单应性 HomographyHomographies

相机把三维世界压成二维图像，深度丢了，一般无法反过来。但有一种特殊情形，反过来完全没有歧义：被看的点都落在同一个平面上。地面、桌面、墙、棋盘格、一页文档、AprilTag 的黑白方块，都是平面。同一个平面被两台相机拍到，两幅图像里对应点之间的关系就是一个 3×3 矩阵 \(H\)：\(\lambda\tilde{\mathbf x}'=H\tilde{\mathbf x}\)，这个矩阵叫单应性（homography）。

另一种同样"免费"的情形是相机只转动、不平移（原地环顾、云台转动、手机拍全景）。这时场景有多深都无所谓，两幅图像也只差一个单应性。这个 3×3 矩阵撑起了机器人视觉里一长串应用：

• 鸟瞰图（逆透视映射）：把前视相机的地面像素映成俯视的度量地图，用于车道线、AMR 导航和可通行区域（FIG 5）；
• 障碍物检测：地面上的像素服从 \(H\)，不服从的就是"高出地面"的东西（FIG 2）；
• 全景拼接：旋转相机拍的多张图合成一张（FIG 3）；
• 平面标记的位姿：AprilTag / ArUco 的四个角点 → \(H\) → 标记相对相机的位姿；文档与白板的矫正也是同一件事；
• 平面标定：10.4 节中每张棋盘格照片给出一个 \(H\)，Zhang 方法从多个 \(H\) 里解出内参 \(K\)；
• SLAM 初始化：场景接近平面或相机几乎纯旋转时，对极几何（11.1 节）退化，要改用 \(H\)。

本节先用四个手柄建立直觉（FIG 1），看 DLT 如何从四个点对解出矩阵；再从几何推出 \(H=K'(R+\mathbf t\mathbf n\T/d)K^{-1}\) 并让平面之外的点"露馅"（FIG 2）；然后看纯旋转（FIG 3）、噪声与归一化（FIG 4）；最后在一个合成的公路场景上做鸟瞰图（FIG 5）。

A camera squashes the 3D world into 2D and the depth is lost, so in general you cannot go back. One special case has no ambiguity at all: all the points you look at lie on one plane. The ground, a tabletop, a wall, a checkerboard, a page of a document, the black-and-white square of an AprilTag are all planes. When two cameras see the same plane, the relation between corresponding points in the two images is a single 3×3 matrix \(H\): \(\lambda\tilde{\mathbf x}'=H\tilde{\mathbf x}\). That matrix is called a homography.

A second case is just as "free": the camera only rotates and does not translate (looking around from one spot, a pan-tilt head, a phone taking a panorama). Then it does not matter how deep the scene is: the two images again differ by a homography. This one 3×3 matrix carries a long list of robot-vision applications:

• Bird's-eye view (inverse perspective mapping): turn the ground pixels of a forward camera into a top-down metric map for lane lines, AMR navigation and drivable area (FIG 5);
• Obstacle detection: pixels on the ground obey \(H\), the ones that do not stand above the ground (FIG 2);
• Panorama stitching: images from a rotating camera merge into one (FIG 3);
• Pose of a planar marker: the four corners of an AprilTag / ArUco → \(H\) → the marker's pose relative to the camera; document and whiteboard rectification is the same computation;
• Planar calibration: in Lesson 10.4 every checkerboard photo yields one \(H\), and Zhang's method recovers the intrinsics \(K\) from several of them;
• SLAM initialization: when the scene is nearly planar or the camera nearly purely rotates, epipolar geometry (Lesson 11.1) degenerates and \(H\) takes over.

We first build intuition with four handles (FIG 1) and watch the DLT solve the matrix from four point pairs; then derive \(H=K'(R+\mathbf t\mathbf n\T/d)K^{-1}\) from the geometry and let points off the plane give themselves away (FIG 2); then pure rotation (FIG 3), noise and normalization (FIG 4); finally a bird's-eye view of a synthetic road (FIG 5).

