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本帖最后由 阿帕奇 于 2011-5-25 08:03 编辑
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9 Y5 J0 e- { q0 [: B) S; Bx=(128*sin(180/18 + atan((16*cos(180/18) - 15)/(16*sin(180/18))))*((16*cos(180/18) - 15)^2/(256*sin(180/18)^2) + 1)^(1/2)*(sin(180/18)*sin(k)^2 + cos(180/18)*sin(180/18)*cos(k)*(cos(k) - 1)))/3 + (128*cos(180/18 + atan((16*cos(180/18) - 15)/(16*sin(180/18))))*((16*cos(180/18) - 15)^2/(256*sin(180/18)^2) + 1)^(1/2)*(cos(k)*(cos(k) - sin(180/18)^2*cos(k)^2*(cos(k) - 1)) + sin(k)*(cos(180/18)*sin(k) - sin(180/18)^2*cos(k)*sin(k)*(cos(k) - 1))))/3
- K, s: I2 i9 ?, f& H' y* w% Py=- (128*sin(180/18 + atan((16*cos(180/18) - 15)/(16*sin(180/18))))*(sin(180/18)*cos(k)*sin(k) - cos(180/18)*sin(180/18)*sin(k)*(cos(k) - 1))*((16*cos(180/18) - 15)^2/(256*sin(180/18)^2) + 1)^(1/2))/3 - (128*cos(180/18 + atan((16*cos(180/18) - 15)/(16*sin(180/18))))*((16*cos(180/18) - 15)^2/(256*sin(180/18)^2) + 1)^(1/2)*(cos(k)*(cos(k)*sin(k)*(cos(k) - 1)*sin(180/18)^2 + cos(180/18)*sin(k)) - sin(k)*(cos(k) - sin(180/18)^2*sin(k)^2*(cos(k) - 1))))/36 ^& p& t v) B: w5 m- \. F J
z=(128*cos(180/18 + atan((16*cos(180/18) - 15)/(16*sin(180/18))))*((16*cos(180/18) - 15)^2/(256*sin(180/18)^2) + 1)^(1/2)*(cos(k)*(sin(180/18)*sin(k)^2 - cos(180/18)*sin(180/18)*cos(k)*(cos(k) - 1)) - sin(k)*(sin(180/18)*cos(k)*sin(k) + cos(180/18)*sin(180/18)*sin(k)*(cos(k) - 1))))/3 - (128*sin(180/18 + atan((16*cos(180/18) - 15)/(16*sin(180/18))))*(cos(k) - cos(180/18)^2*(cos(k) - 1))*((16*cos(180/18) - 15)^2/(256*sin(180/18)^2) + 1)^(1/2))/3
: l+ F D Q. x) b这个方程在ug当中画出曲线 2 h- v# c: _0 Y. j- x$ s" A- ~" z
k是变量 从0到360度* q! F3 y9 a) _# W, H" |
希望高人解答 |
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