3

我在第一象限有两个点 A(X,Y) 和 B(P,Q)。还有一点 C(L,M)。如何在顺时针方向找到CA和CB之间的角度?

我进行了很多搜索,所有解决方案都使用了 atan2() 但它找到了原点相对于 x 轴的角度。

可以假设 C 和 A 是固定的。 B 可以在第一象限的任何地方

可以假设 C 和 A 是固定的。B 可以在第一象限的任何地方。角度必须是顺时针方向并且在 0-360(或 0 到 360-1)范围内。

我在 C/C++ 中这样做。

编辑:为每个请求添加代码。这有点不同,因为我陷入了一个概念并且需要对此进行澄清。如果点 x,y 位于 50,50 和 P 之间,则该函数应该返回。P 是相对于 CA 的角度。

bool isInsideAngle(long double x,long double y, long double p)
{   
    if((atan2(y,x) >= atan2(50,100)) && (atan2(y,x) <= (p * PI / 50)))
    {
        // cout<<"YES!";
        //   cout<<" atan2(y,x) = " <<atan2(y,x)*180/PI<<endl;
        //   cout<<" atan2(50,50) = " <<atan2(50,100)*180/PI<<endl;
        //   cout<<" (p * PI / 50) = "<<(p * PI / 50)*180/PI<<endl;
        return true;
    }

    else
        return false;

 }
4

3 回答 3

5

如何在顺时针方向找到CA和CB之间的角度?

// The atan2 functions return arctan y/x in the interval [−π , +π] radians
double Dir_C_to_A = atan2(Ay - Cy, Ax - Cx);
double Dir_C_to_B = atan2(By - Cy, Bx - Cx);
double Angle_ACB = Dir_C_to_A - Dir_C_to_B;

// Handle wrap around
const double Pi = acos(-1);  // or use some π constant
if (Angle_ACB > Pi) Angle_ACB -= 2*Pi;
else if (Angle_ACB < -Pi) Angle_ACB += 2*Pi;

// Answer is in the range of [-pi...pi]
return Angle_ACB;
于 2017-01-07T05:23:40.117 回答
2

由于 chux 已经回答了您的问题并且您接受了他的回答,因此这里有另一种方法可以从数学和编程的角度来看待它。这可以应用于任何三个不同的相对点。


数学- 这将在 2D 空间中完成,但可以应用于任何维度空间。

  • 点数将由小写字母捐赠,向量为大写字母。
  • 点积将表示为*
  • 矢量的长度与其大小相同。
  • 点将与其轴分量 (x,y) 一起显示
  • 向量将与其向量分量 (i,j) 一起显示
  • Theta 将是两个向量之间的角度,Phi 将是外角

> **抽象** - 向量的规则和方程 > 我们有三个点 `a(x,y)`、`b(x,y)` 和 `c(x,y)` > 从这些我们可以构造两个向量 `A` & `B` > `A = ca` 和 `B = cb` > `A = ` > `B = ` > 现在我们已经定义了两个向量 `A` 和 `B` 让我们找到 `Theta` . > 为了找到它们之间的 `cos(angle)`,我们需要知道 > `A` 和 `B` 的大小以及它们之间的 `Dot Product`。

MagnitudeLength向量sqrt( (i^2) + (j^2) )
Dot ProductA*B = (Ai x Bi) + (Aj x Bj)
Cos(Angle) = (A*B) / (magA * magB)
Theta = acos( Cos(Angle) )
Phi = 360 - Theta


证明- 一个例子,点是a(5,7), b(3,5)&c(5,3)

A = < 5-5, 7-3 > = < 0, 4 >
B = < 3-5, 5-3 > = < -2, 2 >

magA = sqrt( 0^2 + 4^2 ) = 4
magB = sqrt( (-2)^2 + 2^2 ) = 2sqrt(2)

cosAngle = A*B / (magA * magB)
cosAngle = (0*-2 + 4*2) = 8 / ( 4 x 2sqrt(2) ) = 0.7071067812

