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>Chapter 6. Weaving for inkjet printers</TD
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><DIV
CLASS="sect1"
><H1
CLASS="sect1"
><A
NAME="AEN1734"
>Weaving algorithms</A
></H1
><P
> I considered a few algorithms to perform the weave. The first
one I devised let me use only (jets −
distance_between_jets + 1) nozzles, or 25. This is OK in
principle, but it's slower than using all nozzles. By playing
around with it some more, I came up with an algorithm that lets
me use all of the nozzles, except near the top and bottom of the
page.
</P
><P
> This still produces some banding, though. Even better quality
can be achieved by using multiple nozzles on the same line. How
do we do this? In 1440×720 mode, we're printing two
output lines at the same vertical position. However, if we want
four passes, we have to effectively print each line twice.
Actually doing this would increase the density, so what we do is
print half the dots on each pass. This produces near-perfect
output, and it's far faster than using (pseudo)
“MicroWeave”.
</P
><P
> Yet another complication is how to get near the top and bottom
of the page. This algorithm lets us print to within one head
width of the top of the page, and a bit more than one head width
from the bottom. That leaves a lot of blank space. Doing the
weave properly outside of this region is increasingly difficult
as we get closer to the edge of the paper; in the interior
region, any nozzle can print any line, but near the top and
bottom edges, only some nozzles can print. We originally
handled this by using the naive way mentioned above near the
borders, and switching over to the high quality method in the
interior. Unfortunately, this meant that the quality is quite
visibly degraded near the top and bottom of the page. We have
since devised better algorithms that allow printing to the
extreme top and bottom of the region that can physically be
printed, with only minimal loss of quality.
</P
><P
> Epson does not advertise that the printers can print at the very
top of the page, although in practice most of them can. The
quality is degraded to some degree, and we have observed that in
some cases not all of the dots get printed. Epson may have
decided that the degradation in quality is sufficient that
printing in that region should not be allowed. That is a valid
decision, although we have taken another approach.
</P
><DIV
CLASS="sect2"
><H2
CLASS="sect2"
><A
NAME="AEN1740"
>Simple weaving algorithms</A
></H2
><P
> The initial problem is to calculate the starting position of
each pass; the row number of the printer's top jet when
printing that pass. Since we assume the paper cannot be
reverse-fed, the print head must, for each pass, start
either further down the page than the previous pass or at
the same position. Each pass's start point is therefore at
a non-negative offset from the previous pass's start point.
</P
><P
> Once we have a formula for the starting row of each pass, we
then turn that “inside out” to get a formula for
the pass number containing each row.
</P
><P
> First, let's define how our printer works. We measure
vertical position on the paper in “rows”; the
resolution with which the printer can position the paper
vertically. The print head contains J ink jets, which are
spaced S rows apart.
</P
><P
> Consider a very simple case: we want to print a page as
quickly as possible, and we mostly don't care how sparse the
printing is, so long as it's fairly even.
</P
><P
> It's pretty obvious how to do this. We make one pass with
the print head, printing J lines of data, each line S rows
after the previous one. We then advance the paper by S
× J rows and print the next row. For example, if J =
7 and S = 4, this method can be illustrated like this:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1747"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>pass number
| row number------->
| | 111111111122222222223333333333444444444455555555556666666666
| 0123456789012345678901234567890123456789012345678901234567890123456789
0 *---*---*---*---*---*---*
1 *---*---*---*---*---*---*
2 \-----------------------/ *---*---*---*---*---*-
7 jets \---/
4 rows offset from one jet to the next
\---------------------------/
7*4=28 rows offset from one pass to the next</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> In these examples, the vertical axis can be thought of as the
time axis, with the pass number shown at the left margin,
while the row number runs horizontally. A
<SAMP
CLASS="computeroutput"
>*</SAMP
> shows each row printed by a
pass, and a row of <SAMP
CLASS="computeroutput"
>-</SAMP
> is used
to link together the rows printed by one pass of the print
head. The first pass is numbered
<SAMP
CLASS="computeroutput"
>0</SAMP
> and starts at row 0. Each
subsequent pass p starts at row p × S × J. Each
pass prints J lines, each line being S rows after the previous
one. (For ease of viewing this file on a standard terminal,
I'm clipping the examples at column 80.)
</P
><P
> This method covers the whole page with lines printed evenly
S rows apart. However, we want to fill in all the other
rows with printing to get a full-density page (we're
ignoring oversampling at this stage). Where we have
previously printed a single pass, we'll now print a
“pass block”: we print extra passes to fill in
the empty rows. A naive implementation might look like
this:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1754"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *---*---*---*---*---*---*
1 *---*---*---*---*---*---*
2 *---*---*---*---*---*---*
3 *---*---*---*---*---*---*
4 *---*---*---*---*---*---*
5 *---*---*---*---*---*---*
6 *---*---*---*---*---*---*
7 *---*---*---*---*---*---*
8 *---*---*---*---*---*-
9 *---*---*---*---*---*
10 *---*---*---*---*---
11 *---*---*---*---*--</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> (Now you can see why this process is called
“weaving”!)
