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Math was never the problem. The explanation was.

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Can you reverse the spin of this tetrahedron?

Can you reverse the spin of this tetrahedron?

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Can you reverse the spin of these snakes?

Can you reverse the spin of these snakes?

670,426 просмотров

A turnstile antenna is two crossed dipoles driven 90° apart, one dipole p turning at ω about its mast. Its far field E ∝ r̂ × (r̂ × p̈(t − r/c)) has phase ωt − kr − φ at every latitude, so each wavefront is the spiral r = (ωt − φ − 2πn)/k. Along the mast E ∝ (1/r)[cos(kr − ωt) ŷ + sin(kr − ωt) x̂] keeps its length and turns once per period, and B = (1/c) r̂ × E turns with it. In the plane of the cross the same wave is linearly polarized.

A turnstile antenna is two crossed dipoles driven 90° apart, one dipole p turning at ω about its mast. Its far field E ∝ r̂ × (r̂ × p̈(t − r/c)) has phase ωt − kr − φ at every latitude, so each wavefront is the spiral r = (ωt − φ − 2πn)/k. Along the mast E ∝ (1/r)[cos(kr − ωt) ŷ + sin(kr − ωt) x̂] keeps its length and turns once per period, and B = (1/c) r̂ × E turns with it. In the plane of the cross the same wave is linearly polarized.

81,717 просмотров

Can you reverse the spin of this ambiguous sphere with your mind?

Can you reverse the spin of this ambiguous sphere with your mind?

265,958 просмотров

Can you reverse the spin of this trefoil knot?

Can you reverse the spin of this trefoil knot?

202,859 просмотров

An electric dipole p(t) = p₀ sin(ωt) ẑ oscillating at a point radiates a field that, far away, falls off only as 1/r. Its radiation term is E = (μ₀/4π) p̈(t − r/c) sin θ / r along θ̂, with B = r̂ × E / c along φ̂, so E, B and the direction of travel r̂ are mutually perpendicular. The intensity goes as sin²θ / r²: zero along the axis, strongest at the equator. Every sphere around the source carries the same mean power, p₀²ω⁴ / (12πε₀c³).

An electric dipole p(t) = p₀ sin(ωt) ẑ oscillating at a point radiates a field that, far away, falls off only as 1/r. Its radiation term is E = (μ₀/4π) p̈(t − r/c) sin θ / r along θ̂, with B = r̂ × E / c along φ̂, so E, B and the direction of travel r̂ are mutually perpendicular. The intensity goes as sin²θ / r²: zero along the axis, strongest at the equator. Every sphere around the source carries the same mean power, p₀²ω⁴ / (12πε₀c³).

91,589 просмотров

Can you reverse the spin of this hyperbolic paraboloid?

Can you reverse the spin of this hyperbolic paraboloid?

89,098 просмотров

Can you reverse the spin of this hourglass?

Can you reverse the spin of this hourglass?

87,157 просмотров

A Gaussian wave packet with energy E = 12.5 meets a square barrier of height V₀ = 20 and width 0.29, with ħ = m = 1. Inside the barrier the wave decays as e^(−κx), κ = √(2m(V₀ − E))/ħ, so the wall leaks T ~ e^(−2κw). Almost none of the packet has energy above V₀, yet 32.7 % of the probability ends up on the far side. The same exponential sets the current in a scanning tunnelling microscope and the half-life of alpha decay.

A Gaussian wave packet with energy E = 12.5 meets a square barrier of height V₀ = 20 and width 0.29, with ħ = m = 1. Inside the barrier the wave decays as e^(−κx), κ = √(2m(V₀ − E))/ħ, so the wall leaks T ~ e^(−2κw). Almost none of the packet has energy above V₀, yet 32.7 % of the probability ends up on the far side. The same exponential sets the current in a scanning tunnelling microscope and the half-life of alpha decay.

28,158 просмотров

A double integral is the limit of a sum of prisms: cut [0, π]² into n × n squares of area ΔA = (π/n)² and raise each to the height of z = sin³x sin³y at its centre. The volume is (∫₀^π sin³x dx)² = (4/3)² = 16/9 ≈ 1.7777778. The prisms give 1.7602152 at n = 4, 1.7768141 at n = 8 and 1.7777776 at n = 64, always from below. Since sin³x = (3 sin x − sin 3x)/4, the h² errors of the two waves cancel, and the gap shrinks like 7π⁴/(180n⁴).

