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.
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”.
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.
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.
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.
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.
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.
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.
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:
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 |
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 * shows each row printed by a pass, and a row of - is used to link together the rows printed by one pass of the print head. The first pass is numbered 0 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.)
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:
0 *---*---*---*---*---*---* 1 *---*---*---*---*---*---* 2 *---*---*---*---*---*---* 3 *---*---*---*---*---*---* 4 *---*---*---*---*---*---* 5 *---*---*---*---*---*---* 6 *---*---*---*---*---*---* 7 *---*---*---*---*---*---* 8 *---*---*---*---*---*- 9 *---*---*---*---*---* 10 *---*---*---*---*--- 11 *---*---*---*---*-- |
(Now you can see why this process is called “weaving”!)
This simple weave pattern prints every row, but will give conspicuous banding patterns for the reasons discussed above.
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.
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.
Here's what this “perfect” weave looks like:
start of full weave
|
0 *---*---*---*---*---*---*
1 *---*---*---*---*---*---*
2 *---*---*---*---*---*---*
3 *---*---*---*---*---*---*
4 *---*---*---*---*---*---*
5 *---*---*---*---*---*---*
6 *---*---*---*---*---*---*
7 *---*---*---*---*---*---*
8 *---*---*---*---*---*-
9 *---*---*---*--
10 *---*---
11 * |
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:
For the moment, we will ignore this problem with the weave. We'll consider later how to fill in the missing rows.
Let's look at a few more examples of perfect weaves:
S = 2, J = 7, start = (2−1) × (7−1) = 6:
starting row of full weave
|
0 *-*-*-*-*-*-*
1 *-*-*-*-*-*-*
2 *-*-*-*-*-*-*
3 *-*-*-*-*-*-*
4 *-*-*-*-*-*-*
5 *-*-*-*-*-*-*
6 *-*-*-*-*-*-*
7 *-*-*-*-*-*-* |
S = 7, J = 2, start = 6:
start
|
0 *------*
1 *------*
2 *------*
3 *------*
4 *------*
5 *------*
6 *------*
7 *------*
8 *------*
9 *------* |
S = 4, J = 13, start = 36:
start
|
0 *---*---*---*---*---*---*---*---*---*---*---*---*
1 *---*---*---*---*---*---*---*---*---*---*---*---*
2 *---*---*---*---*---*---*---*---*---*---*---*---*
3 *---*---*---*---*---*---*---*---*---*--
4 *---*---*---*---*---*---*-
5 *---*---*---* |
S = 13, J = 4, start = 36:
start
|
0 *------------*------------*------------*
1 *------------*------------*------------*
2 *------------*------------*------------*
3 *------------*------------*------------*
4 *------------*------------*------------*
5 *------------*------------*------------*
6 *------------*------------*------------*
7 *------------*------------*------------*
8 *------------*------------*------------*
9 *------------*------------*------------*
10 *------------*------------*-----------
11 *------------*------------*-------
12 *------------*------------*---
13 *------------*------------
14 *------------*--------
15 *------------*----
16 *------------*
17 *---------
18 *-----
19 *- |
S = 8, J = 5, start = 28:
start
|
0 *-------*-------*-------*-------*
1 *-------*-------*-------*-------*
2 *-------*-------*-------*-------*
3 *-------*-------*-------*-------*
4 *-------*-------*-------*-------*
5 *-------*-------*-------*-------*
6 *-------*-------*-------*-------*
7 *-------*-------*-------*-------*
8 *-------*-------*-------*-------*
9 *-------*-------*-------*-------*
10 *-------*-------*-------*---
11 *-------*-------*------
12 *-------*-------*-
13 *-------*----
14 *-------
15 *-- |
S = 9, J = 5, start = 32:
start
|
0 *--------*--------*--------*--------*
1 *--------*--------*--------*--------*
2 *--------*--------*--------*--------*
3 *--------*--------*--------*--------*
4 *--------*--------*--------*--------*
5 *--------*--------*--------*--------*
6 *--------*--------*--------*--------*
7 *--------*--------*--------*--------*
8 *--------*--------*--------*--------*
9 *--------*--------*--------*-----
10 *--------*--------*--------*
11 *--------*--------*----
12 *--------*--------
13 *--------*---
14 *-------
15 *-- |
S = 6, J = 7, start = 30:
start
|
0 *-----*-----*-----*-----*-----*-----*
1 *-----*-----*-----*-----*-----*-----*
2 *-----*-----*-----*-----*-----*-----*
3 *-----*-----*-----*-----*-----*-----*
4 *-----*-----*-----*-----*-----*-----*
5 *-----*-----*-----*-----*-----*-----*
6 *-----*-----*-----*-----*-----*-----
7 *-----*-----*-----*-----*----
8 *-----*-----*-----*---
9 *-----*-----*--
10 *-----*-
11 * |
A perfect weave is not possible in all cases. Let's look at another example:
S = 6, J = 4:
0 *-----*-----*-----*
1 *-----*-----*-----*
2 *-----*-----*-----*
3 *-----*-----*-----*
4 ^ *-^---*-----*-----*
5 | ^ | *-^---*-----*-----*
OUCH! ^ | ^
| | |
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:
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:
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 ^ in our diagrams, and correct rows by *:
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.