想象把一张印好的图平放在桌上，从斜上方拍照：矩形变成了梯形，近大远小，直线仍是直线，平行线却可能汇聚。左图是原图，右图是你"拍到"的样子。拖动右图的四个角，每个角对应原图的一个角。四个点对 \((\mathbf x_i\mapsto\mathbf x_i')\) 恰好够解一个 \(H\)：每个点对给两个方程，\(H\) 有 9 个元素，但整体乘一个常数不改变映射，只剩 8 个自由度，\(4\times2=8\)。页面上每一帧都在运行真正的 DLT 算法（见下面的数学部分），再用 \(H^{-1}\) 反向映射每个输出像素到原图并做双线性采样，得到右图。

注意 \(H\) 的列有几何意义：第 1、2 列是原图 \(x\) 轴、\(y\) 轴方向无穷远点的像，也就是消失点（10.2 节）；第 3 列是原点的像。当四边形变成平行四边形，\(H\) 的最后一行变成 \((0,0,1)\)，消失点跑到无穷远，映射退化为仿射。

Lay a printed picture flat on a table and photograph it from an oblique angle: the rectangle becomes a trapezoid, near parts are larger, straight lines stay straight and parallel lines may converge. The left image is the original, the right image is "what the camera sees". Drag the four corners of the right image; each corresponds to one corner of the original. Four pairs \((\mathbf x_i\mapsto\mathbf x_i')\) are exactly enough for one \(H\): each pair gives two equations, \(H\) has nine entries, but an overall scale changes nothing, so only 8 degrees of freedom remain, and \(4\times2=8\). Every frame runs the real DLT algorithm (derived in the math section), then maps every output pixel backwards through \(H^{-1}\) into the original and samples it bilinearly to produce the right image.

The columns of \(H\) have a geometric meaning: columns 1 and 2 are the images of the points at infinity of the original's \(x\) and \(y\) axes, that is, vanishing points (Lesson 10.2); column 3 is the image of the origin. When the quadrilateral becomes a parallelogram the last row of \(H\) becomes \((0,0,1)\), the vanishing points run off to infinity, and the map degenerates to an affine one.

FIG 1 里的 \(H\) 是"拟合"出来的。真实的相机对里，它由几何完全决定。设相机 1 坐标系中有平面 \(\mathbf n\T\mathbf X_1=d\)（\(\mathbf n\) 单位法向，\(d\) 是相机 1 光心到平面的距离），相机 2 与相机 1 的关系为 \(\mathbf X_2=R\mathbf X_1+\mathbf t\)。则（见下面的数学部分） $$H=K'\Big(R+\frac{\hl{t}{\mathbf t}\,\hl{n}{\mathbf n}\T}{\hl{d}{d}}\Big)K^{-1}.$$ 下面的实验里，相机 1、2 看着一块平面（可以倾斜），平面中心立着一根杆。你可以移动相机 2、倾斜平面、改变杆高，页面做三件事：用上面的公式由几何算出 \(H\)；用平面上 25 个点的投影，用 DLT 独立解出 \(H\)；然后用 \(H\) 把相机 1 的点搬到相机 2。平面上的点严丝合缝（误差 \(10^{-12}\) 像素量级）；杆顶不在平面上，被 \(H\) 搬到的位置是错的，这个误差就是视差。

几何解释在 3D 视图里：\(H\) 把相机 1 的射线"压"到平面上，再从相机 2 看这个平面点。杆顶的射线先落在平面的 \(\mathbf Q\)（紫色），\(H\) 预测相机 2 在 \(\mathbf Q\) 的位置看到杆顶，而真正的杆顶在别处。这就是"地面单应性残差"检测障碍物的原理：地面像素残差为零，高出地面的物体残差正比于高度和基线。

