How Do Tool Path Algorithms Decide Which Direction to Print a Closed-Loop Polygon
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I understand how slicer programs create sets of closed-loop polygons to print on a layer-by-layer basis. For a given closed loop polygon which needs to be printed, the tool path generator will know the coordinates and how those coordinates are connected to each other, such that traversing a set of segments in that order will bring the extruder head back to the first coordinate to complete the closed loop.
My question is: By what mechanism does the tool path generator decide which direction to traverse the closed loop? As it is a loop, that loop could be printed "clockwise" or "counter-clockwise", as it were. Any details, and links to further explanations of how some of the big-name slicer programs determine this is much appreciated.
slicing
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add a comment |
$begingroup$
I understand how slicer programs create sets of closed-loop polygons to print on a layer-by-layer basis. For a given closed loop polygon which needs to be printed, the tool path generator will know the coordinates and how those coordinates are connected to each other, such that traversing a set of segments in that order will bring the extruder head back to the first coordinate to complete the closed loop.
My question is: By what mechanism does the tool path generator decide which direction to traverse the closed loop? As it is a loop, that loop could be printed "clockwise" or "counter-clockwise", as it were. Any details, and links to further explanations of how some of the big-name slicer programs determine this is much appreciated.
slicing
$endgroup$
add a comment |
$begingroup$
I understand how slicer programs create sets of closed-loop polygons to print on a layer-by-layer basis. For a given closed loop polygon which needs to be printed, the tool path generator will know the coordinates and how those coordinates are connected to each other, such that traversing a set of segments in that order will bring the extruder head back to the first coordinate to complete the closed loop.
My question is: By what mechanism does the tool path generator decide which direction to traverse the closed loop? As it is a loop, that loop could be printed "clockwise" or "counter-clockwise", as it were. Any details, and links to further explanations of how some of the big-name slicer programs determine this is much appreciated.
slicing
$endgroup$
I understand how slicer programs create sets of closed-loop polygons to print on a layer-by-layer basis. For a given closed loop polygon which needs to be printed, the tool path generator will know the coordinates and how those coordinates are connected to each other, such that traversing a set of segments in that order will bring the extruder head back to the first coordinate to complete the closed loop.
My question is: By what mechanism does the tool path generator decide which direction to traverse the closed loop? As it is a loop, that loop could be printed "clockwise" or "counter-clockwise", as it were. Any details, and links to further explanations of how some of the big-name slicer programs determine this is much appreciated.
slicing
slicing
edited 4 hours ago
Thetravellingfool
asked 6 hours ago
ThetravellingfoolThetravellingfool
1186
1186
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add a comment |
1 Answer
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$begingroup$
Math
In math, there is a way how a path is to be followed, and that is usually counterclockwise:
Assuming a perimeter path of a circle with $r=1$ around $(2,2)$, then the path can be defined as $f(p)={{cos(p)+2}choose{sin(p)+2}}$ - where $p$ is the path parameter, in this case an angle of 0 to 360°, and just increasing the angle rotates right hand around. If we had the same path but a different starting point, a shift by $theta$, then the path would read $f(p)={{cos(p+theta)+2}choose{sin(p+theta)+2}}$. So math is usually counterclockwise.
Slicers
Now, any slicer is applied math. As far as I can tell, Cura does slice with the result of a clockwise path, which is always performed. However, that does not invalidate the math at all, since solving a path with $p=0°to360°$ and solving it $p=360°to0°$ results in producing the exact same print, just opposed print direction of the perimeter.
As long as the perimeter of an object is solved as being done as one closed loop, the perimeter will need to have just one, prescribed direction. This direction will be clockwise or counterclockwise depending on how the slicer exactly solves its calculations. Since both directions are equally valid, it is a programmer's decision. Some options might be:
- Solving the path correctly counterclockwise and putting the slicing into a FILO memory, resulting in a clockwise operation starting from the last solved point.
- Solving counterclockwise and saving into FIFO, running counterclockwise.
- Solving deliberately clockwise for arbitrary reasons.
Conclusion
Slicers for 3D printing have a hard-coded direction that is followed when producing gcode. Any and all perimeters will be printed, starting from some arbitrary point, into that direction. In the end, it is a choice of the programmer of the slicing engine that determines if the path is to be run down "forward" or "backward" in mathematical sense.
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$begingroup$
Assuming the circle begins at θ = 0, doesn't increasing the angle through 360 rotate counterclockwise?
$endgroup$
– Thetravellingfool
4 hours ago
$begingroup$
@Thetravellingfool good remark, had to look it up, I mixed it up, fixed
$endgroup$
– Trish
4 hours ago
$begingroup$
Thank you for your contribution to the discussion Trish. Unfortunately, my question still stands.
