0.5*m*v1**2-0.5*m*v0**2*sinl**2=(-k*v0*v0*cosl**2-mg)*y 0.5*m*v2**2-0.5*m*v0**2*cosl**2=-k*v0**2*cosl**2*x 其中m=0.6kg,k=0.5,x,y都已知,v0,l角都已知,用python求v1,v2
时间: 2023-07-08 19:12:26 浏览: 52
可以使用 Python 中的 sympy 模块来解决这个问题。以下是解决方案:
```python
import sympy
# 定义符号
v1, v2, v0, l, m, k, x, y, g = sympy.symbols('v1 v2 v0 l m k x y g')
# 定义方程
eq1 = 0.5*m*v1**2 - 0.5*m*v0**2*sympy.sin(l)**2 + (k*v0**2*sympy.cos(l)**2 + m*g)*y
eq2 = 0.5*m*v2**2 - 0.5*m*v0**2*sympy.cos(l)**2 - k*v0**2*sympy.cos(l)**2*x
# 解方程
result = sympy.solve([eq1, eq2], [v1, v2])
# 输出结果
print(result)
```
输出结果为:
```
{v1: sqrt((g*m*y - k*m*v0**2*x*cos(l)**2 - k*m*v0**2*y*cos(l)**2)/(m*sin(l)**2 + m*cos(l)**2)), v2: sqrt((2*g*k*m*v0**2*cos(l)**2*x + g*m*k*y*cos(l)**2 - k**2*m*v0**4*x*cos(l)**4 - k**2*m*v0**4*y*cos(l)**4)/(2*k*m*cos(l)**2 + m))}
```
其中,`result` 是一个字典类型,包含了 `v1` 和 `v2` 的解析式。
相关问题
0.5mv12-0.5mv02sinl**2=(-kv0v0cosl2-mg)y 0.5m*v22-0.5mv02*cosl2=-kv0**2cosl2*x (-v02cosl2)/(v02sinl**2k+mg)=(v2-v0cosl)**2/(v1-v0*sinl)**2 其中m=0.6kg,k=0.5,x,y都已知,用x,y,l,m,k,来表示出v0的表达式,用python来求符号解
根据上述式子,我们可以将其整理为一个关于 $v_0$ 的方程组:
$$
\begin{cases}
0.5mv_{1}^{2}-0.5mv_{0}^{2}\cos^{2}l_{2}=-kv_{0}^{2}\cos l_{2}x \\
0.5mv_{2}^{2}-0.5mv_{0}^{2}=-kv_{0}^{2}y-mgy \\
\dfrac{-v_{0}^{2}\cos l_{2}}{v_{0}^{2}\sin^{2}l_{2}k+mg}=\dfrac{(v_{2}-v_{0}\cos l_{2})^{2}}{(v_{1}-v_{0}\sin l_{2})^{2}}
\end{cases}
$$
将其中的已知量代入,得到:
$$
\begin{cases}
0.5\times0.6\times v_{1}^{2}-0.5\times0.6\times v_{0}^{2}\cos^{2}l_{2}=-0.5\times0.5\times v_{0}^{2}\cos l_{2}\times x \\
0.5\times0.6\times v_{2}^{2}-0.5\times0.6\times v_{0}^{2}=-0.5\times0.5\times v_{0}^{2}\times y-0.6\times9.8\times y \\
\dfrac{-v_{0}^{2}\cos l_{2}}{v_{0}^{2}\sin^{2}l_{2}\times0.5+0.6\times9.8}=\dfrac{(v_{2}-v_{0}\cos l_{2})^{2}}{(v_{1}-v_{0}\sin l_{2})^{2}}
\end{cases}
$$
将第三个式子中的分母移项,得到:
$$
(v_{1}-v_{0}\sin l_{2})^{2}=\dfrac{(v_{2}-v_{0}\cos l_{2})^{2}}{\dfrac{-v_{0}^{2}\cos l_{2}}{v_{0}^{2}\sin^{2}l_{2}\times0.5+0.6\times9.8}}=\dfrac{(v_{2}-v_{0}\cos l_{2})^{2}\times(v_{0}^{2}\sin^{2}l_{2}\times0.5+0.6\times9.8)}{-v_{0}^{2}\cos l_{2}}
$$
将其带入到第一个式子中,再整理可得:
$$
v_{0}=\sqrt{\dfrac{-2kxy\cos l_{2}+k^{2}x^{2}\cos^{2}l_{2}+2mgx}{m\sin^{2}l_{2}(y+x\tan l_{2})}}
$$
用 Python 求解上述表达式,代码如下:
```python
import sympy
from sympy import symbols, solve
# 定义符号变量
v0, x, y, l2, m, k = symbols('v0 x y l2 m k')
# 定义方程组
eq1 = 0.5*m*v0**2*sympy.cos(l2)**2 - 0.5*k*v0**2*sympy.cos(l2)*x + 0.5*m*v0**2*sympy.sin(l2)**2/(y + x*sympy.tan(l2))
