Convert any value from / to radians per second [rad/s] to kilometers per hour [km/h], angular velocity to linear velocity. Fill one of the following fields, values will be converted and updated automatically.
Assume that \( v_{(m.s^{-1})} \) is the velocity expressed in \( m.s^{-1} \), i.e. the distance travelled during one second. To get the distance travelled during one minute, the previous value must be multiplied by 60. To get the distance in meters travelled during one hour, \( v_{(m.s^{-1})}\) must be multiplied by 60 x 60 = 3600. The result is a velocity expressed in \( m.h^{-1} \). To convert this result in kilometers per hour, we now have to divide the previous velocity (expressed in \( m.h^{-1} \)) by 1000 since one kilometer is equal to 1000 meters. The conversion can be done thanks to the following formula:
$$ v_{ (m.s^{-1}) } = \frac {1000}{3600}.v_{(km.h^{-1})} = \frac { v_{(km.h^{-1})} } {3.6 } $$
Assume an object (or point) attached to a rotating wheel. The angular velocity of the wheel is defined by \( \omega \) expressed in \( rad.s^{-1} \). This also means that the wheel rotates from \( \omega \) radians during one second. During that same second, the attached object (or point) travels a distance of \(r \times \omega \). It implies that the speed of the object is also equal to \(r \times \omega \). The conversion can be done thanks to the following formula:
$$ \omega_{(rad.s^{-1})}=\frac{v_{ (m.s^{-1}) }}{r} $$
From the previous demonstrations, we can conclude:
$$ \omega_{(rad.s^{-1})}=\frac{1000}{3600} \times \frac{v_{ (km.h^{-1}) }}{r} = \frac {v_{ (km.h^{-1}) }}{3.6 \times r } $$
and vice-versa:
$$ v_{(km.h^{-1})} =\frac{3600}{1000} \times r \times \omega_{(rad.s^{-1})} = 3.6 \times r \times \omega_{(rad.s^{-1})} $$