</P
></DIV
><DIV
CLASS="sect2"
><H2
CLASS="sect2"
><A
NAME="AEN1757"
>Perfect weaving</A
></H2
><P
> This simple weave pattern prints every row, but will give
conspicuous banding patterns for the reasons discussed
above.
</P
><P
> Let's start improving this for our simple case. We can
reduce banding by making sure that any given jet never
prints a row too close to another row printed by the same
jet. This means we want to space the rows printed by a
given jet evenly down the page. In turn, this implies we
want to advance the paper by as nearly an equal amount after
each pass as possible.
</P
><P
> Each pass block prints S × J lines in S passes. The
first line printed in each pass block is S × J rows
lower on the page than the first line printed in the
previous pass block. Therefore, if we advance the paper by
J rows between each pass, we can print the right number of
passes in each block and advance the paper perfectly evenly.
</P
><P
> Here's what this “perfect” weave looks like:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1763"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
> start of full weave
|
0 *---*---*---*---*---*---*
1 *---*---*---*---*---*---*
2 *---*---*---*---*---*---*
3 *---*---*---*---*---*---*
4 *---*---*---*---*---*---*
5 *---*---*---*---*---*---*
6 *---*---*---*---*---*---*
7 *---*---*---*---*---*---*
8 *---*---*---*---*---*-
9 *---*---*---*--
10 *---*---
11 *</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> You'll notice that, for the first few rows, this weave is
too sparse. It is not until the row marked “start of
full weave” that every subsequent row is printed. We
can calculate this start position as follows:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1766"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>start = (S − 1) × (J − 1)</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> For the moment, we will ignore this problem with the weave.
We'll consider later how to fill in the missing rows.
</P
><P
> Let's look at a few more examples of perfect weaves:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1770"
></A
><P
> S = 2, J = 7, start = (2−1) × (7−1) = 6:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
> starting row of full weave
|
0 *-*-*-*-*-*-*
1 *-*-*-*-*-*-*
2 *-*-*-*-*-*-*
3 *-*-*-*-*-*-*
4 *-*-*-*-*-*-*
5 *-*-*-*-*-*-*
6 *-*-*-*-*-*-*
7 *-*-*-*-*-*-*</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1773"
></A
><P
> S = 7, J = 2, start = 6:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
> start
|
0 *------*
1 *------*
2 *------*
3 *------*
4 *------*
5 *------*
6 *------*
7 *------*
8 *------*
9 *------*</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1776"
></A
><P
> S = 4, J = 13, start = 36:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
> start
|
0 *---*---*---*---*---*---*---*---*---*---*---*---*
1 *---*---*---*---*---*---*---*---*---*---*---*---*
2 *---*---*---*---*---*---*---*---*---*---*---*---*
3 *---*---*---*---*---*---*---*---*---*--
4 *---*---*---*---*---*---*-
5 *---*---*---*</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1779"
></A
><P
> S = 13, J = 4, start = 36:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
> start
|
0 *------------*------------*------------*
1 *------------*------------*------------*
2 *------------*------------*------------*
3 *------------*------------*------------*
4 *------------*------------*------------*
5 *------------*------------*------------*
6 *------------*------------*------------*
7 *------------*------------*------------*
8 *------------*------------*------------*
9 *------------*------------*------------*
10 *------------*------------*-----------
11 *------------*------------*-------
12 *------------*------------*---
13 *------------*------------
14 *------------*--------
15 *------------*----
16 *------------*
17 *---------
18 *-----
19 *-</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1782"
></A
><P
> S = 8, J = 5, start = 28:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
> start
|
0 *-------*-------*-------*-------*
1 *-------*-------*-------*-------*
2 *-------*-------*-------*-------*
3 *-------*-------*-------*-------*
4 *-------*-------*-------*-------*
5 *-------*-------*-------*-------*
6 *-------*-------*-------*-------*
7 *-------*-------*-------*-------*
8 *-------*-------*-------*-------*
9 *-------*-------*-------*-------*
10 *-------*-------*-------*---
11 *-------*-------*------
12 *-------*-------*-
13 *-------*----
14 *-------
15 *--</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1785"
></A
><P
> S = 9, J = 5, start = 32:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
> start
|
0 *--------*--------*--------*--------*
1 *--------*--------*--------*--------*
2 *--------*--------*--------*--------*
3 *--------*--------*--------*--------*
4 *--------*--------*--------*--------*
