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Helical coordinate on a cone

I could not find the equations for a helix on a cone, so I derive them below. An example of results is Figure 2.

 
trycart
trycart
Figure 2
Helix on a cone. Top shows the embedded helix. Bottom hides it. Apex angle $\phi=1/2$.


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Define $\phi$ to be the angle from the axis of the cone to its surface. I call this the apex angle. I discovered by accident that certain apex angles give interesting patterns in the crossline directions while most do not. I found these patterns to be insensitive to the choice of N and show them here for about N=1400. Another value of apex angle with an interesting pattern is $\phi=1/3$.It gives the charming pattern in Figure 3. (I've seen this pattern before on a party hat. I attributed it to an ingenious artist. Now I realize that like all mathematics, this art existed before the big bang.)

 
YinYang
YinYang
Figure 3
Helix on a cone. Top shows the embedded helix. Bottom hides it. Apex angle equals one third radian.


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The radius r of the cone divided by its altitude z has a ratio $r/z=\tan\phi$where $\phi$ is the angle from the axis of the cone to its surface. The area of a circle is $\pi R^2$.The surface area of a cone is $\pi R^2/\sin\phi$.Dividing the area into N cells, the surface area per cell is $\pi R^2/(N\sin\phi)$.Taking any cell area to be square, the length of a side is
\begin{displaymath}
\Delta s \quad =\quad R\ \sqrt{\pi \over N \sin\phi }\end{displaymath} (3)
The number of points running one cycle around a rim is
\begin{displaymath}
N_{\rm rim} \quad =\quad{2\pi r \over \Delta s}
 \quad =\quad{2r\over R} \sqrt{N \pi \sin\phi}\end{displaymath} (4)
On one cycle around the rim, the radius must change by $\Delta r=\Delta h\sin\phi$where the hypotenuse $\Delta h$ of the triangle lies on the surface of the cone. Thus, for each mesh point going around the rim
\begin{displaymath}
\Delta r \quad =\quad{ \Delta h\ \sin\phi \over N_{\rm rim} }\end{displaymath} (5)
Since we want ``square'' mesh points, $\Delta h$ must equal $\Delta s$, hence
\begin{displaymath}
\Delta r \quad =\quad{ R^2 \over 2rN}\end{displaymath} (6)
The algorithm starts on the rim at $\theta =0$ and r=R. At each step, update r and $\theta$ with $-\Delta r$ and $\Delta\theta=r\Delta s$.Stop before $r-\Delta r$ becomes negative.


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Next: FILLING VOLUMES Up: Claerbout: Helical meshes on Previous: Helical coordinate on a
Stanford Exploration Project
4/20/1999