A double integral is the limit of a sum of prisms: cut [0, π]² into n × n squares of area ΔA = (π/n)² and raise each to the height of z = sin³x sin³y at its centre. The volume is (∫₀^π sin³x dx)² = (4/3)² = 16/9 ≈ 1.7777778. The prisms give 1.7602152 at n = 4, 1.7768141 at n = 8 and 1.7777776 at n = 64, always from below. Since sin³x = (3 sin x − sin 3x)/4, the h² errors of the two waves cancel, and the gap shrinks like 7π⁴/(180n⁴).

22,893 просмотров

The curve y = ∛(x²) + 0.9 sin(kx)√(3 − x²) is a sine wave on |x| ≤ √3, carried by the cusp ∛(x²) and scaled by a semicircle. Since |sin(kx)| ≤ 1, it never leaves the band between ∛(x²) − 0.9√(3 − x²) and ∛(x²) + 0.9√(3 − x²). Those two envelopes are the heart: lobes peaking at 2.27 near x = ±1.03, a point at −0.9√3 ≈ −1.56 on the axis. Raising k only packs the wave tighter. At k = 100 it crosses the band about 110 times.

The curve y = ∛(x²) + 0.9 sin(kx)√(3 − x²) is a sine wave on |x| ≤ √3, carried by the cusp ∛(x²) and scaled by a semicircle. Since |sin(kx)| ≤ 1, it never leaves the band between ∛(x²) − 0.9√(3 − x²) and ∛(x²) + 0.9√(3 − x²). Those two envelopes are the heart: lobes peaking at 2.27 near x = ±1.03, a point at −0.9√3 ≈ −1.56 on the axis. Raising k only packs the wave tighter. At k = 100 it crosses the band about 110 times.

18,271 просмотров

Can you reverse the spin of this Penrose triangle?

Can you reverse the spin of this Penrose triangle?

50,239 просмотров

Can you reverse the spin of this hyperboloid?

Can you reverse the spin of this hyperboloid?

43,585 просмотров

A bar magnet spinning at angular speed Ω is a magnetic dipole m turning in its own plane, and it radiates at f = Ω/2π. In that plane E points straight out of it, E ∝ m̈ × r̂ at the retarded time t − r/c, so each far crest lies on θ = φ(t − r/c) + π/2: two spiral arms of pitch λ = 2πc/Ω. The whine is 1024f, ten octaves up. In empty space J = 0, so in ∇ × B = μ₀ε₀ ∂E/∂t + μ₀J only the changing E keeps B alive. The radiated power grows as Ω⁴.

A bar magnet spinning at angular speed Ω is a magnetic dipole m turning in its own plane, and it radiates at f = Ω/2π. In that plane E points straight out of it, E ∝ m̈ × r̂ at the retarded time t − r/c, so each far crest lies on θ = φ(t − r/c) + π/2: two spiral arms of pitch λ = 2πc/Ω. The whine is 1024f, ten octaves up. In empty space J = 0, so in ∇ × B = μ₀ε₀ ∂E/∂t + μ₀J only the changing E keeps B alive. The radiated power grows as Ω⁴.

23,468 просмотров

Can you reverse the spin of this sphere? Took me more than 10 seconds.

Can you reverse the spin of this sphere? Took me more than 10 seconds.

40,784 просмотров

Can you reverse the spin of this T Block?

Can you reverse the spin of this T Block?

30,017 просмотров

A sawtooth, a square and a triangle wave, each summed from its first ten nonzero Fourier terms, f(t) = ∑ bₙ sin(nt). Each term is a circle of radius |bₙ| turning n times per period. The sawtooth has bₙ = 2(−1)ⁿ⁺¹/(nπ), the square 4/(nπ) for odd n, the triangle ±8/(n²π²) for odd n. At ten terms, the square peaks at 1.18 and the sawtooth at 1.09; the triangle stays within 0.021 of exact. A jump forces coefficients like 1/n, a corner only 1/n².

A sawtooth, a square and a triangle wave, each summed from its first ten nonzero Fourier terms, f(t) = ∑ bₙ sin(nt). Each term is a circle of radius |bₙ| turning n times per period. The sawtooth has bₙ = 2(−1)ⁿ⁺¹/(nπ), the square 4/(nπ) for odd n, the triangle ±8/(n²π²) for odd n. At ten terms, the square peaks at 1.18 and the sawtooth at 1.09; the triangle stays within 0.021 of exact. A jump forces coefficients like 1/n, a corner only 1/n².

