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Stephen Beckham
10-29-2009, 1:31 PM
I was wondering if anyone knows how to calculate how much "area" is used by a graphic. Maybe even break it down by how much area per color?

Application is more along the lines of sublimation more than engraving, but I figure there are more Corel Guru's over here.

If so, I guess it could be used to apply how much etching will happen as well.

Thanks for anyone with knowledge, advice or suggestions.

Steve

Lee DeRaud
10-29-2009, 1:52 PM
I'd hate to try to do it inside CorelDRAW, but one (very rough) approach might be export the printable area as a bitmap, convert the colors to gray levels, and look at the histogram in the contrast enhancement. It's probably normalized to the predominant color though.

Worst case, cobble up a small filter program to brute-force scan the BMP and count pixels of a certain value and divide that by the total size. VBA under PhotoPaint maybe?

George D Gabert
10-29-2009, 2:19 PM
I have not tried this but a friend recommended this program. It is shareware but does require a fee after trial period.

http://www.oberonplace.com/products/curveworks/index.htm

GDG

Lee DeRaud
10-29-2009, 2:56 PM
If so, I guess it could be used to apply how much etching will happen as well.Not as useful as it sounds: there's almost zero correlation between the area engraved and the cost of running the laser.

(Warning: off-topic information of dubious usefulness follows.)
What matters for a laser is total time, which is nearly impossible to estimate with any accuracy. Consider:

1. A vertical stripe 1/8" wide and 8" top to bottom takes much longer to engrave than a 1" solid square. (both are 1 sqare inch in area.)
2. That same stripe turned horizontally engraves much faster on most lasers.
3. If you duplicate the stripe in example #1 and put the duplicates 6" apart, it takes the same amount of time to engrave them as a solid square 6" wide by 8" tall. (On the lasers I know about anyway. There may very well be a laser device driver out there smart enough to split that into two separate jobs, but I've never heard of one.)

I could go on all day. :p

There have been several attempts to handle the estimation problem; I have yet to see a really robust one.

Gary Hair
10-29-2009, 3:49 PM
there's almost zero correlation between the area engraved and the cost of running the laser.

The only correlation between the two exists when you have a solid area and it is taller than it is wide, or square - other than that, all bets are off.

This issue is magnified by slow speeds but can be negligable at high speeds and is also dependant on the DPI you are rastering. If you raster an image at 100% speed and 125 DPI then you'll get pretty close times whether you run an image vertical or horizontal - but again, this depends on your laser and the speed it runs.

Gary

Lee DeRaud
10-29-2009, 4:09 PM
The only correlation between the two exists when you have a solid area and it is taller than it is wide, or square - other than that, all bets are off.

This issue is magnified by slow speeds but can be negligable at high speeds and is also dependant on the DPI you are rastering. If you raster an image at 100% speed and 125 DPI then you'll get pretty close times whether you run an image vertical or horizontal - but again, this depends on your laser and the speed it runs.

GaryTrue. The effects of things like my example #3 tend to dominate anyway.

It just seems odd that the drivers tend to do a decent job of optimizing order/direction while vectoring (where "dead" time is done by full-speed movement anyway), yet they treat all rastering as if it's solid within the bounding envelope.