The \(H\) in FIG 1 was "fitted". For a real camera pair it is completely determined by the geometry. Let the plane be \(\mathbf n\T\mathbf X_1=d\) in camera 1's frame (\(\mathbf n\) a unit normal, \(d\) the distance from camera 1's centre to the plane) and let camera 2 be related by \(\mathbf X_2=R\mathbf X_1+\mathbf t\). Then (derived in the math section) $$H=K'\Big(R+\frac{\hl{t}{\mathbf t}\,\hl{n}{\mathbf n}\T}{\hl{d}{d}}\Big)K^{-1}.$$ In the lab below, cameras 1 and 2 look at a plane (which you can tilt) with a pole standing at its middle. Move camera 2, tilt the plane, change the pole height; the page does three things: it computes \(H\) from the geometry with the formula above; it solves \(H\) independently with the DLT from the projections of 25 points on the plane; and it uses \(H\) to carry points from camera 1 to camera 2. Points on the plane match perfectly (errors of order \(10^{-12}\) px); the pole tip is not on the plane, \(H\) carries it to the wrong place, and that error is the parallax.

The 3D view shows why: \(H\) "pushes" camera 1's ray onto the plane and then looks at that plane point from camera 2. The ray through the pole tip first meets the plane at \(\mathbf Q\) (purple); \(H\) predicts that camera 2 sees the tip where it sees \(\mathbf Q\), but the real tip is elsewhere. That is the principle of detecting obstacles by the ground-homography residual: ground pixels have zero residual, objects above the ground have a residual proportional to their height and to the baseline.

让 \(\mathbf t=\mathbf 0\)，公式里的 \(\mathbf t\mathbf n\T/d\) 整项消失：\(H=K'RK^{-1}\)，\(\mathbf n\) 和 \(d\) 都不出现，场景可以是任意的三维结构。直观地说，绕光心旋转时每条射线只是转了个方向，物体多远都沿同一条射线成像。下面的场景里有 60 个深度从 3 m 到 25 m 的点。滑块让相机绕竖直轴转 \(\theta\)，并（或不）向右平移 \(t_x\)。紫色圆圈是用 DLT 在所有点上拟合的最优单应性的预测；纯旋转时它与真实点重合，哪怕深度差了 8 倍；一旦平移，近处的点移动得更多（\(\approx f\,t_x/Z\)），任何一个 \(H\) 都解释不了。

反过来，由 \(H\) 还能读出旋转：\(R=K^{-1}HK\)（先把 \(H\) 乘上 \(\det^{-1/3}\) 使 \(\det=1\)），再用 \(\theta=\arccos\frac{\operatorname{tr}R-1}2\) 或 5.6 节的 \(\Log\)。全景拼接就是这样估计相邻图像的相对旋转的（Brown–Lowe，2007）。

Set \(\mathbf t=\mathbf 0\) and the whole term \(\mathbf t\mathbf n\T/d\) disappears: \(H=K'RK^{-1}\); neither \(\mathbf n\) nor \(d\) appears, and the scene may have any 3D structure. Intuitively, when the camera rotates about its centre each ray merely changes direction, and a point images along the same ray however far away it is. The scene below has 60 points at depths from 3 m to 25 m. The sliders rotate the camera about the vertical axis by \(\theta\) and (or not) translate it to the right by \(t_x\). The purple circles are the predictions of the best homography fitted to all points with the DLT; under pure rotation they coincide with the true points even though depths differ by a factor of eight. Once you translate, near points move more (\(\approx f\,t_x/Z\)) and no single \(H\) explains the image.

Conversely, \(H\) reveals the rotation: \(R=K^{-1}HK\) (after scaling \(H\) by \(\det^{-1/3}\) so that \(\det=1\)), then \(\theta=\arccos\frac{\operatorname{tr}R-1}2\) or the \(\Log\) of Lesson 5.6. This is how panorama stitching estimates the relative rotation of neighbouring images (Brown–Lowe, 2007).

