How is the Formula for Torsion Derived?

[thechunk]

Is anyone familiar with the derivation for this formula for torsion.
[tex]
\tau = \frac {\left( \begin{array}{ccc} \dot{x} & \ddot{x} & \dddot{x}\\\dot{y}& \ddot{y}& \ddot{y} \\\dot{z} & \ddot{z} & \dddot{z}\end{array} \right)} {|v \times a|^2}
[/tex]
I know of expressing torsion as [tex] \tau = -\frac {dB} {dS} \cdot N [/itex], but I do not know how to derive the former. My teacher said it could be derived with knowledge from our multivariable class however my textbook reads that the derivation is found in more advanced texts. The numerator in the first formula looks like [tex] (v \times a) \cdot a' [/itex], but I do not know where to go from there. Any ideas?

 

[johny_doe]

Anyone know the answer to this question?

I'm totally lost on it.

 

[tacman]

The formula for torsion, as you have mentioned, can be expressed as [tex] \tau = -\frac {dB} {dS} \cdot N [/itex], where B is the binormal vector, S is the arc length parameter, and N is the normal vector. This formula is derived using the Frenet-Serret formulas, which describe the behavior of a curve in three-dimensional space.

To derive the formula [tex] \tau = \frac {\left( \begin{array}{ccc} \dot{x} & \ddot{x} & \dddot{x}\\\dot{y}& \ddot{y}& \ddot{y} \\\dot{z} & \ddot{z} & \dddot{z}\end{array} \right)} {|v \times a|^2} [/itex], we can start with the definition of torsion, which is the rate of change of the binormal vector with respect to arc length:

[tex] \tau = \frac {d \vec{B}} {dS} [/itex]

Using the Frenet-Serret formulas, we can express the derivative of the binormal vector as:

[tex] \frac {d \vec{B}} {dS} = \kappa \vec{T} + \tau \vec{N} [/itex]

Where [tex] \kappa [/itex] is the curvature of the curve and [tex] \vec{T} [/itex] is the tangent vector. We can also express the tangent vector and the normal vector in terms of the velocity and acceleration vectors:

[tex] \vec{T} = \frac {\vec{v}} {|v|} [/itex]

[tex] \vec{N} = \frac {\vec{v} \times \vec{a}} {|v \times \vec{a}|} [/itex]

Substituting these expressions into the derivative of the binormal vector, we get:

[tex] \frac {d \vec{B}} {dS} = \kappa \frac {\vec{v}} {|v|} + \tau \frac {\vec{v} \times \vec{a}} {|v \times \vec{a}|} [/itex]

Taking the magnitude of both sides and rearranging, we get:

[tex] \tau