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diff --git a/doc/developer/reference-html/x1734.html b/doc/developer/reference-html/x1734.html index 8c52bfd..dfe36a3 100644 --- a/doc/developer/reference-html/x1734.html +++ b/doc/developer/reference-html/x1734.html @@ -1,14 +1,14 @@ -<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN""http://www.w3.org/TR/html4/loose.dtd"> +<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> <HTML ><HEAD ><TITLE >Weaving algorithms</TITLE ><META NAME="GENERATOR" -CONTENT="Modular DocBook HTML Stylesheet Version 1.79"><LINK +CONTENT="Modular DocBook HTML Stylesheet Version 1.7"><LINK REL="HOME" TITLE="The Developer's Guide to Gutenprint" -HREF="index.html"><LINK +HREF="book1.html"><LINK REL="UP" TITLE="Weaving for inkjet printers" HREF="c1717.html"><LINK @@ -17,14 +17,11 @@ TITLE="Weaving for inkjet printers" HREF="c1717.html"><LINK REL="NEXT" TITLE="Dithering" -HREF="c1968.html"></HEAD +HREF="c1968.html"><META +http-equiv="Content-Type" +content="text/html; charset=UTF-8"></HEAD ><BODY CLASS="sect1" -BGCOLOR="#FFFFFF" -TEXT="#000000" -LINK="#0000FF" -VLINK="#840084" -ALINK="#0000FF" ><DIV CLASS="NAVHEADER" ><TABLE @@ -74,11 +71,11 @@ CLASS="sect1" CLASS="sect1" ><A NAME="AEN1734" ->6.2. Weaving algorithms</A +>Weaving algorithms</A ></H1 ><P > I considered a few algorithms to perform the weave. The first - one I devised let me use only (jets − + 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 @@ -88,13 +85,13 @@ NAME="AEN1734" ><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 1440720 mode, we're printing two + 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”. + “MicroWeave”. </P ><P > Yet another complication is how to get near the top and bottom @@ -128,7 +125,7 @@ CLASS="sect2" CLASS="sect2" ><A NAME="AEN1740" ->6.2.1. Simple weaving algorithms</A +>Simple weaving algorithms</A ></H2 ><P > The initial problem is to calculate the starting position of @@ -141,12 +138,12 @@ NAME="AEN1740" </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 + 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 + 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. @@ -160,7 +157,7 @@ NAME="AEN1740" > 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 = + × J rows and print the next row. For example, if J = 7 and S = 4, this method can be illustrated like this: </P ><DIV @@ -170,6 +167,12 @@ CLASS="informalexample" ><A NAME="AEN1747" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >pass number @@ -183,6 +186,9 @@ CLASS="screen" 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 @@ -204,7 +210,7 @@ CLASS="computeroutput" CLASS="computeroutput" >0</SAMP > and starts at row 0. Each - subsequent pass p starts at row p S J. 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.) @@ -215,7 +221,7 @@ CLASS="computeroutput" 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 + “pass block”: we print extra passes to fill in the empty rows. A naive implementation might look like this: </P @@ -226,6 +232,12 @@ CLASS="informalexample" ><A NAME="AEN1754" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *---*---*---*---*---*---* @@ -240,12 +252,15 @@ CLASS="screen" 9 *---*---*---*---*---* 10 *---*---*---*---*--- 11 *---*---*---*---*--</PRE +></TD +></TR +></TABLE ><P ></P ></DIV ><P > (Now you can see why this process is called - “weaving”!) + “weaving”!) </P ></DIV ><DIV @@ -254,7 +269,7 @@ CLASS="sect2" CLASS="sect2" ><A NAME="AEN1757" ->6.2.2. Perfect weaving</A +>Perfect weaving</A ></H2 ><P > This simple weave pattern prints every row, but will give @@ -271,15 +286,15 @@ NAME="AEN1757" 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 +> 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: +> Here's what this “perfect” weave looks like: </P ><DIV CLASS="informalexample" @@ -288,6 +303,12 @@ CLASS="informalexample" ><A NAME="AEN1763" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" > start of full weave @@ -304,13 +325,16 @@ CLASS="screen" 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 + 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 @@ -320,9 +344,18 @@ CLASS="informalexample" ><A NAME="AEN1766" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" ->start = (S − 1) (J − 1)</PRE +>start = (S − 1) × (J − 1)</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -341,8 +374,14 @@ CLASS="informalexample" NAME="AEN1770" ></A ><P -> S = 2, J = 7, start = (2−1) (7−1) = 6: +> 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 @@ -355,6 +394,9 @@ CLASS="screen" 5 *-*-*-*-*-*-* 6 *-*-*-*-*-*-* 7 *-*-*-*-*-*-*</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -368,6 +410,12 @@ NAME="AEN1773" ><P > S = 7, J = 2, start = 6: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" > start @@ -382,6 +430,9 @@ CLASS="screen" 7 *------* 8 *------* 9 *------*</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -395,6 +446,12 @@ NAME="AEN1776" ><P > S = 4, J = 13, start = 36: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" > start @@ -405,6 +462,9 @@ CLASS="screen" 3 *---*---*---*---*---*---*---*---*---*-- 4 *---*---*---*---*---*---*- 5 *---*---*---*</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -418,6 +478,12 @@ NAME="AEN1779" ><P > S = 13, J = 4, start = 36: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" > start @@ -442,6 +508,9 @@ CLASS="screen" 17 *--------- 18 *----- 19 *-</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -455,6 +524,12 @@ NAME="AEN1782" ><P > S = 8, J = 5, start = 28: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" > start @@ -475,6 +550,9 @@ CLASS="screen" 13 *-------*---- 14 *------- 15 *--</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -488,6 +566,12 @@ NAME="AEN1785" ><P > S = 9, J = 5, start = 32: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" > start @@ -508,6 +592,9 @@ CLASS="screen" 13 *--------*--- 14 *------- 15 *--</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -521,6 +608,12 @@ NAME="AEN1788" ><P > S = 6, J = 7, start = 30: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" > start @@ -537,6 +630,9 @@ CLASS="screen" 9 *-----*-----*-- 10 *-----*- 11 *</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -547,7 +643,7 @@ CLASS="sect2" CLASS="sect2" ><A NAME="AEN1791" ->6.2.3. Weaving collisions</A +>Weaving collisions</A ></H2 ><P > A perfect weave is not possible in all cases. Let's look at @@ -563,6 +659,12 @@ NAME="AEN1794" ><P > S = 6, J = 4: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *-----*-----*-----* @@ -573,6 +675,9 @@ CLASS="screen" 5 | ^ | *-^---*-----*-----* OUCH! ^ | ^ | |</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -589,9 +694,18 @@ CLASS="informalexample" ><A NAME="AEN1798" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" ->row(p, j) = (p J) + (j S)</PRE +>row(p, j) = (p × J) + (j × S)</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -606,10 +720,19 @@ CLASS="informalexample" ><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 +>row(0, 2) = (0 × 4) + (2 × 6) = 12 +row(3, 0) = (3 × 4) + (0 × 6) = 12</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -636,6 +759,12 @@ NAME="AEN1806" ><P > S = 6, J = 4: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *-----*-----*-----* @@ -644,6 +773,9 @@ CLASS="screen" 3 ^-----^-----*-----* 4 ^-----^-----*-----* 5 ^-----^-----*-----*</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -654,7 +786,7 @@ CLASS="sect2" CLASS="sect2" ><A NAME="AEN1809" ->6.2.4. What makes a “perfect” weave?</A +>What makes a “perfect” weave?</A ></H2 ><P > So what causes the perfect weave cases to be perfect, and @@ -730,9 +862,9 @@ TYPE="1" ></LI ></OL ><P -> These repeated subtractions can be done with C's <TT +> These repeated subtractions can be done with C's <VAR CLASS="literal" ->%</TT +>%</VAR > operator, so we can write this in C as follows: </P @@ -743,6 +875,12 @@ CLASS="informalexample" ><A NAME="AEN1836" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="programlisting" >unsigned int @@ -757,18 +895,21 @@ gcd(unsigned int x, unsigned int y) } return y; }</PRE +></TD +></TR +></TABLE ><P ></P ></DIV ><P -> <TT +> <VAR CLASS="literal" ->gcd(S,J)</TT +>gcd(S,J)</VAR > will feature quite prominently in our weaving algorithm. </P ><P -> If 0 ≤ j < J, there should only be a single pair (p, j) +> 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).) @@ -783,6 +924,12 @@ NAME="AEN1841" ><P > S = 8, J = 6, G = 2: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *-------*-------*-------*-------*-------* @@ -791,6 +938,9 @@ CLASS="screen" 3 *-------*-------*-------*-------*-------* 4 ^-------^-------^-------*-------*-------* 5 ^-------^-------^-------*-------*-------*</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -812,6 +962,12 @@ CLASS="informalexample" ><A NAME="AEN1846" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *-------*-------* - - - @@ -820,6 +976,9 @@ CLASS="screen" 3 *-------*-------* - - - 4 *-------*-------* - - - 5 *-------*-------* - - -</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -834,6 +993,12 @@ CLASS="informalexample" ><A NAME="AEN1849" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *-------*-------* - - - @@ -844,6 +1009,9 @@ CLASS="screen" 5 *-------*-------* - - - ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ These rows need filling in.