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BBC Micro Elite

Moving: MVEIT (Part 5 of 9)

Name: MVEIT (Part 5 of 9) [View in context] Type: Subroutine Category: Moving Summary: Move current ship: Rotate ship's location by our pitch and roll Deep dive: Rotating the universe
This routine has multiple stages. This stage does the following: * Rotate the ship's location in space by the amount of pitch and roll of our ship. See below for a deeper explanation of this routine
LDX ALP1 \ Fetch the magnitude of the current roll into X, so \ if the roll angle is alpha, X contains |alpha| LDA INWK \ Set P = ~x_lo (i.e. with all its bits flipped) so that EOR #%11111111 \ we can pass x_lo to MLTU2 below) STA P LDA INWK+1 \ Set A = x_hi JSR MLTU2-2 \ Set (A P+1 P) = (A ~P) * X \ = (x_hi x_lo) * alpha STA P+2 \ Store the high byte of the result in P+2, so we now \ have: \ \ P(2 1 0) = (x_hi x_lo) * alpha LDA ALP2+1 \ Fetch the flipped sign of the current roll angle alpha EOR INWK+2 \ from ALP2+1 and EOR with byte #2 (x_sign), so if the \ flipped roll angle and x_sign have the same sign, A \ will be positive, else it will be negative. So A will \ contain the sign bit of x_sign * flipped alpha sign, \ which is the opposite to the sign of the above result, \ so we now have: \ \ (A P+2 P+1) = - (x_sign x_hi x_lo) * alpha / 256 LDX #3 \ Set (A P+2 P+1) = (y_sign y_hi y_lo) + (A P+2 P+1) JSR MVT6 \ = y - x * alpha / 256 STA K2+3 \ Set K2(3) = A = the sign of the result LDA P+1 \ Set K2(1) = P+1, the low byte of the result STA K2+1 EOR #%11111111 \ Set P = ~K2+1 (i.e. with all its bits flipped) so STA P \ that we can pass K2+1 to MLTU2 below) LDA P+2 \ Set K2(2) = A = P+2 STA K2+2 \ So we now have result 1 above: \ \ K2(3 2 1) = (A P+2 P+1) \ = y - x * alpha / 256 LDX BET1 \ Fetch the magnitude of the current pitch into X, so \ if the pitch angle is beta, X contains |beta| JSR MLTU2-2 \ Set (A P+1 P) = (A ~P) * X \ = K2(2 1) * beta STA P+2 \ Store the high byte of the result in P+2, so we now \ have: \ \ P(2 1 0) = K2(2 1) * beta LDA K2+3 \ Fetch the sign of the above result in K(3 2 1) from EOR BET2 \ K2+3 and EOR with BET2, the sign of the current pitch \ rate, so if the pitch and K(3 2 1) have the same sign, \ A will be positive, else it will be negative. So A \ will contain the sign bit of K(3 2 1) * beta, which is \ the same as the sign of the above result, so we now \ have: \ \ (A P+2 P+1) = K2(3 2 1) * beta / 256 LDX #6 \ Set (A P+2 P+1) = (z_sign z_hi z_lo) + (A P+2 P+1) JSR MVT6 \ = z + K2 * beta / 256 STA INWK+8 \ Set z_sign = A = the sign of the result LDA P+1 \ Set z_lo = P+1, the low byte of the result STA INWK+6 EOR #%11111111 \ Set P = ~z_lo (i.e. with all its bits flipped) so that STA P \ we can pass z_lo to MLTU2 below) LDA P+2 \ Set z_hi = P+2 STA INWK+7 \ So we now have result 2 above: \ \ (z_sign z_hi z_lo) = (A P+2 P+1) \ = z + K2 * beta / 256 JSR MLTU2 \ MLTU2 doesn't change Q, and Q was set to beta in \ the previous call to MLTU2, so this call does: \ \ (A P+1 P) = (A ~P) * Q \ = (z_hi z_lo) * beta STA P+2 \ Set P+2 = A = the high byte of the result, so we \ now have: \ \ P(2 1 0) = (z_hi z_lo) * beta LDA K2+3 \ Set y_sign = K2+3 STA INWK+5 EOR BET2 \ EOR y_sign with BET2, the sign of the current pitch EOR INWK+8 \ rate, and z_sign. If the result is positive jump to BPL MV43 \ MV43, otherwise this means beta * z and y have \ different signs, i.e. P(2 1) and K2(3 2 1) have \ different signs, so we need to add them in order to \ calculate K2(2 1) - P(2 1) LDA P+1 \ Set (y_hi y_lo) = K2(2 1) + P(2 1) ADC K2+1 STA INWK+3 LDA P+2 ADC K2+2 STA INWK+4 JMP MV44 \ Jump to MV44 to continue the calculation .MV43 LDA K2+1 \ Reversing the logic above, we need to subtract P(2 1) SBC P+1 \ and K2(3 2 1) to calculate K2(2 1) - P(2 1), so this STA INWK+3 \ sets (y_hi y_lo) = K2(2 1) - P(2 1) LDA K2+2 SBC P+2 STA INWK+4 BCS MV44 \ If the above subtraction did not underflow, then \ jump to MV44, otherwise we need to negate the result LDA #1 \ Negate (y_sign y_hi y_lo) using two's complement, SBC INWK+3 \ first doing the low bytes: STA INWK+3 \ \ y_lo = 1 - y_lo LDA #0 \ Then the high bytes: SBC INWK+4 \ STA INWK+4 \ y_hi = 0 - y_hi LDA INWK+5 \ And finally flip the sign in y_sign EOR #%10000000 STA INWK+5 .MV44 \ So we now have result 3 above: \ \ (y_sign y_hi y_lo) = K2(2 1) - P(2 1) \ = K2 - beta * z LDX ALP1 \ Fetch the magnitude of the current roll into X, so \ if the roll angle is alpha, X contains |alpha| LDA INWK+3 \ Set P = ~y_lo (i.e. with all its bits flipped) so that EOR #&FF \ we can pass y_lo to MLTU2 below) STA P LDA INWK+4 \ Set A = y_hi JSR MLTU2-2 \ Set (A P+1 P) = (A ~P) * X \ = (y_hi y_lo) * alpha STA P+2 \ Store the high byte of the result in P+2, so we now \ have: \ \ P(2 1 0) = (y_hi y_lo) * alpha LDA ALP2 \ Fetch the correct sign of the current roll angle alpha EOR INWK+5 \ from ALP2 and EOR with byte #5 (y_sign), so if the \ correct roll angle and y_sign have the same sign, A \ will be positive, else it will be negative. So A will \ contain the sign bit of x_sign * correct alpha sign, \ which is the same as the sign of the above result, \ so we now have: \ \ (A P+2 P+1) = (y_sign y_hi y_lo) * alpha / 256 LDX #0 \ Set (A P+2 P+1) = (x_sign x_hi x_lo) + (A P+2 P+1) JSR MVT6 \ = x + y * alpha / 256 STA INWK+2 \ Set x_sign = A = the sign of the result LDA P+2 \ Set x_hi = P+2, the high byte of the result STA INWK+1 LDA P+1 \ Set x_lo = P+1, the low byte of the result STA INWK \ So we now have result 4 above: \ \ x = x + alpha * y \ \ and the rotation of (x, y, z) is done