p******************************** * * * sTEPHEN jUDD * * gEORGE tAYLOR * * sTARTED: 7/11/94 * * fINISHED: 7/19/94 * * * * wELL, IF ALL GOES WELL THIS * * PROGRAM WILL ROTATE A CUBE. * * * * tHIS PROGRAM IS INTENDED TO * * ACCOMPANY THE ARTICLE IN * * c=hACKING, jULY 94 ISSUE. * * fOR DETAILS ON THIS PROGRAM, * * READ THE ARTICLE! * * * * wRITE TO US! * * * * UN(BEE)MO * * * * VI * * N(IN)G * * ARE(TH * * E)YOU(O * * NLY) * * * * ASL(ROSE)EEP * * E.E.CUMMINGS * * * * p.s. tHIS WAS WRITTEN USING * * mERLIN 128. wITH A * * LITTLE MODIFICATION IT * * WILL WORK FINE WITH * * mERLIN 64. iF YOU * * DON'T HAVE EITHER... * * WELL, WE ALL HAVE OUR * * LITTLE FAULTS. * ******************************** org $1000 * cONSTANTS buff1 equ $3000 ;fIRST CHARACTER SET buff2 equ $3800 ;sECOND CHARACTER SET buffer equ $a3 ;pRESUMABLY THE TAPE WON'T BE RUNNING x1 equ $fb ;pOINTS FOR DRAWING A LINE y1 equ $fc ;tHESE ZERO PAGE ADDRESSES x2 equ $fd ;DON'T CONFLICT WITH basic y2 equ $fe dx equ $f9 dy equ $fa temp1 equ $fb ;oF COURSE, COULD CONFLICT WITH X1 temp2 equ $fc ;tEMPORARY VARIABLES acc equ $fb ;tHESE FOUR VARIABLES ARE USED aux equ $fc ;BY THE MULTIPLICATION ROUTINE ext equ $fd rem equ $fe ztemp equ $02 ;uSED FOR BUFFER SWAP. dON'T TOUCH. angmax equ 120 ;tHERE ARE 2*PI/ANGMAX ANGLES offset equ 6 ;fLOAT OFFSET: X=XACTUAL*2^OFFSET * vic vmcsb equ $d018 bkgnd equ $d020 border equ $d021 sstart equ 1344 ;ROW 9 IN SCREEN MEMORY AT 1024 * kERNAL chrout equ $ffd2 getin equ $ffe4 *** mACROS move mac lda ]1 sta ]2 <<< getkey mac ;wAIT FOR A KEYPRESS wait jsr getin cmp #00 beq wait <<< debug mac ;pRINT A CHARACTER do 0 ;dON'T ASSEMBLE lda #]1 jsr chrout >>> getkey ;aND WAIT TO CONTINUE cmp #'S' ;mY SECRECT SWITCH KEY bne l1 jsr cleanup jmp done l1 cmp #'X' ;mY SECRET ABORT KEY bne done jmp cleanup fin done <<< debuga mac do 0 lda ]1 sta 1024 fin donea <<< setbuf mac ;pUT BUFFERS WHERE THEY CAN BE HURT lda #00 sta buffer lda ztemp ;ZTEMP CONTAINS THE HIGH BYTE HERE sta buffer+1 <<< *------------------------------- lda #$00 sta bkgnd sta border lda vmcsb and #%00001111 ;sCREEN MEMORY TO 1024 ora #%00010000 sta vmcsb ldy #00 lda #ttext sta temp2 jmp title ttext hex 9305111111 ;CLEAR SCREEN, WHITE, CRSR DN txt ' CUBE3D',0D,0D txt ' BY',0D hex 9f ;CYAN txt ' STEPHEN JUDD' hex 99 txt ' GEORGE TAYLOR',0D,0D hex 9b txt ' CHECK OUT THE JULY 94 ISSUE OF',0D hex 96 txt ' C=HACKING' hex 9b txt ' FOR MORE DETAILS!',0D hex 0d1d1d9e12 txt 'F1/F2',92 txt ' - INC/DEC X-ROTATION',0D hex 1d1d12 txt 'F3/F4',92 txt ' - INC/DEC Y-ROTATION',0D hex 1d1d12 txt 'F5/F6',92 txt ' - INC/DEC Z-ROTATION',0D hex 1d1d12 txt 'F7',92 txt ' RESETS',0D txt ' PRESS Q TO QUIT',0D hex 0d05 txt ' PRESS ANY KEY TO BEGIN',0D hex 00 title lda (temp1),y beq :cont jsr chrout iny bne title inc temp2 jmp title :cont >>> getkey **** sET UP TABLES(?) * tABLES ARE CURRENTLY SET UP IN basic * AND BY THE ASSEMBLER. tables **** cLEAR SCREEN AND SET UP "BITMAP" setup lda #147 jsr chrout lda #sstart ;rOW 9 sta temp1+1 ;sstart POINTS TO ROW 9 lda #00 ldy #00 ldx #00 ;X WILL COUNT 16 ROWS FOR US clc :loop sta (temp1),y iny adc #16 bcc :loop clc lda temp1 adc #40 ;nEED TO ADD 40 TO THE BASE POINTER sta temp1 ;tO JUMP TO THE NEXT ROW lda temp1+1 adc #00 ;tAKE CARE OF CARRIES sta temp1+1 ldy #00 inx txa ;x IS ALSO AN INDEX INTO THE CHARACTER NUMBER cpx #16 bne :loop ;nEED TO DO IT 16 TIMES >>> debug,'2' **** sET UP BUFFERS lda #buff1 sta buffer+1 sta ztemp ;ZTEMP WILL MAKE LIFE SIMPLE FOR US lda vmcsb and #%11110001 ;sTART HERE SO THAT SWAP BUFFERS WILL WORK RIGHT ora #%00001110 sta vmcsb **** sET UP INITIAL VALUES init lda #00 sta dsx sta dsy sta dsz sta sx sta sy sta sz >>> debug,'4' *------------------------------- * mAIN LOOP **** gET KEYPRESS main kpress jsr getin cmp #133 ;f1? bne :f2 lda dsx cmp #angmax/2 ;nO MORE THAN PI beq :cont inc dsx ;OTHERWISE INCREASE X-ROTATION jmp :cont :f2 cmp #137 ;f2? bne :f3 lda dsx beq :cont dec dsx jmp :cont :f3 cmp #134 bne :f4 lda dsy cmp #angmax/2 beq :cont inc dsy ;iNCREASE Y-ROTATION jmp :cont :f4 cmp #138 bne :f5 lda dsy beq :cont dec dsy jmp :cont :f5 cmp #135 bne :f6 lda dsz cmp #angmax/2 beq :cont inc dsz ;Z-ROTATION jmp :cont :f6 cmp #139 bne :f7 lda dsz beq :cont dec dsz jmp :cont :f7 cmp #136 bne :q jmp init :q cmp #'Q' ;Q QUITS bne :cont jmp cleanup :cont * >>> debug,'5' **** uPDATE ANGLES update clc lda sx adc dsx cmp #angmax ;aRE WE >= MAXIMUM ANGLE? bcc :cont1 sbc #angmax :iF SO, RESET :cont1 sta sx clc lda sy adc dsy cmp #angmax bcc :cont2 sbc #angmax ;sAME DEAL :cont2 sta sy