Texture for 9-9 project[3]

Change Externals matrixPermalink

• 相机中心点
  • 9-9/1-3/
    • $A_{amap}=(120.394685,36.06414)$
    • $A_{cams}=(120.389564,36.063963)$
    • $A_{model}=(79.1888,45.546)$meter
• 模型原点
  • O
    • $O_{amap}=(120.394637,36.064383)$
    • $O_{cams}=o_{amap} + (A_{cams} - A_{amap})$

• 其中
  • 经纬转换
    • Formula 1
      • how-to-convert-latitude-or-longitude-to-meters Stackoverflow
      • $$\begin{aligned} \frac{latitude}{degree}&=111320m/\degree\\ \frac{longitude}{degree}&=40075km\frac{\cos(lat)}{360}=40075km\frac{\cos(36.064\frac{\pi}{180})}{360}\\ &=90013.62031373613m/\degree \end{aligned}$$

          f_lai = 111320
          f_lon = 40075*math.cos(math.pi*36.04/180)/360
        

    • Formula 2
      • SSR WGS84
      • 
      • Geographic coordinate system - Wikipedia
      • $$\begin{aligned} \frac{latitude}{degree}&=111132.92-559.82\cos(2lat)+1.175\cos(4lat) - 0.0023\cos(6lat)\\ &=110959.719m/\degree\\ \frac{longitude}{degree}&=111412.84\cos(lat)-93.5\cos(3lat)+0.118\cos(5lat)\\ &=90118.10182968706m/\degree\\ \end{aligned}$$

      c = lambda x:math.cos(x/180*math.pi)
      f_lai = lambda l:111132.92-559.82*c(2*l) + 1.175*c(4*l)-0.0023*c(6*l)
      f_lon = lambda l:111412.84-93.5*c(3*l) + 0.118*c(5*l)
      

      • laititude	laititude_meter_per_degree	longitude_meter_per_degree
        0$\degree$	110574.2727	111319.458
        36.04$\degree$	110959.71900894716	90118.10182968706
        45$\degree$	111131.745	78846.80572069259
        60$\degree$	111412.24020000001	55799.979000000014
        90$\degree$	111693.9173	0

• wrong f ox3888(x+3,y+20)
• $(111320,71546)$

• f1 ox3895(x+90,y+10)
• $(111320,90013.62031373613)$

• f2 ox3895(x+80-0.18,y+6)
• 系数变更对精度影响不是很大，f2比f1在80米尺度上精确0.18米
• $(110959.71900894716,90118.10182968706)$