Theta = acos( cosAngle ) = 45 degrees
Phi = 360 - 45 = 315

Phiclockwise angle您在左侧第一张图中所要求的,
其中Theta是任意两个向量之间的角度。


唯一剩下的就是现在将这些方程式应用于您选择的编程语言。

笔记- 这只会找到两个向量之间的角度 Theta 并从 360 中减去 Theta 以找到 Phi 将为您提供这两个向量周围的外角。这并不包含或暗示角度本身的任何旋转方向。这不区分顺时针或逆时针。用户必须自己计算。这只是使用 3 个点或 2 个向量的基本属性和操作来找到它们之间的角度,这只是整个问题的一个步骤。如果参考上面的证明,内角是45度,外角是315度;如果将点 b 更改为 (7,5) 而不是点 b 在矢量 A 上反射,则输出将是完全相同的值:两个矢量之间的角度为 45 度,外角为 315。这不知道你在哪个方向旋转;除非您考虑使用和携带余弦函数的符号,如果它是 - 或 +,那么它可能会有所不同,但记住 Trig cosine 在不同的象限中也是 + 和 - 。


编程 - C++

主文件

#include <iostream>
#include "Vector2.h"

int main() {   
    // We can assume that points and vectors are similar except that points
    // Don't have a direction where vectors do.
    Vector2 pointA( 5, 7 );
    Vector2 pointB( 3, 5 );
    Vector2 pointC( 5, 3 );

    Vector2 vec1 = pointA - pointC;
    Vector2 vec2 = pointB - pointC;

    // This is the actual angle
    float ThetaA = vec1.getAngle( vec2, false, false );     

    // Other Option
    float ThetaB = vec1.getCosAngle( vec2, false );
    ThetaB = acos( ThetaB );
    ThetaB = Math::radian2Degree( ThetaB );
    
    // Same as other option above which this is already being done in getAngle()
    float ThetaC = Math::radian2Degree( acos( vec1.getCosAngle( vec2, false ) ) );

    std::cout << "ThetaA = " << ThetaA << std::endl;
    std::cout << "ThetaB = " << ThetaB << std::endl;
    std::cout << "ThetaC = " << ThetaC << std::endl;

    float Phi = 360.0f - ThetaA;
    std::cout << "Phi = " << Phi << std::endl;
 
    return 0;
}

输出

ThetaA = 45
ThetaB = 45
ThetaC = 45
Phi = 315

主要的简短版本

#include <iostream>
#include "Vector2.h" 

int main() {
    Vector2 pointA( 5, 7 );
    Vector2 pointB( 3, 5 );
    Vector2 pointC( 5, 3 );

    Vector2 vec1 = pointA - pointC;
    Vector2 vec2 = pointB - pointC;

    float angle = vec1.getAngle( vec2, false, false );
    float clockwiseAngle = 360 - angle;
  
    std::cout << "Theta " << angle << std::endl;
    std::cout << "Phi " << clockwiwseAngle << std::endl;
   
    return 0;
}

输出

Theta = 45
Phi = 315

这就是你如何从左边第一个图中的 3 个点找到顺时针角度的方法。至于右边的图表,只需找到角度,不要从 360 中减去它。使用的类如下。


矢量2.h

#ifndef VECTOR2_H
#define VECTOR2_H

#include "GeneralMath.h"

class Vector2 { 

public:
    union {
        float m_f2[2];
        struct {
            float m_fX;
            float m_fY;
        };  
    };

    // Constructors
    inline Vector2();
    inline Vector2( float x, float y );
    inline Vector2( float *pfv );

    // Destructor
    ~Vector2(){}

    // Operators
    inline Vector2  operator+( const Vector2 &v2 ) const;
    inline Vector2  operator+() const;
    inline Vector2& operator+=( const Vector2 &v2 );
    inline Vector2  operator-( const Vector2 &v2 ) const;
    inline Vector2  operator-() const;
    inline Vector2& operator-=( const Vector2 &v2 );
    inline Vector2  operator*( const float &fValue ) const;
    inline Vector2& operator*=( const float &fValue );
    inline Vector2  operator/( const float &fValue ) const;
    inline Vector2& operator/=( const float &fValue );  