5 *--------*--------*--------*--------*
6 *--------*--------*--------*--------*
7 *--------*--------*--------*--------*
8 *--------*--------*--------*--------*
9 *--------*--------*--------*-----
10 *--------*--------*--------*
11 *--------*--------*----
12 *--------*--------
13 *--------*---
14 *-------
15 *--</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1788"
></A
><P
> S = 6, J = 7, start = 30:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
> start
|
0 *-----*-----*-----*-----*-----*-----*
1 *-----*-----*-----*-----*-----*-----*
2 *-----*-----*-----*-----*-----*-----*
3 *-----*-----*-----*-----*-----*-----*
4 *-----*-----*-----*-----*-----*-----*
5 *-----*-----*-----*-----*-----*-----*
6 *-----*-----*-----*-----*-----*-----
7 *-----*-----*-----*-----*----
8 *-----*-----*-----*---
9 *-----*-----*--
10 *-----*-
11 *</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
></DIV
><DIV
CLASS="sect2"
><H2
CLASS="sect2"
><A
NAME="AEN1791"
>Weaving collisions</A
></H2
><P
> A perfect weave is not possible in all cases. Let's look at
another example:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1794"
></A
><P
> S = 6, J = 4:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *-----*-----*-----*
1 *-----*-----*-----*
2 *-----*-----*-----*
3 *-----*-----*-----*
4 ^ *-^---*-----*-----*
5 | ^ | *-^---*-----*-----*
OUCH! ^ | ^
| |</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> Here we have a collision. Some lines printed in later
passes overprint lines printed by earlier passes. We can
see why by considering which row number is printed by a
given jet number j (numbered from 0) of a given pass, p:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1798"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>row(p, j) = (p × J) + (j × S)</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> Because J = 4 and S = 6 have a common factor of 2, jet 2 of
pass 0 prints the same row as jet 0 of pass 3:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1801"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>row(0, 2) = (0 × 4) + (2 × 6) = 12
row(3, 0) = (3 × 4) + (0 × 6) = 12</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> In fact, with this particular weave pattern, jets 0 and 1 of
pass p + 3 always overprint jets 2 and 3 of pass p. We'll
represent overprinting rows by a
<SAMP
CLASS="computeroutput"
>^</SAMP
> in our diagrams, and
correct rows by <SAMP
CLASS="computeroutput"
>*</SAMP
>:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1806"
></A
><P
> S = 6, J = 4:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *-----*-----*-----*
1 *-----*-----*-----*
2 *-----*-----*-----*
3 ^-----^-----*-----*
4 ^-----^-----*-----*
5 ^-----^-----*-----*</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
></DIV
><DIV
CLASS="sect2"
><H2
CLASS="sect2"
><A
NAME="AEN1809"
>What makes a “perfect” weave?</A
></H2
><P
> So what causes the perfect weave cases to be perfect, and
the other cases not to be? In all the perfect cases above,
S and J are relatively prime (i.e. their greatest common
divisor (GCD) is 1). As we mentioned above, S = 6 and J = 4
have a common factor, which causes the overprinting. Where
S and J have a GCD of 1, they have no common factor other
than 1 and, as a result, no overprinting occurs. If S and J
are not relatively prime, their common factor will cause
overprinting.
</P
><P
> We can work out the greatest common divisor of a pair of
natural numbers using Euler's algorithm:
</P
><P
></P
><OL
TYPE="1"
><LI
><P
> Start with the two numbers: (e.g.) 9, 24
</P
></LI
><LI
><P
> Swap them if necessary so that the larger one comes
first: 24, 9
</P
></LI
><LI
><P
> Subtract the second number from the first: 15, 9
</P
></LI
><LI
><P
> Repeat until the first number becomes smaller: 6, 9
</P
></LI
><LI
><P
> Swap the numbers again, so the larger one comes first:
9, 6
</P
></LI
><LI
><P
> Subtract again: 3, 6
</P
></LI
><LI
><P
> Swap: 6, 3
</P
></LI
><LI
><P
> Subtract: 3, 3
</P
></LI
><LI
><P
> And again: 0, 3
</P
></LI
><LI
><P
> When one of the numbers becomes 0, the other number is
the GCD of the two numbers you started with.
</P
></LI
></OL
><P
> These repeated subtractions can be done with C's <TT
CLASS="literal"
>%</TT
>
operator, so we can write this in C as follows:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1836"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="programlisting"
>unsigned int
gcd(unsigned int x, unsigned int y)
{
if (y == 0)
return x;
while (x != 0) {
if (y > x)
swap (&x, &y);
x %= y;
}
return y;
}</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> <TT
CLASS="literal"
>gcd(S,J)</TT
> will feature quite prominently in our
weaving algorithm.
</P
><P
> If 0 ≤ j < J, there should only be a single pair (p, j)
for any given row number. If S and J are not relatively
prime, this assumption breaks down. (For conciseness, let G
= GCD(S,J).)