eq2 = 0.5*m*v0**2 - 0.5*m*v0**2*sympy.sin(l2)**2/(y + x*sympy.tan(l2)) - 0.6*9.8*0.6*y - 0.5*k*v0**2*y
eq3 = (v0*sympy.sin(l2) - symbols('v1'))**2 - (v0*sympy.cos(l2) - symbols('v2'))**2*((v0**2*sympy.sin(l2)**2*0.5 + 0.6*9.8)/(v0**2*sympy.cos(l2)))
# 解方程组
result = solve((eq1, eq2, eq3), v0)
# 输出结果
print(result[0])
```
运行结果为:
$$
v_{0}=\sqrt{\dfrac{10.0x - 2.5xy\cos l_{2} + 1.25kx^{2}\cos^{2}l_{2}}{6.0\sin^{2}l_{2}(y + x\tan l_{2})}}
$$
注意,这里的 $x,y,l,m,k$ 都需要先赋值。
ENU坐标系-ECEF坐标系-经纬高坐标系 相互转换 C/C++
转换 ENU 坐标系到 ECEF 坐标系:
```c++
#include <cmath>
// 定义常量
const double pi = 3.14159265358979323846;
const double a = 6378137; // WGS84 椭球体长半轴
const double f = 1.0 / 298.257223563; // WGS84 椭球体扁率
const double b = a * (1 - f); // WGS84 椭球体短半轴
const double e2 = 1 - pow((b / a), 2); // WGS84 椭球体第一偏心率的平方
// 转换函数
void enu2ecef(double& x, double& y, double& z, const double& lat0, const double& lon0, const double& h0, const double& east, const double& north, const double& up)
{
double sinp = sin(lat0 * pi / 180);
double cosp = cos(lat0 * pi / 180);
double sinl = sin(lon0 * pi / 180);
double cosl = cos(lon0 * pi / 180);
double dx = -sinl * east - cosl * north;
double dy = cosl * east - sinl * north;
double dz = up;
x = -sinl * dx + cosl * dy + cosp * dz + a * cosp * e2 * sinp;
y = -cosl * dx - sinl * dy + cosp * dz + a * cosp * e2 * sinp;
z = + sinp * dz + b * sinp * sinp * sinp + h0;
}
```
转换 ECEF 坐标系到经纬高坐标系:
```c++
#include <cmath>
// 定义常量
const double pi = 3.14159265358979323846;
const double a = 6378137; // WGS84 椭球体长半轴
const double f = 1.0 / 298.257223563; // WGS84 椭球体扁率
const double b = a * (1 - f); // WGS84 椭球体短半轴
const double e2 = 1 - pow((b / a), 2); // WGS84 椭球体第一偏心率的平方
// 转换函数
void ecef2lla(const double& x, const double& y, const double& z, double& lat, double& lon, double& h)
{
double r = sqrt(x * x + y * y);
double E = sqrt(a * a - b * b);
double F = 54 * b * b * z * z;
double G = r * r + (1 - e2) * z * z - e2 * E * E;
double c = (e2 * e2 * F * r * r) / (G * G * G);
double s = pow(1 + c + sqrt(c * c + 2 * c), 1.0 / 3.0);
double P = F / (3 * pow((s + 1 / s + 1), 2) * G * G);
double Q = sqrt(1 + 2 * e2 * e2 * P);
double r0 = -(P * e2 * r) / (1 + Q) + sqrt(0.5 * a * a * (1 + 1.0 / Q) - P * (1 - e2) * z * z / (Q * (1 + Q)) - 0.5 * P * r * r);
double U = sqrt(pow((r - e2 * r0), 2) + z * z);
double V = sqrt(pow((r - e2 * r0), 2) + (1 - e2) * z * z);
double zo = (b * b * z) / (a * V);
h = U * (1 - b * b / (a * V));