    // Functions
    inline void     normalize();
    inline void     zero();
    inline bool     isZero() const;
    inline float    dot( const Vector2 v2 ) const;
    inline float    length2() const;
    inline float    length() const;
    inline float    getCosAngle( const Vector2 &v2, const bool bNormalized = false );
    inline float    getAngle( const Vector2 &v2, const bool bNormalized = false, bool bRadians = true );

    // Pre Multiple Vector By A Scalar
    inline friend Vector2 Vector2::operator*( const float &fValue, const Vector2 v2 ) {         
        return Vector2( fValue*v2.m_fX, fValue*v2.m_fY );    
    } // operator*  

    // Pre Divide Vector By A Scalar
    inline friend Vector2 Vector2::operator/( const float &fValue, const Vector2 v2 ) {
        Vector2 vec2;
        if ( Math::isZero( v2.m_fX ) ) {
            vec2.m_fX = 0.0f;
        } else {
            vec2.m_fX = fValue / v2.m_fX;
        }

        if ( Math::isZero( v2.m_fY ) ) {
            vec2.m_fY = 0.0f;
        } else {
            vec2.m_fY = fValue / v2.m_fY;
        }

        return vec2;    
    } // operator/

}; // Vector2

inline Vector2::Vector2() : m_fX( 0.0f ), m_fY( 0.0f ) {
} // Vector2

inline Vector2::Vector2( float x, float y ) : m_fX( x ), m_fY( y ) {
} // Vector2

inline Vector2::Vector2( float *pfv ) {
    m_fX = pfv[0];
    m_fY = pfv[1];
} // Vector2

// Unary + Operator
inline Vector2 Vector2::operator+() const {
    return *this;
} // operator+

// Binary + Take This Vector And Add Another Vector To It
inline Vector2 Vector2::operator+( const Vector2 &v2 ) const {
    return Vector2( m_fX + v2.m_fX, m_fY + v2.m_fY );
} // operator+

// Add Two Vectors Together
inline Vector2 &Vector2::operator+=( const Vector2 &v2 ) {
    m_fX += v2.m_fX;
    m_fY += v2.m_fY;
    return *this;
} // operator+=

// Unary - Operator: Negate Each Value
inline Vector2 Vector2::operator-() const {
    return Vector2( -m_fX, -m_fY );
} // operator-

// Unary - Take This Vector And Subtract Another Vector From It
inline Vector2 Vector2::operator-( const Vector2 &v2 ) const {
    return Vector2( m_fX - v2.m_fX, m_fY - v2.m_fY );
} // operator-

// Subtract Two Vectors From Each Other
inline Vector2 &Vector2::operator-=( const Vector2 &v2 ) {
    m_fX -= v2.m_fX;
    m_fY -= v2.m_fY;
    return *this;
} // operator-=

// Post Multiply Vector By A Scalar
inline Vector2 Vector2::operator*( const float &fValue ) const {
    return Vector2( m_fX * fValue, m_fY * fValue );
} // operator*

// Multiply This Vector By A Scalar
inline Vector2& Vector2::operator*=( const float &fValue ) {    
    m_fX *= fValue;
    m_fY *= fValue;    
    return *this;    
} // operator*

// Post Divide Vector By A Scalar
inline Vector2 Vector2::operator/( const float &fValue ) const {    
    Vector2 vec2;
    if ( Math::isZero( fValue ) ) {
        vec2.m_fX = 0.0f;
        vec2.m_fY = 0.0f;
    } else {
        float fValue_Inv = 1/fValue;
        vec2.m_fX = vec2.m_fX * fValue_Inv;
        vec2.m_fY = vec2.m_fY * fValue_Inv;
    }    
    return vec2;
} // operator/

// Divide This Vector By A Scalar Value
inline Vector2& Vector2::operator/=( const float &fValue ) {    
    if ( Math::isZero( fValue ) ) {
        m_fX = 0.0f;
        m_fY = 0.0f;
    } else {
        float fValue_Inv = 1/fValue;
        m_fX *= fValue_Inv;
        m_fY *= fValue_Inv;
    }    
    return *this;
} // operator/=