$endgroup$
– Thetravellingfool
4 hours ago
add a comment |
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1 Answer
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$begingroup$
Math
In math, there is a way how a path is to be followed, and that is usually counterclockwise:
Assuming a perimeter path of a circle with $r=1$ around $(2,2)$, then the path can be defined as $f(p)={{cos(p)+2}choose{sin(p)+2}}$ - where $p$ is the path parameter, in this case an angle of 0 to 360°, and just increasing the angle rotates right hand around. If we had the same path but a different starting point, a shift by $theta$, then the path would read $f(p)={{cos(p+theta)+2}choose{sin(p+theta)+2}}$. So math is usually counterclockwise.
Slicers
Now, any slicer is applied math. As far as I can tell, Cura does slice with the result of a clockwise path, which is always performed. However, that does not invalidate the math at all, since solving a path with $p=0°to360°$ and solving it $p=360°to0°$ results in producing the exact same print, just opposed print direction of the perimeter.
As long as the perimeter of an object is solved as being done as one closed loop, the perimeter will need to have just one, prescribed direction. This direction will be clockwise or counterclockwise depending on how the slicer exactly solves its calculations. Since both directions are equally valid, it is a programmer's decision. Some options might be:
- Solving the path correctly counterclockwise and putting the slicing into a FILO memory, resulting in a clockwise operation starting from the last solved point.
- Solving counterclockwise and saving into FIFO, running counterclockwise.
- Solving deliberately clockwise for arbitrary reasons.
Conclusion
Slicers for 3D printing have a hard-coded direction that is followed when producing gcode. Any and all perimeters will be printed, starting from some arbitrary point, into that direction. In the end, it is a choice of the programmer of the slicing engine that determines if the path is to be run down "forward" or "backward" in mathematical sense.
$endgroup$
$begingroup$
Assuming the circle begins at θ = 0, doesn't increasing the angle through 360 rotate counterclockwise?
$endgroup$
– Thetravellingfool
4 hours ago
$begingroup$
@Thetravellingfool good remark, had to look it up, I mixed it up, fixed
$endgroup$
– Trish
4 hours ago
$begingroup$
Thank you for your contribution to the discussion Trish. Unfortunately, my question still stands.
$endgroup$
– Thetravellingfool
4 hours ago
add a comment |
$begingroup$
Math
In math, there is a way how a path is to be followed, and that is usually counterclockwise:
Assuming a perimeter path of a circle with $r=1$ around $(2,2)$, then the path can be defined as $f(p)={{cos(p)+2}choose{sin(p)+2}}$ - where $p$ is the path parameter, in this case an angle of 0 to 360°, and just increasing the angle rotates right hand around. If we had the same path but a different starting point, a shift by $theta$, then the path would read $f(p)={{cos(p+theta)+2}choose{sin(p+theta)+2}}$. So math is usually counterclockwise.
Slicers
Now, any slicer is applied math. As far as I can tell, Cura does slice with the result of a clockwise path, which is always performed. However, that does not invalidate the math at all, since solving a path with $p=0°to360°$ and solving it $p=360°to0°$ results in producing the exact same print, just opposed print direction of the perimeter.
As long as the perimeter of an object is solved as being done as one closed loop, the perimeter will need to have just one, prescribed direction. This direction will be clockwise or counterclockwise depending on how the slicer exactly solves its calculations. Since both directions are equally valid, it is a programmer's decision. Some options might be:
- Solving the path correctly counterclockwise and putting the slicing into a FILO memory, resulting in a clockwise operation starting from the last solved point.
- Solving counterclockwise and saving into FIFO, running counterclockwise.
- Solving deliberately clockwise for arbitrary reasons.
Conclusion
Slicers for 3D printing have a hard-coded direction that is followed when producing gcode. Any and all perimeters will be printed, starting from some arbitrary point, into that direction. In the end, it is a choice of the programmer of the slicing engine that determines if the path is to be run down "forward" or "backward" in mathematical sense.
$endgroup$
$begingroup$
Assuming the circle begins at θ = 0, doesn't increasing the angle through 360 rotate counterclockwise?
$endgroup$
– Thetravellingfool
4 hours ago
$begingroup$
@Thetravellingfool good remark, had to look it up, I mixed it up, fixed
$endgroup$
– Trish
4 hours ago
$begingroup$
Thank you for your contribution to the discussion Trish. Unfortunately, my question still stands.
$endgroup$
– Thetravellingfool
4 hours ago
add a comment |
$begingroup$
Math
In math, there is a way how a path is to be followed, and that is usually counterclockwise:
Assuming a perimeter path of a circle with $r=1$ around $(2,2)$, then the path can be defined as $f(p)={{cos(p)+2}choose{sin(p)+2}}$ - where $p$ is the path parameter, in this case an angle of 0 to 360°, and just increasing the angle rotates right hand around. If we had the same path but a different starting point, a shift by $theta$, then the path would read $f(p)={{cos(p+theta)+2}choose{sin(p+theta)+2}}$. So math is usually counterclockwise.
