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    <p>Your question is unclear on a couple of points, so I have to make some assumptions:</p> <ol> <li>During each time interval, translation speed and rotational velocity are constant. <li>You know the values of these variables in every time interval (and you know rotational velocity in usable units, like radians per second, not just "very left"). <li>You know initial heading. <li> You can maintain enough precision that roundoff error is not a problem. </ol> <p>Given that, <b>there is an exact solution</b>. First the easy part:</p> <pre><code>delta_angle = omega * delta_t </code></pre> <p>Where omega is the angular velocity. The distance traveled (maybe along a curve) is</p> <pre><code>dist = speed * delta_t </code></pre> <p>and the radius of the curve is</p> <pre><code>radius = dist / delta_angle </code></pre> <p>(This gets huge when angular velocity is near zero-- we'll deal with that in a moment.) <b>If angle (at the beginning of the interval) is zero</b>, defined as pointing in the +x direction, then the translation in the interval is easy, and we'll call it deta_x_0 and delta_y_0:</p> <pre><code>delta_x_0 = radius * sin(delta_angle) delta_y_0 = radius * (1 - cos(delta_angle)) </code></pre> <p>Since we want to be able to deal with very small delta_angle and very large radius, we'll expand sin and cos, and use this only when angular velocity is close to zero:</p> <pre><code>dx0 = r * sin(da) = (dist/da) * [ da - da^3/3! + da^5/5! - ...] = dist * [ 1 - da^2/3! + da^4/5! - ...] dy0 = r * (1-cos(da)) = (dist/da) * [ da^2/2! - da^4/4! + da^6/6! - ...] = dist * [ da/2! - da^3/4! + da^5/6! - ...] </code></pre> <p><b>But angle generally isn't equal to zero, so we have to rotate these displacements</b>:</p> <pre><code>dx = cos(angle) * dx0 - sin(angle) * dy0 dy = sin(angle) * dx0 - cos(angle) * dy0 </code></pre>
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    1. CO+1: Nice, I like it. Could possibly do with a little more description on that very last step i.e. that your analysis holds at each time step in a local x-y frame where x is in the current direction and so you are just accounting for the current angle at that timestep.
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    2. CO@Troubadour: you're right, I could have made that clearer-- but I don't want to get too verbose. I think putting a couple of lines in boldface will do...
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