真实数据有噪声，点数多于 4，还有错配。下面把图像 2 中的点加上高斯噪声 \(\sigma\)，分别用Hartley 归一化的 DLT（青色）和直接用像素坐标的 DLT（红色）估计 \(H\)，用整幅 640×480 图像上一个 9×7 测试网格上的均方根误差（灰色是真值）评价。有三个旋钮：点数、点占图像的比例（"分布"）、噪声。要看到的现象是：点铺满图像时两者差不多；点挤在一个小区域且离坐标原点很远时，未归一化的 DLT 条件数 \(\kappa(A)\) 暴涨，误差大一个数量级；而不论哪个，外推到没有点的区域都是危险的。"加外点"按钮把个别点偏移 50 像素量级，最小二乘立刻被拖偏；"RANSAC"用 4 点采样、重复拟合找出内点（11.5 节详细讲）。

Real data are noisy, come in more than four pairs, and contain mismatches. Below, Gaussian noise \(\sigma\) is added to the points of image 2 and \(H\) is estimated twice: by the Hartley-normalized DLT (cyan) and by the DLT on raw pixel coordinates (red). Both are scored by the RMS error on a 9×7 test grid over the whole 640×480 image (grey is the truth). Three knobs: number of points, the fraction of the image they occupy ("spread") and the noise. What to see: when points fill the image the two agree closely; when points are packed into a small region far from the coordinate origin, the raw DLT's condition number \(\kappa(A)\) explodes and its error is an order of magnitude larger; and in either case extrapolating to regions without points is dangerous. "Add outlier" shifts individual points by about 50 px and least squares is pulled off at once; "RANSAC" samples 4 points, fits repeatedly and keeps the inliers (Lesson 11.5 treats it in detail).

用 10.2 节的齐次坐标 \(\tilde{\mathbf x}=(x,y,1)\T\)，单应性是一个可逆的 3×3 矩阵作用在 \(\mathbb P^2\) 上：

In the homogeneous coordinates \(\tilde{\mathbf x}=(x,y,1)\T\) of Lesson 10.2, a homography is an invertible 3×3 matrix acting on \(\mathbb P^2\):

\(cH\)（\(c\neq0\)）给出同一个映射，所以 9 个元素只有 8 个自由度，每个点对提供 2 个方程，至少 4 个点对。分母 \(w=h_{31}x+h_{32}y+h_{33}\) 就是 FIG 1 里的"\(w\) 比值"，除以 \(w\) 这一步让映射成为非线性的。\(H\) 的最后一行为 \((0,0,1)\) 时是仿射（6 个自由度，3 个点对，平行线保持平行）；再要求左上 2×2 是旋转乘缩放时是相似变换（4 个自由度，2 个点对）。复合就是矩阵乘法：\(H_{13}=H_{23}H_{12}\)，反向是 \(H^{-1}\)。

\(cH\) (\(c\neq0\)) gives the same map, so the nine entries carry only 8 degrees of freedom; each point pair supplies 2 equations, so at least 4 pairs are needed. The denominator \(w=h_{31}x+h_{32}y+h_{33}\) is FIG 1's "\(w\) ratio", and dividing by \(w\) is what makes the map nonlinear. If the last row of \(H\) is \((0,0,1)\) the map is affine (6 DoF, 3 pairs, parallelism preserved); if the upper-left 2×2 is additionally a rotation times a scale it is a similarity (4 DoF, 2 pairs). Composition is matrix multiplication, \(H_{13}=H_{23}H_{12}\), and the reverse direction is \(H^{-1}\).

\(\lambda\) 未知，所以不写 \(\tilde{\mathbf x}'=H\tilde{\mathbf x}\)，而写"两个向量平行"：\(\tilde{\mathbf x}'\times H\tilde{\mathbf x}=\mathbf 0\)（叉积 1.8 节）。记 \(\tilde{\mathbf x}'=(u',v',1)\T\)，\(H\) 的三行为 \(\mathbf h_1\T,\mathbf h_2\T,\mathbf h_3\T\)，展开：

\(\lambda\) is unknown, so instead of \(\tilde{\mathbf x}'=H\tilde{\mathbf x}\) we state that the two vectors are parallel: \(\tilde{\mathbf x}'\times H\tilde{\mathbf x}=\mathbf 0\) (cross product, Lesson 1.8). Write \(\tilde{\mathbf x}'=(u',v',1)\T\) and let \(\mathbf h_1\T,\mathbf h_2\T,\mathbf h_3\T\) be the rows of \(H\). Expanding:

