# User:Peter Mercator/Math snippets

User:Peter Mercator/Sandbox

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$\phi$ $\lambda$ $\alpha$

$k(\lambda,\,\phi,\,\alpha)=\lim_{Q\to P}\frac{P'Q'}{PQ},$

$a\delta\phi$ $a$ $(a\cos\phi)\delta\lambda$ $(a\cos\phi)$

$\delta x=a\delta\lambda$   $\delta y$

horizontal scale factor   $\quad k\;=\;\frac{\delta x}{a\cos\phi\,\delta\lambda\,}=\,\sec\phi\qquad\qquad{}$
vertical scale factor     $\quad h\;=\;\frac{\delta y}{a\delta\phi\,}=\frac{y'(\phi)}{a}$
$x = a\lambda \qquad\qquad y = a\phi,$

$\pi/180$) $[{-}\pi,\pi]$ $\phi$ $[{-}\pi/2,\pi/2]$.

$y'(\phi)=1$

horizontal scale, $\quad k\;=\;\frac{\delta x}{a\cos\phi\,\delta\lambda\,}=\,\sec\phi\qquad\qquad{}$       vertical scale $\quad h\;=\;\frac{\delta y}{a\delta\phi\,}=\,1$

$y$-direction $x$-direction $2\pi a\cos\phi$$\sec\phi$ $2\pi a$

$x = a\lambda \qquad\qquad y = a\ln \left(\tan \left(\frac{\pi}{4} + \frac{\phi}{2} \right) \right)$
horizontal scale,  $\quad k\;=\;\frac{\delta x}{a\cos\phi\,\delta\lambda\,}=\,\sec\phi\qquad\qquad{}$
vertical scale     $\quad h\;=\;\frac{\delta y}{a\delta\phi\,}=\,\sec\phi$

Lambert $x = a\lambda \qquad\qquad y = a\sin\phi$

horizontal scale,  $\quad k\;=\;\frac{\delta x}{a\cos\phi\,\delta\lambda\,}=\,\sec\phi\qquad\qquad{}$
vertical scale     $\quad h\;=\;\frac{\delta y}{a\delta\phi\,}=\,\cos\phi$

40,000 km

$1

$x = 0.9996a\lambda \qquad\qquad y = 0.9996a\ln \left(\tan \left(\frac{\pi}{4} + \frac{\phi}{2} \right) \right).$
$\text{(a)}\quad \tan\alpha=\frac{a\cos\phi\,\delta\lambda}{a\,\delta\phi},$
$\text{(b)}\quad \tan\beta=\frac{\delta x}{\delta y} =\frac{a\delta \lambda}{\delta y},$
$\text{(c)}\quad \tan\beta=\frac{a\sec\phi}{y'(\phi)} \tan\alpha.\,$
$\mu_{\alpha}=\lim_{Q\to P}\frac{P'Q'}{PQ} = \lim_{Q\to P}\frac{\sqrt{\delta x^2 +\delta y^2}} {\sqrt{ a^2\, \delta\phi^2+a^2\cos^2\!\phi\, \delta\lambda^2}}.$
$\mu_\alpha(\phi) = \sec\phi \left[\frac{\sin\alpha}{\sin\beta}\right].$