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Drawing planets: PLS22

[BBC Master version]

Name: PLS22 [Show more] Type: Subroutine Category: Drawing planets Summary: Draw an ellipse or half-ellipse Deep dive: Drawing ellipses Drawing meridians and equators Drawing craters
Context: See this subroutine in context in the source code References: This subroutine is called as follows: * PL9 (Part 3 of 3) calls PLS22

Draw an ellipse or half-ellipse, to be used for the planet's equator and meridian (in which case we draw half an ellipse), or crater (in which case we draw a full ellipse). The ellipse is defined by a centre point, plus two conjugate radius vectors, u and v, where: u = [ u_x ] v = [ v_x ] [ u_y ] [ v_y ] The individual components of these 2D vectors (i.e. u_x, u_y etc.) are 16-bit sign-magnitude numbers, where the high bytes contain only the sign bit (in bit 7), with bits 0 to 6 being clear. This means that as we store u_x as (XX16 K2), for example, we know that |u_x| = K2. This routine calls BLINE to draw each line segment in the ellipse, passing the coordinates as follows: K6(1 0) = K3(1 0) + u_x * cos(CNT2) + v_x * sin(CNT2) K6(3 2) = K4(1 0) - u_y * cos(CNT2) - v_y * sin(CNT2) The y-coordinates are negated because BLINE expects pixel coordinates but the u and v vectors are extracted from the orientation vector. The y-axis runs in the opposite direction in 3D space to that on the screen, so we need to negate the 3D space coordinates before we can combine them with the ellipse's centre coordinates.
Arguments: K(1 0) The planet's radius K3(1 0) The pixel x-coordinate of the centre of the ellipse K4(1 0) The pixel y-coordinate of the centre of the ellipse (XX16 K2) The x-component of u (i.e. u_x), where XX16 contains just the sign of the sign-magnitude number (XX16+1 K2+1) The y-component of u (i.e. u_y), where XX16+1 contains just the sign of the sign-magnitude number (XX16+2 K2+2) The x-component of v (i.e. v_x), where XX16+2 contains just the sign of the sign-magnitude number (XX16+3 K2+3) The y-component of v (i.e. v_y), where XX16+3 contains just the sign of the sign-magnitude number TGT The number of segments to draw: * 32 for a half ellipse (a meridian) * 64 for a full ellipse (a crater) CNT2 The starting segment for drawing the half-ellipse
.PLS22 LDX #0 \ Set CNT = 0 STX CNT DEX \ Set FLAG = &FF to reset the ball line heap in the call STX FLAG \ to the BLINE routine below .PLL4 LDA CNT2 \ Set X = CNT2 mod 32 AND #31 \ TAX \ So X is the starting segment, reduced to the range 0 \ to 32, so as there are 64 segments in the circle, this \ reduces the starting angle to 0 to 180 degrees, so we \ can use X as an index into the sine table (which only \ contains values for segments 0 to 31) \ \ Also, because CNT2 mod 32 is in the range 0 to 180 \ degrees, we know that sin(CNT2 mod 32) is always \ positive, or to put it another way: \ \ sin(CNT2 mod 32) = |sin(CNT2)| LDA SNE,X \ Set Q = sin(X) STA Q \ = sin(CNT2 mod 32) \ = |sin(CNT2)| LDA K2+2 \ Set A = K2+2 \ = |v_x| JSR FMLTU \ Set R = A * Q / 256 STA R \ = |v_x| * |sin(CNT2)| LDA K2+3 \ Set A = K2+3 \ = |v_y| JSR FMLTU \ Set K = A * Q / 256 STA K \ = |v_y| * |sin(CNT2)| LDX CNT2 \ If CNT2 >= 33 then this sets the C flag, otherwise CPX #33 \ it's clear, so this means that: \ \ * C is clear if the segment starts in the first half \ of the circle, 0 to 180 degrees \ \ * C is set if the segment starts in the second half \ of the circle, 180 to 360 degrees \ \ In other words, the C flag contains the sign bit for \ sin(CNT2), which is positive for 0 to 180 degrees \ and negative for 180 to 360 degrees LDA #0 \ Shift the C flag into the sign bit of XX16+5, so ROR A \ XX16+5 has the correct sign for sin(CNT2) STA XX16+5 \ \ Because we set the following above: \ \ K = |v_y| * |sin(CNT2)| \ R = |v_x| * |sin(CNT2)| \ \ we can add XX16+5 as the high byte to give us the \ following: \ \ (XX16+5 K) = |v_y| * sin(CNT2) \ (XX16+5 R) = |v_x| * sin(CNT2) LDA CNT2 \ Set X = (CNT2 + 16) mod 32 CLC \ ADC #16 \ So we can use X as a lookup index into the SNE table AND #31 \ to get the cosine (as there are 16 segments in a TAX \ quarter-circle) \ \ Also, because the sine table only contains positive \ values, we know that sin((CNT2 + 16) mod 32) will \ always be positive, or to put it another way: \ \ sin((CNT2 + 16) mod 32) = |cos(CNT2)| LDA SNE,X \ Set Q = sin(X) STA Q \ = sin((CNT2 + 16) mod 32) \ = |cos(CNT2)| LDA K2+1 \ Set A = K2+1 \ = |u_y| JSR FMLTU \ Set K+2 = A * Q / 256 STA K+2 \ = |u_y| * |cos(CNT2)| LDA K2 \ Set A = K2 \ = |u_x| JSR FMLTU \ Set P = A * Q / 256 STA P \ = |u_x| * |cos(CNT2)| \ \ The call to FMLTU also sets the C flag, so in the \ following, ADC #15 adds 16 rather than 15 LDA CNT2 \ If (CNT2 + 16) mod 64 >= 33 then this sets the C flag, ADC #15 \ otherwise it's clear, so this means that: AND #63 \ CMP #33 \ * C is clear if the segment starts in the first or \ last quarter of the circle, 0 to 90 degrees or 270 \ to 360 degrees \ \ * C is set if the segment starts in the second or \ third quarter of the circle, 90 to 270 degrees \ \ In other words, the C flag contains the sign bit for \ cos(CNT2), which is positive for 0 to 90 degrees or \ 270 to 360 degrees, and negative for 90 to 270 degrees LDA #0 \ Shift the C flag into the sign bit of XX16+4, so: ROR A \ XX16+4 has the correct sign for cos(CNT2) STA XX16+4 \ \ Because we set the following above: \ \ K+2 = |u_y| * |cos(CNT2)| \ P = |u_x| * |cos(CNT2)| \ \ we can add XX16+4 as the high byte to give us the \ following: \ \ (XX16+4 K+2) = |u_y| * cos(CNT2) \ (XX16+4 P) = |u_x| * cos(CNT2) LDA XX16+5 \ Set S = the sign of XX16+2 * XX16+5 EOR XX16+2 \ = the sign of v_x * XX16+5 STA S \ \ So because we set this above: \ \ (XX16+5 R) = |v_x| * sin(CNT2) \ \ we now have this: \ \ (S R) = v_x * sin(CNT2) LDA XX16+4 \ Set A = the sign of XX16 * XX16+4 EOR XX16 \ = the sign of u_x * XX16+4 \ \ So because we set this above: \ \ (XX16+4 P) = |u_x| * cos(CNT2) \ \ we now have this: \ \ (A P) = u_x * cos(CNT2) JSR ADD \ Set (A X) = (A P) + (S R) \ = u_x * cos(CNT2) + v_x * sin(CNT2) STA T \ Store the high byte in T, so the result is now: \ \ (T X) = u_x * cos(CNT2) + v_x * sin(CNT2) BPL PL42 \ If the result is positive, jump down to PL42 TXA \ The result is negative, so we need to negate the EOR #%11111111 \ magnitude using two's complement, first doing the low CLC \ byte in X ADC #1 TAX LDA T \ And then the high byte in T, making sure to leave the EOR #%01111111 \ sign bit alone ADC #0 STA T .PL42 TXA \ Set K6(1 0) = K3(1 0) + (T X) ADC K3 \ STA K6 \ starting with the low bytes LDA T \ And then doing the high bytes, so we now get: ADC K3+1 \ STA K6+1 \ K6(1 0) = K3(1 0) + (T X) \ = K3(1 0) + u_x * cos(CNT2) \ + v_x * sin(CNT2) \ \ K3(1 0) is the x-coordinate of the centre of the \ ellipse, so we now have the correct x-coordinate for \ our ellipse segment that we can pass to BLINE below LDA K \ Set R = K = |v_y| * sin(CNT2) STA R LDA XX16+5 \ Set S = the sign of XX16+3 * XX16+5 EOR XX16+3 \ = the sign of v_y * XX16+5 STA S \ \ So because we set this above: \ \ (XX16+5 K) = |v_y| * sin(CNT2) \ \ and we just set R = K, we now have this: \ \ (S R) = v_y * sin(CNT2) LDA K+2 \ Set P = K+2 = |u_y| * cos(CNT2) STA P LDA XX16+4 \ Set A = the sign of XX16+1 * XX16+4 EOR XX16+1 \ = the sign of u_y * XX16+4 \ \ So because we set this above: \ \ (XX16+4 K+2) = |u_y| * cos(CNT2) \ \ and we just set P = K+2, we now have this: \ \ (A P) = u_y * cos(CNT2) JSR ADD \ Set (A X) = (A P) + (S R) \ = u_y * cos(CNT2) + v_y * sin(CNT2) EOR #%10000000 \ Store the negated high byte in T, so the result is STA T \ now: \ \ (T X) = - u_y * cos(CNT2) - v_y * sin(CNT2) \ \ This negation is necessary because BLINE expects us \ to pass pixel coordinates, where y-coordinates get \ larger as we go down the screen; u_y and v_y, on the \ other hand, are extracted from the orientation \ vectors, where y-coordinates get larger as we go up \ in space, so to rectify this we need to negate the \ result in (T X) before we can add it to the \ y-coordinate of the ellipse's centre in BLINE BPL PL43 \ If the result is positive, jump down to PL43 TXA \ The result is negative, so we need to negate the EOR #%11111111 \ magnitude using two's complement, first doing the low CLC \ byte in X ADC #1 TAX LDA T \ And then the high byte in T, making sure to leave the EOR #%01111111 \ sign bit alone ADC #0 STA T .PL43 \ We now call BLINE to draw the ellipse line segment \ \ The first few instructions of BLINE do the following: \ \ K6(3 2) = K4(1 0) + (T X) \ \ which gives: \ \ K6(3 2) = K4(1 0) - u_y * cos(CNT2) \ - v_y * sin(CNT2) \ \ K4(1 0) is the pixel y-coordinate of the centre of the \ ellipse, so this gives us the correct y-coordinate for \ our ellipse segment (we already calculated the \ x-coordinate in K3(1 0) above) JSR BLINE \ Call BLINE to draw this segment, which also returns \ the updated value of CNT in A CMP TGT \ If CNT > TGT then jump to PL40 to stop drawing the BEQ P%+4 \ ellipse (which is how we draw half-ellipses) BCS PL40 LDA CNT2 \ Set CNT2 = (CNT2 + STP) mod 64 CLC ADC STP AND #63 STA CNT2 JMP PLL4 \ Jump back to PLL4 to draw the next segment .PL40 RTS \ Return from the subroutine