第三行等于前两行的线性组合（\(u'\)×第一行 + \(v'\)×第二行 \(=-\)第三行），所以每个点对只有 2 个独立方程，这就是"8 = 4×2"。把 \(n\) 个点对的 \(A_i\) 叠成 \(A\in\R^{2n\times9}\)：\(A\mathbf h=\mathbf 0\)。恰好 4 个点（任意三点不共线）时 \(A\) 秩为 8，零空间（1.6 节）是一维的，\(\mathbf h\) 唯一（相差一个常数）；\(n>4\) 时方程无精确解，取

The third row is a linear combination of the first two (\(u'\times\)row 1 \(+\,v'\times\)row 2 \(=-\)row 3), so every pair contributes only 2 independent equations: that is the "8 = 4×2". Stack the \(A_i\) of \(n\) pairs into \(A\in\R^{2n\times9}\): \(A\mathbf h=\mathbf 0\). With exactly 4 points (no three collinear) \(A\) has rank 8, its null space (Lesson 1.6) is one-dimensional and \(\mathbf h\) is unique up to scale; for \(n>4\) there is no exact solution and we take

（1.12 节的 SVD。）若三点在两幅图里都共线，\(A\) 的秩降到 7，\(\sigma_8\approx0\)，零空间多出一维，\(H\) 不唯一；只在一幅图里共线时 \(A\) 仍满秩，但解出的 \(H\) 是奇异的（FIG 1 读数里的 \(\sigma_3(H)/\sigma_1(H)\)）。\(\|A\mathbf h\|\) 叫代数误差，它没有几何含义，只是方便；本节末尾会回到几何误差。

(The SVD of Lesson 1.12.) If three points are collinear in both images, the rank drops to 7, \(\sigma_8\approx0\), the null space gains a dimension and \(H\) is not unique; if they are collinear in one image only, \(A\) keeps full rank but the solution \(H\) is singular (the \(\sigma_3(H)/\sigma_1(H)\) readout of FIG 1). \(\|A\mathbf h\|\) is the algebraic error: it has no geometric meaning and is merely convenient; we return to geometric errors at the end of this section.

像素坐标在百的量级，\(A\) 的一行里同时出现 \(1\)、\(x\sim10^2\)、\(u'x\sim10^4\) 三种尺度，\(\kappa(A)\) 高达 \(10^6\) 以上，SVD 的最小奇异向量对噪声极其敏感（FIG 4 读数）。办法：对两幅图各做一次相似变换 \(T,T'\)——平移使质心在原点，缩放使到原点的平均距离是 \(\sqrt2\)——在归一化坐标里解出 \(\hat H\)，再换回来：

Pixel coordinates are in the hundreds, so one row of \(A\) mixes the scales \(1\), \(x\sim10^2\) and \(u'x\sim10^4\); \(\kappa(A)\) exceeds \(10^6\), and the smallest singular vector reacts violently to noise (see FIG 4's readout). The remedy: apply a similarity \(T,T'\) to each image, translating the centroid to the origin and scaling the mean distance from the origin to \(\sqrt2\); solve for \(\hat H\) in normalized coordinates, then transform back:

它不增加点数，也不改变理想数据的解（4 个无噪声点时两种 DLT 一样准），只改善条件数，代价是几行代码。Hartley（1997）为八点算法提出，同一思想适用于所有 DLT。

It adds no points and does not change the solution for ideal data (with four noise-free points both DLTs are equally exact); it only improves the conditioning, at the cost of a few lines of code. Hartley (1997) proposed it for the eight-point algorithm, and the same idea applies to every DLT.

