Solve a mark. Find its place.

The dots are not equally spaced: each position is calculated from its equation’s result. ◆ identifies the first equation; it is a clue, not the 12 o’clock mark. Use the 18-entry list for the catalogue order and the illustrative inputs.

Choose Oblique 3-D view to see the three vector components. Face-on makes the equation circle clear but hides the normal component; projection or rotation can also overlap positions or hide a component.

Show the position rule

For catalogue entry j, counted from 0 to 17 (the listed entry number minus one), calculate its result using the listed inputs. Express it as a dimensionless number v: divide angles by 1 rad; multiply an angle per metre by 1 m and divide by 1 rad. The other displayed results are already dimensionless. For complex v, use both its real and imaginary parts.

q = Re(v) + Im(v)
f = q − floor(q)
p = (j + f)/18 turns clockwise after 12
αⱼ = 360°p = 20°(j + f)

Thus the mark represents 12p hours after 12. The clock numerically evaluates the documented source constructions using the listed fixed inputs. It uses equivalent normalizations or documented intermediate expressions where applicable, with integration, root finding and series summation as needed. The exact worked results remain available for comparison. The choice of this encoding and the one-metre reference is part of the artwork, not a radar measurement. The fractional-part operation intentionally discards whole-number cycles, and Re(v) + Im(v) can map different complex values to the same number.

To locate 12, solve the ◆ equation (j = 0), find its p, and move backwards by 360p degrees in the original clock plane. That recovered direction defines the H axis of the single time vector. Its M axis points 90° counterclockwise in the reference plane; its S axis is the right-handed plane normal. The twelve-hour face alone does not distinguish AM from PM.

Reveal a worked first position

The first result is 1 + √2. Its fractional part is √2 − 1, so its mark is (√2 − 1)/18 of a turn after 12: about 16 minutes 34 seconds along the hour-hand cycle. It is not the 12 mark itself.

Read the projected 3-D vector (spoiler)

The single vector uses three coordinates: H = ((hour mod 12) + minute/60 + second/3600)/12, M = (minute + second/60)/60, and S = second/60. H points toward the hidden 12 direction, M is 90° counterclockwise in the reference plane, and S is its right-handed normal. Tilt the whole vector, then project it.

The colored H, M and S segments are projected head-to-tail. To keep the vector visible, its display gain changes with the time: g = min(12, 1.08/√(H² + M² + S²)), taking g = 12 when all three coordinates are zero. The same gain scales every component and preserves the vector’s direction. For each segment, divide its screen displacement by g times the displayed clock radius R, reversing the screen y direction, to obtain normalized d. Then, with its projected unit axis b, recover its coordinate with (d · b)/(b · b). The × value shown beside the vector is the gain, also given below; it is needed to recover the coordinate magnitudes from the image. Then seconds = round(60S), minutes = round(60M − seconds/60), hours = round(12H − minutes/60 − seconds/3600). The white arrow alone generally cannot recover three coordinates; the separate segments provide the extra information.

For a manual calculation, solve ◆ to obtain its encoded angle α◆, calculate θ and ψ from the date, and use the view angles below. Form A = P Rₓ Rᵧ C D T [eₓ eᵧ]. Subtract the displayed center from the ◆ coordinates, invert A to recover its reference-plane direction, then rotate that direction counterclockwise by α◆ to recover the hidden 12 axis. If A is too close to singular, this direct method cannot establish the reference reliably. Use Freeze time to hold a live instant. The shown tilted 3-D vector comes from the time used to draw the model; the recovered time comes from the model’s unrounded SVG segment endpoints, not measurements of screenshot pixels.

Show the tilt and projection formulas

Let g = (cos ψ, sin ψ), d = gx·x + gy·y, c = cos θ, and s = sin θ. For a vector (x,y,z) in the reference frame, the rigid tilt is:

T(x,y,z) = (x + (c−1)d gx − s z gx,
y + (c−1)d gy − s z gy,
s d + c z)

For the projection display, D(x,y,z) = (x,y,0); for the clock-plane display, D is the identity. The selected base camera C is T⁻¹ for face-on, or the identity for overhead and oblique views. The view applies yaw and pitch after that. With P(x,y,z) = (x,y), the projected component axes are bᵢ = P Rₓ(pitch) Rᵧ(yaw) C D T eᵢ. The illustrative separation changes positions but cancels from the component differences.

