MATLAB的答案

帮助三角方程的解决方案。谢谢!

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杰克董
杰克董 2021年7月21日
评论道: 杰克董2021年7月22日
嗨Matlab专家,
  • 我建立了“脚本”来计算R1, R2从最初的输入:X = 0.9425;Z = sin (X) + 2 = 2.809;β= 0.531和结果为R1, R2是显示如下。
ans =
7.7733501156970570278303966915701
ans =
11.261695645672471499665334704332
  • 现在,我想用计算R1, R2值输入到“逆脚本”向后计算的初始值X, Z,β。期望是X = 0.9425;Z = sin (X) + 2 = 2.809;β= 0.531作为初始。但是,返回β值是50.796485621即使这是真的结果三角学的但这不是我想要的值(0.531)。β,我认为我们应该分配一些条件,以便它可以返回预期的值0.531,但我不知道请帮助我。谢谢!
ans =
0.94245504646393978078105437144795
ans =
2.8090036222785815735410394206806
ans =
50.796485621630240525665530786852
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向前脚本计算 R1, R2:
信谊X Y Zαβ伽马
函数=π/ 180;
theta_b = 15 *函数;
theta_p = 15 *函数;
R_b = 1;
R_p = 0.9;
Q0 = 1;%初始腿的长度
p = [X, Y, Z];
B1 = [R_b * cos(15 *函数);R_b * sin(15 *函数);0];
B2 = [R_b * cos(105 *函数);R_b *罪(105 *函数);0];
B3 = [R_b * cos(135 *函数);R_b *罪(135 *函数);0];
B4 = [R_b * cos(225 *函数);R_b *罪(225 *函数);0];
B5 = [R_b * cos(255 *函数);R_b *罪(255 *函数);0];
B6 = [R_b * cos(345 *函数);R_b *罪(345 *函数);0];
:Bx1系列= B1 (1);
:Bx2 = B2 (1);
:Bx3 = B3 (1);
:Bx4 = B4 (1);
:Bx5 = B5 (1);
:Bx6 = B6 (1);
:By1 = B1 (2);
By2 = B2 (2);
By3 = B3 (2);
By4 = B4 (2);
欢迎光临= B5 (2:);
6□= B6 (2);
Bz1 = B1 (3);
Bz2 =获取B2 (3);
Bz3 = B3 (3);
Bz4 = B4 (3);
Bz5 = B5 (3);
Bz6 = B6 (3);
P1 = [R_p * cos(75 *函数);R_b *罪(75 *函数);0];
P2 = [R_p * cos(165 *函数);R_b *罪(165 *函数);0];
P3 = [R_p * cos(195 *函数);R_b *罪(195 *函数);0];
P4 = [R_p * cos(285 *函数);R_b *罪(285 *函数);0];
P5 = [R_p * cos(315 *函数);R_b *罪(315 *函数);0];
P6 = [R_p * cos(405 *函数);R_b *罪(405 *函数);0];
:Px1 = P1 (1);
:Px2 = P2 (1);
:Px3 = P3 (1);
:Px4 = P4 (1);
:Px5 = P5 (1);
:Px6 = P6 (1);
Py1 = P1 (2);
:Py2 = P2 (2);
Py3 = P3 (2);
Py4 = P4 (2);
Py5 = P5 (2);
Py6 = P6 (2);
Pz1 = P1 (3);
Pz2 = P2 (3);
Pz3 = P3 (3);
Pz4 = P4 (3);
Pz5 = P5 (3);
Pz6 = P6 (3);
X = 0.9425;
β= 0.531;
Y = 0;
伽马= 0;
α= 0;
Z = sin (X) + 2;
R1 = (X + ((cos(伽马)* cos(β))* Px1) + (((cos(伽马)* sin(β)* sin(α))——(sin(伽马)* cos(α)))* Py1) + (((cos(伽马)* sin(β)* cos(α))+(罪(伽马)* sin(α)))* Pz1) bx1系列)^ 2 + (Y + ((sin(伽马)* cos(β))* Px1) +(((罪(伽马)* sin(β)* sin(α))+ (cos(伽马)* cos(α)))* Py1) +(((罪(伽马)* sin(β)* cos(α))——(cos(伽马)* sin(α)))* Pz1) by1) ^ 2 + (Z + ((sin(β))* Px1) + ((cos(β)* sin(α))* Py1) + ((cos(β)* cos(α))* Pz1) -Bz1) ^ 2;
L1 = R1 ^ 0.5 q0处;