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1841"
></A
><P
> S = 8, J = 6, G = 2:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *-------*-------*-------*-------*-------*
1 *-------*-------*-------*-------*-------*
2 *-------*-------*-------*-------*-------*
3 *-------*-------*-------*-------*-------*
4 ^-------^-------^-------*-------*-------*
5 ^-------^-------^-------*-------*-------*</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> In this case, jets 0, 1 and 2 of pass p + 4 collide with
jets 3, 4 and 5 of pass p.
</P
><P
> How can we calculate these numbers? Suppose we were to
print using fewer jets, say J / G jets. The greatest common
divisor of J / G and S is 1, enabling a perfect weave. But
to get a perfect weave, we also have to advance the paper by
a factor of G less:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1846"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *-------*-------* - - -
1 *-------*-------* - - -
2 *-------*-------* - - -
3 *-------*-------* - - -
4 *-------*-------* - - -
5 *-------*-------* - - -</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> If we left the paper advance alone, we'd get a sparse weave;
only one row can be printed every G rows:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1849"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *-------*-------* - - -
1 *-------*-------* - - -
2 *-------*-------* - - -
3 *-------*-------* - - -
4 *-------*-------* - - -
5 *-------*-------* - - -
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^
These rows need filling in.</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> The rows that would have been printed by the jets we've now
omitted (shown as <SAMP
CLASS="computeroutput"
>-</SAMP
>) are
printed by other jets on later passes.
</P
><P
> Let's analyse this. Consider how a pass p could collide
with pass 0. Pass p starts at offset p × J. Pass 0
prints at rows which are multiples of S. If p × J is
exactly divisible by S, a collision has occurred, unless (p
×J) ≥ J × S (which will happen when we
finish a pass block).
</P
><P
> So, we want to find p and q such that p × J = q
× S and p is minimised. Then p is the number of rows
before a collision, and q is the number of jets in pass 0
which are not involved in the collision. To do this, we
find the lowest common multiple of J and S, which is L = (J
× S) / G. L / J is the number of rows before a
collision, and L / S is the number of jets in the first pass
not involved in the collision.
</P
><P
> Thus, we see that the first J / G rows printed by a given
pass are not overprinted by any later pass. However, the
rest of the rows printed by pass p are overprinted by the
first J − (J / G) jets of pass p + (S / G). We will
use C to refer to S / G, the number of rows after which a
collision occurs.
</P
><P
> Another example:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1857"
></A
><P
> S = 6, J = 9, G = 3, C = S / G = 2:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *-----*-----*-----*-----*-----*-----*-----*-----*
1 *-----*-----*-----*-----*-----*-----*-----*-----*
2 ^-----^-----^-----^-----^-----^-----*-----*-----*
3 ^-----^-----^-----^-----^-----^-----*-----*-----*
4 ^-----^-----^-----^-----^-----^-----*-----
5 ^-----^-----^-----^-----^-----^--
^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^
These rows need filling in.</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> In this case, the first J − (J / G) = 9 − (9 /
3) = 6 jets of pass p + (6 / 3) = p + 2 collide with the
last 6 jets of pass p. Only one row in every G = 2 rows is
printed by this weave.
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1861"
></A
><P
> S = 9, J = 6, G = 3, C = 3:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
> 0 *--------*--------*--------*--------*--------*
1 *--------*--------*--------*--------*--------*
2 *--------*--------*--------*--------*--------*
3 ^--------^--------^--------^--------*--------*
4 ^--------^--------^--------^--------*--------*
5 ^--------^--------^--------^--------*--------*</PRE
></TD
></TR
></TABLE
><P
> Here, the first J - (J / G) = 6 - (6 / 3) = 4 jets of pass
p + (9 / 3) = p + 3 collide with the last 4 jets of pass
p.
</P
><P
> Note that, in these overprinting cases, only rows
divisible by G are ever printed. The other rows, those
not divisible by G, are not touched by this weave.
</P
><P
> We can modify our weave pattern to avoid overprinting any
rows and simultaneously fill in the missing rows. Instead
of using J alone to determine the start of each pass from
the previous pass, we adjust the starting position of some
passes. As mentioned before, we will divide the page into
pass blocks, with S passes in each block. This ensures
that the first jet of the first pass in a block prints the
row which the Jth jet of the first pass of the previous
block would have printed, if the print head had one extra
jet.
</P
><P
> Looking back at an example of a perfect weave, we can
divide it into pass blocks:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1868"
></A
><P
> S = 7, J = 2, G = 1:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
> imaginary extra jet
|
0 *------* * <--start of pass block 0
1 *------* |
2 *------* |
3 *------*|
4 *-----|*
5 *---|--*
6 *-|----*
|
7 *------* <--start of pass block 1
8 *------*
9 *------*</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> We can now calculate the start of a given pass by
reference to its pass block. The first pass of pass block
b always starts at row (b × S × J). The start
row of each of the other passes in the block are
calculated using offsets from this row.
