Verdure
2017-07-21, 20:42
请问大侠:
附件中F2与F3实际是同一个表达式,怎么用maple把F2化成 F3
restart;
F1 := 2*Fh*tanb;
2 Fh tanb
Fh := Kh*(L0-Lh);
Kh (L0 - Lh)
tanb := (hid-x)/(b-Lh);
hid - x
-------
b - Lh
Lh := b-sqrt(a^2-(hid-x)^2);
(1/2)
/ 2 2\
b - \a - (hid - x) /
F1;
/ (1/2)\
| / 2 2\ |
2 Kh \L0 - b + \a - (hid - x) / / (hid - x)
------------------------------------------------
(1/2)
/ 2 2\
\a - (hid - x) /
F2 := subs(a = r1*L0, b = r2*L0, F1);
/ (1/2)\
| / 2 2 2\ |
2 Kh \L0 - r2 L0 + \r1 L0 - (hid - x) / / (hid - x)
---------------------------------------------------------
(1/2)
/ 2 2 2\
\r1 L0 - (hid - x) /
F3 := 2*Kh*(1/sqrt(r1^2-((hid-x)/L0)^2)-r2/sqrt(r1^2-((hid-x)/L0)^2)+1)*(hid-x);
/ 1 r2 \
2 Kh |----------------------- - ----------------------- + 1| (hid
| (1/2) (1/2) |
|/ 2\ / 2\ |
|| 2 (hid - x) | | 2 (hid - x) | |
||r1 - ----------| |r1 - ----------| |
|| 2 | | 2 | |
\\ L0 / \ L0 / /
- x)
附件中F2与F3实际是同一个表达式,怎么用maple把F2化成 F3
restart;
F1 := 2*Fh*tanb;
2 Fh tanb
Fh := Kh*(L0-Lh);
Kh (L0 - Lh)
tanb := (hid-x)/(b-Lh);
hid - x
-------
b - Lh
Lh := b-sqrt(a^2-(hid-x)^2);
(1/2)
/ 2 2\
b - \a - (hid - x) /
F1;
/ (1/2)\
| / 2 2\ |
2 Kh \L0 - b + \a - (hid - x) / / (hid - x)
------------------------------------------------
(1/2)
/ 2 2\
\a - (hid - x) /
F2 := subs(a = r1*L0, b = r2*L0, F1);
/ (1/2)\
| / 2 2 2\ |
2 Kh \L0 - r2 L0 + \r1 L0 - (hid - x) / / (hid - x)
---------------------------------------------------------
(1/2)
/ 2 2 2\
\r1 L0 - (hid - x) /
F3 := 2*Kh*(1/sqrt(r1^2-((hid-x)/L0)^2)-r2/sqrt(r1^2-((hid-x)/L0)^2)+1)*(hid-x);
/ 1 r2 \
2 Kh |----------------------- - ----------------------- + 1| (hid
| (1/2) (1/2) |
|/ 2\ / 2\ |
|| 2 (hid - x) | | 2 (hid - x) | |
||r1 - ----------| |r1 - ----------| |
|| 2 | | 2 | |
\\ L0 / \ L0 / /
- x)