// Make The Length Of This Vector Equal To One
inline void Vector2::normalize() {    
    float fMag;
    fMag = sqrt( m_fX * m_fX + m_fY * m_fY );
    if ( fMag <= Math::ZERO ) {
        m_fX = 0.0f;
        m_fY = 0.0f;
        return;
    }

    fMag = 1/fMag;
    m_fX *= fMag;
    m_fY *= fMag;    
} // normalize

// Return True if Vector Is ( 0, 0 )
inline bool Vector2::isZero() const {
    if ( Math::isZero( m_fX ) && Math::isZero( m_fY ) ) {
        return true;
    } else {
        return false;
    }
} // isZero

// Set Vector To ( 0, 0 )
inline void Vector2::zero() {
    m_fX = 0.0f;
    m_fY = 0.0f;
} // zero

// Return The Length Of This Vector
inline float Vector2::length() const {
    return sqrtf( m_fX * m_fX + m_fY * m_fY );
} // length

// Return The Length Of This Vector
inline float Vector2::length2() const {
    return ( m_fX * m_fX + m_fY * m_fY );
} // length2

// Return The Dot Product Between THIS Vector And Another Vector
inline float Vector2::dot( const Vector2 v2 ) const {
    return ( m_fX * v2.m_fX + m_fY * v2.m_fY );
} // dot

// Returns The cos(Angle) Value Between THIS Vector And Vector v2.
// This Is Less Expensive Than Using GetAngle()
inline float Vector2::getCosAngle( const Vector2 &v2, const bool bNormalized ) {    
    // A . B = |A||B|cos(angle)
    // -> cos-1((A.B)/(|A||B|))

    float fMagA = length();
    if ( fMagA <= Math::ZERO ) {
        // This (A) Is An Invalid Vector
        return 0;
    }

    float fValue = 0.0f;

    if ( bNormalized ) {
        // v2 Already Normalized
        fValue = dot(v2) / fMagA;
    } else {
        // v2 Not Normalized
        float fMagB = v2.length();
        if ( fMagB <= Math::ZERO ) {
            // B Is An Invalid Vector
            return 0;
        }
        fValue = dot(v2) / ( fMagA * fMagB );
    }

    // Correct Value Due To Rounding Problems
    Math::constrain( -1.0f, 1.0f, fValue );

    return fValue;

} // getCosAngle

// Returns The Angle Between THIS Vector And Vector v2 In RADIANS
inline float Vector2::getAngle( const Vector2 &v2, const bool bNormalized, bool bRadians ) {    
    // A . B = |A||B|cos(angle)
    // -> cos-1((A.B)/(|A||B|))

    if ( bRadians ) {
        return acos( getCosAngle( v2, bNormalized ) );
    } else {
        // Convert To Degrees
        return Math::radian2Degree( acos( getCosAngle( v2, bNormalized ) ) );
    }    
} // GetAngle

#endif // VECTOR2_H

通用数学.h

#ifndef GENERALMATH_H
#define GENERALMATH_H

#include <math.h>

class Math {
public:
    
    static const float PI;
    static const float PI_HALVES;
    static const float PI_THIRDS;
    static const float PI_FOURTHS;
    static const float PI_SIXTHS;
    static const float PI_2;
    static const float PI_INVx180;
    static const float PI_DIV180;
    static const float PI_INV;
    static const float ZERO;

    Math();

    inline static bool  isZero( float fValue );
    inline static float sign( float fValue );

    inline static int   randomRange( int iMin, int iMax );
    inline static float randomRange( float fMin, float fMax );
    
    inline static float degree2Radian( float fDegrees );
    inline static float radian2Degree( float fRadians );
    inline static float correctAngle( float fAngle, bool bDegrees, float fAngleStart = 0.0f );
    inline static float mapValue( float fMinY, float fMaxY, float fMinX, float fMaxX, float fValueX );

    template<class T>
    inline static void constrain( T min, T max, T &value );

    template<class T>
    inline static void swap( T &value1, T &value2 );

}; // Math

// Convert Angle In Degrees To Radians
inline float Math::degree2Radian( float fDegrees ) {
    return fDegrees * PI_DIV180;
} // degree2Radian

// Convert Angle In Radians To Degrees
inline float Math::radian2Degree( float fRadians ) {
    return fRadians * PI_INVx180;
} // radian2Degree