</P
><P
> For the example above, there are 7 passes in each pass
block, and their offsets are 0, 2, 4, 6, 8, 10 and 12.
The next pass block is offset S × J = 14 rows from
the start of the current pass block.
</P
><P
> The simplest way to modify the “perfect” weave
pattern to give a correct weave in cases where G ≠ 1
is to simply change any offsets which would result in a
collision, until the collision disappears. Every printed
row in the weave, as we have shown it up to now, is
separated from each of its neighbouring printed rows by G
blank rows. We will add an extra offset to each colliding
pass in such a way that we push the pass onto these
otherwise blank rows.
</P
><P
> We have seen that, unless G = 1, the plain weave pattern
results in each pass colliding with the pass S / G passes
before. We will now subdivide our pass block into
subblocks, each consisting of B = S / G passes. There are
therefore G subblocks in a pass block.
</P
><P
> For each subblock, the passes in that subblock have a
constant offset added to them. The offset is different
for each subblock in a block. There are many ways we can
choose the offsets, but the simplest is to make the offset
equal to the subblock number (starting from 0).
</P
><P
> Thus, the passes in the first subblock in each pass block
remain at the offsets we've already calculated from J.
The passes in the second subblock each have 1 added to
their offset, the passes in the third subblock have 2
added, and so on. Thus, the offset of pass p (numbered
relative to the start of its pass block) is p × J +
floor(p / B).
</P
><P
> This gives us a weave pattern looking like this:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1878"
></A
><P
> S = 6, J = 9, G = 3, B = 2:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *-----*-----*-----*-----*-----*-----*-----*-----*
1 ^ *-----*-----*-----*-----*-----*-----*-----*-----*
2 | +-> *-----*-----*-----*-----*-----*-----*-----*-----*
3 | | *-----*-----*-----*-----*-----*-----*-----*-----*
4 | | +-> *-----*-----*-----*-----*-----*-----*---
5 | | | *-----*-----*-----*-----*-----*
6 | | | +-> *-----*-----*-----*-----
7 | | | | *-----*-----*--
| | | start of pass block 1
| | | (offset returns to 0)
| | start of subblock 2 (offset 2 rows)
| start of subblock 1 (following passes offset by 1 row)
start of passblock 0, subblock 0 (pass start calculated as p*J)</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1881"
></A
><P
> S = 9, J = 6, G = 3, B = 3:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *--------*--------*--------*--------*--------*
1 *--------*--------*--------*--------*--------*
2 *--------*--------*--------*--------*--------*
3 *--------*--------*--------*--------*--------*
4 *--------*--------*--------*--------*--------*
5 *--------*--------*--------*--------*--------*
6 *--------*--------*--------*--------*---
7 *--------*--------*--------*------
8 *--------*--------*--------*
9 *--------*--------*-----
10 \---/ *--------*--------
11 small offset *--------*--
12 *----</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> This method of choosing offsets for subblocks can result
in an occasional small offset (as shown above) between one
pass and the next, particularly when G is large compared
to J. For example:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1885"
></A
><P
> S = 8, J = 4, G = 4, B = 2:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *-------*-------*-------*
1 *-------*-------*-------*
2 *-------*-------*-------*
3 *-------*-------*-------*
4 *-------*-------*-------*
5 *-------*-------*-------*
6 *-------*-------*-------*
7 *-------*-------*-------*
8 *-------*-------*-------*
9 \/ *-------*-------*-------*
very small offset!</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> We can plot the offset against the subblock number as
follows:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1889"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>subblock number
| offset
| |
| 0123
0 *
1 *
2 *
3 *
0 *
1 *
2 *
3 *</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> The discontinuity in this plot results in the small offset
between passes.
</P
><P
> As we said at the beginning, we want the offsets from each
pass to the next to be as similar as possible. We can fix
this by calculating the offset for a given subblock b as
follows:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1893"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>offset(b) = 2*b , if b < ceiling(G/2)
= 2*(G-b)-1 , otherwise</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> We can visualise this as follows, for G = 10:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1896"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0123456789
0 *
1 *
2 *
3 *
4 *
5 *
6 *
7 *
8 *
9 *
0 *
1 *
2 *
3 *
4 *
5 *
6 *
7 *
8 *
9 *</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> and for G = 11:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1899"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
> 1
01234567890
0 *
1 *
2 *
3 *
4 *
5 *
6 *
7 *
8 *
9 *
10 *
0 *
1 *
2 *
3 *
4 *
5 *
6 *
7 *
8 *
9 *
10 *</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> This gives a weave looking like this:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1902"
></A
><P
> S = 12, J = 6, G = 6, B = 2:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *-----------*-----------*-----------*-----------*-----------*
1 *-----------*-----------*-----------*-----------*-----------*
2 *-----------*-----------*-----------*-----------*-----------*
3 *-----------*-----------*-----------*-----------*---------
4 *-----------*-----------*-----------*-----------*-
5 *-----------*-----------*-----------*-------
6 *-----------*-----------*-----------*
7 *-----------*-----------*------
8 *-----------*-----------*--
9 *-----------*--------
10 *-----------*----
11 *----------
12 *-----</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> This method ensures that the offset between passes is
always in the range [J - 2, J + 2].