We can work out the greatest common divisor of a pair of natural numbers using Euler's algorithm:
Start with the two numbers: (e.g.) 9, 24
Swap them if necessary so that the larger one comes first: 24, 9
Subtract the second number from the first: 15, 9
Repeat until the first number becomes smaller: 6, 9
Swap the numbers again, so the larger one comes first: 9, 6
Subtract again: 3, 6
Swap: 6, 3
Subtract: 3, 3
And again: 0, 3
When one of the numbers becomes 0, the other number is the GCD of the two numbers you started with.
These repeated subtractions can be done with C's % operator, so we can write this in C as follows:
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;
} |
gcd(S,J) will feature quite prominently in our weaving algorithm.
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).)
S = 8, J = 6, G = 2:
0 *-------*-------*-------*-------*-------* 1 *-------*-------*-------*-------*-------* 2 *-------*-------*-------*-------*-------* 3 *-------*-------*-------*-------*-------* 4 ^-------^-------^-------*-------*-------* 5 ^-------^-------^-------*-------*-------* |
In this case, jets 0, 1 and 2 of pass p + 4 collide with jets 3, 4 and 5 of pass 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:
0 *-------*-------* - - - 1 *-------*-------* - - - 2 *-------*-------* - - - 3 *-------*-------* - - - 4 *-------*-------* - - - 5 *-------*-------* - - - |
If we left the paper advance alone, we'd get a sparse weave; only one row can be printed every G rows:
0 *-------*-------* - - -
1 *-------*-------* - - -
2 *-------*-------* - - -
3 *-------*-------* - - -
4 *-------*-------* - - -
5 *-------*-------* - - -
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^
These rows need filling in. |
The rows that would have been printed by the jets we've now omitted (shown as -) are printed by other jets on later passes.
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).
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.
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.
Another example:
S = 6, J = 9, G = 3, C = S / G = 2:
0 *-----*-----*-----*-----*-----*-----*-----*-----*
1 *-----*-----*-----*-----*-----*-----*-----*-----*
2 ^-----^-----^-----^-----^-----^-----*-----*-----*
3 ^-----^-----^-----^-----^-----^-----*-----*-----*
4 ^-----^-----^-----^-----^-----^-----*-----
5 ^-----^-----^-----^-----^-----^--
^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^
These rows need filling in. |
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.
S = 9, J = 6, G = 3, C = 3:
0 *--------*--------*--------*--------*--------* 1 *--------*--------*--------*--------*--------* 2 *--------*--------*--------*--------*--------* 3 ^--------^--------^--------^--------*--------* 4 ^--------^--------^--------^--------*--------* 5 ^--------^--------^--------^--------*--------* |
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.
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.
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.
Looking back at an example of a perfect weave, we can divide it into pass blocks:
S = 7, J = 2, G = 1:
imaginary extra jet
|
0 *------* * <--start of pass block 0
1 *------* |
2 *------* |
3 *------*|
4 *-----|*
5 *---|--*
6 *-|----*
|
7 *------* <--start of pass block 1
8 *------*
9 *------* |
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.
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.
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.
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.
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).
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).
This gives us a weave pattern looking like this:
S = 6, J = 9, G = 3, B = 2:
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) |
S = 9, J = 6, G = 3, B = 3:
0 *--------*--------*--------*--------*--------* 1 *--------*--------*--------*--------*--------* 2 *--------*--------*--------*--------*--------* 3 *--------*--------*--------*--------*--------* 4 *--------*--------*--------*--------*--------* 5 *--------*--------*--------*--------*--------* 6 *--------*--------*--------*--------*--- 7 *--------*--------*--------*------ 8 *--------*--------*--------* 9 *--------*--------*----- 10 \---/ *--------*-------- 11 small offset *--------*-- 12 *---- |
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:
S = 8, J = 4, G = 4, B = 2:
0 *-------*-------*-------*
1 *-------*-------*-------*
2 *-------*-------*-------*
3 *-------*-------*-------*
4 *-------*-------*-------*
5 *-------*-------*-------*
6 *-------*-------*-------*
7 *-------*-------*-------*
8 *-------*-------*-------*
9 \/ *-------*-------*-------*
very small offset! |
We can plot the offset against the subblock number as follows:
The discontinuity in this plot results in the small offset between passes.
