mashuai9633
2015-01-13, 22:44
http://g.hiphotos.baidu.com/zhidao/pic/item/b03533fa828ba61ece1bb5b44234970a314e59b7.jpg
计算这个方程,其中a0=15°,k=0.000422,q=0.04,Fa=22,Z=17,Db=6.85;
我编写的MATLAB程序如下
a0=15*pi/180;
k=0.000422;
q=0.04;
Fa=22;
Z=17;
Db=6.85;
b=(k/q)*(Fa/(Z*Db))^(2/3);
a=solve('(cos(a0)/cos(a*pi/180)-1)*(sin(a*pi/180))^(2/3)=b','a')
结果是
b =
0.0035
a =
(Dom::ImageSet(-(180*(acos(z1) + 2*pi*k))/pi, [k, z1], [Z_, RootOf(z^5 - 3*z^4*cos(a0) - z^3*(- 3*cos(a0)^2 + b^3 + 1) + z^2*(3*cos(a0) - cos(a0)^3) - 3*z*cos(a0)^2 + cos(a0)^3, z)]) intersect (Dom::ImageSet((180*(asinh(x)*i + 2*pi*k))/pi, [k, x], [Z_, Dom::Interval(0, Inf)]) union Dom::ImageSet(180*k, k, Z_) union Dom::ImageSet((180*(pi - asin(x*i + y) + 2*pi*k))/pi, [k, x, y], [Z_, R_, Dom::Interval(0, Inf)]) union Dom::ImageSet(-(180*(pi + asinh(x)*i + 2*pi*k))/pi, [k, x], [Z_, Dom::Interval(0, Inf)]) union Dom::ImageSet((180*(asin(x*i + y) + 2*pi*k))/pi, [k, x, y], [Z_, R_, Dom::Interval(0, Inf)])) union Dom::ImageSet((180*(acos(z1) + 2*pi*k))/pi, [k, z1], [Z_, RootOf(z^5 - 3*z^4*cos(a0) - z^3*(- 3*cos(a0)^2 + b^3 + 1) + z^2*(3*cos(a0) - cos(a0)^3) - 3*z*cos(a0)^2 + cos(a0)^3, z)]) intersect (Dom::ImageSet((180*(asinh(x)*i + 2*pi*k))/pi, [k, x], [Z_, Dom::Interval(0, Inf)]) union Dom::ImageSet(180*k, k, Z_) union Dom::ImageSet((180*(pi - asin(x*i + y) + 2*pi*k))/pi, [k, x, y], [Z_, R_, Dom::Interval(0, Inf)]) union Dom::ImageSet(-(180*(pi + asinh(x)*i + 2*pi*k))/pi, [k, x], [Z_, Dom::Interval(0, Inf)]) union Dom::ImageSet((180*(asin(x*i + y) + 2*pi*k))/pi, [k, x, y], [Z_, R_, Dom::Interval(0, Inf)]))) minus (Dom::ImageSet(-(180*(a0 + 2*pi*k))/pi, k, Z_) union Dom::ImageSet((180*(a0 + 2*pi*k))/pi, k, Z_) union Dom::ImageSet(180*k + 90, k, Z_))
我想要的是a的数值解,但是现在的结果确实这样的,如何解决这个问题。
计算这个方程,其中a0=15°,k=0.000422,q=0.04,Fa=22,Z=17,Db=6.85;
我编写的MATLAB程序如下
a0=15*pi/180;
k=0.000422;
q=0.04;
Fa=22;
Z=17;
Db=6.85;
b=(k/q)*(Fa/(Z*Db))^(2/3);
a=solve('(cos(a0)/cos(a*pi/180)-1)*(sin(a*pi/180))^(2/3)=b','a')
结果是
b =
0.0035
a =
(Dom::ImageSet(-(180*(acos(z1) + 2*pi*k))/pi, [k, z1], [Z_, RootOf(z^5 - 3*z^4*cos(a0) - z^3*(- 3*cos(a0)^2 + b^3 + 1) + z^2*(3*cos(a0) - cos(a0)^3) - 3*z*cos(a0)^2 + cos(a0)^3, z)]) intersect (Dom::ImageSet((180*(asinh(x)*i + 2*pi*k))/pi, [k, x], [Z_, Dom::Interval(0, Inf)]) union Dom::ImageSet(180*k, k, Z_) union Dom::ImageSet((180*(pi - asin(x*i + y) + 2*pi*k))/pi, [k, x, y], [Z_, R_, Dom::Interval(0, Inf)]) union Dom::ImageSet(-(180*(pi + asinh(x)*i + 2*pi*k))/pi, [k, x], [Z_, Dom::Interval(0, Inf)]) union Dom::ImageSet((180*(asin(x*i + y) + 2*pi*k))/pi, [k, x, y], [Z_, R_, Dom::Interval(0, Inf)])) union Dom::ImageSet((180*(acos(z1) + 2*pi*k))/pi, [k, z1], [Z_, RootOf(z^5 - 3*z^4*cos(a0) - z^3*(- 3*cos(a0)^2 + b^3 + 1) + z^2*(3*cos(a0) - cos(a0)^3) - 3*z*cos(a0)^2 + cos(a0)^3, z)]) intersect (Dom::ImageSet((180*(asinh(x)*i + 2*pi*k))/pi, [k, x], [Z_, Dom::Interval(0, Inf)]) union Dom::ImageSet(180*k, k, Z_) union Dom::ImageSet((180*(pi - asin(x*i + y) + 2*pi*k))/pi, [k, x, y], [Z_, R_, Dom::Interval(0, Inf)]) union Dom::ImageSet(-(180*(pi + asinh(x)*i + 2*pi*k))/pi, [k, x], [Z_, Dom::Interval(0, Inf)]) union Dom::ImageSet((180*(asin(x*i + y) + 2*pi*k))/pi, [k, x, y], [Z_, R_, Dom::Interval(0, Inf)]))) minus (Dom::ImageSet(-(180*(a0 + 2*pi*k))/pi, k, Z_) union Dom::ImageSet((180*(a0 + 2*pi*k))/pi, k, Z_) union Dom::ImageSet(180*k + 90, k, Z_))
我想要的是a的数值解,但是现在的结果确实这样的,如何解决这个问题。