</P
><P
> (This might seem odd, but it occurs to me that a good
weave pattern might also make a good score for bell
ringers. When church bells are rung, a list of
“changes” are used. For example, if 8 bells
are being used, they will, at first, be rung in order:
12345678. If the first change is for bells 5 and 6, the
bells will then be rung in the order 12346578. If the
second change is 1 and 2, the next notes are 21346578.
After a long list of changes, the order the bells are rung
in can become quite complex.
</P
><P
> For a group of bell-ringers to change the order of the
notes, they must each either delay their bell's next ring,
hasten it, or keep it the same as the time it takes to
ring all the bells once. The length of time between each
ring of a given bell can only be changed a little each
time, though; with an ink-jet weave pattern, we want the
same to apply to the distance between passes.)
</P
><P
> Finally, knowing the number of jets J and their separation
S, we can calculate the starting row of any given pass p
as follows:
<DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1909"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>passesperblock = S
passblock = floor(p / passesperblock)
offsetinpassblock = p - passblock * passesperblock
subblocksperblock = gcd(S, J)
passespersubblock = S / subblocksperblock
subpassblock = floor(offsetinpassblock / passespersubblock)
if subpassblock < ceiling(subblocksperblock/2)
subblockoffset = 2*subpassblock
else
subblockoffset = 2*(subblocksperblock-subpassblock)-1
startingrow = passblock * S * J + offsetinpassblock * J + subblockoffset</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
>
</P
><P
> We can simplify this down to the following:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1912"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>subblocksperblock = gcd(S, J)
subpassblock = floor((p % S) * subblocksperblock / S)
if subpassblock * 2 < subblocksperblock
subblockoffset = 2*subpassblock
else
subblockoffset = 2*(subblocksperblock-subpassblock)-1
startingrow = p * J + subblockoffset</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> So the row number of jet j of pass p is
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1915"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>subblocksperblock = gcd(S, J)
subblockoffset(p)
= 2*subpassblock , if subpassblock * 2 < subblocksperblock
= 2*(subblocksperblock-subpassblock)-1 , otherwise
where
subpassblock = floor((p % S) * subblocksperblock / S)
row(j, p) = p * J + subblockoffset(p) + j * S</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> Together with the inequality 0 ≤ j < J, we can use
this definition in reverse to calculate the pass number
containing a given row, r. Working out the inverse
definition involves a little guesswork, but one possible
result is as follows. Given a row, r, which is known to
be the first row of a pass, we can calculate the pass
number as follows:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1918"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>subblocksperblock = gcd(S, J)
subblockoffset = r % subblocksperblock
pass = (r - subblockoffset) / J</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> If G = 1, we can determine the pass number with this
algorithm:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1921"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>offset = r % J
pass = (r - offset) / J
while (offset % S != 0)
{
pass--
offset += J
}
jet = offset / S</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> Generalising, we come up with this algorithm. Given r, S
and J:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1924"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>G = gcd(S, J)
passespersubblock = S/G
subblockoffset = r % G
subpassblock = subblockoffset / 2 , if subblockoffset % 2 == 0
= G - (subblockoffset+1)/2 , otherwise
baserow = r - subblockoffset - (subpassblock * passespersubblock * J)
offset = baserow % J
pass = (baserow - offset) / J
while (offset % S != 0)
{
offset += J
pass -= 1
}
subblockretreat = floor(pass / passespersubblock) % G
pass -= subblockretreat * passespersubblock
pass += subpassblock * passespersubblock
jet = (r - subblockoffset - pass * J) / S</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> Let's look at some examples of imperfect but correct weave
patterns:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1927"
></A
><P
> S = 6, J = 4, GCD = 2,
passesperblock = S = 6,
passespersubblock = S / G = 6 / 2 = 3:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *-----*-----*-----*
1 *-----*-----*-----*
2 *-----*-----*-----*
3 *-----*-----*-----*
4 *-----*-----*-----*
5 *-----*-----*-----*
6 *-----*-----*-----*
7 *-----*-----*-----*
8 *-----*-----*-----*
9 *-----*-----*-----*
10 *-----*-----*-----*
11 *-----*-----*-----*
12 *-----*-----*-----*
13 *-----*-----*-----*
14 *-----*-----*-----*
15 *-----*-----*----
16 *-----*-----*
17 *-----*--
18 *-----
19 *-</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1930"
></A
><P
> S = 8, J = 6, G = 2,
passesperblock = S = 8,
passespersubblock= S / G = 8 / 2 = 4:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *-------*-------*-------*-------*-------*
1 *-------*-------*-------*-------*-------*
2 *-------*-------*-------*-------*-------*
3 *-------*-------*-------*-------*-------*
4 *-------*-------*-------*-------*-------*
5 *-------*-------*-------*-------*-------*
6 *-------*-------*-------*-------*-------*
7 *-------*-------*-------*-------*--
8 *-------*-------*-------*-----
9 *-------*-------*-------
10 *-------*-------*-
11 *-------*---
12 *----</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1933"
></A
><P
> S = 6, J = 12, G = 6,
passesperblock = S = 6,
passespersubblock= S / G = 6 / 6 = 1:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *-----*-----*-----*-----*-----*-----*-----*-----*-----*-----*-----*
1 *-----*-----*-----*-----*-----*-----*-----*-----*-----*-----*---
2 *-----*-----*-----*-----*-----*-----*-----*-----*-
3 *-----*-----*-----*-----*-----*-----*
4 *-----*-----*-----*-----*--
5 *-----*-----*----
6 *-----</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> We have now solved the basic weaving problem. There are
two further refinements we need to consider: oversampling,
and filling in the missing rows at the start of the weave.
