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    <p>Some thoughts about this problem. I suggest there was initial Bezier curve <code>P0</code>-<code>P3</code> with control points <code>P1</code> and <code>P2</code></p> <p><img src="https://i.stack.imgur.com/ncF9A.gif" alt="enter image description here"></p> <p>Let's make two subdivisions at parameters ta and tb. We have now two subcurves (in yellow) - <code>P0</code>-<code>PA3</code> and <code>PB0</code>-<code>P3</code>. Blue interval is lost. <code>PA1</code> and <code>PB2</code> - known control points. We have to find unknown <code>P1</code> and <code>P2</code>.</p> <p>Some equations</p> <p>Initial curve:</p> <pre><code>C = P0*(1-t)^3+3*P1(1-t)^2*t+3*P2*(1-t)*t^2+P3*t^3 </code></pre> <p>Endpoints:</p> <pre><code>PA3 = P0*(1-ta)^3+3*P1*(1-ta)^2*ta+3*P2*(1-ta)*ta^2+P3*ta^3 PB0 = P0*(1-tb)^3+3*P1*(1-tb)^2*tb+3*P2*(1-tb)*tb^2+P3*tb^3 </code></pre> <p>Control points of small curves</p> <pre><code>PA1 = P0*(1-ta)+P1*ta =&gt; P1*ta = PA1 – P0*(1-ta) PB2 = P2*(1-tb)+P3*tb =&gt; P2(1-tb) = PB2 – P3*tb </code></pre> <p>Now substitute unknown points in PA3 equation:</p> <pre><code>**PA3***(1-tb) = **P0***(1-ta)^3*(1-tb)+3*(1-ta)^2*(1-tb)*(**PA1** – **P0***(1-ta))+3*(1-ta)*ta^2*( **PB2** – **P3***tb)+**P3***ta^3*(1-tb) </code></pre> <p>(some multiplication signs have been lost due to SO formatting)</p> <p>This is vector equation, it contains two scalar equations for two unknowns <code>ta</code> and <code>tb</code></p> <pre><code>PA3X*(1-tb) = P0X*(1-ta)^3*(1-tb)+3*(1-ta)^2*(1-tb)*(PA1X – P0X*(1-ta))+3*(1-ta)*ta^2*( PB2X – P3X*tb)+P3X*ta^3*(1-tb) PA3Y*(1-tb) = P0Y*(1-ta)^3*(1-tb)+3*(1-ta)^2*(1-tb)*(PA1Y – P0Y*(1-ta))+3*(1-ta)*ta^2*( PB2Y – P3Y*tb)+P3Y*ta^3*(1-tb) </code></pre> <p>This system might be solved both numerically and analytically (indeed Maple solves it with very-very big cubic formula :( ) </p> <p>If we have points with some error, that makes sense to build overdetermined equation system for some points (<code>PA3</code>, <code>PB0</code>, <code>PA2</code>, <code>PB1</code>) and solve it numerically to minimize deviations.</p>
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