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Maths (Geometry): ARCTAN

[NES version, Bank 1]

Name: ARCTAN [Show more] Type: Subroutine Category: Maths (Geometry) Summary: Calculate A = arctan(P / Q) Deep dive: The sine, cosine and arctan tables
Context: See this subroutine in context in the source code References: This subroutine is called as follows: * PLS4 calls ARCTAN

Calculate the following: A = arctan(P / Q) In other words, this finds the angle in the right-angled triangle where the opposite side to angle A is length P and the adjacent side to angle A has length Q, so: tan(A) = P / Q The result in A is an integer representing the angle in radians. The routine returns values in the range 0 to 128, which covers 0 to 180 degrees (or 0 to PI radians).
.ARCTAN LDA P ; Set T1 = P EOR Q, which will have the sign of P * Q EOR Q STA T1 LDA Q ; If Q = 0, jump to AR2 to return a right angle BEQ AR2 ASL A ; Set Q = |Q| * 2 (this is a quick way of clearing the STA Q ; sign bit, and we don't need to shift right again as we ; only ever use this value in the division with |P| * 2, ; which we set next) LDA P ; Set A = |P| * 2 ASL A CMP Q ; If A >= Q, i.e. |P| > |Q|, jump to AR1 to swap P BCS AR1 ; and Q around, so we can still use the lookup table JSR ARS1 ; Call ARS1 to set the following from the lookup table: ; ; A = arctan(A / Q) ; = arctan(|P / Q|) SEC ; Set the C flag so the SBC instruction in AR3 will be ; correct, should we jump there .AR4 LDX T1 ; If T1 is negative, i.e. P and Q have different signs, BMI AR3 ; jump down to AR3 to return arctan(-|P / Q|) RTS ; Otherwise P and Q have the same sign, so our result is ; correct and we can return from the subroutine .AR1 ; We want to calculate arctan(t) where |t| > 1, so we ; can use the calculation described in the documentation ; for the ACT table, i.e. 64 - arctan(1 / t) LDX Q ; Swap the values in Q and P, using the fact that we STA Q ; called AR1 with A = P STX P ; TXA ; This also sets A = P (which now contains the original ; argument |Q|) JSR ARS1 ; Call ARS1 to set the following from the lookup table: ; ; A = arctan(A / Q) ; = arctan(|Q / P|) ; = arctan(1 / |P / Q|) STA T ; Set T = 64 - T LDA #64 SBC T BCS AR4 ; Jump to AR4 to continue the calculation (this BCS is ; effectively a JMP as the subtraction will never ; underflow, as ARS1 returns values in the range 0-31) .AR2 ; If we get here then Q = 0, so tan(A) = infinity and ; A is a right angle, or 0.25 of a circle. We allocate ; 255 to a full circle, so we should return 63 for a ; right angle LDA #63 ; Set A to 63, to represent a right angle RTS ; Return from the subroutine .AR3 ; A contains arctan(|P / Q|) but P and Q have different ; signs, so we need to return arctan(-|P / Q|), using ; the calculation described in the documentation for the ; ACT table, i.e. 128 - A STA T ; Set A = 128 - A LDA #128 ; SBC T ; The subtraction will work because we did a SEC before ; calling AR3 RTS ; Return from the subroutine .ARS1 ; This routine fetches arctan(A / Q) from the ACT table, ; so A will be set to an integer in the range 0 to 31 ; that represents an angle from 0 to 45 degrees (or 0 to ; PI / 4 radians) JSR LL28 ; Call LL28 to calculate: ; ; R = 256 * A / Q LDA R ; Set X = R / 8 LSR A ; = 32 * A / Q LSR A ; LSR A ; so X has the value t * 32 where t = A / Q, which is TAX ; what we need to look up values in the ACT table LDA ACT,X ; Fetch ACT+X from the ACT table into A, so now: ; ; A = value in ACT + X ; = value in ACT + (32 * A / Q) ; = arctan(A / Q) RTS ; Return from the subroutine