13,928 просмотров

A disk of radius a rolls on a table at tilt θ, its contact point circling at rate Ω. Gravity against the gyroscopic torque fixes Ω² = 4g/(a sin θ), with energy (3/2)Mga sin θ. Rolling resistance μ drains it as dθ/dt = −(2/3)μΩ, so near the end θ√θ falls linearly to zero: a finite-time stop. With a = 0.40 m and μ = 0.0086, θ drops from 70° to 1.4° in 15 s while Ω climbs from 1.63 to 10.08 Hz. The disk itself turns ever slower, at Ω sin θ.

A disk of radius a rolls on a table at tilt θ, its contact point circling at rate Ω. Gravity against the gyroscopic torque fixes Ω² = 4g/(a sin θ), with energy (3/2)Mga sin θ. Rolling resistance μ drains it as dθ/dt = −(2/3)μΩ, so near the end θ√θ falls linearly to zero: a finite-time stop. With a = 0.40 m and μ = 0.0086, θ drops from 70° to 1.4° in 15 s while Ω climbs from 1.63 to 10.08 Hz. The disk itself turns ever slower, at Ω sin θ.

19,220 просмотров

Can you reverse the spin of this Klein bottle?

Can you reverse the spin of this Klein bottle?

29,181 просмотров

A radius turns at a steady rate inside a regular hexagon, a decagon and a circle, and the height y = r(θ) sin θ of its tip is traced against time. For an n-gon of circumradius R with a vertex at θ = 0, r(θ) = R cos(π/n) / cos((θ mod 2π/n) − π/n). Each edge is one piece of the wave; the flat top edge is a plateau at R cos(π/n), 0.866R for n = 6, 0.951R for n = 10. As n grows the plateaus narrow and rise to R, and the wave tends to R sin θ.

A radius turns at a steady rate inside a regular hexagon, a decagon and a circle, and the height y = r(θ) sin θ of its tip is traced against time. For an n-gon of circumradius R with a vertex at θ = 0, r(θ) = R cos(π/n) / cos((θ mod 2π/n) − π/n). Each edge is one piece of the wave; the flat top edge is a plateau at R cos(π/n), 0.866R for n = 6, 0.951R for n = 10. As n grows the plateaus narrow and rise to R, and the wave tends to R sin θ.

11,905 просмотров

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Each coordinate of a closed outline has its own real Fourier series: x(t) = ∑ (aₖ cos kt + bₖ sin kt), and the same for y. A cosine and sine of frequency k make one arm of length √(aₖ² + bₖ²) turning k times per lap. One chain sets x, one sets y, and the pen sits where their guides cross. With N harmonics per axis the knight is off by 30.2 % of its height at N = 1, 0.70 % at N = 50. At N = 1 both coordinates are pure sinusoids: an ellipse.

Each coordinate of a closed outline has its own real Fourier series: x(t) = ∑ (aₖ cos kt + bₖ sin kt), and the same for y. A cosine and sine of frequency k make one arm of length √(aₖ² + bₖ²) turning k times per lap. One chain sets x, one sets y, and the pen sits where their guides cross. With N harmonics per axis the knight is off by 30.2 % of its height at N = 1, 0.70 % at N = 50. At N = 1 both coordinates are pure sinusoids: an ellipse.

1,370,261 просмотров • 9 дней назад

Joseph Fourier’s 1823 engraving, traced as 57 ink shapes joined by thin bridges into one closed outline, is a sum of circles: z(t) = ∑ cₖ eⁱᵏᵗ, the cₖ from its discrete Fourier transform. With the N largest circles the widest gap to the outline is 42.0 % of the portrait’s height at N = 1, 3.16 % at 128 and 0.92 % at 512, and the circles shrink like k^−1.47. The tip’s distance |z(t)| plays as pitch.

Joseph Fourier’s 1823 engraving, traced as 57 ink shapes joined by thin bridges into one closed outline, is a sum of circles: z(t) = ∑ cₖ eⁱᵏᵗ, the cₖ from its discrete Fourier transform. With the N largest circles the widest gap to the outline is 42.0 % of the portrait’s height at N = 1, 3.16 % at 128 and 0.92 % at 512, and the circles shrink like k^−1.47. The tip’s distance |z(t)| plays as pitch.

