Inertial Resistance on Circular Arc

Geometric Setup

We model a dancer turning along a circular arc with a radius 'R' centered at the origin (0, 0). The dancer's path and movement are defined as follows:

Arc Position at Angle θ:

$arc_{point}(θ) = (R * cos(θ), R * sin(θ))$

Velocity Vector (Tangent to the Arc)

The velocity vector ($\vec{v}$) is the derivative of the position with respect to the angle: $$\vec{v}θ=\dfrac{d}{θd} (R \cos(θ), R \sin(θ))$$ Carrying out the derivative: $$\vec{v}θ = (-R * \sin(θ), R * \cos(θ))$$

Normalized Velocity Direction:

To get a unit vector in the direction of motion, divide by the magnitude:

$$\hat{v}θ = \dfrac{\vec{v}(θ)}{\left\|\vec{v}(θ)\right\|}$$ Since the magnitude is$\left|\vec{v}(θ)\right| = R$

$$\hat{v}θ = (-sin(θ), cos(θ))$$

🚀 Inertial Resistance

We assume the initial motion was in the +z direction, i.e., vector:

initial_direction = (0, 1)

Resistance Vector (difference between original and new velocity):

recall that the normalized velocity vector tangent to the arc: $$unit\_velocity(θ) = (-sin(θ), cos(θ))$$ and that:$$initial\_direction = (0, 1)$$the Resistance vector is defined as the _difference_ between the current direction and the initial direction:$$resistance\_vector(θ) = unit\_velocity(θ) - initial\_direction$$ plugging in: unit_velocity:$$= (-sin(θ), cos(θ))$$ and initial_direction: $$= (0, 1)$$ We have: $$resistance\_vector(θ) = (-sin(θ) - 0, cos(θ) - 1)$$

Magnitude of Resistance:

The magnitude of the resistance_vector(θ) (which is the Force we need to overcome):

$$\left\|resistance\right\| = \sqrt{(-sin(θ) - 0)^2 - (cos(θ) - 1)^2}$$ Expanding: $$= \sqrt{sin^2(θ) + cos^2(θ) - 2cos(θ)+1)}$$= \sqrt{2 - 2cos(θ)}$$ which becomes:

$$\left\|resistance\right\| = 2sin\dfrac{θ}{2}$$

Inflection Point Detection

The inflection point is where the direction of travel diverges from the original by 45 degrees (π/4 radians).

Angle Between Tangent and Original Vector:

dot_product = $unit\_velocity(θ) • (0, 1) = cos(θ)$ deviation_angle = $acos(θ)) = θ$

Inflection is reached when:$$θ > \dfrac{π}{4} ≈ 0.7854\, radians$$which is approximately 45^o

This shows that the inertial force becomes significant when θ exceeds 45°, and the resistance magnitude is: $2 x sin(\dfrac{π}{8}) ≈ 0.765$

✅ Summary

Resistance increases as the deviation from the original direction increases
*Magnitude is governed by $2 sin(\dfrac{θ}{2})$**
Inflection begins at 45° which is the optimal point to, for example take a step out of a turn.