This Idea is half-baked because binary data-displays are not common, and so the market would be small, and so most companies would not bother trying to make it. Ah. well.

There is a fairly common mechanical digit-display (in clock radios) which consists of a horizontal axis around which ten short
plates rotate. Each side of one plate has half a numeral on it, so two plates, one above the axis and one below the axis, suffice to present an entire digit to the viewer. The mechanism rotates the plates to let the topmost drop to the bottom, exposing two fresh sides-of-plates, presenting the next numeral.

Okay, in this device there is just one plate, circular, and the axis is vertical, crossing the plate's diameter. The edge of the circular plate is colored to appropriate visibility. When viewed face-on, you see a zero. When rotated one-quarter-turn, you see the colored edge of the circular plate, which is vertical and so looks like a "one". When rotated another quarter-turn, you see the back side of the plate, which is also a zero. When rotated another quarter-turn, you see "one" again. This Idea allows more rapid switching between the two states, than if a zero was on one side of the plate, and a one was on the opposite side.

The only binary displays I can think of are traffic lights, drive access lights and the like, where simply turning the light on and off takes care of things.

But I imagine you could design 3D objects that would represent different digits or letters in different profiles (indeed, I've seen sculptures like that), allowing you to display more than just binary.

Place it in a box then cut the power. Open the box and examine the display, zero, one, or the perceived width of an angled (mid-spin) zero. Voila! - A qbit observed directly without collapsing the wave.

How about an array of disks that spin 90 degrees (1/4 turn) about their access. A 1 is painted on in vertical stripes, and a zero is painted on the same side in horizontal stripes. A transparent screen with vertical mask bars is placed over the top and when the plate is viewed through it, shows either a big, slightly stripy looking serif 1, or an equally stripy looking serif 0.

Q1 - 25 Marks: Vernon makes a freize of these to run around his kitchen wall and once complete has it zeroed and then sets it up to display a number (in binary) that increases by one every second. After pressing go, Vernon subsequently waits aproximately 2 and a half years before it all resets to zero.

Assuming a diameter of 5cm for each disk, and a clearance of 1/2cm either side, what are the dimensions of Vernon's kitchen?

Assuming that the freize takes up the entire perimeter of the kitchen walls, unhindered by obstacles such as doorways etc but including a small gap to indicate the start and finish point, the length, width and height are calculated as follows:

2 and half years is approximated to 903 days (365+365+183). This is equivalent to some 78019200 seconds. This time in seconds can be represented in hexidecimal format as 4A67A80, or 0010 1010 0110 0111 1010 1000 0000 binary. So 28 digits are required to represent the binary time of 2.5 years. The total perimeter is thus 28 x 5cm (digit faces) + 29 x 0.5cm (clearance spaces**) = 154.5cm

**assuming that between any 2 faces there is a total of 0.5 cm space. For 0.5cm each this figure becomes 28 x 2 x 0.5cm.

Let us assume a further clearance space of 5.5cm between the start and finish digit for the installment of a control unit. The total perimiter is therefore: 160cm.

Assuming that the room is rectangular and has length 45cm, it must have a width of 35cm. We can either assume the height to be on a proportional basis, such as 20% of the length (and therefore 9cm) or we can assume a standard height of, say, 9ft - approximately 274cm. Let us assume the latter.

The volume is then 2.74 * .45 * .35 m^3 = 0.432m^3 Vernon has a very small kitchen.

From this we can conclude that, at a rate of 3 minutes per calculation, I have just wasted 15 critical minutes of my time. ** Aye Carumba!! **

<Warning JinBish - do not read> more pointless calculations ahead </Warning JinBish - do not read>

Part 2a: I forgot to mention that Vernon's kitchen is conical in shape, with an angle of 85° between the floor and wall. Assuming that both Vernon, and the floorplan of his kitchen are both of 'reasonable size', how long would the wire running from the device to a wall-mounted, shoulder-height, 'Reset Counter' button have to be?

For bonus points, calculate how long before the time is up before someone would come into the kitchen and say 'What's this button for?' immediately before pressing the button and causing Vernon to have to remain in his kitchen for another 2.5 years.

<Too late!>
Let us assume a reasonable floor area of 3.14m^2. The spherical floor therefore has a radius: sqrt{3.14 / (pi)} = 1
As the room is a conical shape the height can be worked out thus: Tan 85 = height/radius
11.43 = height/1
height = 11.43m

Now, the clock itself is at the point in the room where the circumference of the cone is 1.545m. The radius at this point is given by: 1.545 = pi.2.r
r = 1.545/2pi
r = 0.246m
The distance to the pinnacle of the cone is then:
Tan 85 = height/0.246 =>
height = 0.246m/11.45 = 0.021m
Therefore, the height of the clock from the floor is 11.45m - 0.021m = 11.24m.
The average light switch is at a height of approximate height of 1.5m, giving a length of cable required as 9.74m.
</Too late!>

How would you account for parralax? The '1' digit would appear as a straight line when viewed from directly head-on. If you viewed the digit from an angle the '1' would appear to be curved, due to it being drawn around the edge of a disc. Have I got this correct?

Heh, I was wondering when somebody else would think of that. One solution could be to make the "0" more of an oval than a circle. Another could be to thicken the rotating plate, so that uppermost and lowermost parts, curving away, become a little less obvious by comparison. And finally, we could just consider this problem to be another reason why this Idea is half-baked!

Put an end to that. just squeeze the digit in the middle like a coffe can when you make a pouring spout to empty the last of it. The zero will never look perfectly round anyway, nor will its counterpart digit one ever look perfectly uniform in parallax.

There are just 10 things wrong with this: 01) LEDs already do this just fine; and 10) Saturn has been doing this ever since it had rings (I think Saturn is displaying 1 now). Other than that, I give the idea a Crescent, which appears as a 3/4 formed '0' when viewed at the proper angle 'C'. Your mileage may vary.

[lurch], good show! I guess the only difference I can think of is that that RR signal only oscillates (implied by the simplicity of the mechanism), while the one I was imagining would fully rotate, in quarter-turn stages. --No, I can think of another difference; that one is probably intended to be highly-seen/not-seen (thinness of metal disc from distance), while I was wanting the edge-on view to be quite visible.

This idea makes me imagine a huge Babbage difference engine, but using "Vernon discs" to indicate all numbers in binary. It would clank and blow huge clouds of steam from its boiler as it muliplied 100101 by 10110.

Sorry to cause big problems, [Jinbish] and [zen_tom]... But if Vernon was waiting for all the digits to reset to 0s, wouldn't the whole "clock" get to all 1s then reset? since binary goes on powers of 2, you'd have:
67,108,863 seconds (27 digits), which is less than 2.5 years (about 2.13 years) OR
134,217,727 seconds (28 digits) which is about 4.25 years.

That was a Vernon idea? What happened to the other twenty pages? :) [note to Vernon: your ideas are sometimes interesting, but your verbosity is legendary]

The links are broken, but railroad signals have indeed used the principle of a 90-degree rotating indicator (including ones with a vertical axis) for a long long time.

[emjay]: You have a very good point. The solution is the same as every other digital clock - a reset mechanism that sets the outputs of all the gates to '0' when the magic total of 0010 1010 0110 0111 1010 1000 0000 is reached.

It does appear, though, that the 2 left-most digits are redundant. Therefore we must re-calcuate *everything*...