• 交比不变性证明


    在这里插入图片描述

    A C B C / A D B D 为四个有序点的交比,求证 A C ⋅ B D B C ⋅ A D = a c ⋅ b d b c ⋅ a d 。 \frac{AC }{BC} / \frac{AD }{BD} 为四个有序点的交比,求证 \frac{AC \cdot BD}{BC \cdot AD} = \frac{ac \cdot bd}{bc \cdot ad}。 BCAC/BDAD为四个有序点的交比,求证BCADACBD=bcadacbd
    借助三角形的面积证明。
    世界坐标中:
    S Δ P A C = 1 2 ⋅ A C ⋅ H = 1 2 ⋅ P A ⋅ P C ⋅ sin ⁡ ∠ A P C ⇒ A C = P A ⋅ P C H ⋅ sin ⁡ ∠ A P C S Δ P B C = 1 2 ⋅ B C ⋅ H = 1 2 ⋅ P B ⋅ P C ⋅ sin ⁡ ∠ B P C ⇒ B C = P B ⋅ P C H ⋅ sin ⁡ ∠ B P C S Δ P B D = 1 2 ⋅ B D ⋅ H = 1 2 ⋅ P B ⋅ P D ⋅ sin ⁡ ∠ B P D ⇒ B D = P B ⋅ P D H ⋅ sin ⁡ ∠ B P D S Δ P A D = 1 2 ⋅ A D ⋅ H = 1 2 ⋅ P A ⋅ P D ⋅ sin ⁡ ∠ A P D ⇒ A D = P A ⋅ P D H ⋅ sin ⁡ ∠ A P D S_{\Delta PAC} = \frac{1}{2} \cdot AC \cdot H = \frac{1}{2} \cdot PA \cdot PC \cdot \sin{\angle APC} \\ \Rightarrow AC = \frac{PA \cdot PC}{H} \cdot \sin{\angle APC} \\ {}\\ S_{\Delta PBC} = \frac{1}{2} \cdot BC \cdot H = \frac{1}{2} \cdot PB \cdot PC \cdot \sin{\angle BPC} \\ \Rightarrow BC = \frac{PB \cdot PC}{H} \cdot \sin{\angle BPC}\\ {}\\ S_{\Delta PBD} = \frac{1}{2} \cdot BD \cdot H = \frac{1}{2} \cdot PB \cdot PD \cdot \sin{\angle BPD} \\ \Rightarrow BD = \frac{PB \cdot PD}{H} \cdot \sin{\angle BPD}\\ {}\\ S_{\Delta PAD} = \frac{1}{2} \cdot AD \cdot H = \frac{1}{2} \cdot PA \cdot PD \cdot \sin{\angle APD} \\\Rightarrow AD = \frac{PA \cdot PD}{H} \cdot \sin{\angle APD} SΔPAC=21ACH=21PAPCsinAPCAC=HPAPCsinAPCSΔPBC=21BCH=21PBPCsinBPCBC=HPBPCsinBPCSΔPBD=21BDH=21PBPDsinBPDBD=HPBPDsinBPDSΔPAD=21ADH=21PAPDsinAPDAD=HPAPDsinAPD
    A C ⋅ B D B C ⋅ A D = P A ⋅ P B ⋅ P C ⋅ P D H 2 ⋅ sin ⁡ ∠ A P C ⋅ sin ⁡ ∠ B P D P A ⋅ P B ⋅ P C ⋅ P D H 2 ⋅ sin ⁡ ∠ B P C ⋅ sin ⁡ ∠ A P D = sin ⁡ ∠ A P C ⋅ sin ⁡ ∠ B P D sin ⁡ ∠ B P C ⋅ sin ⁡ ∠ A P D

    ACBDBCAD=PAPBPCPDH2sinAPCsinBPDPAPBPCPDH2sinBPCsinAPD=sinAPCsinBPDsinBPCsinAPD" role="presentation" style="position: relative;">ACBDBCAD=PAPBPCPDH2sinAPCsinBPDPAPBPCPDH2sinBPCsinAPD=sinAPCsinBPDsinBPCsinAPD
    BCADACBD=H2PAPBPCPDsinBPCsinAPDH2PAPBPCPDsinAPCsinBPD=sinBPCsinAPDsinAPCsinBPD

