Objective c 在飞机上拦截日出
我想计算飞机与日出或日落之间最接近的预计时间,记住: 红线是飞机上的大圆轨道。 蓝色的圆圈是飞机。 1-已知太阳赤纬度(纬度)和交叉经度,加上日出半径约为5450海里,因此日出可以显示为一个圆心和半径已知的圆 2-我使用了2D矢量代码,因为大圆路径不能应用于XY平面,所以该代码不起作用 2-飞机在弯曲的大圆轨道上飞行,纬度变化不是线性的,如果纬度变化不是恒定的,我如何使用飞机速度作为速度矢量Objective c 在飞机上拦截日出,objective-c,swift,great-circle,Objective C,Swift,Great Circle,我想计算飞机与日出或日落之间最接近的预计时间,记住: 红线是飞机上的大圆轨道。 蓝色的圆圈是飞机。 1-已知太阳赤纬度(纬度)和交叉经度,加上日出半径约为5450海里,因此日出可以显示为一个圆心和半径已知的圆 2-我使用了2D矢量代码,因为大圆路径不能应用于XY平面,所以该代码不起作用 2-飞机在弯曲的大圆轨道上飞行,纬度变化不是线性的,如果纬度变化不是恒定的,我如何使用飞机速度作为速度矢量 /// Va - Velocity of circle A. Va = new Vect
/// Va - Velocity of circle A.
Va = new Vector2(450, 0);
我用的是c代码
我要找的就是算法。以及如何将飞机速度表示为Velocity Vector2D或Vector3D。我接受任何语言的答案,JAVA、JS、Objective C、Swift、,
/// Calculate the time of closest approach of two moving circles. Also determine if the circles collide.
///
/// Input:
/// Pa - Position of circle A.
/// Pb - Position of circle B.
/// Va - Velocity of circle A.
/// Vb - Velocity of circle B.
/// Ra - Radius of circle A.
/// Rb - Radius of circle B.
// Set up the initial position, velocity, and size of the circles.
Pa = new Vector2(150, 250);
Pb = new Vector2(600, 400);
Va = new Vector2(450, 0);
Vb = new Vector2(-100, -250);
Ra = 60;
Rb = 30;
/// Returns:
/// collision - Returns True if a collision occured, else False.
/// The method returns the time to impact if collision=true, else it returns the time of closest approach.
public float TimeOfClosestApproach(Vector2 Pa, Vector2 Pb, Vector2 Va, Vector2 Vb, float Ra, float Rb, out bool collision)
{
Vector2 Pab = Pa - Pb;
Vector2 Vab = Va - Vb;
float a = Vector2.Dot(Vab, Vab);
float b = 2 * Vector2.Dot(Pab, Vab);
float c = Vector2.Dot(Pab, Pab) - (Ra + Rb) * (Ra + Rb);
// The quadratic discriminant.
float discriminant = b * b - 4 * a * c;
// Case 1:
// If the discriminant is negative, then there are no real roots, so there is no collision. The time of
// closest approach is then given by the average of the imaginary roots, which is: t = -b / 2a
float t;
if (discriminant < 0)
{
t = -b / (2 * a);
collision = false;
}
else
{
// Case 2 and 3:
// If the discriminant is zero, then there is exactly one real root, meaning that the circles just grazed each other. If the
// discriminant is positive, then there are two real roots, meaning that the circles penetrate each other. In that case, the
// smallest of the two roots is the initial time of impact. We handle these two cases identically.
float t0 = (-b + (float)Math.Sqrt(discriminant)) / (2 * a);
float t1 = (-b - (float)Math.Sqrt(discriminant)) / (2 * a);
t = Math.Min(t0, t1);
// We also have to check if the time to impact is negative. If it is negative, then that means that the collision
// occured in the past. Since we're only concerned about future events, we say that no collision occurs if t < 0.
if (t < 0)
collision = false;
else
collision = true;
}
// Finally, if the time is negative, then set it to zero, because, again, we want this function to respond only to future events.
if (t < 0)
t = 0;
return t;
}
JAVA , JS, Objective-C , Swift , C#.