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:
We can visualise this as follows, for G = 10:
and for G = 11:
1 01234567890 0 * 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 0 * 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * |
This gives a weave looking like this:
S = 12, J = 6, G = 6, B = 2:
0 *-----------*-----------*-----------*-----------*-----------* 1 *-----------*-----------*-----------*-----------*-----------* 2 *-----------*-----------*-----------*-----------*-----------* 3 *-----------*-----------*-----------*-----------*--------- 4 *-----------*-----------*-----------*-----------*- 5 *-----------*-----------*-----------*------- 6 *-----------*-----------*-----------* 7 *-----------*-----------*------ 8 *-----------*-----------*-- 9 *-----------*-------- 10 *-----------*---- 11 *---------- 12 *----- |
This method ensures that the offset between passes is always in the range [J - 2, J + 2].
(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.
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.)
Finally, knowing the number of jets J and their separation S, we can calculate the starting row of any given pass p as follows:
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 |
We can simplify this down to the following:
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 |
So the row number of jet j of pass p is
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 |
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:
subblocksperblock = gcd(S, J) subblockoffset = r % subblocksperblock pass = (r - subblockoffset) / J |
If G = 1, we can determine the pass number with this algorithm:
offset = r % J
pass = (r - offset) / J
while (offset % S != 0)
{
pass--
offset += J
}
jet = offset / S |
Generalising, we come up with this algorithm. Given r, S and J:
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 |
Let's look at some examples of imperfect but correct weave patterns:
S = 6, J = 4, GCD = 2, passesperblock = S = 6, passespersubblock = S / G = 6 / 2 = 3:
0 *-----*-----*-----* 1 *-----*-----*-----* 2 *-----*-----*-----* 3 *-----*-----*-----* 4 *-----*-----*-----* 5 *-----*-----*-----* 6 *-----*-----*-----* 7 *-----*-----*-----* 8 *-----*-----*-----* 9 *-----*-----*-----* 10 *-----*-----*-----* 11 *-----*-----*-----* 12 *-----*-----*-----* 13 *-----*-----*-----* 14 *-----*-----*-----* 15 *-----*-----*---- 16 *-----*-----* 17 *-----*-- 18 *----- 19 *- |
S = 8, J = 6, G = 2, passesperblock = S = 8, passespersubblock= S / G = 8 / 2 = 4:
0 *-------*-------*-------*-------*-------* 1 *-------*-------*-------*-------*-------* 2 *-------*-------*-------*-------*-------* 3 *-------*-------*-------*-------*-------* 4 *-------*-------*-------*-------*-------* 5 *-------*-------*-------*-------*-------* 6 *-------*-------*-------*-------*-------* 7 *-------*-------*-------*-------*-- 8 *-------*-------*-------*----- 9 *-------*-------*------- 10 *-------*-------*- 11 *-------*--- 12 *---- |
S = 6, J = 12, G = 6, passesperblock = S = 6, passespersubblock= S / G = 6 / 6 = 1:
0 *-----*-----*-----*-----*-----*-----*-----*-----*-----*-----*-----* 1 *-----*-----*-----*-----*-----*-----*-----*-----*-----*-----*--- 2 *-----*-----*-----*-----*-----*-----*-----*-----*- 3 *-----*-----*-----*-----*-----*-----* 4 *-----*-----*-----*-----*-- 5 *-----*-----*---- 6 *----- |
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.
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.
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.)
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.
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”).
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.
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.
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:
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:
0 *---*---*---*---*---*---*---*---*---* 1 *---*---*---*---*---*---*---*---*---* 2 *---*---*---*---*---*---*---*---*---* 3 *---*---*---*---*---*---*---*---*---* 4 *---*---*---*---*---*---*---*---*---* 5 *---*---*---*---*---*---*---*---*---* 6 *---*---*---*---*---*---*---*---*---* 7 *---*---*---*---*---*---*---*---*---* 8 *---*---*---*---*---*---*---*---*---* 9 *---*---*---*---*---*---*---*---* 10 *---*---*---*---*---*---*--- 11 *---*---*---*---*---*-- 12 *---*---*---*---*- 13 *---*---*---* 14 *---*--- 15 *-- |
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 *s with integers in the range [0…H−1].
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:
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-- |
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:
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- |
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.
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
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--- |
We can calculate the starting row and subpass number of a given pass in this scheme as follows:
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 |
So the row number of jet j of pass p is
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 |
To be continued…