Involute function

The involute function is used to calculate involute gears . The involute function is defined as:

${\displaystyle \operatorname {inv} (\alpha )=\tan(\alpha )-\alpha \quad {\text{mit}}\quad -{\frac {\pi }{2}}<\alpha <{\frac {\pi }{2}}}$

Example:

${\displaystyle \operatorname {inv} (20^{\circ })=\tan(20^{\circ })-{\frac {20^{\circ }\cdot \pi }{180^{\circ }}}=\tan(20^{\circ })-{\frac {\pi }{9}}=0{,}014904384\ldots =\operatorname {inv} \left({\frac {\pi }{9}}\right)}$

See also involute .

Inverse function

The inverse function of the involute function is referred to below with . She is one in a very defined, analytic function that is strictly increasing and whose function graph point symmetrical to (0,0) and amount by limited (that is similar to the real arc tangent function ). The values ​​of this inverse function of the involute function can be determined efficiently iteratively . From the series expansion of the involute function ${\displaystyle \operatorname {inv} ^{-1}}$ ${\displaystyle \mathbb {R} }$ ${\displaystyle \pi /2}$

${\displaystyle \operatorname {inv} (\alpha )=\tan(\alpha )-\alpha ={\frac {1}{3}}\alpha ^{3}+{\frac {2}{15}}\alpha ^{5}+{\frac {17}{315}}\alpha ^{7}+\dots }$

it can be deduced that for the inverse involute function

${\displaystyle \alpha =\operatorname {inv} ^{-1}(\tan(\alpha )-\alpha )={\sqrt[{3}]{3\ \operatorname {inv} (\alpha )}}-{\frac {2}{5}}\operatorname {inv} (\alpha )+\Theta \left(\operatorname {inv} (\alpha )^{\frac {5}{3}}\right)\approx {\sqrt[{3}]{3\ \operatorname {inv} (\alpha )}}-{\frac {2}{5}}\operatorname {inv} (\alpha )}$

is an acceptable approximation if is sufficiently small. With the help of Newton's method , this approximate value can be further improved for: ${\displaystyle |\operatorname {inv} (\alpha )|}$ ${\displaystyle \alpha }$

${\displaystyle \alpha _{i+1}=\alpha _{i}+{\frac {\operatorname {inv} (\alpha )-\tan(\alpha _{i})+\alpha _{i}}{\tan(\alpha _{i})^{2}}}\quad {\text{und}}\quad \alpha _{0}={\sqrt[{3}]{3\ \operatorname {inv} (\alpha )}}-{\frac {2}{5}}\operatorname {inv} (\alpha )}$

Is , you should choose as a starting value so that the above Newton method also converges. ${\displaystyle |\operatorname {inv} (\alpha )|>2}$ ${\displaystyle \alpha _{0}=\arctan(\operatorname {inv} (\alpha ))}$