相机 2 的坐标 \(\mathbf X_2=R\mathbf X_1+\mathbf t\)（与 5.3 节一致：\(R,\mathbf t\) 是相机 1 → 相机 2 的相对位姿），平面 \(\mathbf n\T\mathbf X_1=d\)，\(\mathbf n\) 是相机 1 坐标系里的单位法向，\(d>0\)。平面上的点满足 \(\mathbf n\T\mathbf X_1/d=1\)，所以可以把 \(\mathbf t\) 乘上这个"1"：

Camera 2's coordinates are \(\mathbf X_2=R\mathbf X_1+\mathbf t\) (as in Lesson 5.3: \(R,\mathbf t\) is the relative pose camera 1 → camera 2) and the plane is \(\mathbf n\T\mathbf X_1=d\), \(\mathbf n\) a unit normal in camera 1's frame, \(d>0\). Points on the plane satisfy \(\mathbf n\T\mathbf X_1/d=1\), so \(\mathbf t\) may be multiplied by this "1":

用的是 \(\mathbf X_1=\lambda K^{-1}\tilde{\mathbf x}_1\)，\(\lambda\) 被"\(\sim\)"吞掉了。几点说明。

• 符号：这里平面写成 \(\mathbf n\T\mathbf X=d\)，所以是加号。Hartley–Zisserman 把平面写成 \(\mathbf n\T\mathbf X+d=0\)，得到 \(R-\mathbf t\mathbf n\T/d\)。FIG 2 对加号做了数值验证；换号会差几百像素。
• 只出现 \(\mathbf t/d\)：把整个场景和基线同时放大 2 倍，\(H\) 不变，这就是单目的尺度不确定。
• 欧氏单应性：\(K'^{-1}HK=R+\mathbf t\mathbf n\T/d\) 只含运动与平面。分解它可得 \((R,\mathbf t/d,\mathbf n)\)，一般有至多 4 组数学解，要求点在两个相机前方后通常剩 2 组，第三个视角或已知法向再消歧（Malis–Vargas 2007；OpenCV 的 decomposeHomographyMat）。
• 平面外的点：若 \(\mathbf n\T\mathbf X_1=d'\neq d\)，则 \(\mathbf X_2=(R+\mathbf t\mathbf n\T/d)\mathbf X_1+\mathbf t\,(1-d'/d)\)。多出来的 \(\mathbf t(1-d'/d)\) 就是视差：正比于基线 \(\mathbf t\) 和"离平面有多远"，方向沿着极线（指向极点，11.1 节）。\(\mathbf t=\mathbf 0\) 时为零。

This uses \(\mathbf X_1=\lambda K^{-1}\tilde{\mathbf x}_1\), with \(\lambda\) swallowed by "\(\sim\)". A few remarks.

• Sign: the plane is written \(\mathbf n\T\mathbf X=d\) here, hence the plus. Hartley–Zisserman write the plane as \(\mathbf n\T\mathbf X+d=0\) and obtain \(R-\mathbf t\mathbf n\T/d\). FIG 2 verifies the plus sign numerically; the wrong sign is off by hundreds of pixels.
• Only \(\mathbf t/d\) appears: scaling the scene and the baseline by 2 leaves \(H\) unchanged, which is the scale ambiguity of monocular vision.
• Euclidean homography: \(K'^{-1}HK=R+\mathbf t\mathbf n\T/d\) contains only motion and plane. Decomposing it yields \((R,\mathbf t/d,\mathbf n)\), with up to 4 mathematical solutions, usually 2 after requiring the points to be in front of both cameras, and a third view or a known normal resolves the rest (Malis–Vargas 2007; OpenCV's decomposeHomographyMat).
• Points off the plane: if \(\mathbf n\T\mathbf X_1=d'\neq d\), then \(\mathbf X_2=(R+\mathbf t\mathbf n\T/d)\mathbf X_1+\mathbf t\,(1-d'/d)\). The extra \(\mathbf t(1-d'/d)\) is the parallax: proportional to the baseline \(\mathbf t\) and to the distance from the plane, directed along the epipolar line (towards the epipole, Lesson 11.1). It vanishes for \(\mathbf t=\mathbf 0\).