Slicers
Now, any slicer is applied math. As far as I can tell, Cura does slice with the result of a clockwise path, which is always performed. However, that does not invalidate the math at all, since solving a path with $p=0°to360°$ and solving it $p=360°to0°$ results in producing the exact same print, just opposed print direction of the perimeter.
As long as the perimeter of an object is solved as being done as one closed loop, the perimeter will need to have just one, prescribed direction. This direction will be clockwise or counterclockwise depending on how the slicer exactly solves its calculations. Since both directions are equally valid, it is a programmer's decision. Some options might be:
- Solving the path correctly counterclockwise and putting the slicing into a FILO memory, resulting in a clockwise operation starting from the last solved point.
- Solving counterclockwise and saving into FIFO, running counterclockwise.
- Solving deliberately clockwise for arbitrary reasons.
Conclusion
Slicers for 3D printing have a hard-coded direction that is followed when producing gcode. Any and all perimeters will be printed, starting from some arbitrary point, into that direction. In the end, it is a choice of the programmer of the slicing engine that determines if the path is to be run down "forward" or "backward" in mathematical sense.
$endgroup$
Math
In math, there is a way how a path is to be followed, and that is usually counterclockwise:
Assuming a perimeter path of a circle with $r=1$ around $(2,2)$, then the path can be defined as $f(p)={{cos(p)+2}choose{sin(p)+2}}$ - where $p$ is the path parameter, in this case an angle of 0 to 360°, and just increasing the angle rotates right hand around. If we had the same path but a different starting point, a shift by $theta$, then the path would read $f(p)={{cos(p+theta)+2}choose{sin(p+theta)+2}}$. So math is usually counterclockwise.
Slicers
Now, any slicer is applied math. As far as I can tell, Cura does slice with the result of a clockwise path, which is always performed. However, that does not invalidate the math at all, since solving a path with $p=0°to360°$ and solving it $p=360°to0°$ results in producing the exact same print, just opposed print direction of the perimeter.
As long as the perimeter of an object is solved as being done as one closed loop, the perimeter will need to have just one, prescribed direction. This direction will be clockwise or counterclockwise depending on how the slicer exactly solves its calculations. Since both directions are equally valid, it is a programmer's decision. Some options might be:
- Solving the path correctly counterclockwise and putting the slicing into a FILO memory, resulting in a clockwise operation starting from the last solved point.
- Solving counterclockwise and saving into FIFO, running counterclockwise.
- Solving deliberately clockwise for arbitrary reasons.
Conclusion
Slicers for 3D printing have a hard-coded direction that is followed when producing gcode. Any and all perimeters will be printed, starting from some arbitrary point, into that direction. In the end, it is a choice of the programmer of the slicing engine that determines if the path is to be run down "forward" or "backward" in mathematical sense.
edited 1 min ago
answered 5 hours ago
TrishTrish
5,076938
5,076938
$begingroup$
Assuming the circle begins at θ = 0, doesn't increasing the angle through 360 rotate counterclockwise?
$endgroup$
– Thetravellingfool
4 hours ago
$begingroup$
@Thetravellingfool good remark, had to look it up, I mixed it up, fixed
$endgroup$
– Trish
4 hours ago
$begingroup$
Thank you for your contribution to the discussion Trish. Unfortunately, my question still stands.
$endgroup$
– Thetravellingfool
4 hours ago
add a comment |
$begingroup$
Assuming the circle begins at θ = 0, doesn't increasing the angle through 360 rotate counterclockwise?
$endgroup$
– Thetravellingfool
4 hours ago
$begingroup$
@Thetravellingfool good remark, had to look it up, I mixed it up, fixed
$endgroup$
– Trish
4 hours ago
$begingroup$
Thank you for your contribution to the discussion Trish. Unfortunately, my question still stands.
$endgroup$
– Thetravellingfool
4 hours ago
$begingroup$
Assuming the circle begins at θ = 0, doesn't increasing the angle through 360 rotate counterclockwise?
$endgroup$
– Thetravellingfool
4 hours ago
$begingroup$
Assuming the circle begins at θ = 0, doesn't increasing the angle through 360 rotate counterclockwise?
$endgroup$
– Thetravellingfool
4 hours ago
$begingroup$
@Thetravellingfool good remark, had to look it up, I mixed it up, fixed
$endgroup$
– Trish
4 hours ago
$begingroup$
@Thetravellingfool good remark, had to look it up, I mixed it up, fixed
$endgroup$
– Trish
4 hours ago
$begingroup$
Thank you for your contribution to the discussion Trish. Unfortunately, my question still stands.
$endgroup$
– Thetravellingfool
4 hours ago
$begingroup$
Thank you for your contribution to the discussion Trish. Unfortunately, my question still stands.
$endgroup$
– Thetravellingfool
4 hours ago
add a comment |
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