^{n}, is it true that the centroid of P (the average of all the points in P) is equal to the average of the points on the boundary of P? Intuitively it would appear to be so, but just wanted to make sure.

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Given a convex set P in R^{n}, is it true that the centroid of P (the average of all the points in P) is equal to the average of the points on the boundary of P? Intuitively it would appear to be so, but just wanted to make sure.

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

No.

An isosceles triangle, with vertices at (±b, 0) and (0, h), has centroid (0, h/3). Its edges have lengths 2b and √(b² + h²), and centroids (0, 0) and (±b/2, h/2). So the triangle's boundary has centroid

( 2b(0,0) + √(b² + h²)(b/2,h/2) + √(b² + h²)(-b/2,h/2) ) / ( 2b + √(b² + h²) + √(b² + h²) )

= √(b² + h²)(0, h) / ( 2b + 2√(b² + h²) )

= (0, h√(b² + h²) / (2b + 2√(b² + h²)) )

≠ (0, h/3).

(If you try to set them equal, you get h² = 3b², indicating a regular triangle.)

Now let's try a trapezoid, with vertices at (±(b + a), 0) and (±b, h). It can be split into two right triangles and a rectangle; this gives the centroid

( ah/2 (b + a/3, h/3) + ah/2 (-(b + a/3), h/3) + 2bh (0, h/2) ) / ( ah/2 + ah/2 + 2bh )

= (0, ah²/3 + bh²) / (ah + 2bh)

= (0, h/3 (a + 3b)/(a + 2b) ).

But the centroid of its vertices is (0, h/2).

So we conclude that the centroids of the various k-dimensional skeletons (or frames, or whatever they're called) are generally unrelated. It really is a coincidence that, for an n-dimensional simplex, the n-frame and the 0-frame (vertices) have the same centroid.

An isosceles triangle, with vertices at (±b, 0) and (0, h), has centroid (0, h/3). Its edges have lengths 2b and √(b² + h²), and centroids (0, 0) and (±b/2, h/2). So the triangle's boundary has centroid

( 2b(0,0) + √(b² + h²)(b/2,h/2) + √(b² + h²)(-b/2,h/2) ) / ( 2b + √(b² + h²) + √(b² + h²) )

= √(b² + h²)(0, h) / ( 2b + 2√(b² + h²) )

= (0, h√(b² + h²) / (2b + 2√(b² + h²)) )

≠ (0, h/3).

(If you try to set them equal, you get h² = 3b², indicating a regular triangle.)

Now let's try a trapezoid, with vertices at (±(b + a), 0) and (±b, h). It can be split into two right triangles and a rectangle; this gives the centroid

( ah/2 (b + a/3, h/3) + ah/2 (-(b + a/3), h/3) + 2bh (0, h/2) ) / ( ah/2 + ah/2 + 2bh )

= (0, ah²/3 + bh²) / (ah + 2bh)

= (0, h/3 (a + 3b)/(a + 2b) ).

But the centroid of its vertices is (0, h/2).

So we conclude that the centroids of the various k-dimensional skeletons (or frames, or whatever they're called) are generally unrelated. It really is a coincidence that, for an n-dimensional simplex, the n-frame and the 0-frame (vertices) have the same centroid.

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

Interesting. So basically, the average of vertices, the average of edges, ... the average of n-dimensional elements, are in general all different?

Edit: nevermind, that's exactly what you said.

Edit: nevermind, that's exactly what you said.

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

But we might guess that, for an n-dimensional simplex, the k-frame and the (n - k)-frame have the same centroid, for any k.

This is obviously true in 1D or 2D. So let's try a 3D simplex, with not too much symmetry: a triangular pyramid. We already know that the 0-frame and the 3-frame have the same centroid; we want to know whether the 1-frame (edges) and the 2-frame (faces) have the same centroid.

Say the three base edges have length a, and the three rising edges have length b. The height (in its own plane) of the base triangle is √3/2 a. The height (in its own plane) of a lateral face is l = √(b² - (1/2 a)²) = √(b² - 1/4 a²). The 3D height of the pyramid is h = √(l² - (√3/6 a)²) = √(b² - 1/3 a²); notice that this requires b > 1/√3 a. The area of the base is √3/4 a², and the area of a lateral face is 1/2 a l.