</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -857,19 +1025,19 @@ CLASS="computeroutput" </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 + 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 + ×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 +> 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 + × 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 @@ -877,7 +1045,7 @@ CLASS="computeroutput" > 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 + 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 @@ -894,6 +1062,12 @@ NAME="AEN1857" ><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 *-----*-----*-----*-----*-----*-----*-----*-----* @@ -904,11 +1078,14 @@ CLASS="screen" 5 ^-----^-----^-----^-----^-----^-- ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ ^^ These rows need filling in.</PRE +></TD +></TR +></TABLE ><P ></P ></DIV ><P -> In this case, the first J − (J / G) = 9 − (9 / +> 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. @@ -923,6 +1100,12 @@ NAME="AEN1861" ><P > S = 9, J = 6, G = 3, C = 3: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" > 0 *--------*--------*--------*--------*--------* @@ -931,6 +1114,9 @@ CLASS="screen" 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 @@ -967,6 +1153,12 @@ NAME="AEN1868" ><P > S = 7, J = 2, G = 1: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" > imaginary extra jet @@ -982,25 +1174,28 @@ CLASS="screen" 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 + 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 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 +> 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 @@ -1029,7 +1224,7 @@ CLASS="screen" 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 + + relative to the start of its pass block) is p × J + floor(p / B). </P ><P @@ -1045,6 +1240,12 @@ NAME="AEN1878" ><P > S = 6, J = 9, G = 3, B = 2: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *-----*-----*-----*-----*-----*-----*-----*-----* @@ -1060,6 +1261,9 @@ CLASS="screen" | | 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 @@ -1073,6 +1277,12 @@ NAME="AEN1881" ><P > S = 9, J = 6, G = 3, B = 3: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *--------*--------*--------*--------*--------* @@ -1088,6 +1298,9 @@ CLASS="screen" 10 \---/ *--------*-------- 11 small offset *--------*-- 12 *----</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1107,6 +1320,12 @@ NAME="AEN1885" ><P > S = 8, J = 4, G = 4, B = 2: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *-------*-------*-------* @@ -1120,6 +1339,9 @@ CLASS="screen" 8 *-------*-------*-------* 9 \/ *-------*-------*-------* very small offset!</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1134,6 +1356,12 @@ CLASS="informalexample" ><A NAME="AEN1889" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >subblock number @@ -1148,6 +1376,9 @@ CLASS="screen" 1 * 2 * 3 *</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1168,10 +1399,19 @@ CLASS="informalexample" ><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 @@ -1185,6 +1425,12 @@ CLASS="informalexample" ><A NAME="AEN1896" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0123456789 @@ -1208,6 +1454,9 @@ CLASS="screen" 7 * 8 * 9 *</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1221,6 +1470,12 @@ CLASS="informalexample" ><A NAME="AEN1899" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" > 1 @@ -1247,6 +1502,9 @@ CLASS="screen" 8 * 9 * 10 *</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1263,6 +1521,12 @@ NAME="AEN1902" ><P > S = 12, J = 6, G = 6, B = 2: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *-----------*-----------*-----------*-----------*-----------* @@ -1278,6 +1542,9 @@ CLASS="screen" 10 *-----------*---- 11 *---------- 12 *-----</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1289,7 +1556,7 @@ CLASS="screen" > (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 + “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 @@ -1317,6 +1584,12 @@ CLASS="informalexample" ><A NAME="AEN1909" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >passesperblock = S @@ -1330,6 +1603,9 @@ if subpassblock < ceiling(subblocksperblock/2) else subblockoffset = 2*(subblocksperblock-subpassblock)-1 startingrow = passblock * S * J + offsetinpassblock * J + subblockoffset</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1345,6 +1621,12 @@ CLASS="informalexample" ><A NAME="AEN1912" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >subblocksperblock = gcd(S, J) @@ -1354,6 +1636,9 @@ if subpassblock * 2 < subblocksperblock else subblockoffset = 2*(subblocksperblock-subpassblock)-1 startingrow = p * J + subblockoffset</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1367,6 +1652,12 @@ CLASS="informalexample" ><A NAME="AEN1915" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >subblocksperblock = gcd(S, J) @@ -1378,11 +1669,14 @@ subblockoffset(p) 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 +> 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 @@ -1397,11 +1691,20 @@ CLASS="informalexample" ><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 @@ -1416,6 +1719,12 @@ CLASS="informalexample" ><A NAME="AEN1921" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >offset = r % J @@ -1426,6 +1735,9 @@ pass-- offset += J } jet = offset / S</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1440,6 +1752,12 @@ CLASS="informalexample" ><A NAME="AEN1924" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >G = gcd(S, J) @@ -1459,6 +1777,9 @@ subblockretreat = floor(pass / passespersubblock) % G pass -= subblockretreat * passespersubblock pass += subpassblock * passespersubblock jet = (r - subblockoffset - pass * J) / S</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1478,6 +1799,12 @@ NAME="AEN1927" passesperblock = S = 6, passespersubblock = S / G = 6 / 2 = 3: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *-----*-----*-----* @@ -1500,6 +1827,9 @@ CLASS="screen" 17 *-----*-- 18 *----- 19 *-</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1515,6 +1845,12 @@ NAME="AEN1930" passesperblock = S = 8, passespersubblock= S / G = 8 / 2 = 4: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *-------*-------*-------*-------*-------* @@ -1530,6 +1866,9 @@ CLASS="screen" 10 *-------*-------*- 11 *-------*--- 12 *----</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1545,6 +1884,12 @@ NAME="AEN1933" passesperblock = S = 6, passespersubblock= S / G = 6 / 6 = 1: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *-----*-----*-----*-----*-----*-----*-----*-----*-----*-----*-----* @@ -1554,6 +1899,9 @@ CLASS="screen" 4 *-----*-----*-----*-----*-- 5 *-----*-----*---- 6 *-----</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1572,7 +1920,7 @@ CLASS="sect2" CLASS="sect2" ><A NAME="AEN1937" ->6.2.5. Oversampling</A +>Oversampling</A ></H2 ><P > By oversampling, we mean printing on the same row more than @@ -1589,7 +1937,7 @@ NAME="AEN1937" 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 + 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 @@ -1600,18 +1948,18 @@ NAME="AEN1937" ><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 + 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 + 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”). + for horizontal oversampling (the “extra + oversampling”). </P ><P > It is necessary for all the lines printed by a given pass to @@ -1629,10 +1977,10 @@ NAME="AEN1937" 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 (SJ) / H rows. We therefore name a group of H - pass blocks a “band”. Printing one band - involves advancing the paper SJ rows, as a pass + 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 @@ -1654,6 +2002,12 @@ NAME="AEN1946" passesperblock = S = 4, passespersubblock = S/G = 4/1 = 4: </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >0 *---*---*---*---*---*---*---*---*---* @@ -1672,6 +2026,9 @@ CLASS="screen" 13 *---*---*---* 14 *---*--- 15 *--</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1683,7 +2040,7 @@ CLASS="screen" CLASS="computeroutput" >*</SAMP >s - with integers in the range [0…H−1]. + with integers in the range [0…H−1]. </P ><P > Overprinting occurs once per pass block, so we can simply @@ -1698,6 +2055,12 @@ CLASS="informalexample" ><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 @@ -1716,6 +2079,9 @@ CLASS="screen" 13 1---1---1---1 14 1---1--- 15 1--</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1732,6 +2098,12 @@ NAME="AEN1954" 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 @@ -1747,6 +2119,9 @@ CLASS="screen" 10 0---0---0---0---0 11 0---0---0-- 12 1---1-</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1770,6 +2145,12 @@ NAME="AEN1958" passesperblock = S = 4, passespersubblock = S/G = 4/1 = 4 </P +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >Band 0: @@ -1790,6 +2171,9 @@ Band 1: 12 1---1---1---1- 13 1---1---1 14 1---</PRE +></TD +></TR +></TABLE ><P ></P ></DIV @@ -1804,6 +2188,12 @@ CLASS="informalexample" ><A NAME="AEN1962" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >A = floor(J / H) @@ -1817,6 +2207,9 @@ 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 @@ -1829,6 +2222,12 @@ CLASS="informalexample" ><A NAME="AEN1965" ></A +><TABLE +BORDER="0" +BGCOLOR="#E0E0E0" +WIDTH="100%" +><TR +><TD ><PRE CLASS="screen" >A = floor(J / H) @@ -1845,8 +2244,11 @@ 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… +> To be continued… </P ><P ></P @@ -1880,7 +2282,7 @@ WIDTH="34%" ALIGN="center" VALIGN="top" ><A -HREF="index.html" +HREF="book1.html" ACCESSKEY="H" >Home</A ></TD |