clc lda sz adc dsz cmp #angmax bcc :cont3 sbc #angmax :cont3 sta sz **** rOTATE COORDINATES rotate *** fIRST, CALCULATE T1,T2,...,T10 ** tWO MACROS TO SIMPLIFY OUR LIFE adda mac ;aDD TWO ANGLES TOGETHER clc lda ]1 adc ]2 cmp #angmax ;iS THE SUM > 2*PI? bcc done sbc #angmax ;iF SO, SUBTRACT 2*PI done <<< suba mac ;sUBTRACT TWO ANGLES sec lda ]1 sbc ]2 bcs done adc #angmax ;oOPS, WE NEED TO ADD 2*PI done <<< ** nOW CALCULATE T1,T2,ETC. >>> suba,sy;sz sta t1 ;T1=SY-SZ >>> adda,sy;sz sta t2 ;T2=SY+SZ >>> adda,sx;sz sta t3 ;T3=SX+SZ >>> suba,sx;sz sta t4 ;T4=SX-SZ >>> adda,sx;t2 sta t5 ;T5=SX+T2 >>> suba,sx;t1 sta t6 ;T6=SX-T1 >>> adda,sx;t1 sta t7 ;T7=SX+T1 >>> suba,t2;sx sta t8 ;T8=T2-SX >>> suba,sy;sx sta t9 ;T9=SY-SX >>> adda,sx;sy sta t10 ;T10=SX+SY * eT VOILA! *** nEXT, CALCULATE a,b,c,...,i ** aNOTHER USEFUL LITTLE MACRO div2 mac ;dIVIDE A SIGNED NUMBER BY 2 ;iT IS ASSUMED THAT THE NUMBER bpl pos ;IS IN THE ACCUMULATOR clc eor #$ff ;wE NEED TO UN-NEGATIVE THE NUMBER adc #01 ;BY TAKING IT'S COMPLEMENT lsr ;DIVIDE BY TWO clc eor #$ff adc #01 ;mAKE IT NEGATIVE AGAIN jmp donediv pos lsr ;nUMBER IS POSITIVE donediv <<< mul2 mac ;mULTIPLY A SIGNED NUMBER BY 2 bpl posm clc eor #$ff adc #$01 asl clc eor #$ff adc #$01 jmp donemul posm asl donemul <<< ** nOTE THAT WE ARE CURRENTLY MAKING A MINOR LEAP ** OF FAITH THAT NO OVERFLOWS WILL OCCUR. :calca clc ldx t1 lda cos,x ldx t2 adc cos,x sta a11 ;a=(COS(T1)+COS(T2))/2 :calcb ldx t1 lda sin,x sec ldx t2 sbc sin,x sta b12 ;b=(SIN(T1)-SIN(T2))/2 :calcc ldx sy lda sin,x >>> mul2 sta c13 ;c=SIN(SY) :calcd sec ldx t8 lda cos,x ldx t7 sbc cos,x sec ldx t5 sbc cos,x clc ldx t6 adc cos,x ;dI=(COS(T8)-COS(T7)+COS(T6)-COS(T5))/2 >>> div2 clc ldx t3 adc sin,x sec ldx t4 sbc sin,x sta d21 ;d=(SIN(T3)-SIN(T4)+dI)/2 :calce sec ldx t5 lda sin,x ldx t6 sbc sin,x sec ldx t7 sbc sin,x sec ldx t8 sbc sin,x ;eI=(SIN(T5)-SIN(T6)-SIN(T7)-SIN(T8))/2 >>> div2 clc ldx t3 adc cos,x clc ldx t4 adc cos,x sta e22 ;e=(COS(T3)+COS(T4)+eI)/2 :calcf ldx t9 lda sin,x sec ldx t10 sbc sin,x sta f23 ;f=(SIN(T9)-SIN(T10))/2 :calcg ldx t6 lda sin,x sec ldx t8 sbc sin,x sec ldx t7 sbc sin,x sec ldx t5 sbc sin,x ;gI=(SIN(T6)-SIN(T8)-SIN(T7)-SIN(T5))/2 >>> div2 clc ldx t4 adc cos,x sec ldx t3 sbc cos,x sta g31 ;g=(COS(T4)-COS(T3)+gI)/2 >>> debuga,g31 >>> debug,'G' :calch clc ldx t6 lda cos,x ldx t7 adc cos,x sec ldx t5 sbc cos,x sec ldx t8 sbc cos,x ;hI=(COS(T6)+COS(T7)-COS(T5)-COS(T8))/2 >>> div2 clc ldx t3 adc sin,x clc ldx t4 adc sin,x sta h32 ;h=(SIN(T3)+SIN(T4)+hI)/2 :whew clc ldx t9 lda cos,x ldx t10 adc cos,x sta i33 ;i=(COS(T9)+COS(T10))/2 ** iT'S ALL DOWNHILL FROM HERE. ** rOTATE, PROJECT, AND STORE THE POINTS downhill lda a11 ;tHIS IS GETTING TO BE A REAL MESS sta ta lda b12 ;tHE REASON THIS IS DONE sta tb ;IS TO MAKE THE CODE A LITTLE lda c13 ;EASIER TO READ (AND DEBUG!) sta tc lda d21 ;tHESE ARE ALL TEMPORARY LOCATIONS sta td ;uSED BY THE PROJECTION SUBROUTINE. lda e22 sta te ;oTHERWISE, THERE WOULD BE EIGHT lda f23 ;LONG ROUTINES HERE. sta tf lda g31 ;bUT IT WOULD BE SIGNIFICANTLY FASTER sta tg lda h32 sta th lda i33 sta ti * a NEAT MACRO neg mac ;cHANGE THE SIGN OF A TWO'S COMPLEMENT clc lda ]1 ;NUMBER. eor #$ff adc #$01 <<< * p1=[1 1 1] jsr project ;uNROLL THIS WHOLE THING ldx tx1 ;(SORRY ABOUT THESE TWO LINES) ldy ty1 ;(SEE project FOR REASON WHY) stx p1x ;fOR A PRETTY BIG SPEED INCREASE! sty p1y * p2=[1 -1 1] >>> neg,b12 ;cHANGE THESE ELEMENTS sta tb >>> neg,e22 ;sINCE Y IS NOW -1 sta te >>> neg,h32 sta th jsr project ldx tx1 ldy ty1 stx p2x sty p2y * p3=[-1 -1 1] >>> neg,a11 sta ta >>> neg,d21 sta td >>> neg,g31 sta tg jsr project ldx tx1 ldy ty1 stx p3x sty p3y * p4=[-1 1 1] lda b12 sta tb lda e22 sta te lda h32 sta th jsr project ldx tx1 ldy ty1 stx p4x sty p4y * p8=[-1 1 -1] >>> neg,c13 sta tc >>> neg,f23 sta tf >>> neg,i33 sta ti jsr project ldx tx1 ldy ty1 stx p8x sty p8y * p7=[-1 -1 -1] >>> neg,b12 sta tb >>> neg,e22 sta te >>> neg,h32 sta th jsr project ldx tx1 ldy ty1 stx p7x sty p7y * p6=[1 -1 -1] lda a11 sta ta lda d21 sta td lda g31 sta tg jsr project ldx tx1 ldy ty1 stx p6x sty p6y * p5=[1 1 -1] lda b12 sta tb lda e22 sta te lda h32 sta th jsr project ldx tx1 ldy ty1 stx p5x sty p5y **** cLEAR BUFFER >>> setbuf clrbuf lda #$00 ;pRETTY STRAIGHTFORWARD, ldx #$08 ;i THINK ldy #$00 :loop sta (buffer),y iny bne :loop inc buffer+1 dex bne :loop lda buffer+1 **** fINALLY, DRAW THE LINES. lda p1x ;[1 1 1] sta tx1 lda p1y sta ty1 lda p2x ;[1 -1 1] sta tx2 lda p2y sta ty2 jsr draw ;fIRST LINE lda p3x ;[-1 -1 1] sta tx1 lda p3y