On the length of a slanted circular arc

Question

Consider two concentric circles, of radius $R$ and $R+\mathrm{d}R$ respectively. What is the length of the "crooked" arc in the figure? Suppose this arc subtends a generic angle $\theta$.

Answer 1

I'm assuming from the title that the slanted arc is circular. Now, as shown in the figure below, extend outward normal to the skewed arc from point C and seek the intersection (point A') with the extension of line AB. This is the origin of skewed arc. Let's say this has a length $R+\delta R$ and an included angle $\theta+\delta\theta$. Now observe the chord $c$. This has a length given by

$$c=2(R+\delta R)\sin\left(\frac{\theta+\delta\theta}{2}\right)$$

For triangle A'BC, we have the law of cosines

$$ \cos(\theta+\delta\theta)=\frac{R^2+(R+\delta R)^2-(R+dR)^2}{2\delta r(R+\delta R)} $$

At this point we have two equations in three unknowns, $\delta R, \delta\theta, \text{ and } c$. But $c$ is known from triangle ABC. From the law of cosines again,

$$ c^2=R^2+(R+dR)^2-2R(R+dR)\cos(\theta). $$

You now have two equations in two unknowns. It looks to me to require a numerical solution. However, it may be possible to achieve an approximate analytical solution if indeed $d \text{ and } \delta$ indicate small quantities. In that case you can simplify the $\sin \text{ and } \cos$ terms by Taylor expansions and discard higher-order terms. Of course, the length of the skewed arc is $s=(R+\delta R)\cdot(\theta+\delta\theta)$

Can you take it from here?