For a sphere of radius 1, the azimuthal angle , the polar angle (defined here to correspond to latitude), and
Cartesian unit vectors , , and can be used to write the radius vector as :\mathbf{r}(\lambda,\varphi) = (\cos{\lambda} \cdot \cos{\varphi}) \mathbf{i} + (\sin{\lambda} \cdot \cos{\varphi}) \mathbf{j} + (\sin{\varphi}) \mathbf{k} \, .
Orthogonal unit vectors in the azimuthal and polar directions of the sphere can be written :\begin{align} \boldsymbol{\hat\lambda}(\lambda,\varphi) &= \sec{\varphi} \frac{\partial\mathbf{r}}{\partial\lambda} = (-\sin{\lambda}) \mathbf{i} + (\cos{\lambda}) \mathbf{j} \, , \\[8pt] \boldsymbol{\hat\varphi}(\lambda,\varphi) &= \frac{\partial\mathbf{r}}{\partial\varphi} = (-\cos{\lambda} \cdot \sin{\varphi}) \mathbf{i} + (-\sin{\lambda} \cdot \sin{\varphi}) \mathbf{j} + (\cos{\varphi}) \mathbf{k} \, , \end{align} which have the
scalar products :\boldsymbol{\hat\lambda} \cdot \boldsymbol{\hat\varphi} = \boldsymbol{\hat\lambda} \cdot \mathbf{r} = \boldsymbol{\hat\varphi} \cdot \mathbf{r} = 0 \, . for constant traces out a parallel of latitude, while for constant traces out a meridian of longitude, and together they generate a plane tangent to the sphere. The unit vector :\mathbf{\boldsymbol{\hat\beta}}(\lambda,\varphi) = (\sin{\beta}) \boldsymbol{\hat\lambda} + (\cos{\beta}) \boldsymbol{\hat\varphi} has a constant angle with the unit vector for any and , since their scalar product is :\boldsymbol{\hat\beta} \cdot \boldsymbol{\hat\varphi} = \cos{\beta} \, . A loxodrome is defined as a curve on the sphere that has a constant angle with all meridians of longitude, and therefore must be parallel to the unit vector . As a result, a differential length along the loxodrome will produce a differential displacement :\begin{align} d\mathbf{r} &= \boldsymbol{\hat\beta} \, ds \\[8px] \frac{\partial\mathbf{r}}{\partial\lambda} \, d\lambda + \frac{\partial\mathbf{r}}{\partial\varphi} \, d\varphi &= \bigl((\sin{\beta}) \, \boldsymbol{\hat\lambda} + (\cos{\beta}) \, \boldsymbol{\hat\varphi}\bigr) ds \\[8px] (\cos{\varphi}) \, d\lambda \, \boldsymbol{\hat\lambda} + d\varphi \, \boldsymbol{\hat\varphi} &= (\sin{\beta}) \, ds \, \boldsymbol{\hat\lambda} + (\cos{\beta}) \, ds \, \boldsymbol{\hat\varphi} \\[8px] ds &= \frac{\cos{\varphi} }{\sin{\beta}} \, d\lambda = \frac{d\varphi}{\cos{\beta}} \\[8px] \frac{d\lambda}{d\varphi} &= \tan{\beta} \cdot \sec{\varphi} \\[8px] \lambda(\varphi\,|\,\beta,\lambda_0,\varphi_0) &= \tan\beta \cdot \big( \operatorname{gd}^{-1}\varphi - \operatorname{gd}^{-1}\varphi_0 \big) + \lambda_0 \\[8px] \varphi(\lambda\,|\,\beta,\lambda_0,\varphi_0) &= \operatorname{gd} \big((\lambda - \lambda_0) \cot\beta + \operatorname{gd}^{-1}\varphi_0\big) \end{align} where \operatorname{gd} and \operatorname{gd}^{-1} are the
Gudermannian function and its inverse, \operatorname{gd}\psi = \arctan(\sinh\psi), \operatorname{gd}^{-1}\varphi = \operatorname{arsinh}(\tan\varphi), and \operatorname{arsinh} is the
inverse hyperbolic sine. With this relationship between and , the radius vector becomes a parametric function of one variable, tracing out the loxodrome on the sphere: :\mathbf{r}(\lambda\,|\,\beta,\lambda_0,\varphi_0) = \big(\cos{\lambda} \cdot \operatorname{sech} \psi \big) \mathbf{i} + \big(\sin{\lambda} \cdot \operatorname{sech}\psi\big) \mathbf{j} + \big(\tanh\psi\big) \mathbf{k} \, , where :\psi \equiv (\lambda - \lambda_0) \cot\beta + \operatorname{gd}^{-1}\varphi_0 = \operatorname{gd}^{-1}\varphi is the
isometric latitude. In the Rhumb line, as the latitude tends to the poles, , , the isometric latitude , and longitude increases without bound, circling the sphere ever so fast in a spiral towards the pole, while tending to a finite total arc length Δ given by :\Delta s = R \, \big|(\pm\pi/2 - \varphi_0) \cdot \sec \beta\big|
Connection to the Mercator projection Let be the longitude of a point on the sphere, and its latitude. Then, if we define the map coordinates of the
Mercator projection as :\begin{align} x &= \lambda - \lambda_0 \, , \\ y &= \operatorname{gd}^{-1}\varphi = \operatorname{arsinh}(\tan\varphi)\, , \end{align} a loxodrome with constant
bearing from true north will be a straight line, since (using the expression in the previous section) :y = m x with a slope :m=\cot\beta\,. Finding the loxodromes between two given points can be done graphically on a Mercator map, or by solving a nonlinear system of two equations in the two unknowns and . There are infinitely many solutions; the shortest one is that which covers the actual longitude difference, i.e. does not make extra revolutions, and does not go "the wrong way around". The distance between two points , measured along a loxodrome, is simply the absolute value of the
secant of the bearing (azimuth) times the north–south distance (except for
circles of latitude for which the distance becomes infinite): :\Delta s = R \, \big|(\varphi - \varphi_0)\cdot \sec \beta \big| where is one of the
earth average radii. ==Application==