sta ty1 jsr draw ;sECOND LINE lda p4x ;[-1 1 1] sta tx2 lda p4y sta ty2 jsr draw ;tHIRD LINE lda p1x ;[1 1 1] sta tx1 lda p1y sta ty1 jsr draw ;fOURTH LINE. oNE FACE DONE. lda p5x ;[1 1 -1] sta tx2 lda p5y sta ty2 jsr draw ;fIVE lda p6x ;[1 -1 -1] sta tx1 lda p6y sta ty1 jsr draw ;sIX lda p2x ;[1 -1 1] sta tx2 lda p2y sta ty2 jsr draw ;sEVEN lda p7x ;[-1 -1 -1] sta tx2 lda p7y sta ty2 jsr draw ;eIGHT lda p3x ;[-1 -1 1] sta tx1 lda p3y sta ty1 jsr draw ;nINE lda p8x ;[-1 1 -1] sta tx1 lda p8y sta ty1 jsr draw ;tEN lda p4x ;[-1 1 1] sta tx2 lda p4y sta ty2 jsr draw ;eLEVEN lda p5x ;[1 1 -1] sta tx2 lda p5y sta ty2 jsr draw ;tWELVE! **** sWAP BUFFERS swapbuf lda vmcsb eor #$02 ;pRETTY TRICKY, EH? sta vmcsb lda #$08 eor ztemp ;ZTEMP=HIGH BYTE JUST FLIPS sta ztemp ;BETWEEN $30 AND $38 jmp main ;aROUND AND AROUND WE GO... *------------------------------- * tHIS SUBROUTINE CALCULATES THE PROJECTION OF x AND y project clc lda tg adc th clc adc ti ;tHIS IS ROTATED z clc adc #128 ;wE ARE GOING TO TAKE 128+Z * >>> debuga,aux * >>> debug,'A' tax ;nOW IT IS READY FOR INDEXING lda zdiv,x ;tABLE OF -D/Z sta aux ;tHIS IS FOR THE PROJECTION sta rem ;mULTIPLY CAN CLOBBER aux clc lda ta adc tb clc adc tc sta acc ;tHIS IS ROTATED X jsr smult ;sIGNED MULTIPLY acc*aux/2^offset clc lda acc :cont1 adc #64 ;oFFSET THE COORDINATE * sEE BELOW FOR THE REASON WHY THIS * NEXT INSTRUCTION IS COMMENTED OUT * tax ;nOW x IS X! sta tx1 clc ;dO THE WHOLE THING AGAIN FOR y lda rem sta aux lda td adc te clc adc tf sta acc ;tHIS IS ROTATED Y jsr smult ;sIGNED MULTIPLY acc*aux/2^offset clc lda acc :cont2 adc #64 ;oFFSET THE COORDINATE * fOR SOME COMPLETELY UNKNOWN REASON TO ME * THE INSTRUCTION BELOW DOESN'T WORK. sOMEHOW * THE rts IS MODIFYING x AND y??? * tay ;sTORE IN y sta ty1 rts ;i HOPE TO HECK THIS WORKS. *------------------------------- * smult: 8-BIT SIGNED (SORT-OF) MULTIPLY * * acc*aux/2^offset -> [acc, ext] 16-BIT RESULT LO,HI * * nOTE THAT THIS ROUTINE DIVIDES THE END RESULT BY 2^offset * yUP, ANOTHER MACRO. divoff mac ;dIVIDE BY THE FLOAT OFFSET lup offset ;rEPEAT OFFSET TIMES lsr ;a CONTAINS HIGH BYTE ror acc ;acc IS LOW BYTE --^ <<< smult clc lda acc ;fIRST, IS THE RESULT POSITIVE OR NEGATIVE? eor aux bmi :neg lda acc ;tHEY ARE EITHER BOTH NEGATIVE