纯旋转：\(d\to\infty\)（平面在无穷远）与 \(\mathbf t=\mathbf 0\) 效果相同，\(H_\infty=K'RK^{-1}\)。相机绕光心转 \(\theta\)，\(K\) 相同时，图像中心移动约 \(f\tan\theta\) 像素（小角时 \(\approx f\theta\)），与深度无关。地面：世界平面 \(Z=0\) 上的点 \(\mathbf X=(X,Y,0)\)，\(\mathbf x_c=R\mathbf X+\mathbf t=\mathbf r_1X+\mathbf r_2Y+\mathbf t\)，\(R\) 的第三列 \(\mathbf r_3\) 乘 0 消失，所以 \(H_g=K[\mathbf r_1\ \mathbf r_2\ \mathbf t]\)，这里 \((R,\mathbf t)\) 是世界 → 相机的外参 \({}^cT_w\)（10.3 节）。对高 \(h\)、俯仰角 \(\alpha\) 向前看的相机（世界 \(X\) 向右、\(Y\) 向前、\(Z\) 向上）：

Pure rotation: \(d\to\infty\) (the plane at infinity) acts like \(\mathbf t=\mathbf 0\), giving \(H_\infty=K'RK^{-1}\). If the camera rotates by \(\theta\) about its centre, with equal \(K\), the image centre moves by about \(f\tan\theta\) pixels (\(\approx f\theta\) for small angles), independent of depth. The ground: a point \(\mathbf X=(X,Y,0)\) of the world plane \(Z=0\) has \(\mathbf x_c=R\mathbf X+\mathbf t=\mathbf r_1X+\mathbf r_2Y+\mathbf t\); the third column \(\mathbf r_3\) is multiplied by 0 and drops out, so \(H_g=K[\mathbf r_1\ \mathbf r_2\ \mathbf t]\), where \((R,\mathbf t)\) are the extrinsics world → camera \({}^cT_w\) (Lesson 10.3). For a forward-looking camera at height \(h\), pitched down by \(\alpha\) (world \(X\) right, \(Y\) forward, \(Z\) up):

\(Y\to\infty\) 得到地平线 \(v=c_y-f\tan\alpha\)。反过来，从 \(H\) 求相机位姿（平面目标的位姿，AprilTag 与 10.4 节标定的第一步）：\(K^{-1}H_g\propto[\mathbf r_1\ \mathbf r_2\ \mathbf t]\)，取 \(\lambda=1/\|K^{-1}\mathbf h_1\|\) 使 \(\mathbf r_1\) 为单位向量，\(\mathbf r_3=\mathbf r_1\times\mathbf r_2\)，再用 SVD 把 \([\mathbf r_1\ \mathbf r_2\ \mathbf r_3]\) 投影到最近的旋转矩阵（\(R=UV\T\)，1.12 节），选符号使 \(t_z>0\)（目标在相机前方）。FIG 5 的读数用 4 个点的 DLT 恢复出相机的高度和俯仰角，就是这一段。有噪声时更好的做法是 PnP（11.4 节）或迭代优化。

\(Y\to\infty\) gives the horizon \(v=c_y-f\tan\alpha\). Conversely, recovering the camera pose from \(H\) (the pose of a planar target, the first step of AprilTag and of Lesson 10.4's calibration): \(K^{-1}H_g\propto[\mathbf r_1\ \mathbf r_2\ \mathbf t]\); take \(\lambda=1/\|K^{-1}\mathbf h_1\|\) so that \(\mathbf r_1\) is a unit vector, set \(\mathbf r_3=\mathbf r_1\times\mathbf r_2\), project \([\mathbf r_1\ \mathbf r_2\ \mathbf r_3]\) onto the nearest rotation by SVD (\(R=UV\T\), Lesson 1.12) and pick the sign with \(t_z>0\) (the target is in front of the camera). FIG 5's readout recovers the camera height and pitch from a 4-point DLT with exactly this recipe. With noisy data PnP (Lesson 11.4) or iterative refinement does better.

