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Reconstruction of the 2000 Wrotham formation | ||
| 1. | ![]() |
Start with constructing a equilateral triangle. |
| 2. | ![]() |
Draw three circles centered at each corner of the triangle and tangent to the opposite sides. |
| 3. | ![]() |
Remove the parts of the circles, that lie outside the triangle. |
| 4. | ![]() |
Draw circles centered at the midpoint of each side of the triangle, and tangent to the larger circles. |
| 5. | ![]() |
From the same midpoints, draw smaller circles, touching the other two circles. |
| 6. | ![]() |
As with step 2, draw circles from each corner of the triangle, that touch the latter smaller circles at the far side. |
| 7. | ![]() |
Again, remove from the last drawn circles all parts outside the triangle. |
| 8. | ![]() |
Remove from all circles introduced in steps 4 and 5, all parts inside the triangle; then, remove the triangle. |
| 9. | ![]() |
Finally, remove repeatedly two of the four cross-lines, such that the woven-like knot pattern appears. |
| 10. | ![]() |
The final reconstruction fits neatly. |
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I tried to find mathematical expressions with a graphical representation similar to the pattern of the Wrotham formation. I found the following quite complicated formulas. |
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xi = r1 cos φi |
(1) |
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| yi = r1 sin φi | (2) | |
| ui = xi + r2 cos –a·φi | (3) | |
| vi = yi + r2 sin –a·φi | (4) | |
| pi = xi + c·r2 cos –a·φi | (5) | |
| qi = yi + c·r2 sin –a·φi | (6) | |
| With the following parameter setting: | ||
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a = 2 | ||
| b = 2.6 | ||
| c = 0.55 | ||
| r2 = b·r1 | ||
| φi = 0,...,2π | ||
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the set of formula's will generate this pattern: |
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| To understand what happens, look at the following pictures: | ||
| 11. | ![]() |
Formula's (1) and (2) describe a point A (xi,yi) "walking" around a circle with radius r1. Each position of point A is determined by the value of φ; φ runs counter-clockwise. |
| 12. | ![]() |
Another circle with radius r2 (= 2.6 times r1) has its center in point A. An other point B (ui,vi) is walking around this second circle, in the opposite direction, and two times as fast (formula's (3) and (4)). |
| 13. | ![]() |
Following the path of point B results in the outer
loop of the Celtic knot pattern. The last two formula's (5) and (6) generate the inner loop. It is the path of a point C (pi,qi), that is just some distance (0.55 times r2) from point A on the same ray as point B. |
| Copyright © 2000, Zef Damen, The Netherlands | ||