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    <p>This answer does not (as far as I see) favor one of the methods you mentioned in your question, but is what I would do in this situation.</p> <p>I would calculate the normalized normal (or binormal) of the curve. Let's say I take the normalized normal and have it as a function of t (<code>N(t)</code>). With this I would write a helper function to calculate the offset point P:</p> <pre><code>P(t, o) = B(t) + o * N(t) </code></pre> <p>Where <code>o</code> means the signed offset of the curve in normal direction.</p> <p>Given this function one would simply calculate the points to the left and right of the curve by:</p> <pre><code>Points = [P(t, -w), P(t, w), P(t + s, -w), P(t + s, w)] </code></pre> <p>Where <code>w</code> is the width of the curve you want to achieve.</p> <p>Then connect these points via two triangles.</p> <p>For use in a triangle strip this would mean the indices:</p> <pre><code>0 1 2 3 </code></pre> <p><strong>Edit</strong></p> <p>To do some work with the curve one would generally calculate the Frenet frame.</p> <p>This is a set of 3 vectors (Tangent, Normal, Binormal) that gives the orientation in a curve at a given parameter value (t).</p> <p>The Frenet frame is given by:</p> <pre><code>unit tangent = B'(t) / || B'(t) || unit binormal = (B'(t) x B''(t)) / || B'(t) x B''(t) || unit normal = unit binormal x unit tangent </code></pre> <p>In this example <code>x</code> denotes the cross product of two vectors and <code>|| v ||</code> means the length (or norm) of the enclosed vector <code>v</code>.</p> <p>As you can see you need the first (<code>B'(t)</code>) and the second (<code>B''(t)</code>) derivative of the curve.</p>
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    1. COIt's quite alright if the answer is a departure from the methods I mentioned, I like that :) The logic seems like a great way to go, for simplicity sake. I am still somewhat struggling with normalizing, however. In this case, I'm specifically not sure what the normal of the curve is referring to. I hate to ask you to elaborate a bit on that, but I think it will help me grasp it further. Really appreciate it!
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    2. CO@JavierOtero added a paragraph about the normal. I hope it makes sense now.
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    3. COI'll keep you updated on how it goes, so far it's making sense :)
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