R2 = (X + ((cos(伽马)* cos(β))* Px2) + (((cos(伽马)* sin(β)* sin(α))——(sin(伽马)* cos(α)))* Py2) + (((cos(伽马)* sin(β)* cos(α))+(罪(伽马)* sin(α)))* Pz2) -Bx2) ^ 2 + (Y + ((sin(伽马)* cos(β))* Px2) +(((罪(伽马)* sin(β)* sin(α))+ (cos(伽马)* cos(α)))* Py2) +(((罪(伽马)* sin(β)* cos(α))——(cos(伽马)* sin(α)))* Pz2) -By2) ^ 2 + (Z + ((sin(β))* Px2) + ((cos(β)* sin(α))* Py2) + ((cos(β)* cos(α))* Pz2) -Bz2) ^ 2;
L2 R2 = ^ 0.5 q0处;
vpa (R1。”)
vpa (R2。”)
逆脚本
信谊X Y Zαβ伽马
函数=π/ 180;
theta_b = 15 *函数;
theta_p = 15 *函数;
R_b = 1;
R_p = 0.9;
Q0 = 1;%初始腿的长度
p = [X, Y, Z];
B1 = [R_b * cos(15 *函数);R_b * sin(15 *函数);0];
B2 = [R_b * cos(105 *函数);R_b *罪(105 *函数);0];
B3 = [R_b * cos(135 *函数);R_b *罪(135 *函数);0];
B4 = [R_b * cos(225 *函数);R_b *罪(225 *函数);0];
B5 = [R_b * cos(255 *函数);R_b *罪(255 *函数);0];
B6 = [R_b * cos(345 *函数);R_b *罪(345 *函数);0];
:Bx1系列= B1 (1);
:Bx2 = B2 (1);
:Bx3 = B3 (1);
:Bx4 = B4 (1);
:Bx5 = B5 (1);
:Bx6 = B6 (1);
:By1 = B1 (2);
By2 = B2 (2);
By3 = B3 (2);
By4 = B4 (2);
欢迎光临= B5 (2:);
6□= B6 (2);
Bz1 = B1 (3);
Bz2 =获取B2 (3);
Bz3 = B3 (3);
Bz4 = B4 (3);
Bz5 = B5 (3);
Bz6 = B6 (3);
P1 = [R_p * cos(75 *函数);R_b *罪(75 *函数);0];
P2 = [R_p * cos(165 *函数);R_b *罪(165 *函数);0];
P3 = [R_p * cos(195 *函数);R_b *罪(195 *函数);0];
P4 = [R_p * cos(285 *函数);R_b *罪(285 *函数);0];
P5 = [R_p * cos(315 *函数);R_b *罪(315 *函数);0];
P6 = [R_p * cos(405 *函数);R_b *罪(405 *函数);0];
:Px1 = P1 (1);
:Px2 = P2 (1);
:Px3 = P3 (1);
:Px4 = P4 (1);
:Px5 = P5 (1);
:Px6 = P6 (1);
Py1 = P1 (2);
:Py2 = P2 (2);
Py3 = P3 (2);
Py4 = P4 (2);
Py5 = P5 (2);
Py6 = P6 (2);
Pz1 = P1 (3);
Pz2 = P2 (3);
Pz3 = P3 (3);
Pz4 = P4 (3);
Pz5 = P5 (3);
Pz6 = P6 (3);
Y = 0;
伽马= 0;
α= 0;
R1 = (X + ((cos(伽马)* cos(β))* Px1) + (((cos(伽马)* sin(β)* sin(α))——(sin(伽马)* cos(α)))* Py1) + (((cos(伽马)* sin(β)* cos(α))+(罪(伽马)* sin(α)))* Pz1) bx1系列)^ 2 + (Y + ((sin(伽马)* cos(β))* Px1) +(((罪(伽马)* sin(β)* sin(α))+ (cos(伽马)* cos(α)))* Py1) +(((罪(伽马)* sin(β)* cos(α))——(cos(伽马)* sin(α)))* Pz1) by1) ^ 2 + (Z + ((sin(β))* Px1) + ((cos(β)* sin(α))* Py1) + ((cos(β)* cos(α))* Pz1) -Bz1) ^ 2;
L1 = R1 ^ 0.5 q0处;
R2 = (X + ((cos(伽马)* cos(β))* Px2) + (((cos(伽马)* sin(β)* sin(α))——(sin(伽马)* cos(α)))* Py2) + (((cos(伽马)* sin(β)* cos(α))+(罪(伽马)* sin(α)))* Pz2) -Bx2) ^ 2 + (Y + ((sin(伽马)* cos(β))* Px2) +(((罪(伽马)* sin(β)* sin(α))+ (cos(伽马)* cos(α)))* Py2) +(((罪(伽马)* sin(β)* cos(α))——(cos(伽马)* sin(α)))* Pz2) -By2) ^ 2 + (Z + ((sin(β))* Px2) + ((cos(β)* sin(α))* Py2) + ((cos(β)* cos(α))* Pz2) -Bz2) ^ 2;
L2 R2 = ^ 0.5 q0处;
R2 = vpasolve (R1 = = 7.7731884, = = 11.2615, Z = = sin (X) + 2);
vpa (A.X”)
vpa (A.Z”)
vpa (A.beta”)

