0
$\begingroup$

In Cartesian printers, the extruder is moving along the X axis, Y axis and/or Z axis. Every axes has its own resolution, this is the minimum length that is possible to achieve in that axis and is related with the number of steps that the stepper motors can do.

enter image description here enter image description here

So, if we trace the solid figure with the minimum volume that the printer can achieve based on its axis resolutions we obtained some sort of a cuboid

The length of the cuboid in X is the resolution of the printer in the X axis, the same logic applies for Y axis and Z axis.

In the figure it is possible to see 27 (3x3x3) cuboids genarated:

enter image description here

if the resolution of the printer in X axis is 0.1 mm, in the Y axis: 0.1 mm and in the Z axis: 0.1 mm, the volume of these cuboids is going to be 0.1 mm x 0.1 mm x 0.1 mm

Now, consider a Delta printer:

enter image description here

What is the solid figure with the minimun volume that can be traced with it?

| improve this question | |
$\endgroup$
1
$\begingroup$

I think you misunderstand. It's the nozzle width and extruder step size, not the axis step sizes, that limit detail. Positioning resolution on a typical printer is on the order of 0.01 mm, but nozzle size is at least 0.1 mm and typically 0.4 mm. Also, lack of perfect rigidity in the mechanical parts will produce gradually increasing error as you try to go smaller and smaller in detail. So, it really makes no difference to the achievable detail whether the printer is a Cartesian or a Delta.

| improve this answer | |
$\endgroup$
  • $\begingroup$ yes, I think you right, but either way, I think it would be interesting to see what is the minimum solid that can be created with the positioning resolution, just to use it as a theoretical model $\endgroup$ – DieDauphin Jun 25 at 6:42
  • $\begingroup$ You can work out the shape traced out by stepping each motor +- 1 with some math, but it's not translation-invariant; it will be shaped differently depending on where it's located. For this reason, assuming you want to be moving effectively in an XY plane to print layers, a delta printer really needs to have motor steps significantly smaller than the order of magnitude of the moves you're trying to produce. $\endgroup$ – R.. GitHub STOP HELPING ICE Jun 25 at 21:59
  • $\begingroup$ In theory you could slice in the nonlinear space of the delta axes, but you'd effectively be doing "z-hops" to get around before the outer edges of your print built up to a level where the center of the build area was reachable, and it doesn't seem like a very practical or smart thing to do. $\endgroup$ – R.. GitHub STOP HELPING ICE Jun 25 at 21:59
  • $\begingroup$ I want to work out the shape traced out by stepping each motor +- 1 with some math. But I don't have a Delta Printer and I don't understand how they work .. $\endgroup$ – DieDauphin Jun 26 at 0:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.