Arc Compensation

Reprapper Nophead has pointed out that when the RepRap extrude head moves it lays down material equally on either side of its centre line. But what an arc is being created, less material is needed on the inside of the arc than the outside. This page describes the calculation that's needed to compensate for that.

Suppose RepRap is making a circular hole of radius R. The extrude head lays down a track of thickness t. The head has to move at an offset radius r to get the edge of the hole in the right place. For big values of R:

    r ≈ R + 0.5t.

Indeed for straight lines (i.e. R is infinite), that is exactly what you want. But for small R the hole will come out too small because more material is laid down in the inside area, A_i than is needed. So r has to be larger than the simple calculation gives to compensate.

Without loss of generality, consider the entire circle of which the arc along which the centre of the extruder moves is a part. Let the circumference of the circle be length L:

    L = 2πr

Clearly, for a straight line (that is, infinite R):

    A_i = Lt/2 = 2πrt/2 = πrt

For finite values of R:

    A_i = π(r² - R²)

If we equate the A_i values so that r gives the area we would have for a straight line we have

    πrt = π(r² - R²)

The πs cancel, giving (after re-arrangement) the quadratic in r:

    r² - rt - R² = 0

which has the root:

    r = [t + √(t² + 4R²)]/2

(There is also the negative square root which we can ignore.) We can tabulate this formula for some typical values:

As you can see, above R = 6.5 mm, the simple offset calculation of adding half t gives the right answer to within 10 microns. But below that the differences are significant.

-- AdrianBowyer - 26 Jul 2008