We can put the vertices at coordinates (√3/3 a, 0, 0), (-√3/6 a, ±1/2 a, 0), and (0, 0, h). Then the 1-frame's centroid is (0, 0, (3b)/(3a + 3b) 1/2 h), while the 2-frame's centroid is (0, 0, (3/2 a l)/(√3/4 a² + 3/2 a l) 1/3 h). And of course the 0-frame and 3-frame have centroid (0, 0, 1/4 h).

These are not equal in general. But let's equate them and see what happens:

b/(a + b) 1/2 h = l/(√3/2 a + 3 l) h

b (√3/2 a + 3l) = 2(a + b) l

√3/2 ab = (2a - b) l

The left side is positive, while the right side is positive only if b < 2a. Continuing, with x = b/a to simplify:

3/4 a²b² = (4a² - 4ab + b²) (b² - 1/4 a²)

3x² = (4 - 4x + x²) (4x² - 1)

3x² = 4x⁴ - 16x³ + 16x² - x² + 4x - 4

0 = 4x⁴ - 16x³ + 12x² + 4x - 4

Obviously x = 1 is a solution, corresponding to a regular tetrahedron, so we can factor that out:

0 = 4 (x - 1) (x³ - 3x² + 1)

This cubic polynomial has one solution in the relevant interval 1/√3 < x < 2, which happens to be exactly

x = 1 - 2 cos(4π/9) = 1/(2 cos(2π/9)) ≈ 0.6527

Very interesting! Somehow this equal-centroid pyramid is related to the regular enneagon.

This is obviously true in 1D or 2D. So let's try a 3D simplex, with not too much symmetry: a triangular pyramid. We already know that the 0-frame and the 3-frame have the same centroid; we want to know whether the 1-frame (edges) and the 2-frame (faces) have the same centroid.

Say the three base edges have length a, and the three rising edges have length b. The height (in its own plane) of the base triangle is √3/2 a. The height (in its own plane) of a lateral face is l = √(b² - (1/2 a)²) = √(b² - 1/4 a²). The 3D height of the pyramid is h = √(l² - (√3/6 a)²) = √(b² - 1/3 a²); notice that this requires b > 1/√3 a. The area of the base is √3/4 a², and the area of a lateral face is 1/2 a l.

We can put the vertices at coordinates (√3/3 a, 0, 0), (-√3/6 a, ±1/2 a, 0), and (0, 0, h). Then the 1-frame's centroid is (0, 0, (3b)/(3a + 3b) 1/2 h), while the 2-frame's centroid is (0, 0, (3/2 a l)/(√3/4 a² + 3/2 a l) 1/3 h). And of course the 0-frame and 3-frame have centroid (0, 0, 1/4 h).

These are not equal in general. But let's equate them and see what happens:

b/(a + b) 1/2 h = l/(√3/2 a + 3 l) h

b (√3/2 a + 3l) = 2(a + b) l

√3/2 ab = (2a - b) l

The left side is positive, while the right side is positive only if b < 2a. Continuing, with x = b/a to simplify:

3/4 a²b² = (4a² - 4ab + b²) (b² - 1/4 a²)

3x² = (4 - 4x + x²) (4x² - 1)

3x² = 4x⁴ - 16x³ + 16x² - x² + 4x - 4

0 = 4x⁴ - 16x³ + 12x² + 4x - 4

Obviously x = 1 is a solution, corresponding to a regular tetrahedron, so we can factor that out:

0 = 4 (x - 1) (x³ - 3x² + 1)

This cubic polynomial has one solution in the relevant interval 1/√3 < x < 2, which happens to be exactly

x = 1 - 2 cos(4π/9) = 1/(2 cos(2π/9)) ≈ 0.6527

Very interesting! Somehow this equal-centroid pyramid is related to the regular enneagon.

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

Hmm. This is indeed an extremely interesting coincidence. I wonder if there may be some kind of construction involving this equal-centroid pyramid that results in enneagonal symmetry? Perhaps some kind of tiling of space with 11-gonal symmetry, that contains this pyramid as one of the tile shapes?

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Due to crystallographic restriction, there is no tiling of 3D space with 9-fold or 11-fold rotation symmetry.

Indeed, in any dimension, a tiling can't have simple 9-fold rotation symmetry. But a compound rotation is possible! The general condition of crystallographic restriction is that the characteristic polynomial of the transformation matrix needs to have integer coefficients.

