Применение MATHCAD в инженерных расчетах. Панферов А.И - 27 стр.

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27
Рис. 8. Некоторые допустимые определения дискретного аргумента
10 1 2
2
0
2
y
i
x
i
y
i
0
0.603
1.063
1.285
1.245
1
0.657
0.333
0.112
0.015
0
–0.015
–0.112
–0.333
–0.657
–1
=
x
i
2
1.856
1.464
0.933
0.405
0
–0.214
–0.242
–0.155
–0.047
0
–0.047
–0.155
–0.242
–0.214
0
=
r
i
2
1.951
1.809
1.588
1.309
1
0.691
0.412
0.191
0.049
0
0.049
0.191
0.412
0.691
1
=υ
i
0
0.314
0.628
0.942
1.257
1.571
1.885
2.199
2.513
2.827
3.142
3.456
3.77
4.084
4.398
4.712
=
j6
0
5
10
15
=
j5
10
11
12
13
14
15
=
j4
0
2
4
6
8
10
=
j3
0
1
2
3
4
5
6
7
8
9
10
=
j2
10
9
8
7
6
5
4
3
2
1
0
=
j1
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
7
7.5
=
y
i
r
i
sin
υ
i
()
:=
x
i
r
i
cos
υ
i
()
:=
j6 0 5
,
18
..:=
j5 10 15
..:=
r
i
cos
υ
i
()
1
+:=υ
i
2
π
i
N
:=
j4 0 2
,
10
..:=
j3 0 10
..:=
i0N
..:=
N20
:=
j2 10 0
..:=
j1 0 0.5
,
8
..:=