Coordinates here are normalized, with screen y upward. The projection is orthographic; the two displayed camera angles describe the view, not a separately adjustable physical projection direction. A projected unit axis shorter than 8 display pixels is flagged as too small for this direct method. Passing that check does not guarantee individual seconds are visually distinguishable in a screenshot. The clock components contain redundancy, so other methods may sometimes infer a hidden component. AM/PM is not encoded.

Reveal traditional hand positions (spoiler)

This traditional-hand explanation uses the known device or example time and the calculated equation positions; it does not infer time from the displayed image. Choose Clock hands to show these hands on the face.

The angles and percentages describe the original clock plane between the calculated, unevenly spaced marks, not distances on the projected screen. The twelve-hour face alone does not show AM or PM.

From a tilted circle to its projection.

This fixed overhead construction separates the clock plane from its image so you can follow the mapping. The lines connect matching points. Change the date to see it change; the separation is only for the drawing. The faint projected outline on the main clock turns with the scene.

Tilted clock planeOverhead projectionCorresponding points

The day bends the circle.

Every year begins the count anew. Find this date’s place in the year; divide it by thirteen and keep what is left. Which version of the rule belongs to this day?

Count the ordinal day D: January 1 is day 1. Include February 29 when it occurs.

Find the remainder r = D mod 13, from 0 through 12. Choose it below. The choices substitute a value into one rule; they are not thirteen different rules.

Start at z₀ = 0. Use n = 0, 1, …, 12 in order, taking each output as the next input. From z₁₃, find the tilt θ and direction ψ using the expressions below.

Choose your remainder

r = D mod 13

z₀ = 0

z_{n + 1} = 1 + 0.9 z_{n} \exp {i [0.4 - \frac{6}{1 + {| z_{n} |}^{2}} + \frac{π}{7} \sin (\frac{2 π (n + r)}{13})]}

θ = arctan|z₁₃| · ψ = arg(z₁₃)

Use radians for the trigonometric functions and complex argument. θ is the tilt; ψ is the direction of steepest slope. To express θ in degrees, multiply its radian value by 180/π.

Hints for the calendar and calculation

Count days in completed months, then add the day of the month. A leap year has 29 days in February: years divisible by 4 are leap years, except century years unless divisible by 400. January 1 always starts at 1, and z₀ resets to 0 for each date.

If the division is exact, choose 0. To calculate without a complex-number function, write z_n = x_n + i y_n and call the bracketed angle above a_n. Start with x₀ = y₀ = 0, then repeat these equations for n = 0 through 12:

x_n+1 = 1 + 0.9(x_n cos a_n − y_n sin a_n)

y_n+1 = 0.9(x_n sin a_n + y_n cos a_n)

|z_n|² = x_n² + y_n²

After 13 steps, |z₁₃| = √(x₁₃² + y₁₃²) and ψ = atan2(y₁₃, x₁₃).

AN IMAGINARY CLOCK / A VISIBLE PROJECTION

Time, in another plane.

Each expression has a place. The day tilts the plane; the hands move with it. Each solved equation determines its own unequal position around a twelve-hour cycle.

Time zoneDate for tiltHands

◆ MARKS THE FIRST EQUATION · SHORT HAND / HOUR · LONG HAND / MINUTE

The lines connect each equation to its place on the tilted face. Select an equation to see the complete source, exact value and evaluation. Long equations can be scrolled within their labels.

READING THE CLOCK

Equations make the intervals.

Use the position rule above: the fixed inputs give a result, and its dimensionless real and imaginary parts determine the marker’s offset within its catalogue sector. The resulting gaps between markers are unequal. ◆ identifies the first equation and does not itself mark 12.

After finding the hidden 12 direction from a solved equation, read clockwise in the imaginary plane. A full hour-hand turn is 12 hours; a full minute-hand turn is 60 minutes. Projection can compress or reverse apparent screen ordering. The optional reading aid below uses the known clock time, not a measurement from the image.

Reading aid is off.

Explore all 18 exact results

These results and inputs are illustrative mathematical constructions, not radar measurements. Select a clock equation for its complete source and working.

How the date shapes the projection

The ordinal day D gives r = D mod 13. Starting from z₀ = 0, iterate zₙ₊₁ = 1 + 0.9zₙ exp{i[0.4 − 6/(1 + |zₙ|²) + (π/7) sin(2π(n+r)/13)]} thirteen times. The inclination is arctan|z₁₃| and the direction of steepest slope is arg(z₁₃). A rigid rotation followed by orthographic projection applies to both labels and hands. The forcing is an artistic adaptation, not a measurement.

A clock in three circles.

Solve a mark. Find its place.

From a tilted circle to its projection.

The day bends the circle.

Time, in another plane.

Equations make the intervals.