Autonomous quadcopter drones navigate urban delivery corridors by evaluating onboard sensor feedback and tracking pre-computed 3D parametric spline trajectories. Aerospace and computer engineering teams decouple the drone's instantaneous flight speed from its 3D directional heading vector using the principal unit tangent vector T ^ ( t ) \hat{T}(t) T ^ ( t )
An autonomous quadcopter executes an ascending spiral corridor inspection with 3D position vector (in meters) given as a function of time t t t
r β ( t ) = β¨ 4 cos β‘ ( t ) , 4 sin β‘ ( t ) , 3 t β© \vec{r}(t) = \langle 4\cos(t), 4\sin(t), 3t \rangle r ( t ) = β¨ 4 cos ( t ) , 4 sin ( t ) , 3 t β©
Calculate the instantaneous velocity vector v β ( t ) = r β β² ( t ) \vec{v}(t) = \vec{r}'(t) v ( t ) = r β² ( t ) a β ( t ) = r β β² β² ( t ) \vec{a}(t) = \vec{r}''(t) a ( t ) = r β²β² ( t )
Determine the scalar flight speed β₯ v β ( t ) β₯ \|\vec{v}(t)\| β₯ v ( t ) β₯
Compute the principal unit tangent vector T ^ ( t ) \hat{T}(t) T ^ ( t )
Evaluate the velocity v β ( Ο ) \vec{v}(\pi) v ( Ο ) a β ( Ο ) \vec{a}(\pi) a ( Ο ) T ^ ( Ο ) \hat{T}(\pi) T ^ ( Ο ) t = Ο Β seconds t = \pi\text{ seconds} t = Ο Β seconds
Step-by-Step Solution:
Calculate Velocity v β ( t ) \vec{v}(t) v ( t ) a β ( t ) \vec{a}(t) a ( t )
Differentiating component-wise:
v β ( t ) = r β β² ( t ) = β¨ d d t [ 4 cos β‘ ( t ) ] , d d t [ 4 sin β‘ ( t ) ] , d d t [ 3 t ] β© = β¨ β 4 sin β‘ ( t ) , 4 cos β‘ ( t ) , 3 β© Β m/s \vec{v}(t) = \vec{r}'(t) = \left\langle \frac{d}{dt}[4\cos(t)], \frac{d}{dt}[4\sin(t)], \frac{d}{dt}[3t] \right\rangle = \langle -4\sin(t), 4\cos(t), 3 \rangle \text{ m/s} v ( t ) = r β² ( t ) = β¨ d t d β [ 4 cos ( t )] , d t d β [ 4 sin ( t )] , d t d β [ 3 t ] β© = β¨ β 4 sin ( t ) , 4 cos ( t ) , 3 β© Β m/s a β ( t ) = v β β² ( t ) = β¨ β 4 cos β‘ ( t ) , β 4 sin β‘ ( t ) , 0 β© Β m/s 2 \vec{a}(t) = \vec{v}'(t) = \langle -4\cos(t), -4\sin(t), 0 \rangle \text{ m/s}^2 a ( t ) = v β² ( t ) = β¨ β 4 cos ( t ) , β 4 sin ( t ) , 0 β© Β m/s 2 Compute Scalar Flight Speed β₯ v β ( t ) β₯ : \|\vec{v}(t)\|: β₯ v ( t ) β₯ :
β₯ v β ( t ) β₯ = ( β 4 sin β‘ ( t ) ) 2 + ( 4 cos β‘ ( t ) ) 2 + 3 2 = 16 sin β‘ 2 ( t ) + 16 cos β‘ 2 ( t ) + 9 \|\vec{v}(t)\| = \sqrt{(-4\sin(t))^2 + (4\cos(t))^2 + 3^2} = \sqrt{16\sin^2(t) + 16\cos^2(t) + 9} β₯ v ( t ) β₯ = ( β 4 sin ( t ) ) 2 + ( 4 cos ( t ) ) 2 + 3 2 β = 16 sin 2 ( t ) + 16 cos 2 ( t ) + 9 β sin β‘ 2 ( t ) + cos β‘ 2 ( t ) = 1 \sin^2(t) + \cos^2(t) = 1 sin 2 ( t ) + cos 2 ( t ) = 1 β₯ v β ( t ) β₯ = 16 ( 1 ) + 9 = 25 = 5 Β m/s \|\vec{v}(t)\| = \sqrt{16(1) + 9} = \sqrt{25} = 5\text{ m/s} β₯ v ( t ) β₯ = 16 ( 1 ) + 9 β = 25 β = 5 Β m/s 5 Β m/s 5\text{ m/s} 5 Β m/s Compute Principal Unit Tangent Vector T ^ ( t ) \hat{T}(t) T ^ ( t )
Normalize velocity by dividing by speed:
T ^ ( t ) = v β ( t ) β₯ v β ( t ) β₯ = β¨ β 4 sin β‘ ( t ) , 4 cos β‘ ( t ) , 3 β© 5 = β¨ β 4 5 sin β‘ ( t ) , 4 5 cos β‘ ( t ) , 3 5 β© \hat{T}(t) = \frac{\vec{v}(t)}{\|\vec{v}(t)\|} = \frac{\langle -4\sin(t), 4\cos(t), 3 \rangle}{5} = \left\langle -\frac{4}{5}\sin(t), \frac{4}{5}\cos(t), \frac{3}{5} \right\rangle T ^ ( t ) = β₯ v ( t ) β₯ v ( t ) β = 5 β¨ β 4 sin ( t ) , 4 cos ( t ) , 3 β© β = β¨ β 5 4 β sin ( t ) , 5 4 β cos ( t ) , 5 3 β β© Evaluate at t = Ο Β s t = \pi\text{ s} t = Ο Β s
Velocity: v β ( Ο ) = β¨ β 4 sin β‘ ( Ο ) , 4 cos β‘ ( Ο ) , 3 β© = β¨ 0 , β 4 , 3 β© Β m/s \vec{v}(\pi) = \langle -4\sin(\pi), 4\cos(\pi), 3 \rangle = \langle 0, -4, 3 \rangle \text{ m/s} v ( Ο ) = β¨ β 4 sin ( Ο ) , 4 cos ( Ο ) , 3 β© = β¨ 0 , β 4 , 3 β© Β m/s
Acceleration: a β ( Ο ) = β¨ β 4 cos β‘ ( Ο ) , β 4 sin β‘ ( Ο ) , 0 β© = β¨ 4 , 0 , 0 β© Β m/s 2 \vec{a}(\pi) = \langle -4\cos(\pi), -4\sin(\pi), 0 \rangle = \langle 4, 0, 0 \rangle \text{ m/s}^2 a ( Ο ) = β¨ β 4 cos ( Ο ) , β 4 sin ( Ο ) , 0 β© = β¨ 4 , 0 , 0 β© Β m/s 2
Unit Tangent Heading: T ^ ( Ο ) = β¨ 0 , β 4 5 , 3 5 β© \hat{T}(\pi) = \left\langle 0, -\frac{4}{5}, \frac{3}{5} \right\rangle T ^ ( Ο ) = β¨ 0 , β 5 4 β , 5 3 β β©
Meaning of the Answer & Real-Life Application:
Even though the drone maintains a constant speed of 5 Β m/s 5\text{ m/s} 5 Β m/s β₯ a β ( t ) β₯ = 4 Β m/s 2 \|\vec{a}(t)\| = 4\text{ m/s}^2 β₯ a ( t ) β₯ = 4 Β m/s 2 z z z T ^ ( t ) \hat{T}(t) T ^ ( t ) a β ( t ) \vec{a}(t) a ( t )