105,487 просмотров • 4 дней назад

The treble clef 𝄞, its four contours joined by thin bridges and sampled at 4096 points, is a sum of circles: z(t) = ∑ cₖ eⁱᵏᵗ. With the N largest circles the widest gap to the outline falls from 35.7 % of its height at N = 1 to 0.37 % at 256. The tip's distance |z(t)| is played as pitch, floored to C major. One circle keeps |z| fixed: one note. With 64 the tip plays the outline's own note 93 % of the lap; with 256, 98 %.

The treble clef 𝄞, its four contours joined by thin bridges and sampled at 4096 points, is a sum of circles: z(t) = ∑ cₖ eⁱᵏᵗ. With the N largest circles the widest gap to the outline falls from 35.7 % of its height at N = 1 to 0.37 % at 256. The tip's distance |z(t)| is played as pitch, floored to C major. One circle keeps |z| fixed: one note. With 64 the tip plays the outline's own note 93 % of the lap; with 256, 98 %.

189,597 просмотров • 9 дней назад

The curves y = x² and y = 1.1ˣ cross exactly three times: at x = −0.955, at x = 1.051 and at x = 95.717. Between the second and third crossings the square leads, by a lot: 50² = 2500 while 1.1⁵⁰ ≈ 117. The ratio 1.1ˣ/x² bottoms out at x = 2/ln 1.1 ≈ 21 and grows without bound after it, so the third crossing is final. By x = 200, 1.1ˣ ≈ 4 700 · x².

The curves y = x² and y = 1.1ˣ cross exactly three times: at x = −0.955, at x = 1.051 and at x = 95.717. Between the second and third crossings the square leads, by a lot: 50² = 2500 while 1.1⁵⁰ ≈ 117. The ratio 1.1ˣ/x² bottoms out at x = 2/ln 1.1 ≈ 21 and grows without bound after it, so the third crossing is final. By x = 200, 1.1ˣ ≈ 4 700 · x².

47,724 просмотров • 2 дней назад

Give every pixel a Pokemon type, then let it fight its neighbours. Fire beats grass, grass beats water, water beats fire. That loop has no winner, so no colour ever owns the grid. Domains swell, collide, and get eaten from behind.

Give every pixel a Pokemon type, then let it fight its neighbours. Fire beats grass, grass beats water, water beats fire. That loop has no winner, so no colour ever owns the grid. Domains swell, collide, and get eaten from behind.

1,725,973 просмотров • 3 месяцев назад

Each coordinate of the Koch snowflake's outline has its own real Fourier series, x(t) = ∑ (aₖ cos kt + bₖ sin kt), and one chain of arms per axis draws it where their guides cross. The snowflake's six-fold symmetry leaves only k ≡ ±1 (mod 6), so five arms in six are zero, and its mirror leaves x only sines and y only cosines. The error falls from 11.5 % of the height at N = 1 to 0.60 % at N = 193. At N = 1 the two arms are equal: a circle.

Each coordinate of the Koch snowflake's outline has its own real Fourier series, x(t) = ∑ (aₖ cos kt + bₖ sin kt), and one chain of arms per axis draws it where their guides cross. The snowflake's six-fold symmetry leaves only k ≡ ±1 (mod 6), so five arms in six are zero, and its mirror leaves x only sines and y only cosines. The error falls from 11.5 % of the height at N = 1 to 0.60 % at N = 193. At N = 1 the two arms are equal: a circle.

92,000 просмотров • 6 дней назад

The Moore curve of order 4 is a loop of 256 unit steps through every cell of a 16 × 16 grid. Each coordinate has its own real Fourier series, and one chain of arms per axis draws it. The second half of the loop is the first half turned 180°, so every even harmonic is zero, and its mirror leaves x only sines and y only cosines. A curve that wiggles everywhere converges slowly: 32.3 % of the height at N = 1, still 0.47 % at N = 255.

The Moore curve of order 4 is a loop of 256 unit steps through every cell of a 16 × 16 grid. Each coordinate has its own real Fourier series, and one chain of arms per axis draws it. The second half of the loop is the first half turned 180°, so every even harmonic is zero, and its mirror leaves x only sines and y only cosines. A curve that wiggles everywhere converges slowly: 32.3 % of the height at N = 1, still 0.47 % at N = 255.

66,592 просмотров • 5 дней назад

The dumbest way to solve a maze, and it works every time. Release thousands of particles at the entrance and let each random walk. The first to stumble out has traced a legal path. Retrace its steps and the answer is there. Diffusion, not thinking.

The dumbest way to solve a maze, and it works every time. Release thousands of particles at the entrance and let each random walk. The first to stumble out has traced a legal path. Retrace its steps and the answer is there. Diffusion, not thinking.

