考虑到对称性,仅考虑四分之一截面。每级硅钢片对应角度为:
θ
1
、
θ
2
…
…
θ
n
θ_1 、θ_2……θ_n
θ1、θ2……θn 叠片宽度:
B
n
=
r
c
o
s
(
θ
n
)
B_n=rcos(θ_n)
Bn=rcos(θn) 叠片厚度:
A
n
=
r
[
s
i
n
(
θ
n
)
−
s
i
n
(
θ
n
−
1
)
]
(
n
≥
2
)
;
A
1
=
r
s
i
n
(
θ
1
)
A_n=r[sin(θ_n )-sin(θ_{n-1})](n\geq2) ; A_1=rsin(θ_1 )
An=r[sin(θn)−sin(θn−1)](n≥2);A1=rsin(θ1) 有效面积:
S
=
∑
A
n
B
n
=
r
2
[
s
i
n
(
θ
1
)
c
o
s
(
θ
1
)
+
(
s
i
n
(
θ
2
)
−
s
i
n
(
θ
1
)
)
c
o
s
(
θ
2
)
+
⋯
⋯
+
(
s
i
n
(
θ
n
)
−
s
i
n
(
θ
n
−
1
)
)
c
o
s
(
θ
n
)
]
S=∑A_n B_n=r^2 [sin(θ_1 )cos(θ_1 )+(sin(θ_2 )-sin(θ_1 ))cos(θ_2 )+⋯⋯+(sin(θ_n )-sin(θ_{n-1} ))cos(θ_n )]
S=∑AnBn=r2[sin(θ1)cos(θ1)+(sin(θ2)−sin(θ1))cos(θ2)+⋯⋯+(sin(θn)−sin(θn−1))cos(θn)] 利用极值原理求S最大值,分别对
θ
1
、
θ
2
…
…
θ
n
θ_1 、θ_2……θ_n
θ1、θ2……θn求导,令导数为零得:
∂
S
∂
θ
1
=
r
2
(
c
o
s
(
2
θ
1
)
−
c
o
s
(
θ
1
)
c
o
s
(
θ
1
)
)
=
0
\frac{∂S}{∂θ_1}=r^2 (cos(2θ_1 )-cos(θ_1) cos(θ_1) )=0
∂θ1∂S=r2(cos(2θ1)−cos(θ1)cos(θ1))=0
∂
S
∂
θ
2
=
r
2
(
c
o
s
(
2
θ
2
)
+
s
i
n
(
θ
1
)
s
i
n
(
θ
2
)
−
c
o
s
(
θ
2
)
c
o
s
(
θ
3
)
)
=
0
\frac{∂S}{∂θ_2}=r^2 (cos(2θ_2)+sin(θ_1)sin( θ_2)-cos(θ_2) cos(θ_3) )=0
∂θ2∂S=r2(cos(2θ2)+sin(θ1)sin(θ2)−cos(θ2)cos(θ3))=0
……
∂
S
∂
θ
n
−
1
=
r
2
(
c
o
s
(
2
θ
n
−
1
)
+
s
i
n
(
θ
n
−
2
)
s
i
n
(
θ
n
−
1
)
−
c
o
s
(
θ
n
−
1
)
c
o
s
(
θ
n
)
)
=
0
\frac{∂S}{∂θ_{n-1} }=r^2 (cos(2θ_{n-1} )+sin(θ_{n-2})sin(θ_{n-1})-cos(θ_{n-1}) cos(θ_n) )=0
∂θn−1∂S=r2(cos(2θn−1)+sin(θn−2)sin(θn−1)−cos(θn−1)cos(θn))=0 化简得:
c
o
s
(
θ
2
)
=
c
o
s
(
2
θ
1
)
c
o
s
(
θ
1
)
cos(θ_2)=\frac{cos(2θ_1 )}{cos(θ_1)}
cos(θ2)=cos(θ1)cos(2θ1)
c
o
s
(
θ
3
)
=
c
o
s
(
2
θ
2
)
c
o
s
(
θ
2
)
+
s
i
n
(
θ
1
)
t
a
n
(
θ
2
)
cos(θ_3)=\frac{cos(2θ_2)}{cos(θ_2)} +sin(θ_1)tan(θ_2)
cos(θ3)=cos(θ2)cos(2θ2)+sin(θ1)tan(θ2)
……
c
o
s
(
θ
n
)
=
c
o
s
(
2
θ
n
−
1
)
c
o
s
(
θ
n
−
1
)
+
s
i
n
(
θ
n
−
2
)
t
a
n
(
θ
n
−
1
)
cos(θ_n)=\frac{cos(2θ_{n-1})}{cos(θ_{n-1})} +sin(θ_{n-2})tan(θ_{n-1})
cos(θn)=cos(θn−1)cos(2θn−1)+sin(θn−2)tan(θn−1) 从而可得:
A
1
=
r
2
−
B
1
2
A_1=\sqrt{r^2-B_1^2 }
A1=r2−B12
B
2
=
2
B
1
2
−
r
2
B
1
,
A
2
=
r
2
−
B
2
2
B_2=\frac{2B_1^2-r^2}{B_1} ,A_2=\sqrt{r^2-B_2^2 }
B2=B12B12−r2,A2=r2−B22
B
3
=
2
B
2
2
−
r
2
+
A
1
A
2
B
2
,
A
3
=
r
2
−
B
3
2
B_3=\frac{2B_2^2-r^2+A_1 A_2}{B_2} ,A_3=\sqrt{r^2-B_3^2 }
B3=B22B22−r2+A1A2,A3=r2−B32
……
B
n
=
2
B
n
−
1
2
−
r
2
+
A
n
−
2
A
n
−
1
B
n
−
1
,
A
n
=
r
2
−
B
n
2
B_n=\frac{2B_{n-1}^2-r^2+A_{n-2} A_{n-1}}{B_{n-1}} ,A_n=\sqrt{r^2-B_n^2 }
Bn=Bn−12Bn−12−r2+An−2An−1,An=r2−Bn2 一般硅钢片级数大于2,则有:
B
2
=
2
B
1
2
−
r
2
B
1
>
0
B_2=\frac{2B_1^2-r^2}{B_1} >0
B2=B12B12−r2>0 即:
2
2
r
<
B
1
<
r
\frac{\sqrt{2}}{2} r<B_1<r
22r<B1<r 由此确定
B
1
B_1
B1的范围为:
2
2
r
<
B
1
<
r
\frac{\sqrt{2}}{2} r<B_1<r
22r<B1<r 当
B
1
B_1
B1确定后,
B
2
B_2
B2,
B
3
B_3
B3,……
B
n
B_n
Bn的值都可以确定,所以可以在
2
2
r
<
B
1
<
r
\frac{\sqrt{2}}{2} r<B_1<r
22r<B1<r范围内采用迭代的方法,每次迭代
B
1
B_1
B1的值加1(mm),求出最大面积和对应的片宽和叠厚,流程图如下。