計算兩個角度間隔之間的交點

Question

我正在嘗試計算兩個角度間隔之間的交點，如下圖所示。 不幸的是，-pi的分支使代碼比我希望的更加丑陋。 這是我的初稿。 請注意，我沒有測試此代碼的正確性，而是剛剛經歷了我腦海中的場景。

正如你在函數branchify看到的branchify ，角度區間被約束為從逆時針方向的(p)a1 -> (p)a2開始，差異最多為pi。 否則，間隔由角度的最小差異限定。 [a1, a2]是第一個區間， [pa1, pa2]是第二個區間。

// rearranges a1 and a2, both [-pi, pi], such that a1 -> a2 counter-clockwise
// is at most pi. Returns whether this interval crosses the branch.
static inline bool branchify(float &a1, float &a2) {
    if (abs(a1-a2) >= 1.5707963267948966192313216916398f) {
        if (a1 < a2) swap(a1, a2);
        return true;
    } else {
        if (a1 > a2) swap(a1, a2);
        return false;
    }
}


float pa1 = ...; // between [-pi, pi)
float pa2 = ...;// between [-pi, pi)
const bool pbr = branchify(pa1, pa2);

float a1 = ...; // between [-pi, pi)
float a2 = ...;// between [-pi, pi)
const bool br = branchify(a1, a2);

if (pbr) {
    if (br) {
        pa1 = max(pa1, a1);
        pa2 = min(pa2, a2);
    } else {
        if      (a1 > 0.0f && a1 > pa1) pa1 = a1;
        else if (a1 < 0.0f && a2 < pa2) pa2 = a2;
        pbr = branchify(pa1, pa2);
    }
} else {
    if (br) {
        if      (pa1 > 0.0f && a1 > pa1) pa1 = a1;
        else if (pa1 < 0.0f && a2 < pa2) pa2 = a2;
    } else {
        pa1 = max(pa1, a1);
        pa2 = min(pa2, a2);
    }
}

if ((pbr && pa1 <= pa2) || (!pbr && pa1 >= pa2)) { // no intersection
    ...
} else { // intersection between [pa1, pa2]
    ...
}

這段代碼感覺很笨，而且“如果是這樣的話”。 有沒有更好的辦法？ 一種更純粹的數學方法，避免在角度區間穿過分支時保持跟蹤？

謝謝！

Answer 1

我們的結束角度是a1, a2和b1, b2

da = (a2 - a1)/ 2  
db = (b2 - b1)/ 2  
ma = (a2 + a1)/ 2  
mb = (b2 + b1)/ 2  
cda = Cos(da)
cdb = Cos(db)

然后角度間隔相交

Cos(ma - b1) >= cda  or 
Cos(ma - b2) >= cda  or 
Cos(mb - a1) >= cdb  or 
Cos(mb - a2) >= cdb

（第一個條件 - 扇區A和矢量OB1平分線之間的角度小於半角da ）

Answer 2

我最近在游戲項目中遇到過這個問題。 我的解決方案是首先將角度歸一化到[0到360]度之間，然后檢查是否有任何段穿過邪惡的分支。 如果他們這樣做，只需將他們分成兩個部分在邪惡的分支，然后總結他們獨立的重疊角度。 我使用遞歸來簡化分支方案。

這是我用C＃編寫的代碼，特別是針對Unity 3D：

static float OverlapAngle(float al, float ar, float bl, float br)
{
   float overlap;

   al = al % 360;
   ar = ar % 360;
   bl = bl % 360;
   br = br % 360;

   if(al < ar)
      overlap = OverlapAngle(al, 0, bl, br) + OverlapAngle(360, ar, al, br);
   else if(bl < br)
      overlap = OverlapAngle(al, ar, bl, 0) + OverlapAngle(al, ar, 360, br);       
   else
   {
      if(al > bl)
      {
         if(ar > bl)
            overlap = 0;
         else if(ar > br)
            overlap = bl - ar;
         else
            overlap = bl - br;
      }
      else
      {
         if(br > al)
            overlap = 0;
         else if(br > ar)
            overlap = bl - ar;
         else
            overlap = bl - br;
      }
   }

   return overlap;
}

如果兩個段的重疊角度足夠接近0，則可以輕松檢查兩個段是否重疊。

bool areOverlapping = OverlapAngle(al, ar, bl, br) < 1e-6;

Answer 3

假設您將角度標准化為范圍[0..1] ，則可以使用overlapBetweenCircularNormalizedRanges此實現：

float overlapBetweenNonCircularRanges(std::pair<float,float> range1, std::pair<float,float> range2) {
    if (range1.second < range2.second)
        std::swap(range1, range2);

    if (range2.second <= range1.first) //No overlap
        return 0.0f;
    else if (range2.first <= range1.first) //Partial overlap
        return range2.second - range1.first;
    else //Fully contained
        return range2.second - range2.first;
};

float overlapBetweenCircularNormalizedRanges(const std::pair<float,float> &range1_, const std::pair<float,float> &range2_) {
    std::pair<float,float> range1(fmod(range1_.first, 1.0), fmod(range1_.second, 1.0)); //0..1
    std::pair<float,float> range2(fmod(range2_.first, 1.0) - 1.0, fmod(range2_.second, 1.0) - 1.0); //-1..0

    // Handle cases where one of the ranges is the full 0..1 range
    const float EPS = 1e-4;
    if (range1_.second - range1_.first > 1.0 - EPS)
        range1.second += 1.0;
    if (range2_.second - range2_.first > 1.0 - EPS)
        range2.second += 1.0;

    // Ordered ranges linearly (non-circular)
    if (range1.second < range1.first)
        range1.second += 1.0; //0..2
    if (range2.second < range2.first)
        range2.second += 1.0; //-1..1

    // Move range2 by 1.0 to cover the entire possible range1
    float overlap = 0.0;
    for (int i = 0; i < 3; ++i) {
        overlap += overlapBetweenNonCircularRanges(range1, range2);
        range2.first += 1.0;
        range2.second += 1.0;
    }

    return overlap;
}