[LISP]如何用简单的方法求出第四个顶点坐标？

WGMOK

如图为一旋转的线性标注尺寸，

((-1 . <图元名: 7ef6f000>) (0 . "DIMENSION") (330 . <图元名: 7ef53cf8>) (5 .
"B0") (100 . "AcDbEntity") (67 . 0) (410 . "Model") (8 . "0") (100 .
"AcDbDimension") (2 . "*D2") (10 564.706 464.251 0.0) (11 550.206 457.307 0.0)
(12 0.0 0.0 0.0) (70 . 32) (1 . "") (71 . 5) (72 . 1) (41 . 1.0) (42 . 31.8832)
(52 . 0.0) (53 . 0.0) (54 . 0.0) (51 . 0.0) (210 0.0 0.0 1.0) (3 . "ISO-25")
(100 . "AcDbAlignedDimension") (13 546.0 434.534 0.0) (14 578.102 443.653 0.0)
(15 0.0 0.0 0.0) (16 0.0 0.0 0.0) (40 . 0.0) (50 . 0.576623) (100 .
"AcDbRotatedDimension"))

如何用简单的方法求出第四个顶点的坐标？

谢了！

BDYCAD

PT13 和 PT14  在DXF 的 (13 546.0 434.534 0.0) (14 578.102 443.653 0.0)

PT10 就是DXF的(10 564.706 464.251 0.0)

rongyifei

那另外一点是不是(11 550.206 457.307 0.0)?

WGMOK

不是呀，那是标注尺寸的数字（也就是图中数字31.88）中点的值

ZZXXQQ

3. (DEFUN C:DIMPT ()
4. (IF (SETQ S1 (ENTSEL "\nSelect a Dimation :")) (PROGN
5.   (SETQ ENT (ENTGET (CAR S1))
6. PT14 (CDR (ASSOC 14 ENT))
7. PT13 (CDR (ASSOC 13 ENT))
8. PT10 (CDR (ASSOC 10 ENT))
9. ANG (CDR (ASSOC 50 ENT))
10. PT12 (INTERS
11. (POLAR PT10 (IF (EQUAL ANG 0 1E-6) (ANGLE PT14 PT13) ANG) 1)
12. PT10
13. (POLAR PT13 (ANGLE PT14 PT10) 1)
14. PT13
15. nil)
16.   )
17. (COMMAND "POINT" PT12)
18. ))
19. )

(DEFUN C:DIMPT () (IF (SETQ S1 (ENTSEL "\nSelect a Dimation :")) (PROGN   (SETQ ENT (ENTGET (CAR S1)) PT14 (CDR (ASSOC 14 ENT)) PT13 (CDR (ASSOC 13 ENT)) PT10 (CDR (ASSOC 10 ENT)) ANG (CDR (ASSOC 50 ENT)) PT12 (INTERS (POLAR PT10 (IF (EQUAL ANG 0 1E-6) (ANGLE PT14 PT13) ANG) 1) PT10 (POLAR PT13 (ANGLE PT14 PT10) 1) PT13 nil)   ) (COMMAND "POINT" PT12) )) )

WGMOK

谢谢上面的朋友哟，成了

不过不是太理解上面的求PT12代码，得好好体会一下了。