Yes - this is exactly it, thanks very much. I will play about with it until

I can properly use all aspects.

Many thanks!

-----Original Message-----

From: Discuss [mailto:

[hidden email]] On Behalf Of

Parkinbot

Sent: 24 May 2016 00:04

To:

[hidden email]
Subject: Re: [OpenSCAD] Curved linear extrude

Have a look at this code. It uses an exponential funktion.

> pf = 1.004;

>

> for (i=[0:153])

> hull()

> {

> translate([0, 0, i])

> scale(pow(pf, i)*[1,1.19255])

> cylinder(r=161/2, h = .01);

> translate([0, 0, i+1])

> scale(pow(pf, i+1)*[1,1.19255])

> cylinder(r=161/2, h = .01);

> }

in a more general formulation you would write your own function f to

describe the trajectory. E.g.

> step = 10;

> for (i=[0:step:153])

> hull()

> {

> translate([0, 0, i])

> scale(f(i)*[1,1.19255])

> cylinder(r=161/2, h = .01);

> translate([0, 0, i+step])

> scale(f(i+step)*[1,1.19255])

> cylinder(r=161/2, h = .01);

> }

>

> function f(i) = 1+ pow(i/200,2);

<

http://forum.openscad.org/file/n17409/hat.png>

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