744,788 просмотров • 3 месяцев назад

A square wave is a sum of odd sines, f(x) = (4/π) ∑ sin((2n−1)x)/(2n−1), and each term is a circle of radius 4/(π(2n−1)) turning 2n−1 times per period. Chained end to end, the height of the n-th joint is the partial sum Sₙ(x). The overshoot at each jump tends to 0.179, about 9 % of the jump, however many terms are added. At x = π/2 every arm points straight up or down and f = 1, which is Leibniz's series: π/4 = 1 − 1/3 + 1/5 − 1/7 + ⋯.

A square wave is a sum of odd sines, f(x) = (4/π) ∑ sin((2n−1)x)/(2n−1), and each term is a circle of radius 4/(π(2n−1)) turning 2n−1 times per period. Chained end to end, the height of the n-th joint is the partial sum Sₙ(x). The overshoot at each jump tends to 0.179, about 9 % of the jump, however many terms are added. At x = π/2 every arm points straight up or down and f = 1, which is Leibniz's series: π/4 = 1 − 1/3 + 1/5 − 1/7 + ⋯.

47,203 просмотров • 8 дней назад

The lemniscates of the Mandelbrot set are the curves |pₙ(c)| = 2, with p₁(c) = c and pₙ₊₁ = pₙ² + c. At n = 1 the curve is the circle |c| = 2. Each lies inside the last, and all pass through c = −2, where pₙ = 2 for n ≥ 2. The area inside falls from 4π ≈ 12.57 to 1.92 at n = 12, a curve of degree 4096. They close in on the set: c lies in M exactly when |pₙ(c)| ≤ 2 for every n, since an orbit that leaves the disc of radius 2 escapes.

The lemniscates of the Mandelbrot set are the curves |pₙ(c)| = 2, with p₁(c) = c and pₙ₊₁ = pₙ² + c. At n = 1 the curve is the circle |c| = 2. Each lies inside the last, and all pass through c = −2, where pₙ = 2 for n ≥ 2. The area inside falls from 4π ≈ 12.57 to 1.92 at n = 12, a curve of degree 4096. They close in on the set: c lies in M exactly when |pₙ(c)| ≤ 2 for every n, since an orbit that leaves the disc of radius 2 escapes.

38,435 просмотров • 7 дней назад

The outline of π, sampled at 2048 points, is a sum of circles: z(t) = ∑ cₖ eⁱᵏᵗ, the cₖ from its discrete Fourier transform. Circle k has radius |cₖ| and turns k times per lap. With the N largest, the widest gap to the outline is 41.6 % of the letter's height at N = 1, 1.37 % at 64, 0.36 % at 256. Finitely many circles make a smooth curve, and π has four sharp corners, so no chain is exact. The error stays below the sum of the radii left out.

The outline of π, sampled at 2048 points, is a sum of circles: z(t) = ∑ cₖ eⁱᵏᵗ, the cₖ from its discrete Fourier transform. Circle k has radius |cₖ| and turns k times per lap. With the N largest, the widest gap to the outline is 41.6 % of the letter's height at N = 1, 1.37 % at 64, 0.36 % at 256. Finitely many circles make a smooth curve, and π has four sharp corners, so no chain is exact. The error stays below the sum of the radii left out.

51,299 просмотров • 11 дней назад

A string fixed at both ends vibrates in standing waves sin(nπx/L), with frequencies fₙ = n·f₁ = (n/2L)√(T/μ). On the A string, f₁ = 110 Hz, the first eight are A2, A3, E4, A4, C#5, E5, G5, A5. Six sit within 2 cents of a piano key; the 5th is 14 cents flat and the 7th, 770 Hz, is 31 cents flat of G5. A pluck at p = 1/5 is the sum of all modes with bₙ = 2h sin(nπp)/(n²π²p(1−p)). Since sin(π) = 0, it has no 5th harmonic.

A string fixed at both ends vibrates in standing waves sin(nπx/L), with frequencies fₙ = n·f₁ = (n/2L)√(T/μ). On the A string, f₁ = 110 Hz, the first eight are A2, A3, E4, A4, C#5, E5, G5, A5. Six sit within 2 cents of a piano key; the 5th is 14 cents flat and the 7th, 770 Hz, is 31 cents flat of G5. A pluck at p = 1/5 is the sum of all modes with bₙ = 2h sin(nπp)/(n²π²p(1−p)). Since sin(π) = 0, it has no 5th harmonic.