Let's consider a triple rotation in 6D, with angles θ₁,θ₂,θ₃ in three orthogonal 2D planes. The rotation matrix has eigenvalues

z₁=exp(-iθ₁), z₂=exp(iθ₁),

z₃=exp(-iθ₂), z₄=exp(iθ₂),

z₅=exp(-iθ₃), z₆=exp(iθ₃),

and the characteristic polynomial is (t - z₁)(t - z₂)(t - z₃)(t - z₄)(t - z₅)(t - z₆). This paper by P. A. Damianou shows that, if the roots z of an integer-coefficient monic polynomial are all on the unit circle |z|=1, then they must be roots of unity; that is, each θ must be a rational fraction of a circle. Furthermore, the polynomial must be a product of cyclotomic polynomials (which are just the minimal polynomials of roots of unity).

(I don't know why Damianou mentions Newton's formulas; the coefficients of f_{k} are not all simple sums of powers of z₁,...,z₆. What matters is that they are at least symmetric polynomials in z₁,...,z₆, and all symmetric polynomials are generated by the elementary symmetric polynomials; since the latter have integer values at (z₁,...,z₆) (being the coefficients of f itself), the former have integer values as well. Then f_{j}=f_{k} for some j≠k because there are finitely many such f.)

So we could have a characteristic polynomial t⁶+t³+1, for a triple rotation by 2π/9,4π/9,8π/9. But I don't know any specific example of a 6D tiling or symmetry group containing this rotation. Crystallographic restriction is a necessary condition, not a sufficient condition, for a tiling to exist with such a symmetry.

Indeed, in any dimension, a tiling can't have simple 9-fold rotation symmetry. But a compound rotation is possible! The general condition of crystallographic restriction is that the characteristic polynomial of the transformation matrix needs to have integer coefficients.

Let's consider a triple rotation in 6D, with angles θ₁,θ₂,θ₃ in three orthogonal 2D planes. The rotation matrix has eigenvalues

z₁=exp(-iθ₁), z₂=exp(iθ₁),

z₃=exp(-iθ₂), z₄=exp(iθ₂),

z₅=exp(-iθ₃), z₆=exp(iθ₃),

and the characteristic polynomial is (t - z₁)(t - z₂)(t - z₃)(t - z₄)(t - z₅)(t - z₆). This paper by P. A. Damianou shows that, if the roots z of an integer-coefficient monic polynomial are all on the unit circle |z|=1, then they must be roots of unity; that is, each θ must be a rational fraction of a circle. Furthermore, the polynomial must be a product of cyclotomic polynomials (which are just the minimal polynomials of roots of unity).

(I don't know why Damianou mentions Newton's formulas; the coefficients of f

So we could have a characteristic polynomial t⁶+t³+1, for a triple rotation by 2π/9,4π/9,8π/9. But I don't know any specific example of a 6D tiling or symmetry group containing this rotation. Crystallographic restriction is a necessary condition, not a sufficient condition, for a tiling to exist with such a symmetry.

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

mr_e_man wrote:So we could have a characteristic polynomial t⁶+t³+1, for a triple rotation by 2π/9,4π/9,8π/9. But I don't know any specific example of a 6D tiling or symmetry group containing this rotation. Crystallographic restriction is a necessary condition, not a sufficient condition, for a tiling to exist with such a symmetry.

Actually I think you can just apply the triple rotation to any vector not aligned with the three planes, to generate other vectors, which can be taken as translation symmetries. This should produce a valid symmetry group.

(If the original vector is aligned with the planes, then the generated vectors would only span a 2D or 4D subspace, not the whole 6D space.)

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

Reading the Wikipedia page on crystallographic restriction, I noticed an interesting claim: The tesseract has a (45°, 135°) double rotation symmetry. Can anyone explain this?

What are the "axes" of such a rotation? Consider the two planes of rotation, and the intersection of one plane with the surface of the tesseract (or some uniform truncation of it). Where are the lines formed by this intersection?

What are the "axes" of such a rotation? Consider the two planes of rotation, and the intersection of one plane with the surface of the tesseract (or some uniform truncation of it). Where are the lines formed by this intersection?

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

Well, that page gives the rotation matrices explicitly. Studying those (and getting confused by one matrix being the inverse of what I expected), I eventually figured it out.