    像素坐标中:
    S Δ P a c = 1 2 ⋅ a c ⋅ h = 1 2 ⋅ P a ⋅ P c ⋅ sin ⁡ ∠ a P c ⇒ a c = P a ⋅ P c h ⋅ sin ⁡ ∠ a P c S Δ P b c = 1 2 ⋅ b c ⋅ h = 1 2 ⋅ P b ⋅ P c ⋅ sin ⁡ ∠ b P c ⇒ b c = P b ⋅ P c h ⋅ sin ⁡ ∠ b P c S Δ P b d = 1 2 ⋅ b d ⋅ h = 1 2 ⋅ P b ⋅ P d ⋅ sin ⁡ ∠ b P d ⇒ b d = P b ⋅ P d h ⋅ sin ⁡ ∠ b P d S Δ P a d = 1 2 ⋅ a d ⋅ h = 1 2 ⋅ P a ⋅ P d ⋅ sin ⁡ ∠ a P d ⇒ a d = P a ⋅ P d h ⋅ sin ⁡ ∠ a P d S_{\Delta Pac} = \frac{1}{2} \cdot ac \cdot h = \frac{1}{2} \cdot Pa \cdot Pc \cdot \sin{\angle aPc} \\ \Rightarrow ac = \frac{Pa \cdot Pc}{h} \cdot \sin{\angle aPc} \\ {}\\ S_{\Delta Pbc} = \frac{1}{2} \cdot bc \cdot h = \frac{1}{2} \cdot Pb \cdot Pc \cdot \sin{\angle bPc} \\ \Rightarrow bc = \frac{Pb \cdot Pc}{h} \cdot \sin{\angle bPc}\\ {}\\ S_{\Delta Pbd} = \frac{1}{2} \cdot bd \cdot h = \frac{1}{2} \cdot Pb \cdot Pd \cdot \sin{\angle bPd} \\ \Rightarrow bd = \frac{Pb \cdot Pd}{h} \cdot \sin{\angle bPd}\\ {}\\ S_{\Delta Pad} = \frac{1}{2} \cdot ad \cdot h = \frac{1}{2} \cdot Pa \cdot Pd \cdot \sin{\angle aPd} \\\Rightarrow ad = \frac{Pa \cdot Pd}{h} \cdot \sin{\angle aPd} SΔPac=21ach=21PaPcsinaPcac=hPaPcsinaPcSΔPbc=21bch=21PbPcsinbPcbc=hPbPcsinbPcSΔPbd=21bdh=21PbPdsinbPdbd=hPbPdsinbPdSΔPad=21adh=21PaPdsinaPdad=hPaPdsinaPd
    a c ⋅ b d b c ⋅ a d = P a ⋅ P b ⋅ P c ⋅ P d h 2 ⋅ sin ⁡ ∠ a P c ⋅ sin ⁡ ∠ b P d P a ⋅ P b ⋅ P c ⋅ P d h 2 ⋅ sin ⁡ ∠ b P c ⋅ sin ⁡ ∠ a P d = sin ⁡ ∠ a P c ⋅ sin ⁡ ∠ b P d sin ⁡ ∠ b P c ⋅ sin ⁡ ∠ a P d

    acbdbcad=PaPbPcPdh2sinaPcsinbPdPaPbPcPdh2sinbPcsinaPd=sinaPcsinbPdsinbPcsinaPd" role="presentation" style="position: relative;">acbdbcad=PaPbPcPdh2sinaPcsinbPdPaPbPcPdh2sinbPcsinaPd=sinaPcsinbPdsinbPcsinaPd
    bcadacbd=h2PaPbPcPdsinbPcsinaPdh2PaPbPcPdsinaPcsinbPd=sinbPcsinaPdsinaPcsinbPd
    ∵ sin ⁡ ∠ A P C = sin ⁡ ∠ a P c sin ⁡ ∠ B P D = sin ⁡ ∠ b P d sin ⁡ ∠ B P C = sin ⁡ ∠ b P c sin ⁡ ∠ A P D = sin ⁡ ∠ a P d ∴ A C ⋅ B D B C ⋅ A D = a c ⋅ b d b c ⋅ a d \begin {aligned} \because \sin{\angle APC} &= \sin{\angle aPc} \\ \sin{\angle BPD} &= \sin{\angle bPd} \\ \sin{\angle BPC} &= \sin{\angle bPc} \\ \sin{\angle APD} &= \sin{\angle aPd} \\ \therefore \frac{AC \cdot BD}{BC \cdot AD} &= \frac{ac \cdot bd}{bc \cdot ad} \end{aligned} sinAPCsinBPDsinBPCsinAPDBCADACBD=sinaPc=sinbPd=sinbPc=sinaPd=bcadacbd

    工业镜头
    参考文献
    工业镜头是运用于工业自动化领域的摄像或摄影镜头,主要作用是进行光学成像

    工业镜头是机器视觉中必不可少的核心基础硬件,对成像的质量有关键影响。可以分为fa镜头(也有称CCTV镜头)、变倍镜头(有手动也有自动)、远心镜头(一般有物方远心和双远心)等。

    基础参数
    1.工作距离WD:指镜头下端到物体表面的距离。

    机器视觉光学系统所需要的空间,不仅要考虑相机镜头组本身的长度,还要留下镜头的工作距离。如果空间无法满足这两者之和,应该立即考虑棱镜改变工作方向,或是重新定制光学系统。

    2.焦距:镜头到成像面的距离。

    3.景深DOF:镜头能够看清的纵深范围。

    景深与焦距的关系是:焦距越长,景深越小;焦距越短,景深越大。

    景深与工作距离的关系是:工作距离越短,镜头离物体越近,景深越小,反之则越大。

    4.视野FOV:指镜头能够看到的最大范围。简单来说就是镜头能够正常进行检测工作的区域。

    对于大面积检测,选择大视野的镜头能够显著地节省检测成本。但是选择时也要兼具分辨率,大视野很容易发生视野边缘灰度值不足的情况,一定要找专业光学定制!

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  • 原文地址:https://blog.csdn.net/weixin_44284128/article/details/127611616