HP 32SII: area of a polygon by coordinates

12292014, 09:19 PM
(This post was last modified: 12292014 09:22 PM by Carlos CM (Mexico).)
Post: #1




HP 32SII: area of a polygon by coordinates
Program to calculate the area of a polygon by coordinates
Code: [LBL] [A] REGISTERS USED: A: area of the polygon N: total of points to enter X: xi Y: yi LABELS USED: A (20 bytes), B (28.5 bytes) NOTES: X<>Y is SWAP Rv is ROLL DOWN INSTRUCTIONS: 1. The polygon must be in the 1st Cuadrant of Cartesian System 2. Must begin/end with P1(0,0) coordinate 3. The order to enter the coordinates is clockwise to normal figures and counter clockwise if the figure has a HOLE. 4. Enter in order the coordinates, P1, P2...Pn1,P1 in this way: [X coordinate] [R/S] [Y coordinate] [R/S] 6. At the end of the last point, P1(0,0) the program display the desired area. EXAMPLE 1. For example, in a square (3x3=9 squared units): [XEQ] [A] N? ***Total of points to enter [5] [R/S] *** In this case 5 points [0] [R/S] [0] [R/S] *** P1(0,0) [0] [R/S] [3] [R/S] *** P2(0,3) [3] [R/S] [3] [R/S] *** P3(3,3) [3] [R/S] [0] [R/S] *** P4(3,0) [0] [R/S] [0] [R/S] *** P1(0,0) A=9 ***AREA OF THE POLYGON EXAMPLE 2. Example with a hole at center of the last square: [XEQ][A] N? ***Total of points to enter [1][1] [R/S] *** In this case 11 points [0] [R/S] [0] [R/S] *** P1 (0,0) (Begin with 0,0) [0] [R/S] [3] [R/S] *** P2 (0,3) [3] [R/S] [3] [R/S] *** P3 (3,3) [3] [R/S] [0] [R/S] *** P4 (3,0) [0] [R/S] [0] [R/S] *** P5 (0,0) (Close the polygon) [1] [R/S] [1] [R/S] *** P6 (1,1) (here the order of the points in counter clockwise) [2] [R/S] [1] [R/S] *** P7 (2,1) [2] [R/S] [2] [R/S] *** P8 (2,2) [1] [R/S] [2] [R/S] *** P9 (1,2) [1] [R/S] [1] [R/S] *** P10(1,1) (close the polygon that generate the hole) [0] [R/S] []0 [R/S] *** P11(0,0) (end with 0,0) A=8 ***AREA OF THE POLYGON RESULT=8 squared units Thanks for your aditional comments. Best Regards 

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