Skip to main content


For a thermal gradient φ\varphi applied to a beam the curvature is

κ=αφ\kappa = \alpha\varphi

So the radius of curvature is

r=1κr = \frac{1}{\kappa}

Assuming a circular arc the rotation at the end is perpendicular to the radial lien of the arc. This means that the angle of the radial line to the beam original configuration is

cosβ=l2rβ=cos1(lκ2)\begin{aligned} \cos\beta &= \frac{\frac{l}{2}}{r} \\ \beta &= \cos^{- 1}\left( \frac{l\kappa}{2} \right) \end{aligned}


ϕ=π2cos1(lκ2)\phi = \frac{\pi}{2} - \cos^{- 1}\left( \frac{l\kappa}{2} \right)
ϕ=π2cos1(lαφ2)\phi = \frac{\pi}{2} - \cos^{- 1}\left( \frac{l\alpha\varphi}{2} \right)