Rotating Koch curve

(FB/Math/BM)

Let a Koch Snowflake (http://en.wikipedia.org/wiki/Koch_snowflake) rotate around a line in its plane through the centre. Assume that the flake was created starting from en equilateral triangel with side length a.
Then the generated body will have infinite surface area since the circumference is infinite but finite volume, depending on the orientation of the rotation axis. Compute the minimal and maximal volume volume.

(Inspired by a FB question on objects with finite/infinite volume/surface area.)

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Rotating Koch curve.

The axis of rotation goes through the centre and I consider only the cases that the axis is either horizontal or vertical.

The curve is generated starting from an equilateral triangle (fig 1) with side length L. At the middle of each side is placed a smaller equilateral triangle with side L/3. Then at the middle of each the 12 segments is placed new equilateral triangles with side L/9. Continuing in this way after an infinity of steps there results the Koch curve, a fractal with (Hausdorff) dimension log 4/log 3 ≈ 1.26. The enclosed are is exactly 8A₀/5 where A₀ is the area of the original triangle, A₀ = L²√3/4.

To find the rotational volume Guldin’s second rule is used, which demands us to find the centre of gravity of the area at one side of the axis. For this start with the green part (fig 2), for simplicity placed in another position (fig 3) and find the vertical distance from the base line (length L) to the centre of gravity. Then there remains a comparatively easy combination of pieces.

For easier reading I consider just this first problem in this part and formulate it as a challenge with some hints.

Let the distance looked for be H. The entire area in fig 3 is exactly similar to each of the four smaller green parts with a length ratio of 1/3. So the vertical distance from the base line to the centre of mass of the two horizontal green parts is H/3 and each has area A/9 where A is the entire area in fig 3. The centre of gravity of the other two green parts can be expressed in L and H using elementary geometry. The red triangle has area A₀/9. – Now, find H using the using the fact that the coordinates of the centre of gravity for a composite body (or, in this case, a surface) is a weigthed sum of the centres of gravity of the parts.

This is really not very hard and no long calculation either.

Web pages:

http://en.wikipedia.org/wiki/Koch_snowflake

http://www.vias.org/comp_geometry/geom_3d_general_rules.html

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The small green parts on the sides of the red triangle (fig 3) have centre of gravity is at vertical distance L/12·√3 + H/3·1/2 from the base line. The red triangle has its centre of gravity at height L/6·√3·1/3. Also A = (8A₀/5 – A₀)/3 = A…₀/5. Then by the weighted sum of the centres of gravity for the parts (three green and the red triangle)

H = [H/3·A/9·2 + (L/12·√3 + H/3·1/2)·A/9·2 + L/6·√3·1/3·5A/9] / [A/9·2 + A/9·2 + 5A/9]

From this H can be solved: H = L√3/18

Notice that we have just used elementary geometry for equilateral triangles and the self-similarity property of the fractal.

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Before going on to the entire ”snow flake” let’s find the volume generated when the surface in fig 3 rotates around the bas line (indeed it is often the fractal upper boundary of this that is named the Koch curve). The area is A = A₀/5 = L²…√3/4·5 = L²√3/20, so by Guldin’s second theorem the volume is 2π·H·A = 2π·L√3/18·L²√3/20 =
= πL³/60. Rather nice!

Assuming, of course, that Guldin’s theorem and the calculation of centre of gravity work as usual even when the boundary is a fractal!

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Next step is to find the centre of gravity of the surface resulting from that in fig 3 when one of the smaller green parts at the base line is removed. This is easy using similarity and weighted addition (or subtraction).

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In fact that is unnecessary. The green part of fig 2 above the horizontal line is composed of three parts similar to the whole surface of fig 3 and one equilateral triangle. The half ”snow flake” above the horizontal line or to the right of… the vertical line is composed of two such pieces and a larger equilateral or isosceles triangle respectively. The centre of gravity for these two halves can be found in one step. (Only the y-coordinate or the x-coordinate respectively is needed for the volume calculation; and also the other one is trivial from symmetry).
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Upper half:

Ycg = { [(L√3/12 – H/6) + (L√3/6 + H/3) + (L/4√3 + H/6)]·A/9·2 + (L√3/12 + L√3/36)·5A/9·2 + L√3/9·20A/9 } / ( A/9·2·3 + 5A/9·2 + 20A/9 ) = (4·H + 27·√3·L)/216 = 245·√3·L/1944
(H = L√3/18)
Volume of rotation = 2π·Ycg·4A₀/5 = 2π·245·…√3·L/1944·L²√3/5 = 49πL³/324.
(A₀ = L²√3/4)

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Right half:

Xcg = { [(L/12 + H√3/6) + (L/6 + L/6) + (5L/12 + H√3/6)]·A/9·2 + (L/4 + L/12)·5A/9·2 + L/9·20A/9 } / ( A/9·2·3 + 5A/9·2 + 20A/9 ) =
= (6·√3·H + 65·L)/324 = 11L/54
(H = L√3/18)
…Volume of rotation = 2π·Xcg·4A₀/9 = 2π·11L/54·L²√3/9 = 11πL³√3/243
(A₀ = L²√3/4)
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Correction for upper half:
4A₀/5 should be 4A₀/9 so the volume is 49πL³/324·5/9 = 245πL³/2916

V(hor ax)/V(ver ax) = 245·√3/396 ≈ 1.07

Annonser

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