OR bpl :cont1 ;BOTH POSITIVE eor #$ff ;iN THIS CASE, MAKE THEM adc #$01 ;BOTH POSITIVE! sta acc >>> neg,aux ;lITTLE MACRO USED EARLIER. :cont1 lda #00 ;mULTIPLY THE TWO NUMBERS ldy #$09 ]loop lsr ;rEAD THE ARTICLE FOR DETAILS. ror acc bcc :mult1 ;oR FIGURE IT OUT YOURSELF! clc adc aux :mult1 dey bne ]loop >>> divoff ;rEMOVE THIS LINE FOR A GENERAL MULTIPLY sta ext rts :neg lda acc ;oNE OF THE TWO IS NEGATIVE bmi :cont2 >>> neg,aux ;oTHERWISE IT'S aux jmp :cont3 :cont2 eor #$ff ;tAKE TWO'S COMPLEMENT adc #$01 sta acc :cont3 lda #00 ;mULTIPLY ldy #$09 ]loop2 lsr ror acc bcc :mult2 clc adc aux :mult2 dey bne ]loop2 >>> divoff ;aGAIN, DIVIDE BY THE OFFSET sta ext lda acc bpl :ok ;sOMETHING IS REALLY WRONG IF THIS IS NEGATIVE. jsr choke :ok eor #$ff ;oTHERISE, EVERYTHING RELEVANT SHOULD adc #$01 ;BE COMPLETELY IN THE LOW BYTE. sta acc rts ;i HOPE... *------------------------------- * gENERAL QUESTIONABLE-VALUE ERROR PROCEDURE choke ldx #00 :loop lda :ctext,x beq :done jsr chrout inx jmp :loop :done rts :ctext hex 0d ;cr txt 'SOMETHING CHOKED :(' hex 0d00 *------------------------------- * dRAWIN' A LINE. a FAHN LAHN. *** sOME USEFUL MACROS plotpx mac ;PLOT A POINT IN X pha ;uSE THIS ONE EVERY TIME lda bitp,x ;x IS INCREASED bpl c1 eor buffer sta buffer bmi c2 inc buffer+1 c2 lda #%10000000 c1 ora (buffer),y sta (buffer),y pla ;nEED TO SAVE a! <<< plotpy mac ;pLOT A POINT IN Y: SIMPLER AND NECESSARY! pha ;uSE THIS ONE WHEN YOU JUST INCREASE y lda bitp,x ;BUT x DOESN'T CHANGE ora (buffer),y sta (buffer),y pla <<< cinit mac ;mACRO TO INITIALIZE THE COUNTER lda ]1 ;DX OR DY lsr eor #$ff ;(nOT TWO'S COMPLEMENT) adc #$01 ;a = 256-DX/2 OR 256-DY/2 <<< ;tHE DX/2 MAKES A NICER LOOKING LINE xstep mac ;mACRO TO TAKE A STEP IN x xloop inx adc dy bcc l1 * dO WE USE iny OR dey HERE? if i,]1 ;iF THE FIRST CHARACTER IS AN 'i' iny else dey fin sbc dx l1 >>> plotpx ;aLWAYS TAKE A STEP IN x cpx x2 bne xloop <<< ystep mac ;sAME THING, BUT FOR y yloop if i,]1 iny else dey clc ;vERY IMPORTANT! fin adc dx bcc l2 inx ;aLWAYS INCREASE x sbc dy >>> plotpx jmp l3 l2 >>> plotpy ;wE ONLY INCREASED y l3 cpy y2 bne yloop <<< **** iNITIAL LINE SETUP draw >>> move,tx1;x1 ;mOVE STUFF INTO ZERO PAGE >>> move,tx2;x2 ;wHERE IT CAN BE MODIFIED >>> move,ty1;y1 >>> move,ty2;y2 >>> setbuf ;nOW WE CAN CLOBBER THE BUFFER sec ;mAKE SURE X1Y1? lda y1 ;oTHERWISE DY=Y1-Y2 sbc y2 :cont2 sta dy cmp dx ;wHO'S BIGGER: DY OR DX? bcs stepiny ;iF DY, WE NEED TO TAKE BIG STEPS IN Y stepinx ldy y1 ;x IS ALREADY SET TO X1 lda bitp,x ;pLOT THE FIRST POINT ora (buffer),y sta (buffer),y >>> cinit,dx ;iNITIALIZE THE COUNTER cpy y2 bcs xdecy ;dO WE STEP FORWARDS OR BACKWARDS IN y? xincy >>> xstep,iny rts xdecy >>> xstep,dey rts stepiny ldy y1 ;wELL, A LITTLE REPETITION NEVER HURT ANYONE lda bitp,x ora (buffer),y sta (buffer),y >>> cinit,dy cpy y2 bcs ydecy yincy >>> ystep,iny rts ydecy >>> ystep,dey rts *------------------------------- * cLEAN UP cleanup lda vmcsb ;sWITCH CHAR ROM BACK IN and #%11110101 ;DEFAULT sta vmcsb rts ;BYE! *------------------------------- * sOME VARIABLES tx1 ds 1 ty1 ds 1 tx2 ds 1 ty2 ds 1 p1x ds 1 ;tHESE ARE TEMPORARY STORAGE p1y ds 1 ;uSED IN PLOTTING THE PROJECTION p2x ds 1 p2y ds 1 ;tHEY ARE HERE SO THAT WE p3x ds 1 ;DON'T HAVE TO RECALCULATE THEM. p3y ds 1 p4x ds 1 ;tHEY MAKE LIFE EASY. p4y ds 1 p5x ds 1 ;wHY ARE YOU LOOKING AT ME LIKE THAT? p5y ds 1 ;dON'T YOU TRUST ME? p6x ds 1 p6y ds 1 ;hAVING ANOTHER CHILD WASN'T MY IDEA. p7x ds 1 p7y ds 1 p8x ds 1 p8y ds 1 dsx ds 1 ;dsx IS THE INCREMENT FOR ROTATING AROUND X dsy ds 1 ;sIMILAR FOR dsy, dsz dsz ds 1 sx ds 1 ;tHESE ARE THE ACTUAL ANGLES IN X Y AND Z sy ds 1 sz ds 1 t1 ds 1 ;tHESE ARE USED IN THE ROTATION t2 ds 1 t3 ds 1 ;sEE THE ARTICLE FOR MORE DETAILS t4 ds 1 t5 ds 1 t6 ds 1 t7 ds 1 t8 ds 1 t9 ds 1 t10 ds 1 a11 ds 1 ;tHESE ARE THE ELEMENTS OF THE ROTATION MATRIX b12 ds 1 ;xyz c13 ds 1 d21 ds 1 ;tHE NUMBER DENOTES (ROW,COLUMN) e22 ds 1 f23 ds 1 g31 ds 1 h32 ds 1 i33 ds 1 ta ds 1 ;tHESE ARE TEMPORARY LOCATIONS tb ds 1 ;FOR USE BY THE PROJECTION ROUTINE tc ds 1 td ds 1 te ds 1 tf ds 1 tg ds 1 th ds 1 ti ds 1 *------------------------------- * sET UP BIT TABLE ds ^ ;cLEAR TO END OF PAGE ;sO THAT TABLES START ON A PAGE BOUNDARY bitp lup 16 ;128 eNTRIES FOR x dfb %10000000 dfb %01000000 dfb %00100000 dfb %00010000 dfb %00001000 dfb %00000100 dfb %00000010 dfb %00000001 --^ sin ;tABLE OF SINES, 120 BYTES cos equ sin+128 ;tABLE OF COSINES ;bOTH OF THESE TRIG TABLES ARE ;CURRENTLY SET UP FROM basic zdiv equ cos+128 ;dIVISION TABLE