DLT 最小化的 \(\|A\mathbf h\|\) 是代数量。真正想要的是像素误差。转移误差 \(\sum_i\|\mathbf x_i'-\pi(H\tilde{\mathbf x}_i)\|^2\)（\(\pi\) 是除以 \(w\)）只算图像 2 的误差；对称转移误差再加上 \(\|\mathbf x_i-\pi(H^{-1}\tilde{\mathbf x}_i')\|^2\)；若两幅图的点都有噪声，最大似然是重投影误差：同时估计 \(H\) 和精确点 \(\hat{\mathbf x}_i,\hat{\mathbf x}_i'\)（Sampson 距离是它的一阶近似）。后两种是非线性最小二乘，用 6.6、6.9 节的高斯–牛顿 / LM，初值取 DLT 结果。OpenCV 的 findHomography 在 RANSAC 之后也会在内点上做 LM 精化。

What the DLT minimizes, \(\|A\mathbf h\|\), is an algebraic quantity; what you really want is pixel error. The transfer error \(\sum_i\|\mathbf x_i'-\pi(H\tilde{\mathbf x}_i)\|^2\) (\(\pi\) divides by \(w\)) counts the error in image 2 only; the symmetric transfer error adds \(\|\mathbf x_i-\pi(H^{-1}\tilde{\mathbf x}_i')\|^2\); if points in both images are noisy, the maximum-likelihood criterion is the reprojection error: estimate \(H\) and the exact points \(\hat{\mathbf x}_i,\hat{\mathbf x}_i'\) together (the Sampson distance is its first-order approximation). The latter two are nonlinear least squares solved by Gauss–Newton / LM (Lessons 6.6, 6.9) started from the DLT result. OpenCV's findHomography also runs an LM refinement on the inliers after RANSAC.

前视相机在路上看到的车道线汇聚到消失点，这对"在地图上走多远、车道多宽"毫无帮助。如果假设地面是平面，则 \(H_g\) 把地面坐标 \((X,Y)\)（米）映到像素，它的逆 \(H_g^{-1}\) 把每个像素映回地面米制坐标。对鸟瞰图的每个输出像素 \((X,Y)\)，用 \(H_g\) 找到它在相机图像中的位置采样，就得到一张俯视的度量图：车道线平行、宽度恒为 3.5 m、距离刻度均匀。这就是反向映射（输出 → 输入），不会留下空洞。下面是一个合成的两车道公路（所有像素由射线与地面求交渲染），相机高 \(h\)、俯仰 \(\alpha\) 可调。\(H_g\) 有两种来源：公式 \(K[\mathbf r_1\ \mathbf r_2\ \mathbf t]\)（已知安装高度和俯仰），或者四个手柄：在图中点出地面已知矩形（车道线与两条横向标线的交点，3.5 m × 10 m）的四个角，用 DLT 求 \(H\)，这是没有标定数据时的做法。

但这只对地面上的东西成立。高出地面的物体（琥珀色箱子）违反了平面假设：射线穿过它的顶部，落在地面上更远的位置，于是箱子被沿着视线方向拖长。侧视图里的相似三角形给出顶端落点 \(d'=d\,\dfrac{h}{h-z_b}\)（\(d\) 是水平距离，\(z_b\) 是高度）。

The lane lines seen by a forward camera converge to a vanishing point, which is no help for "how far to drive" or "how wide is the lane". If we assume the ground is a plane, \(H_g\) maps ground coordinates \((X,Y)\) in metres to pixels, and its inverse \(H_g^{-1}\) maps every pixel back to metric ground coordinates. For every output pixel \((X,Y)\) of the bird's-eye view, \(H_g\) tells where to sample the camera image, producing a top-down metric picture: lane lines parallel, lane width a constant 3.5 m, uniform distance ticks. That is inverse mapping (output → input), so no holes appear. Below is a synthetic two-lane road (every pixel is rendered by intersecting rays with the ground) with adjustable camera height \(h\) and pitch \(\alpha\). \(H_g\) can come from two sources: the formula \(K[\mathbf r_1\ \mathbf r_2\ \mathbf t]\) (mounting height and pitch known) or four handles: click the four corners of a known rectangle on the ground (the crossings of the lane lines with two painted cross-bars, 3.5 m × 10 m) and run the DLT, which is what you do when there is no calibration data.

But this holds only for things on the ground. An object standing above the ground (the amber box) violates the planar assumption: the ray through its top continues to a farther ground point, so the box is smeared along the line of sight. The similar triangles of the side view give the landing point of the top \(d'=d\,\dfrac{h}{h-z_b}\) (\(d\) the horizontal distance, \(z_b\) the height).