# Can the derivative of a spherical curve be calculated in this way?

$$β=arctan [ (1+tan^2 \theta ) K\sin \alpha + tan \theta \sqrt {1+(1+tan^2 \theta )K^2 \sin^2 \alpha} ]$$

This is the general straight line (circle) equation on the sphere. $\alpha$ is longitude, $\beta$ is latitudes, $K$ is the slope of a straight line (circle) and $\theta$ is the distance between the point of the straight line and the equator when alpha equals zero.

diagram

Diagram, the horizontal line (circle) is the equator, and its $K=0$. The $K$ values of other lines (circles) are $-0.3$, and their $\theta$ values are $0.45pi, \ 0.25pi, \ 0, \ 0.$

The calculation formula of the slope of the spherical curve is:

$$K={\sinβ_2-\sinβ_1\over\sinα_2\cosβ_2-\sinα_1\cosβ_1}$$

In the sphere very small area, this formula is approximate to the formula of the slope of the plane curve. So the slope of a plane curve is only a special case of the slope of a spherical curve.

According to this formula, we can find the derivative of the spherical curve and the calculus operation for the sphere.

• Definitions and terminology can be improved. – Narasimham Mar 29 '18 at 19:11
• @Narasimham Do you mean I want to improve? – enbin zheng Mar 29 '18 at 21:14
• I find a bit confusing. About the reference link, great circles have zero slip (geodesics ) which is rate of chage of opening radius ( $r_o$ Clairaut's constant) with respect to axial distance z of For small circles slip $\tan \gamma= d r_o/d z,\, \gamma$ is angle between sphere normal and local arc normal without sphere from Frenet Serret relations. – Narasimham Mar 30 '18 at 4:06
• @Narasimham The link is to say that the curvature of a small circle is equal to zero. Because the slope of the latitudinal circle equals zero. – enbin zheng Mar 30 '18 at 5:41
• I am not able to understand this statement. Small circles have geodesic curvature $\kappa_g\ne 0$ , great circle geodesics have it vanish. Did you mean torsion of small circle in space $=0$ without reference to sphere? – Narasimham Mar 30 '18 at 6:33

You could correlate with the following result that has a slightly different notation ($\varphi$ latitude, $\theta$ longitude, $\psi$ angle small circle makes to longitude, $\alpha$ inclination of small circle to plane of equator).
$$\tan \alpha = \dfrac{\sin \theta \cos \varphi + \cos \theta \tan \psi}{ \sin \varphi}{}$$
• $$\phi =arctan ( (1+tan^2 ψ ) \tan \alpha \sin \theta + tan ψ \sqrt {1+(1+tan^2 ψ ) \tan \alpha^2 \sin^2 \theta} )$$Is that what it means? – enbin zheng Mar 29 '18 at 21:55
• You brought in a new $\phi$ without defining it. Maybe you get better help with your derivation, not just the end result. – Narasimham Mar 30 '18 at 4:14