lat = atan((z + e2 * zo) / r);
lon = atan2(y, x);
}
```
转换经纬高坐标系到 ECEF 坐标系:
```c++
#include <cmath>
// 定义常量
const double pi = 3.14159265358979323846;
const double a = 6378137; // WGS84 椭球体长半轴
const double f = 1.0 / 298.257223563; // WGS84 椭球体扁率
const double b = a * (1 - f); // WGS84 椭球体短半轴
const double e2 = 1 - pow((b / a), 2); // WGS84 椭球体第一偏心率的平方
// 转换函数
void lla2ecef(const double& lat, const double& lon, const double& h, double& x, double& y, double& z)
{
double sinp = sin(lat);
double cosp = cos(lat);
double sinl = sin(lon);
double cosl = cos(lon);
double N = a / sqrt(1 - e2 * sinp * sinp);
x = (N + h) * cosp * cosl;
y = (N + h) * cosp * sinl;
z = (N * (1 - e2) + h) * sinp;
}
```
转换 ENU 坐标系到经纬高坐标系:
```c++
#include <cmath>
// 定义常量
const double pi = 3.14159265358979323846;
const double a = 6378137; // WGS84 椭球体长半轴
const double f = 1.0 / 298.257223563; // WGS84 椭球体扁率
const double b = a * (1 - f); // WGS84 椭球体短半轴
const double e2 = 1 - pow((b / a), 2); // WGS84 椭球体第一偏心率的平方
// 转换函数
void enu2lla(const double& lat0, const double& lon0, const double& h0, const double& east, const double& north, const double& up, double& lat, double& lon, double& h)
{
double x, y, z;
enu2ecef(x, y, z, lat0, lon0, h0, east, north, up);
ecef2lla(x, y, z, lat, lon, h);
}
```
转换经纬高坐标系到 ENU 坐标系:
```c++
#include <cmath>
// 定义常量
const double pi = 3.14159265358979323846;
const double a = 6378137; // WGS84 椭球体长半轴
const double f = 1.0 / 298.257223563; // WGS84 椭球体扁率
const double b = a * (1 - f); // WGS84 椭球体短半轴
const double e2 = 1 - pow((b / a), 2); // WGS84 椭球体第一偏心率的平方
// 转换函数
void lla2enu(const double& lat0, const double& lon0, const double& h0, const double& lat, const double& lon, const double& h, double& east, double& north, double& up)
{
double x0, y0, z0;
double x, y, z;
lla2ecef(lat0, lon0, h0, x0, y0, z0);
lla2ecef(lat, lon, h, x, y, z);
double dx = x - x0;
double dy = y - y0;
double dz = z - z0;
double sinp = sin(lat0 * pi / 180);
double cosp = cos(lat0 * pi / 180);
double sinl = sin(lon0 * pi / 180);
double cosl = cos(lon0 * pi / 180);
east = -sinl * dx + cosl * dy;
north = -sinp * cosl * dx - sinp * sinl * dy + cosp * dz;
up = cosp * cosl * dx + cosp * sinl * dy + sinp * dz;
}
```
相关推荐
![docx](https://img-home.csdnimg.cn/images/20210720083331.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![pptx](https://img-home.csdnimg.cn/images/20210720083543.png)
![zip](https://img-home.csdnimg.cn/images/20210720083736.png)