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楼主 |
发表于 2010-11-25 23:10
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计算不求人里求解渐开线反涵数的N种方法:
一:
inv(80)=4.2750184180222
ainv(4.2750184180222)=80
二:
rote[a,0 to 90,tan(a)-torad(a)-4.2750184180222,13]=80.0000000001455
三:
1: val1=4.2750184180222
2: middle=45
3: pg=13
4: min=0
5: max=90
6: val2=inv(middle)= 0.214602
7: i=0
8: wcycle()
9: max= among[val2>val1,middle,max]= 80.000000
10: min=among[val2<val1, middle,min]= 80.000000
11: middle=(min+max)/2= 80.000000
12: val2=inv(middle)= 4.275018
13: i=i+1= 35
14: if(i>100)
15: exit wcycle
16: end if
17: end wcycle(abs(val2-val1)<=(1/10^pg))
18: angle=middle= 80.000000
前面的行序号是可以不要的.
四:
1: val1=4.2750184180222
2: middle=45
3: pg=13
4: min=0
5: max=90
6: val2=inv(middle)= 0.214602
7: i=0
8: wcycle()
9: if(val2>val1)
10: max=middle= 80.000000
11: else
12: min=middle= 80.000000
13: end if
14: middle=(min+max)/2= 80.000000
15: val2=inv(middle)= 4.275018
16: i=i+1= 35
18: if(i>100)
19: exit wcycle
20: end if
21: end wcycle(abs(val2-val1)<=(1/10^pg))
22: angle=middle= 80.000000
五:
1: val1=4.2750184180222
2: middle=45
3: pg=13
4: min=0
5: max=90
6: val2=inv(middle)= 0.214602
7: i=0
8: wcycle(abs(val2-val1)>(1/10^pg))
9: if(val2>val1)
10: max=middle= 80.000000
11: else
12: min=middle= 80.000000
13: end if
14: middle=(min+max)/2= 80.000000
15: val2=inv(middle)= 4.275018
16: i=i+1= 35
18: if(i>100)
19: exit wcycle
20: end if
21: end wcycle
22: angle=middle= 80.000000
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