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624272884 发表于 2015-5-8 17:17
算过的例子我这里有几个,我以前也发过几个,不过回应的人确实不多,可能我的案例太弱智了
>> syms t q a k A R2 >> x=R2*t*sin(t)+a
x =
a + R2*t*sin(t)
>> y=R2*t*cos(t)
y =
R2*t*cos(t)
>> x1=-x*cos(k*q)-y*sin(k*q)+A*cos(q)
x1 =
A*cos(q) - cos(k*q)*(a + R2*t*sin(t)) - R2*t*sin(k*q)*cos(t)
>> y1=-x*sin(k*q)+y*cos(k*q)+A*sin(q)
y1 =
A*sin(q) - sin(k*q)*(a + R2*t*sin(t)) + R2*t*cos(k*q)*cos(t)
>> diff(x1,t)
ans =
R2*t*sin(k*q)*sin(t) - R2*sin(k*q)*cos(t) - cos(k*q)*(R2*sin(t) + R2*t*cos(t))
>> diff(x1,q)
ans =
k*sin(k*q)*(a + R2*t*sin(t)) - A*sin(q) - R2*k*t*cos(k*q)*cos(t)
>> diff(y1,t)
ans =
R2*cos(k*q)*cos(t) - sin(k*q)*(R2*sin(t) + R2*t*cos(t)) - R2*t*cos(k*q)*sin(t)
>> diff(y1,q)
ans =
A*cos(q) - k*cos(k*q)*(a + R2*t*sin(t)) - R2*k*t*sin(k*q)*cos(t)
>> f1=sym('(A*cos(q) - k*cos(k*q)*(a + R2*t*sin(t)) - R2*k*t*sin(k*q)*cos(t))*(R2*t*sin(k*q)*sin(t) - R2*sin(k*q)*cos(t) - cos(k*q)*(R2*sin(t) + R2*t*cos(t)))-(k*sin(k*q)*(a + R2*t*sin(t)) - A*sin(q) - R2*k*t*cos(k*q)*cos(t))*(R2*cos(k*q)*cos(t) - sin(k*q)*(R2*sin(t) + R2*t*cos(t)) - R2*t*cos(k*q)*sin(t))')
f1 =
(k*cos(k*q)*(a + R2*t*sin(t)) - A*cos(q) + R2*k*t*sin(k*q)*cos(t))*(cos(k*q)*(R2*sin(t) + R2*t*cos(t)) + R2*sin(k*q)*cos(t) - R2*t*sin(k*q)*sin(t)) - (A*sin(q) - k*sin(k*q)*(a + R2*t*sin(t)) + R2*k*t*cos(k*q)*cos(t))*(sin(k*q)*(R2*sin(t) + R2*t*cos(t)) - R2*cos(k*q)*cos(t) + R2*t*cos(k*q)*sin(t))
>> simplify(f1)
ans =
A*R2*sin(q - t - k*q) + R2^2*k*t + R2*a*k*sin(t) - A*R2*t*cos(q - t - k*q) + R2*a*k*t*cos(t)
令 A*R2*sin(q - t - k*q) + R2^2*k*t + R2*a*k*sin(t) - A*R2*t*cos(q - t - k*q) + R2*a*k*t*cos(t)=0
则q的表达式可手算出来
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