In x4o3x3o, consider the secondary type of square, not corresponding to squares in the tesseract. The diagonals of these squares are the double rotation "axes".

Quickfur, maybe you could make an animation of x4o3x3o performing a continuous double rotation, with one rate of rotation being 3 times the other, and with the camera centred on these diagonal lines.

Those 48 double rotations by (45°, 135°) are the least obvious of the tesseract's 384 symmetries. All the others (even including rotoreflections) have "axes" passing through the centres of surtopes of the tesseract, so they're not difficult to find.

In x4o3x3o, consider the secondary type of square, not corresponding to squares in the tesseract. The diagonals of these squares are the double rotation "axes".

Quickfur, maybe you could make an animation of x4o3x3o performing a continuous double rotation, with one rate of rotation being 3 times the other, and with the camera centred on these diagonal lines.

Those 48 double rotations by (45°, 135°) are the least obvious of the tesseract's 384 symmetries. All the others (even including rotoreflections) have "axes" passing through the centres of surtopes of the tesseract, so they're not difficult to find.

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

Wow this is very interesting indeed! I'll have to spend some time to figure out exactly how to orient the tesseract so that it rotates around the requisite planes.

"Axes" of rotation are a 3D-centric notion that's ambiguous in higher dimensions. To specify a rotation you need a plane, i.e., a pair of linearly-independent vectors.

"Axes" of rotation are a 3D-centric notion that's ambiguous in higher dimensions. To specify a rotation you need a plane, i.e., a pair of linearly-independent vectors.

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Yes, by "axis" I meant the intersection of the 2D plane with the 3D surface, which is thus 1D.

The two linearly independent vectors are two vertices of the cantellated tesseract (e.g. the endpoints of one of those diagonal lines: (1, 1, 1+√2, 1+√2), (1+√2, -1, 1, 1+√2)).

The two linearly independent vectors are two vertices of the cantellated tesseract (e.g. the endpoints of one of those diagonal lines: (1, 1, 1+√2, 1+√2), (1+√2, -1, 1, 1+√2)).

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

Gotcha, I'll take a look and see how to orient it correctly.

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

What's the link to the Wikipedia page that claims this symmetry?

I finally found a way to orient the tesseract according to the plane you describe, but I'm not seeing a 45°/135° double rotation symmetry here. Are you sure the reference points you gave are correct? And that the claim is actually valid? Maybe somebody just made it up.

Either that, or I made a huge mistake in my calculations, which is probably the most likely case.

I finally found a way to orient the tesseract according to the plane you describe, but I'm not seeing a 45°/135° double rotation symmetry here. Are you sure the reference points you gave are correct? And that the claim is actually valid? Maybe somebody just made it up.

Either that, or I made a huge mistake in my calculations, which is probably the most likely case.

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

https://en.wikipedia.org/wiki/Crystallo ... dimensions

I am sure it's correct. (But I don't like the specific example they chose; A and B have more negative than positive entries.)

Let's focus on the cantellated tesseract.

We'll equate 4-tuples with 4×1 column matrices. Let w=1+√2, so w²=2w+1.

One vertex is u = (-1, +1, -w, +w).

The next vertex is Au = (-w, -1, -1, +w), which is 45° away from u (measuring the angle at the centre of the polychoron).

The next is A²u = (-w, -w, +1, +1), which is 90° away from u.

Notice that Au = (u + A²u)/√2, so all three vertices (indeed, all eight, up to A⁷u) are in one 2D plane, which is one of the two planes of the double rotation A.

Orthogonal to that plane is a vertex v = (+1, +1, +w, +w).

Then we have Av = (-w, +1, -1, -w) which is 135° away from v,

and A²v = (+w, -w, -1, +1) which is 90° or 270° away from v.

Notice that Av = (-v - A²v)/√2, so again all three vertices are in one 2D plane.

Thus, the four orthogonal vectors u, A²u, v, -A²v make it quite clear that the matrix A is in fact a (45°, 135°) double rotation. Collect these vectors into a change-of-basis matrix C:

Now, to make a continuous double rotation, which has the symmetry A = A(45°) as a special case:

Vary θ from 0° to 45°, and use A(θ) = C D(θ) C⁻¹ (applying it to apacs(1,1,w,w)) to rotate the polychoron.