// Returns An Angle Value That Is Alway Between fAngleStart And fAngleStart + 360
// If Radians Are Used, Then Range Is fAngleStart To fAngleStart + 2PI
inline float Math::correctAngle( float fAngle, bool bDegrees, float fAngleStart ) {    
    if ( bDegrees ) {
        // Using Degrees
        if ( fAngle < fAngleStart ) {
            while ( fAngle < fAngleStart ) {
                fAngle += 360.0f;
            }
        } else if ( fAngle >= (fAngleStart + 360.0f) ) {
            while ( fAngle >= (fAngleStart + 360.0f) ) {
                fAngle -= 360.0f;
            }
        }
        return fAngle;
    } else {
        // Using Radians
        if ( fAngle < fAngleStart ) {
            while ( fAngle < fAngleStart ) {
                fAngle += Math::PI_2;
            }
        } else if ( fAngle >= (fAngleStart + Math::PI_2) ) {
            while ( fAngle >= (fAngleStart + Math::PI_2) ) {
                fAngle -= Math::PI_2;
            }
        }
        return fAngle;
    }    
} // correctAngle

// Tests If Input Value Is Close To Zero
inline bool Math::isZero( float fValue ) {
    if ( (fValue > -ZERO) && (fValue < ZERO) ) {
        return true;
    }
    return false;
} // isZero

// Returns 1 If Value Is Positive, -1 If Value Is Negative Or 0 Otherwise
inline float Math::Sign( float fValue ) {    
    if ( fValue > 0 ) {
        return 1.0f;
    } else if ( fValue < 0 ) {
        return -1.0f;
    }
    return 0;
} // sign

// Return A Random Number Between iMin And iMax Where iMin < iMax
inline int Math::randomRange( int iMin, int iMax ) {    
    if ( iMax < iMin ) {
        swap( iMax, iMin );
    }    
    return (iMin + ((iMax - iMin +1) * rand()) / (RAND_MAX+1) );
    
} // randomRange

// Return A Random Number Between fMin And fMax Where fMin < fMax
inline float Math::randomRange( float fMin, float fMax ) {
    if ( fMax < fMin ) {
        swap( fMax, fMin );
    }    
    return (fMin + (rand()/(float)RAND_MAX)*(fMax-fMin));    
} // randomRange

// Returns The fValueY That Corresponds To A Point On The Line Going From Min To Max
inline float Math::mapValue( float fMinY, float fMaxY, float fMinX, float fMaxX, float fValueX ) {    
    if ( fValueX >= fMaxX ) {
        return fMaxY;
    } else if ( fValueX <= fMinX ) {
        return fMinY;
    } else {
        float fM = (fMaxY - fMinY) / (fMaxX - fMinX);
        float fB = fMaxY - fM * fMaxX;

        return (fM*fValueX + fB);
    }    
} // mapValue

// Constrain a Value To Be Between T min & T max
template<class T>
inline void Math::constrain( T min, T max, T &value ) {    
    if ( value < min ) {
        value = min;
        return;
    }

    if ( value > max ) {
        value = max;
    }    
} // constrain

// Swap Two Values
template<class T>
inline void Math::Swap( T &value1, T &value2 ) {    
    T temp;

    temp   = value1;
    value1 = value2;
    value2 = temp;    
} // swap

#endif // GENERALMATH_H

通用数学.cpp

#include "GeneralMath.h"

const float Math::PI            = 4.0f  * atan(1.0f); // tan(pi/4) = 1 or acos(-1)
const float Math::PI_HALVES     = 0.50f * Math::PI;
const float Math::PI_THIRDS     = Math::PI * 0.3333333333333f;
const float Math::PI_FOURTHS    = 0.25f * Math::PI;
const float Math::PI_SIXTHS     = Math::PI * 0.6666666666667f;
const float Math::PI_2          = 2.00f * Math::PI;
const float Math::PI_DIV180     = Math::PI / 180.0f;
const float Math::PI_INVx180    = 180.0f / Math::PI;
const float Math::PI_INV        = 1.0f / Math::PI;
const float Math::ZERO          = (float)1e-7;