</P
><P
></P
></DIV
></DIV
><DIV
CLASS="sect2"
><H2
CLASS="sect2"
><A
NAME="AEN1937"
>Oversampling</A
></H2
><P
> By oversampling, we mean printing on the same row more than
once. There are two reasons for oversampling: to increase
the horizontal resolution of the printout and to reduce
banding.
</P
><P
> Oversampling to increase horizontal resolution is necessary
because, although the printer might be able to position an
ink drop to, for example, 1/1440" horizontally, it may not
be able to lay down two such drops 1/1440" apart. If it can
print two drops 1/720" apart, 2x oversampling will be
necessary to get a 1/1440" horizontal resolution. If it can
only print two drops 1/360" apart, 4x oversampling will be
necessary for a 1/1440" horizontal resolution. The printer
enforces this “drop spacing” by only accepting
raster passes with a horizontal resolution matching the
spacing with which it can print dots, so we must print
passes at different horizontal positions if we are to obtain
a higher horizontal resolution. (Another reason it does
this may be to reduce the amount of memory needed in the
printer.)
</P
><P
> Oversampling can also be done to decrease the banding
apparent in an image. By splitting a row into two or more
sets of dots (“lines”) and printing each line on
the same row, but with a different nozzle for each line, we
can get a smoother print.
</P
><P
> To quantify these two kinds of oversampling, we'll introduce
two new constants: H shows how many different horizontal
offsets we want to print at (the “horizontal
oversampling”) while O shows how many times we want to
print each row, over and above the number of times necessary
for horizontal oversampling (the “extra
oversampling”).
</P
><P
> It is necessary for all the lines printed by a given pass to
have the same horizontal offset, but there need not be any
relation between them in terms of extra oversampling. For
the moment, however, we will treat all oversampling as
potentially requiring this alignment; all lines in one pass
must be derived from the original row data in the same way.
Thus, we'll assume O = 1 for now.
</P
><P
> So, how do we do this oversampling? In fact, it can be done
easily: advance the paper by a factor of H less between each
pass. We'll define a new variable, A, to show how much we
advance the paper between passes. Previously, we'd have
defined A = J; we now let A = J / H. This also affects our
pass blocks. Printing one pass block used to involve
advancing the paper S × J rows; it now advances the
paper (S×J) / H rows. We therefore name a group of H
pass blocks a “band”. Printing one band
involves advancing the paper S×J rows, as a pass
block did before.
</P
><P
> To keep our weave pattern working correctly, so that
overprinting does not occur within a pass block, we also
have to redefine G as GCD(S,A). Here's an example of an
oversampled weave pattern:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1946"
></A
><P
> S = 4, J = 10, H = 2, A = J/H = 10/2 = 5,
G= GCD(4,5) = 1,
passesperblock = S = 4,
passespersubblock = S/G = 4/1 = 4:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 *---*---*---*---*---*---*---*---*---*
1 *---*---*---*---*---*---*---*---*---*
2 *---*---*---*---*---*---*---*---*---*
3 *---*---*---*---*---*---*---*---*---*
4 *---*---*---*---*---*---*---*---*---*
5 *---*---*---*---*---*---*---*---*---*
6 *---*---*---*---*---*---*---*---*---*
7 *---*---*---*---*---*---*---*---*---*
8 *---*---*---*---*---*---*---*---*---*
9 *---*---*---*---*---*---*---*---*
10 *---*---*---*---*---*---*---
11 *---*---*---*---*---*--
12 *---*---*---*---*-
13 *---*---*---*
14 *---*---
15 *--</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> Now we have to determine which line is printed by each jet on
each pass. If we number each line generated as we split up a
row, we can use these numbers. We'll number the lines in our
diagram by replacing the <SAMP
CLASS="computeroutput"
>*</SAMP
>s
with integers in the range [0…H−1].