...More simply, use D(θ) C⁻¹. Then you shouldn't have to rotate the camera; it should already be looking at those diagonal lines.

But this would put the slow rotation in the x,y plane, and the fast rotation in the z,w plane. I would rather have the slow rotation in the z,w plane. Of course this is easily fixed by a 90° double rotation.

I am sure it's correct. (But I don't like the specific example they chose; A and B have more negative than positive entries.)

Let's focus on the cantellated tesseract.

We'll equate 4-tuples with 4×1 column matrices. Let w=1+√2, so w²=2w+1.

One vertex is u = (-1, +1, -w, +w).

The next vertex is Au = (-w, -1, -1, +w), which is 45° away from u (measuring the angle at the centre of the polychoron).

The next is A²u = (-w, -w, +1, +1), which is 90° away from u.

Notice that Au = (u + A²u)/√2, so all three vertices (indeed, all eight, up to A⁷u) are in one 2D plane, which is one of the two planes of the double rotation A.

Orthogonal to that plane is a vertex v = (+1, +1, +w, +w).

Then we have Av = (-w, +1, -1, -w) which is 135° away from v,

and A²v = (+w, -w, -1, +1) which is 90° or 270° away from v.

Notice that Av = (-v - A²v)/√2, so again all three vertices are in one 2D plane.

Thus, the four orthogonal vectors u, A²u, v, -A²v make it quite clear that the matrix A is in fact a (45°, 135°) double rotation. Collect these vectors into a change-of-basis matrix C:

- Code: Select all
`⌈ -1⌉ ⌈-1 -w +1 -w⌉ ⌈-1 +1 -w +w⌉`

A = |+1 | C = |+1 -w +1 +w| C⁻¹ = (2-√2)/8 |-w -w +1 +1|

| -1 | , |-w +1 +w +1| , |+1 +1 +w +w|

⌊ -1 ⌋ ⌊+w +1 +w -1⌋ ⌊-w +w +1 -1⌋

⌈+1/√2 -1/√2 ⌉

D = C⁻¹AC = |+1/√2 +1/√2 |

| -1/√2 -1/√2|

⌊ +1/√2 -1/√2⌋

Now, to make a continuous double rotation, which has the symmetry A = A(45°) as a special case:

- Code: Select all
`⌈cos(θ) -sin(θ) ⌉`

D(θ) = |sin(θ) cos(θ) |

| cos(3θ) -sin(3θ)|

⌊ sin(3θ) cos(3θ)⌋

Vary θ from 0° to 45°, and use A(θ) = C D(θ) C⁻¹ (applying it to apacs(1,1,w,w)) to rotate the polychoron.

...More simply, use D(θ) C⁻¹. Then you shouldn't have to rotate the camera; it should already be looking at those diagonal lines.

But this would put the slow rotation in the x,y plane, and the fast rotation in the z,w plane. I would rather have the slow rotation in the z,w plane. Of course this is easily fixed by a 90° double rotation.

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

quickfur wrote:I finally found a way to orient the tesseract according to the plane you describe, but I'm not seeing a 45°/135° double rotation symmetry here.

Perhaps you were thinking of two separate simple rotations?

There is no 45° rotation symmetry anywhere.

You have to do both rotations at the same time.

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

Or perhaps you apply a matrix to a vector as v*M, whereas I apply a matrix to a vector as M*v?

In that case, everything should be transposed; so the vertices of the rotating cantellated tesseract are

(Again, w=1+√2, and θ varies from 0° to 45°.)

(Also, I'm ignoring a constant scale factor, 2√(2+√2), coming from the matrix in the middle.)

In that case, everything should be transposed; so the vertices of the rotating cantellated tesseract are

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`⌈-1 -w +1 -w⌉ ⌈ cos(θ) sin(θ) 0 0 ⌉`

|+1 -w +1 +w| |-sin(θ) cos(θ) 0 0 |

apacs[1 1 w w] * |-w +1 +w +1| * | 0 0 cos(3θ) sin(3θ)|

⌊+w +1 +w -1⌋ ⌊ 0 0 -sin(3θ) cos(3θ)⌋

(Again, w=1+√2, and θ varies from 0° to 45°.)

(Also, I'm ignoring a constant scale factor, 2√(2+√2), coming from the matrix in the middle.)

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

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