## Abstract

The characteristics of an all-sky camera with a concave mirror are analyzed. A differential equation for a concave aspheric mirror with constant angular magnification is derived for the general dependence of the camera image height on the camera field angle. This equation is solved in parametric form for the case of a concave mirror with a constant angular magnification. The explicit equations for the shape of the aspheric mirror are given for some particular values of the angular magnification. Parametric equations of the surface shape for sevenfold angular magnification are developed into a power series that is used to analyze the imaging performance of such a mirror. The performance of the concave aspheric mirror is compared with that of a spherical mirror. The minimal camera-to-mirror distance is determined as a function of the blur allowed and the camera lens aperture. Some characteristics of convex mirrors are also presented for comparison.

© 1996 Optical Society of America

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### Equations (31)

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(1)
$$h=\frac{1}{M}\beta ,$$
(2)
$$h=g(\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}}),$$
(3)
$$g(\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}})=\frac{1}{M}\beta .$$
(4)
$$g(\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}})=Ftan\phantom{\rule{0em}{0ex}}(\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}}),$$
(5)
$$g(\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}})\approx F\phantom{\rule{0em}{0ex}}\alpha $$
(6)
$$\beta =2\phantom{\rule{0em}{0ex}}\delta +\alpha $$
(7)
$$\delta =\phi -\alpha .$$
(8)
$$tan\phantom{\rule{0em}{0ex}}(\phantom{\rule{0em}{0ex}}\sigma \phantom{\rule{0em}{0ex}})=-\frac{1}{\frac{\text{d}\phantom{\rule{0em}{0ex}}f(x)}{\text{d}x}}$$
(9)
$$\sigma =90\xb0+\phi ,$$
(10)
$$\frac{\text{d}\phantom{\rule{0em}{0ex}}f(x)}{\text{d}x}=tan\phantom{\rule{0em}{0ex}}(\phantom{\rule{0em}{0ex}}\delta +\alpha \phantom{\rule{0em}{0ex}})\phantom{\rule{0.2em}{0ex}}.$$
(11)
$$\frac{\text{d}\phantom{\rule{0em}{0ex}}f(x)}{\text{d}x}=tan\phantom{\rule{0em}{0ex}}\left(\frac{\phantom{\rule{0em}{0ex}}\beta +\alpha}{2}\phantom{\rule{0em}{0ex}}\right)$$
(12)
$$\frac{\text{d}y}{\text{d}x}=tan\phantom{\rule{0em}{0ex}}\left[\frac{Mg(\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}})+\alpha}{2}\phantom{\rule{0em}{0ex}}\right]\phantom{\rule{0.2em}{0ex}}.$$
(13)
$$\alpha =\text{arctan}\left(\frac{x}{\text{d}-y}\right)\phantom{\rule{0.2em}{0ex}}.$$
(14)
$$\frac{\text{d}y}{\text{d}x}=tan\phantom{\rule{0em}{0ex}}\left[\frac{M-1}{2}\text{arctan}\left(\frac{x}{\text{d}-y}\right)\right].$$
(15)
$$X=\frac{x}{\text{d}}\phantom{\rule{0.2em}{0ex}},$$
(16)
$$Y=\frac{y}{\text{d}}\phantom{\rule{0.2em}{0ex}},$$
(19)
$$N=\frac{M-1}{2}.$$
(20)
$$\frac{\text{d}Z}{Z}=-\frac{tan\phantom{\rule{0em}{0ex}}[(N+1)\text{arctan}(U)]}{1+Utan[(N+1)\text{arctan}(U)]}\text{d}U$$
(21)
$$x(\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}})=\text{d}sin\phantom{\rule{0em}{0ex}}(\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}}){\left[cos\phantom{\rule{0em}{0ex}}\left(\frac{M-1}{2}\alpha \right)\right]}^{2/(M-1)},$$
(22)
$$y(\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}})=\text{d}-\text{d}cos(\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}}){\left[\phantom{\rule{0em}{0ex}}cos\phantom{\rule{0em}{0ex}}\left(\frac{M-1}{2}\alpha \phantom{\rule{0em}{0ex}}\right)\right]}^{2/(M-1)},$$
(23)
$${T}_{N}\left\{\frac{(1-Y)}{{[{X}^{2}+{(1-Y)}^{2}]}^{1/2}}\right\}={[{X}^{2}+{(1-Y)}^{2}]}^{N/2},$$
(24)
$$M=3,\phantom{\rule{1em}{0ex}}Y=1-\frac{1+{(1-4{X}^{2})}^{1/2}}{2},$$
(25)
$$M=5,\phantom{\rule{1em}{0ex}}Y=1-{\left[\frac{1-2{X}^{2}+{(1-8{X}^{2})}^{1/2}}{2}\right]}^{1/2}.$$
(26)
$$\begin{array}{c}M=7,\hfill \\ Y=2{X}^{2}+7{X}^{4}+\frac{140}{3}{X}^{6}+\frac{4187}{15}{X}^{8}+O({X}^{10}),\hfill \end{array}$$
(27)
$$R=\frac{2\text{d}}{m+1},$$
(28)
$$R=\frac{2\text{d}}{m-1},$$
(29)
$$\beta =2\phantom{\rule{0.2em}{0ex}}\text{arcsin}\left(\frac{m-1}{2}sin\alpha \right)+\alpha ,$$
(30)
$$\beta =2\phantom{\rule{0.2em}{0ex}}\text{arcsin}\left(\frac{m+1}{2}sin\alpha \right)-\alpha .$$
(31)
$$\text{DIST}=\frac{\beta -m\phantom{\rule{0em}{0ex}}\alpha}{m\phantom{\rule{0em}{0ex}}\alpha}.$$