Here is a mini study of some of the options using the Scatter Extension under Extensions > Generate from Path > Scatter. I was trying to figure out a better way to control the number of instances that would occur ... based upon the Source Path perimeter length to Destination Path perimeter length ratio ... and the various spacing settings. It appears to be trial and error. (If anyone knows a simple, consistent way to determine this in advance, I would like to know.)
For example, if I wanted the end result to be exactly 5 instances (copies of the Source Path equally spaced apart) - how to determine the exact spacing option to set.
The problem is that the spacing option always uses millimetres even when the objects and the document use pixels. In your example, the semicircle thus is treated as having a length of 264.58mm. The scatter object's path length is irrelevant, only the width matters – or the height if you select 'Pattern is vertical' or 'Rotate pattern 90° clockwise' if you use the latest version. Both values are (I suspect) 50px or 13.229mm in your drawing. The number of instances is then calculated as
path-length / (pattern-width + spacing) + 1
The extra instance is due to the fact that the first instance does not need a spacing. If you select 'Stretch spaces …', the result of the above expression will be rounded to the nearest integer. Otherwise, it is rounded down. To get a specific number of instances, you should first determine your sizes in millimetres and then choose a spacing by computing
path-length / (instances − 1) − pattern-width
For your desired 5 instances that yields: 264.58 / (5 − 1) − 13.229 = 52.916 millimetres of spacing. Because of the rounding, you do not need the exact value, as you can tell from your examples 50 is fine.
the spacing option always uses millimetres even when the objects and the document use pixels...
...For your desired 5 instances that yields: 264.58 / (5 − 1) − 13.229 = 52.916 millimetres of spacing. Because of the rounding, you do not need the exact value, as you can tell from your examples 50 is fine.
Tips, Tricks, and Tutorials Exploring Using Scatter Extension
Inkscape.org
Hi, Inkscapers.
Here is a mini study of some of the options using the Scatter Extension under Extensions > Generate from Path > Scatter. I was trying to figure out a better way to control the number of instances that would occur ... based upon the Source Path perimeter length to Destination Path perimeter length ratio ... and the various spacing settings. It appears to be trial and error. (If anyone knows a simple, consistent way to determine this in advance, I would like to know.)
For example, if I wanted the end result to be exactly 5 instances (copies of the Source Path equally spaced apart) - how to determine the exact spacing option to set.
Meanwhile, see the graphics below.
Have a great day.
Has this extension been removed in 1.3.2 ???
Edit: Name changed to "Distrubute Along Path"
The Scatter extension has been renamed to Distribute Along Path:
Yup - caught a video on it and was scratching my head for a few.
https://youtu.be/POz_6s1r9tA?si=8JhKPEuTxprMliRC
The problem is that the spacing option always uses millimetres even when the objects and the document use pixels. In your example, the semicircle thus is treated as having a length of 264.58mm. The scatter object's path length is irrelevant, only the width matters – or the height if you select 'Pattern is vertical' or 'Rotate pattern 90° clockwise' if you use the latest version. Both values are (I suspect) 50px or 13.229mm in your drawing. The number of instances is then calculated as
path-length / (pattern-width + spacing) + 1
The extra instance is due to the fact that the first instance does not need a spacing. If you select 'Stretch spaces …', the result of the above expression will be rounded to the nearest integer. Otherwise, it is rounded down. To get a specific number of instances, you should first determine your sizes in millimetres and then choose a spacing by computing
path-length / (instances − 1) − pattern-width
For your desired 5 instances that yields: 264.58 / (5 − 1) − 13.229 = 52.916 millimetres of spacing. Because of the rounding, you do not need the exact value, as you can tell from your examples 50 is fine.
Might depend on the document scale...