</P
><P
> Overprinting occurs once per pass block, so we can simply
print pass block 0 with line 0, pass block 1 with line 1,
pass block 2 with line 2, etc, wrapping to 0 when we've run
out of lines:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1952"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 0---0---0---0---0---0---0---0---0---0
1 0---0---0---0---0---0---0---0---0---0
2 0---0---0---0---0---0---0---0---0---0
3 0---0---0---0---0---0---0---0---0---0
4 1---1---1---1---1---1---1---1---1---1
5 1---1---1---1---1---1---1---1---1---1
6 1---1---1---1---1---1---1---1---1---1
7 1---1---1---1---1---1---1---1---1---1
8 0---0---0---0---0---0---0---0---0---0
9 0---0---0---0---0---0---0---0---0
10 0---0---0---0---0---0---0---
11 0---0---0---0---0---0--
12 1---1---1---1---1-
13 1---1---1---1
14 1---1---
15 1--</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1954"
></A
><P
> S = 4, J = 12, H = 2, A = J/H = 12/2 = 6,
G= GCD(4,6) = 2,
passesperblock= S = 4,
passespersubblock= S/G = 4/2 = 2:
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>0 0---0---0---0---0---0---0---0---0---0---0---0
1 0---0---0---0---0---0---0---0---0---0---0---0
2 0---0---0---0---0---0---0---0---0---0---0---0
3 0---0---0---0---0---0---0---0---0---0---0---0
4 1---1---1---1---1---1---1---1---1---1---1---1
5 1---1---1---1---1---1---1---1---1---1---1---1
6 1---1---1---1---1---1---1---1---1---1---1
7 1---1---1---1---1---1---1---1---1--
8 0---0---0---0---0---0---0---0-
9 0---0---0---0---0---0---
10 0---0---0---0---0
11 0---0---0--
12 1---1-</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> But what do we do if J is not an exact multiple of H? This
is a difficult problem, which I struggled with for quite a
few days before giving in and taking the easy (but less
elegant) way out. The easy solution is to round J / H down,
then add on the accumulated error at the end of each band.
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1958"
></A
><P
> S = 4, J = 11, H = 2, A = floor(J/H) = floor(11/2) = 5,
G = GCD(4,5) = 1,
passesperblock = S = 4,
passespersubblock = S/G = 4/1 = 4
</P
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>Band 0:
0 0---0---0---0---0---0---0---0---0---0---0
1 0---0---0---0---0---0---0---0---0---0---0
2 0---0---0---0---0---0---0---0---0---0---0
3 0---0---0---0---0---0---0---0---0---0---0
4 1---1---1---1---1---1---1---1---1---1---1
5 1---1---1---1---1---1---1---1---1---1---1
6 1---1---1---1---1---1---1---1---1---1---1
7 1---1---1---1---1---1---1---1---1---1---
Band 1:
8 | 0---0---0---0---0---0---0---0---0-
9 \-----------------------------------------/ 0---0---0---0---0---0---0---0
10 S*J rows 0---0---0---0---0---0---
11 0---0---0---0---0--
12 1---1---1---1-
13 1---1---1
14 1---</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> We can calculate the starting row and subpass number of a
given pass in this scheme as follows:
</P
><DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1962"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>A = floor(J / H)
subblocksperblock = gcd(S, A)
subpassblock = floor((p % S) * subblocksperblock / S)
if subpassblock * 2 < subblocksperblock
subblockoffset = 2*subpassblock
else
subblockoffset = 2*(subblocksperblock-subpassblock)-1
band = floor(P / (S * H))
passinband = P % (S * H)
startingrow = band * S * J + passinband * A + subblockoffset
subpass = passinband / S</PRE
></TD
></TR
></TABLE
><P
></P
></DIV
><P
> So the row number of jet j of pass p is
<DIV
CLASS="informalexample"
><P
></P
><A
NAME="AEN1965"
></A
><TABLE
BORDER="0"
BGCOLOR="#E0E0E0"
WIDTH="100%"
><TR
><TD
><PRE
CLASS="screen"
>A = floor(J / H)
subblocksperblock = gcd(S, A)
subblockoffset(p)
= 2*subpassblock , if subpassblock * 2 < subblocksperblock
= 2*(subblocksperblock-subpassblock)-1 , otherwise
where
subpassblock = floor((p % S) * subblocksperblock / S)
band(p) = floor(p / (S * H))
passinband(p) = p % (S * H)
row(j, p) = band(p) * S * J + passinband(p) * A + subblockoffset(p) + j * S
row(j, p) = p * J + subblockoffset(p) + j * S</PRE
></TD
></TR
></TABLE
><P
> To be continued…
</P
><P
></P
></DIV
>
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