26,780 просмотров • 8 дней назад

Ten functions y(x) are played as pitch, f = 130.8 · 2ʸ Hz, with x advancing 2.5 units per second, so every graph is four bars of music. A kick is y = −1.6 + 5e^(−kx): a five octave jump that falls back toward 43 Hz. ⌊·⌋ snaps two sines onto steps of 7/12 of an octave, a fifth. arctan(s sin 4πx)/arctan s squares a vibrato as s climbs from 0.5 to 40, and sin(πp²/4) quickens its own notes.

Ten functions y(x) are played as pitch, f = 130.8 · 2ʸ Hz, with x advancing 2.5 units per second, so every graph is four bars of music. A kick is y = −1.6 + 5e^(−kx): a five octave jump that falls back toward 43 Hz. ⌊·⌋ snaps two sines onto steps of 7/12 of an octave, a fifth. arctan(s sin 4πx)/arctan s squares a vibrato as s climbs from 0.5 to 40, and sin(πp²/4) quickens its own notes.

33,662 просмотров • 11 дней назад

The first phrase of Beethoven's Ode to Joy is a single formula: d = ⌊2.27 + 1.2 sin(π(s + 1)/8) − cos(πs/4)⌋ with s = ⌊x⌋ + ½, played at f = 261.63 · 2ʸ Hz. The tune moves only by scale steps, so two harmonics carry its shape and the floor snaps each beat onto a degree: E E F G G F E D C C D E E D D. Then ⌊(12d + 5)/7⌋ = 0, 2, 4, 5, 7, 9, 11 turns degrees into semitones. The major scale itself is a floor: seven notes spread evenly over twelve.

The first phrase of Beethoven's Ode to Joy is a single formula: d = ⌊2.27 + 1.2 sin(π(s + 1)/8) − cos(πs/4)⌋ with s = ⌊x⌋ + ½, played at f = 261.63 · 2ʸ Hz. The tune moves only by scale steps, so two harmonics carry its shape and the floor snaps each beat onto a degree: E E F G G F E D C C D E E D D. Then ⌊(12d + 5)/7⌋ = 0, 2, 4, 5, 7, 9, 11 turns degrees into semitones. The major scale itself is a floor: seven notes spread evenly over twelve.

13,817 просмотров • 9 дней назад

A square wave flipping between ±1 at multiples of π is the limit of f_N(t) = (4/π) ∑ sin((2k+1)t)/(2k+1) for k < N. Only odd harmonics appear, since f(t + π) = −f(t). Term k is a circle of radius (4/π)/(2k+1) turning 2k+1 times as fast as the first; the chain ends at height f_N(t). The sum overshoots each jump. Its first peak, at t = π/(2N), is 1.182 for N = 5 and 1.179 for N = 100, about 9 % of the jump. More terms narrow it, never remove it.

A square wave flipping between ±1 at multiples of π is the limit of f_N(t) = (4/π) ∑ sin((2k+1)t)/(2k+1) for k < N. Only odd harmonics appear, since f(t + π) = −f(t). Term k is a circle of radius (4/π)/(2k+1) turning 2k+1 times as fast as the first; the chain ends at height f_N(t). The sum overshoots each jump. Its first peak, at t = π/(2N), is 1.182 for N = 5 and 1.179 for N = 100, about 9 % of the jump. More terms narrow it, never remove it.

23,614 просмотров • 22 дней назад

Can you reverse the spin of this staircase?

Can you reverse the spin of this staircase?

25,679 просмотров • 1 месяц назад

A polyrhythm is two even pulses sharing one span. One hand turns every 3 s and sounds each corner it crosses: the square's four, every 0.75 s, against a regular n-gon's n, every 3/n s. The two pulses meet gcd(n, 4) times a turn: once for 3:4, 5:4, 7:4, 9:4 and 15:4, twice for 6:4 and 10:4, and on every square beat for 8:4 and 20:4.

A polyrhythm is two even pulses sharing one span. One hand turns every 3 s and sounds each corner it crosses: the square's four, every 0.75 s, against a regular n-gon's n, every 3/n s. The two pulses meet gcd(n, 4) times a turn: once for 3:4, 5:4, 7:4, 9:4 and 15:4, twice for 6:4 and 10:4, and on every square beat for 8:4 and 20:4.

17,238 просмотров • 23 дней назад

Can you reverse the spin of this knot?

Can you reverse the spin of this knot?

17,157 просмотров • 1 месяц назад

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