## Abstract

We propose a new panossramic optical system that provides an additional field of view (FOV) channel without expanding the physical size of a conventional panoramic annular lens (PAL). The two channels are contained within one PAL, their optical paths do not interfere with each other, and the two images are realized on a single image plane. A prototype panoramic lens was developed that provides a 360° × (38–80°) front FOV channel and a 360° × (102–140°) back FOV channel.

© 2017 Optical Society of America

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### Equations (14)

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(1)
$${\theta}_{4}=\mathrm{arc}\mathrm{tan}(\frac{{C}_{y}-{A}_{y}}{{C}_{x}-{A}_{x}})+\mathrm{arc}\mathrm{sin}(({\theta}_{2}-{\theta}_{1})/n).$$
(2)
$$L+\text{dis}(B,O)=\text{dis}(C,O)+\text{dis}(C,A)={L}_{total}.$$
(3)
$$L+\sqrt{{(L\cdot \mathrm{cos}({\theta}_{4})+{A}_{x})}^{2}+{(L\cdot \mathrm{sin}({\theta}_{4})-{A}_{y})}^{2}}=\sqrt{{C}_{x}{}^{2}+{C}_{y}{}^{2}}+\sqrt{{({C}_{x}-{A}_{x})}^{2}+{({C}_{y}-{A}_{y})}^{2}}={L}_{total}.$$
(4)
$$L=\frac{{L}_{{}_{total}}^{2}-({A}_{x}{}^{2}+{A}_{y}{}^{2})}{2\cdot (\mathrm{cos}({\theta}_{4})\cdot {A}_{x}-\mathrm{sin}({\theta}_{4})\cdot {A}_{y}+{L}_{total})}.$$
(5)
$${\theta}_{4}^{\text{'}}={\theta}_{3}^{\text{'}}+\mathrm{arc}\mathrm{sin}(({\theta}_{2}^{\text{'}}-{\theta}_{1}^{\text{'}})/n).$$
(6)
$${\text{R}}_{1}=\text{L}\cdot \mathrm{cos}{\theta}_{3}^{\text{'}}+\text{L,}$$
(7)
$${R}_{2}={B}_{x}+\sqrt{{B}_{x}{}^{2}+{B}_{y}{}^{2}}.$$
(8)
$$\frac{{R}_{1}}{2}-{L}^{\text{'}}\cdot \mathrm{cos}{\theta}_{4}^{\text{'}}=\frac{{({L}^{\text{'}}\cdot \mathrm{sin}{\theta}_{4}^{\text{'}})}^{2}}{2{R}_{1}}.$$
(9)
$${L}^{\text{'}}=\frac{(-\mathrm{cos}{\theta}_{4}^{\text{'}}+1)}{{\text{(}\mathrm{sin}{\theta}_{4}^{\text{'}})}^{2}}\cdot {R}_{1}.$$
(10)
$$\Delta {\text{z}}_{1}=\frac{1}{2{R}_{1}}\cdot ({(L\cdot \mathrm{sin}{\theta}_{3}^{\text{'}})}^{2}-{({L}^{\text{'}}\cdot \mathrm{sin}{\theta}_{4}^{\text{'}})}^{2}),$$
(11)
$$\Delta {\text{z}}_{2}=\frac{1}{2{R}_{2}}\cdot ({(L\cdot \mathrm{sin}{\theta}_{3}^{\text{'}})}^{2}-{({L}^{\text{'}}\cdot \mathrm{sin}{\theta}_{4}^{\text{'}})}^{2}),$$
(12)
$$\text{z}=\frac{c{r}_{g}^{2}}{1+\sqrt{1-(1+k){c}^{2}{r}_{g}^{2}}}+{\displaystyle \sum _{i=1}^{4}{\alpha}_{i}}{r}_{g}^{2i},$$
(13)
$${\text{r}}_{\text{g}}={\text{r}}_{\text{o}}+\sqrt{{x}^{2}+{y}^{2}},$$
(13)
$$\text{z}=\frac{c{r}^{2}}{1+\sqrt{1-(1+k){c}^{2}{r}^{2}}}+{\displaystyle \sum _{i=1}^{4}{\alpha}_{i}}{\text{r}}^{2i},r=\sqrt{{\text{x}}^{2}+{\text{y}}^{2}}.$$