在回答这个问题。

接受的答案

大卫Goodmanson
大卫Goodmanson 2021年7月21日
编辑:大卫Goodmanson 2021年7月21日
嗨,杰克,
幸运的是想出正确的角解算器,除了2 *π模棱两可。所以,解决后你可以用它工作
国防部(50.796485621,2 *π)
ans = 0.5310
1评论
杰克董
杰克董 2021年7月22日
大卫感谢你伟大的答案,但我们可以将它嵌入我的脚本,运行后,我立即得到正确的测试。谢谢!
我尝试过但是失败了!
vpa (A.X”)
vpa (A.Z”)
vpa (A.beta”)
beta1 =国防部(A.beta 2 *π)
- - - - - - >结果
ans =
0.94245504646393978078105437144795
ans =
2.8090036222785815735410394206806
ans =
50.796485621630240525665530786852
beta1 =
国防部(50.796485621630240525665530786852,2 *π)
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脚本
信谊X Y Zαβ伽马
函数=π/ 180;
theta_b = 15 *函数;
theta_p = 15 *函数;
R_b = 1;
R_p = 0.9;
Q0 = 1;%初始腿的长度
p = [X, Y, Z];
B1 = [R_b * cos(15 *函数);R_b * sin(15 *函数);0];
B2 = [R_b * cos(105 *函数);R_b *罪(105 *函数);0];
B3 = [R_b * cos(135 *函数);R_b *罪(135 *函数);0];
B4 = [R_b * cos(225 *函数);R_b *罪(225 *函数);0];
B5 = [R_b * cos(255 *函数);R_b *罪(255 *函数);0];
B6 = [R_b * cos(345 *函数);R_b *罪(345 *函数);0];
:Bx1系列= B1 (1);
:Bx2 = B2 (1);
:Bx3 = B3 (1);
:Bx4 = B4 (1);
:Bx5 = B5 (1);
:Bx6 = B6 (1);
:By1 = B1 (2);
By2 = B2 (2);
By3 = B3 (2);
By4 = B4 (2);
欢迎光临= B5 (2:);
6□= B6 (2);
Bz1 = B1 (3);
Bz2 =获取B2 (3);
Bz3 = B3 (3);
Bz4 = B4 (3);
Bz5 = B5 (3);
Bz6 = B6 (3);
P1 = [R_p * cos(75 *函数);R_b *罪(75 *函数);0];
P2 = [R_p * cos(165 *函数);R_b *罪(165 *函数);0];
P3 = [R_p * cos(195 *函数);R_b *罪(195 *函数);0];
P4 = [R_p * cos(285 *函数);R_b *罪(285 *函数);0];
P5 = [R_p * cos(315 *函数);R_b *罪(315 *函数);0];
P6 = [R_p * cos(405 *函数);R_b *罪(405 *函数);0];
:Px1 = P1 (1);
:Px2 = P2 (1);
:Px3 = P3 (1);
:Px4 = P4 (1);
:Px5 = P5 (1);
:Px6 = P6 (1);
Py1 = P1 (2);
:Py2 = P2 (2);
Py3 = P3 (2);
Py4 = P4 (2);
Py5 = P5 (2);
Py6 = P6 (2);
Pz1 = P1 (3);
Pz2 = P2 (3);
Pz3 = P3 (3);
Pz4 = P4 (3);
Pz5 = P5 (3);
Pz6 = P6 (3);
Y = 0;
伽马= 0;
α= 0;
R1 = (X + ((cos(伽马)* cos(β))* Px1) + (((cos(伽马)* sin(β)* sin(α))——(sin(伽马)* cos(α)))* Py1) + (((cos(伽马)* sin(β)* cos(α))+(罪(伽马)* sin(α)))* Pz1) bx1系列)^ 2 + (Y + ((sin(伽马)* cos(β))* Px1) +(((罪(伽马)* sin(β)* sin(α))+ (cos(伽马)* cos(α)))* Py1) +(((罪(伽马)* sin(β)* cos(α))——(cos(伽马)* sin(α)))* Pz1) by1) ^ 2 + (Z + ((sin(β))* Px1) + ((cos(β)* sin(α))* Py1) + ((cos(β)* cos(α))* Pz1) -Bz1) ^ 2;
L1 = R1 ^ 0.5 q0处;
R2 = (X + ((cos(伽马)* cos(β))* Px2) + (((cos(伽马)* sin(β)* sin(α))——(sin(伽马)* cos(α)))* Py2) + (((cos(伽马)* sin(β)* cos(α))+(罪(伽马)* sin(α)))* Pz2) -Bx2) ^ 2 + (Y + ((sin(伽马)* cos(β))* Px2) +(((罪(伽马)* sin(β)* sin(α))+ (cos(伽马)* cos(α)))* Py2) +(((罪(伽马)* sin(β)* cos(α))——(cos(伽马)* sin(α)))* Pz2) -By2) ^ 2 + (Z + ((sin(β))* Px2) + ((cos(β)* sin(α))* Py2) + ((cos(β)* cos(α))* Pz2) -Bz2) ^ 2;
L2 R2 = ^ 0.5 q0处;
R2 = vpasolve (R1 = = 7.7731884, = = 11.2615, Z = = sin (X) + 2);
vpa (A.X”)
vpa (A.Z”)
vpa (A.beta”)
beta1 =国防部(A.beta 2 *π)

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