This code appears in the following versions (click to see it in the source code):
Code variations between these versions are shown below.
Name: MV40 Type: Subroutine Category: Moving
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.
This variation is blank in the Cassette, Disc (flight), 6502 Second Processor and Master versions.
Arguments: X The type of the planet Other entry points: MV40-1 Contains an RTS
.MV40
As the Electron only has planets and no suns, the MV40 routine only moves the planet by our pitch and roll, and does nothing if asked to move the sun.
This variation is blank in the Cassette, Disc (flight), 6502 Second Processor and Master versions.
TXA \ If bit 0 of X is set, then this is 129, which is the LSR A \ placeholder used to denote that there is no space BCS MV40-1 \ station, so return from the subroutine (as MV40-1 \ contains an RTS)
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
A CLC instruction is omitted from the cassette version in the rotation routine in MV40; it isn't needed, so this claws back one precious byte.
Tap on a block to expand it, and tap it again to revert.
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