Math::Math() {
} // Math
于 2017-01-07T10:30:44.623 回答
1

此解决方案使用向量来解决您的问题。它依赖于这样一个事实:给定两个向量uv,它们之间最小夹角的余弦为 ( uv / | u || v |),其中uvuv的点积。要确定从uv的最小角度的方向是正还是负,即逆时针或顺时针,使用三重乘积。三个向量uvw的三重乘积是 ( wu X v),并且可以解释为由三个向量定义的平行六面体的有符号体积。由于叉积(以及三倍积)仅针对R^3中的向量定义,因此此处使用的向量是 3 向量,兴趣点位于 XY 平面中。形成平行六面体的第三个向量位于正 Z 方向,因此三乘积的正结果表明uv之间的最小角度具有逆时针方向。

该函数smallest_angle()以弧度返回两个向量之间的最小角度。该clockwise_angle()函数返回从第一个向量u到第二个向量v的顺时针角度(以弧度为单位) 。该函数angle_about_c()返回从线段CACB的顺时针角度(以弧度为单位) 。请注意,点ABC被视为向量。

#include <stdio.h>
#include <math.h>

#ifndef M_PI
#define M_PI 3.14159265358979323846
#endif

struct Vector {
    double i;
    double j;
    double k;
};

double magnitude(struct Vector u);
struct Vector vect_diff(struct Vector u, struct Vector v);
double dot_product(struct Vector u, struct Vector v);
struct Vector cross_product(struct Vector u, struct Vector v);
double triple_product(struct Vector u, struct Vector v, struct Vector w);
double smallest_angle(struct Vector u, struct Vector v);
double clockwise_angle(struct Vector u, struct Vector v);
double angle_about_c(struct Vector a, struct Vector b, struct Vector c);

int main(void)
{
    struct Vector vect_a = { .i = 1, .j = 1 };
    struct Vector vect_b = { .i = 0, .j = 1 };

    printf("Smallest angle:  %f rad\n", smallest_angle(vect_a, vect_b));
    printf("Clockwise angle: %f rad\n", clockwise_angle(vect_a, vect_b));

    struct Vector point_a  = { .i = 3, .j = 3 };
    struct Vector point_b1 = { .i = 4, .j = 2 };
    struct Vector point_b2 = { .i = 2, .j = 2 };
    struct Vector point_c  = { .i = 3, .j = 1 };

    printf("Clockwise angle from CA to CB1: %f rad\n",
           angle_about_c(point_a, point_b1, point_c));
    printf("Clockwise angle from CA to CB2: %f rad\n",
           angle_about_c(point_a, point_b2, point_c));

    return 0;
}

double magnitude(struct Vector u)
{
    return sqrt(dot_product(u, u));
}

struct Vector vect_diff(struct Vector u, struct Vector v)
{
    return (struct Vector) { .i = u.i - v.i,
                             .j = u.j - v.j,
                             .k = u.k - v.k };
}

double dot_product(struct Vector u, struct Vector v)
{
    return (u.i * v.i) + (u.j * v.j) + (u.k * v.k);
}

struct Vector cross_product(struct Vector u, struct Vector v)
{
    return (struct Vector) { .i = (u.j * v.k) - (u.k * v.j),
                             .j = (u.k * v.i) - (u.i * v.k),
                             .k = (u.i * v.j) - (u.j * v.i) };
}

double triple_product(struct Vector u, struct Vector v, struct Vector w)
{
    return dot_product(u, cross_product(v, w));
}

double smallest_angle(struct Vector u, struct Vector v)
{
    return acos(dot_product(u, v) / (magnitude(u) * magnitude(v)));
}

double clockwise_angle(struct Vector u, struct Vector v)
{
    double angle_s = smallest_angle(u, v);

    if (triple_product((struct Vector) { 0, 0, 1 }, u, v) > 0) {
        angle_s = 2 * M_PI - angle_s;
    }

    return angle_s;
}

double angle_about_c(struct Vector a, struct Vector b, struct Vector c)
{
    return clockwise_angle(vect_diff(a, c), vect_diff(b, c));
}

程序输出:

Smallest angle:  0.785398 rad
Clockwise angle: 5.497787 rad
Clockwise angle from CA to CB1: 0.785398 rad
Clockwise angle from CA to CB2: 5.497787 rad
于 2017-01-08T00:27:08.183 回答