Scale (map) - Mathematical Addendum

Mathematical Addendum

For normal cylindrical projections the geometry of the infinitesimal elements gives

\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}.

The relationship between the angles and is

\text{(c)}\quad \tan\beta=\frac{a\sec\phi}{y'(\phi)} \tan\alpha.\,

For the Mercator projection giving : angles are preserved. (Hardly surprising since this is the relation used to derive Mercator). For the equidistant and Lambert projections we have and respectively so the relationship between and depends upon the latitude . Denote the point scale at P when the infinitesimal element PQ makes an angle with the meridian by It is given by the ratio of distances:

\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}}.

Setting and substituting and from equations (a) and (b) respectively gives

For the projections other than Mercator we must first calculate from and using equation (c), before we can find . For example the equirectangular projection has so that

If we consider a line of constant slope on the projection both the corresponding value of and the scale factor along the line are complicated functions of . There is no simple way of transferring a general finite separation to a bar scale and obtaining meaningful results.

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