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Elite on the BBC Micro

Moving: MV40 [Cassette version]

Name: MV40 [Show more] Type: Subroutine Category: Moving Summary: Rotate the planet or sun's location in space by the amount of pitch and roll of our ship
Context: See this subroutine in context in the source code Variations: See code variations for this subroutine in the different versions References: This subroutine is called as follows: * MVEIT (Part 2 of 9) calls MV40

We implement this using the same equations as in part 5 of MVEIT, where we rotated the current ship's location by our pitch and roll. Specifically, the calculation is as follows: 1. K2 = y - alpha * x 2. z = z + beta * K2 3. y = K2 - beta * z 4. x = x + alpha * y See the deep dive on "Rotating the universe" for more details on the above.
.MV40 LDA ALPHA \ Set Q = -ALPHA, so Q contains the angle we want to EOR #%10000000 \ roll the planet through (i.e. in the opposite STA Q \ direction to our ship's roll angle alpha) LDA INWK \ Set P(1 0) = (x_hi x_lo) STA P LDA INWK+1 STA P+1 LDA INWK+2 \ Set A = x_sign JSR MULT3 \ Set K(3 2 1 0) = (A P+1 P) * Q \ \ which also means: \ \ K(3 2 1) = (A P+1 P) * Q / 256 \ = x * -alpha / 256 \ = - alpha * x / 256 LDX #3 \ Set K(3 2 1) = (y_sign y_hi y_lo) + K(3 2 1) JSR MVT3 \ = y - alpha * x / 256 LDA K+1 \ Set K2(2 1) = P(1 0) = K(2 1) STA K2+1 STA P LDA K+2 \ Set K2+2 = K+2 STA K2+2 STA P+1 \ Set P+1 = K+2 LDA BETA \ Set Q = beta, the pitch angle of our ship STA Q LDA K+3 \ Set K+3 to K2+3, so now we have result 1 above: STA K2+3 \ \ K2(3 2 1) = K(3 2 1) \ = y - alpha * x / 256 \ We also have: \ \ A = K+3 \ \ P(1 0) = K(2 1) \ \ so combined, these mean: \ \ (A P+1 P) = K(3 2 1) \ = K2(3 2 1) JSR MULT3 \ Set K(3 2 1 0) = (A P+1 P) * Q \ \ which also means: \ \ K(3 2 1) = (A P+1 P) * Q / 256 \ = K2(3 2 1) * beta / 256 \ = beta * K2 / 256 LDX #6 \ K(3 2 1) = (z_sign z_hi z_lo) + K(3 2 1) JSR MVT3 \ = z + beta * K2 / 256 LDA K+1 \ Set P = K+1 STA P STA INWK+6 \ Set z_lo = K+1 LDA K+2 \ Set P+1 = K+2 STA P+1 STA INWK+7 \ Set z_hi = K+2 LDA K+3 \ Set A = z_sign = K+3, so now we have: STA INWK+8 \ \ (z_sign z_hi z_lo) = K(3 2 1) \ = z + beta * K2 / 256 \ So we now have result 2 above: \ \ z = z + beta * K2 EOR #%10000000 \ Flip the sign bit of A to give A = -z_sign JSR MULT3 \ Set K(3 2 1 0) = (A P+1 P) * Q \ = (-z_sign z_hi z_lo) * beta \ = -z * beta LDA K+3 \ Set T to the sign bit of K(3 2 1 0), i.e. to the sign AND #%10000000 \ bit of -z * beta STA T EOR K2+3 \ If K2(3 2 1 0) has a different sign to K(3 2 1 0), BMI MV1 \ then EOR'ing them will produce a 1 in bit 7, so jump \ to MV1 to take this into account \ If we get here, K and K2 have the same sign, so we can \ add them together to get the result we're after, and \ then set the sign afterwards LDA K \ We now do the following sum: \CLC \ ADC K2 \ (A y_hi y_lo -) = K(3 2 1 0) + K2(3 2 1 0) \ \ starting with the low bytes (which we don't keep) \ \ The CLC instruction is commented out in the original \ source. It isn't needed because MULT3 clears the C \ flag, so this is an example of the authors finding \ one more precious byte to save LDA K+1 \ We then do the middle bytes, which go into y_lo ADC K2+1 STA INWK+3 LDA K+2 \ And then the high bytes, which go into y_hi ADC K2+2 STA INWK+4 LDA K+3 \ And then the sign bytes into A, so overall we have the ADC K2+3 \ following, if we drop the low bytes from the result: \ \ (A y_hi y_lo) = (K + K2) / 256 JMP MV2 \ Jump to MV2 to skip the calculation for when K and K2 \ have different signs .MV1 LDA K \ If we get here then K2 and K have different signs, so SEC \ instead of adding, we need to subtract to get the SBC K2 \ result we want, like this: \ \ (A y_hi y_lo -) = K(3 2 1 0) - K2(3 2 1 0) \ \ starting with the low bytes (which we don't keep) LDA K+1 \ We then do the middle bytes, which go into y_lo SBC K2+1 STA INWK+3 LDA K+2 \ And then the high bytes, which go into y_hi SBC K2+2 STA INWK+4 LDA K2+3 \ Now for the sign bytes, so first we extract the sign AND #%01111111 \ byte from K2 without the sign bit, so P = |K2+3| STA P LDA K+3 \ And then we extract the sign byte from K without the AND #%01111111 \ sign bit, so A = |K+3| SBC P \ And finally we subtract the sign bytes, so P = A - P STA P \ By now we have the following, if we drop the low bytes \ from the result: \ \ (A y_hi y_lo) = (K - K2) / 256 \ \ so now we just need to make sure the sign of the \ result is correct BCS MV2 \ If the C flag is set, then the last subtraction above \ didn't underflow and the result is correct, so jump to \ MV2 as we are done with this particular stage LDA #1 \ Otherwise the subtraction above underflowed, as K2 is SBC INWK+3 \ the dominant part of the subtraction, so we need to STA INWK+3 \ negate the result using two's complement, starting \ with the low bytes: \ \ y_lo = 1 - y_lo LDA #0 \ And then the high bytes: SBC INWK+4 \ STA INWK+4 \ y_hi = 0 - y_hi LDA #0 \ And finally the sign bytes: SBC P \ \ A = 0 - P ORA #%10000000 \ We now force the sign bit to be negative, so that the \ final result below gets the opposite sign to K, which \ we want as K2 is the dominant part of the sum .MV2 EOR T \ T contains the sign bit of K, so if K is negative, \ this flips the sign of A STA INWK+5 \ Store A in y_sign \ So we now have result 3 above: \ \ y = K2 + K \ = K2 - beta * z LDA ALPHA \ Set A = alpha STA Q LDA INWK+3 \ Set P(1 0) = (y_hi y_lo) STA P LDA INWK+4 STA P+1 LDA INWK+5 \ Set A = y_sign JSR MULT3 \ Set K(3 2 1 0) = (A P+1 P) * Q \ = (y_sign y_hi y_lo) * alpha \ = y * alpha LDX #0 \ Set K(3 2 1) = (x_sign x_hi x_lo) + K(3 2 1) JSR MVT3 \ = x + y * alpha / 256 LDA K+1 \ Set (x_sign x_hi x_lo) = K(3 2 1) STA INWK \ = x + y * alpha / 256 LDA K+2 STA INWK+1 LDA K+3 STA INWK+2 \ So we now have result 4 above: \ \ x = x + y * alpha JMP MV45 \ We have now finished rotating the planet or sun by \ our pitch and roll, so jump back into the MVEIT \ routine at MV45 to apply all the other movements