阴圆弧推阳包络线
本帖最后由 shouce 于 2015-12-9 23:18 编辑阴圆弧推阳包络线 >> syms r t a b k q i A
>> x2=r*cos(t)+a;y2=r*sin(t)+b;
>> x1=-x2*cos(k*q)-y2*sin(k*q)+A*cos(q);
>> y1=-x2*sin(k*q)+y2*cos(k*q)+A*sin(q);
>> simplify(diff(x1,t))
ans =
r*sin(t - k*q)
>> simplify(diff(y1,t))
ans =
r*cos(t - k*q)
>> simplify(diff(x1,q))
ans =
k*sin(k*q)*(a + r*cos(t)) - A*sin(q) - k*cos(k*q)*(b + r*sin(t))
>> simplify(diff(y1,q))
ans =
A*cos(q) - k*cos(k*q)*(a + r*cos(t)) - k*sin(k*q)*(b + r*sin(t))
>> f=sym('(r*sin(t - k*q))*(A*cos(q) - k*cos(k*q)*(a + r*cos(t)) - k*sin(k*q)*(b + r*sin(t)))-r*cos(t - k*q)*(k*sin(k*q)*(a + r*cos(t)) - A*sin(q) - k*cos(k*q)*(b + r*sin(t)))')
>> simplify(f)
ans =
A*r*sin(q + t - k*q) + b*k*r*cos(t) - a*k*r*sin(t)
f1=sym('A*r*sin(q + t - k*q) + b*k*r*cos(t) - a*k*r*sin(t)=0')
>> pretty(solve(f1,q))
/ / b k cos(t) - a k sin(t) \ \
| t + asin| ----------------------- | |
| \ A / |
| ----------------------------------- |
| k - 1 |
(t + asin((k*(b*cos(t) - a*sin(t)))/A))/i
阳圆弧推导 阴转子包络线 t + asin((k*(b*cos(t) - a*sin(t))/(A*i))
这编程语言后面不加注释语句吗？别人看起来很困难呢 十字背包客 发表于 2015-12-10 09:55 static/image/common/back.gif
这编程语言后面不加注释语句吗？别人看起来很困难呢
你的意见很好我下次改
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