Slic3r path spacing calculations  

RE: Slic3r path spacing calculations

Well, for starters, a trace width of 0.8mm is extremely large if you are using the default 0.4mm nozzle.  It would be appropriate for a 0.6mm nozzle.

Next, why 40% overlap of infill and perimeters?  That's pretty large when the default is usually 25% or less.  It will probably cause some distortion or artifacts in your print.

But to answer your question directly, Slic3r tries to do what you asked.  Your perimeters are 0.8mm each, so that leaves an open infill width of (6 - 1.6) = 4.4mm.  But you asked for 40% overlap, so add 0.4 * 0.8 = 0.32 on each side.  4.4 + 0.62 = 5.02mm total infill width.  At exactly 0.8mm per trace, that would be 6.275 traces.  But instead of laying down 6 full traces and one trace that's 0.275 of specified width, Slic3r will adjust the width of one or several traces upward to get the requested infill. 

RE: Slic3r path spacing calculations

Thanks for your response! I was able to develop this formula accordingly:

Given some original dimension, D, with some layer thickness h, and width w, some overlap % entered in decimal form, the modified part dimension for an integer # of paths, the following equation attached will provide you with the new dimension of the part such that you have an integer number of paths and follow the path spacing as defined by slic3r, and the round down function rounds down to the nearest positive integer.

spacing = width - h(1-pi/4)

I looked at the gcode values and noticed that the extruder value does not change as path spacing increases, thus it would still slightly under extrude as it automatically increased path spacing. This formula is a guarantee that you will have exactly an integer path spacing with the proper extrusion values.

Hope this helps! 

RE: Slic3r path spacing calculations

PS note that width is the extrusion width as entered into slic3r.