Tangent, Normal and Binormal Vectors for Physics of Walking/Marching and Tossing

WMT surface

With thanks to the following calculus III students:
Saad Alhajeri,
Brian K. Andrasko,
Ben P. Drury,
Laura M. Edmonds,
Thadde Holdinghausen,
James E. Huckstep, III,
Jeremy M. Hudson,
Corey A. Luke,
Nicholas A. Offerman,
Achyut Pradhan,
Sarah K. Taylor,
Son T. Tran, and
Brent S. Trunko

Position Of Labled Points On Parabolic Cylinder Surface f(x,y) = -(1/4)x² + 16 Calculations Based On The Given Vector-Valued Function
Point	Time t	X(t)	Y(t)	Z(t)	Comments On The Vector-Valued Function
0	0.000	0.000	0.000	16.000
1	0.050	1.394	2.780	15.518	R(t) = [ 27.8 t, 55.6 t, -193 t² + 16 ]
2	0.100	2.788	5.560	14.070
3*	0.100	3.168	6.320	13.542	X(t) = 27.8 t for projected horizontal 27.8 in/sec tossing motion
4	0.150	4.182	8.340	11.658
5	0.200	5.576	11.120	8.280	Y(t) = 55.6 t for 55.6 in/sec walking/marching speed
6	0.250	6.970	13.900	3.938
7	0.288	8.029	16.013	0.000	Z(t) = -193 t² + 16 is the tossing free fall motion

* Point 3 is the tip of the unit tangent vector originating from point 2.

TNB with Acceleration

Surface Equation:
f(x,y) = -(1/4)x² + 16

Vector Function:
R(t) = [ 27.8 t, 55.6 t, -193 t² + 16 ]

Velocity Vector Function:
V(t) = R'(t) = [ 27.8, 55.6, -386 t ]

Acceleration Vector Function:
A(t) = V'(t) = R"(t) = [ 0, 0, -386 ]

Tangent Vector:
T(t) = [ 0.4472, 0.8944, -6.2095 t ] / ( 1 + 38.5580 t² )^0.5

Normal Vector:
N(t) = [ -2.7770 t, -5.5540 t, -1 ] / ( 1 + 38.5580 t² )^0.5

Binormal Vector:
B(t) = [ -0.8944, 0.4472, 0 ]

Curvature:
k = 0.09989 / ( 1 + 38.5580 t² )^0.5

Moving Trihedral (Tangent, Normal and Binormal) Position Of Vectors' Terminal Points For Each Corresponding Originating Point Listed Above
Point	Time t	TX(t)	TY(t)	TZ(t)	NX(t)	NY(t)	NZ(t)	BX(t)	BY(t)	BZ(t)
0	0.000	0.447	0.894	16.000	0.000	0.000	15.000	-.894	0.447	16.000
1	0.050	1.821	3.634	15.221	1.261	2.515	14.562	0.500	3.227	15.518
2	0.100	3.168	6.320	13.542	2.552	5.088	13.220	1.894	6.007	14.070
4	0.150	4.509	8.994	10.976	3.877	7.730	10.926	3.288	8.787	11.658
5	0.200	5.856	11.681	7.501	5.228	10.423	7.652	4.682	11.567	8.280
6	0.250	7.212	14.384	3.097	6.594	13.148	3.396	6.076	14.347	3.937
7	0.288	8.248	16.449	-.8810	7.639	15.232	-.496	7.135	16.460	0.000

TNB with Acceleration Speed:
|R'(t)| = 62.1627 ( 1 + 38.5580 t² )^0.5

Acceleration magnitude:
|R"(t)| = 386

Units:
. . Time in seconds
. . Distance in inches
. . Velocity in inches per second
. . Acceleration in inches per
. . . . second squared

Arc Length:
s is the definite integral from 0.0 to t of ( 3864.2 + 148996 t² )^0.5 dt so

s = 193 t ( .025935 + t² )^0.5 + 5.005455 arsinh( 6.2095 t)

Flight distance for an object tossed while walking is a little more than 50 inches.

Acceleration in the plane of the Tangent and Normal unit vectors:
A = (2396.8718/(1 + 38.5580 t²)^0.5)T(t) + (386/( 1 + 38.5580 t²)^0.5)N(t)

Give an equation for the plane of this toss?

Give an equation for the path of this toss in that plane then check curvature and arc length values with those presented in three dimensions.

• Gradient of f(x,y) and F(x,y,z) = f(x,y) - z, a Tangent Plane and More

• The Talk Outline

"Out Juggling In My Class" by William V. Thayer

Walking, Marching and Tossing: An Introduction