastly, my phone has 42MB of RAM. Nice round number. 42,000,000 × 8 × 2 = 672,000,000 transistors. A mere and laughable 672 million.

So that’s 20,896,000,000 transistors or just shy of 21 billion. Probably closer to 22 billion if I account for the MP3 player and phone likely having more RAM than is user-exposed.

That’s more than 3 transistors for every person on earth.

• “Atomic Moog 2000” – Coldcut
• “Enola Gay” – OMD
• “Sacrilege” – Mentallo & The Fixer

If you have N mis-focused images taken from the same point of the same source, is it possible through magical mathematical convolutions to reconstruct a better (but not perfectly focused) image of the source? You have multiple equations for the brightness of an output pixel so, as far as my limited understanding goes, getting a “better” image boils down to solving a large number of simultaneous equations for each pixel.

Thinking it through with diagrams, it seems that it would be impossible without knowing the diameter of a certain sized input point on the output image—else you wouldn’t be able to assign an output pixel to a source pixel. Knowing the ratio of focal lengths might help in some indefinable way.