KAAP686 Mathematics and Signal Processing for Biomechanics
Calculus review 3
Vectors in space
1. Vectors
Magnitude & direction
Examples: pos., veloc., accel., force,
Scalar: a number, with no directionality; a one-dimensional vector.
Special unit direction vectors i, j, k point in positive direction along x, y, z axes.
r = r1 i + r2 j + r3 k : specifies a vector by specifying each of its components, along
coordinate axes. If the coefficient is negative, then that component of the vector goes in
the negative direction of the corresponding axis. This specification of a vector does not
imply that it is “at” one place, for example, “rooted” at the origin. Sometimes one will
say that a certain vector, e.g. a force, acts “at” a certain place, but absent such a
statement, nothing can be concluded about the “location” of a vector.
Shorthand: r = (r1 , r2 , r3 )
Adding, subtracting, multiplying by a constant
Magnitude: |r| = sqrt(r12 + r22 + r32 )
Unit direction vectors: r / |r|
Velocity = speed times direction (constant times a unit direction vector)
Distance between 2 points = magnitude of vector from one to other
Point midway between 2 points = “average” of their 2 position vectors
2. Inner products and cross products
These are 2 different ways of “multiplying” vectors.
Inner product
Also known as dot product or scalar product.
Multiply 2 vectors to get a scalar.
u ∙ v = |u| |v| cos θ where θ is angle from u to v
If u = (u1 , u2 , u3) and v = (v1 , v2 , v3), then u ∙ v = u1v1 + u2v2 + u3v3
u∙v=v∙u
Interpretation: u ∙ v is the magnitude of u times the magnitude of the part of v that
is in the direction of u; this can be flipped too (since u ∙ v = v ∙ u): u ∙ v is also the
magnitude of v times the magnitude of the part of u that is in the direction of v.
Application of inner product
The simplicity of computing u ∙ v makes it a simple way to compute the angle
between two vectors in 3-space:
θ = arccos [u ∙ v / (|u| |v|) ]
θ = arccos [(u1v1 + u2v2 + u3v3) / (sqrt(u12 + u22 + u32) sqrt(v12 + v22 + v32 )) ]
Example (inner product)
Work = F ∙ x
Example
Given a force vector F=(Fx, Fy, Fz) and an axis whose direction is specified by
unit vector is a=(ux, uy, uz), find Fa, the component of F in the direction of a.
One might first confirm that a is a unit vector as advertised: check that ux2 + uy2 +
uz2 = 1. Then
Fa = (F ∙ a) a
Note that (F ∙ a) is a scalar. Since a has unit length, the length of Fa is
|Fa| = (F ∙ a)
Cross Product
Multiply 2 vectors to get another vector.
|u × v| = |u| |v| sin θ where θ is angle CCW from u to v
Interpretation: Magnitude of u × v = area of parallelogram spanned by u and v
Direction of u × v is perpendicular to both u and v
(Use right hand rule to find which way: make fingers of RH go the short way
from u into v; the thumb then points in direction of u × v.)
u × v = - (v × u) (same magnitude, opposite direction)
Computing cross products (for 3D vectors): 3x3 arrays and determinants
Determinant of 3x3 matrix has 6 terms, each with 3 factors.
Determinant of 2x2 matrix has 2 terms, each with 2 factors.
Two equiv ways to compute determinant of 3x3 matrix:
1. +, -, + across the top, times the “minors” (minor = determinant of 2x2
matrix left when delete the current row & column)
2. Extend the array by repeat columns 1, 2; draw diagonal arrows;
remember + (down arrows) and – (up arrows)
Examples: torque, angular velocity
Torque (Moment)
Torque and moment are synonymous terms. Torque is the amount of “twisting force” about a
point. Torque is a vector, i.e. it has a magnitude and direction. The magnitude is given by
|τ| = |r| |F| sin θ
τ is the torque vector, F is the force vector, and θ is the angle between r and F. r is the vector
from the point of torque (i.e. the point on the axis of twist) to the point where force is applied.
|τ|, |r|, and |F| indicate the magnitudes of the respective vectors. The direction of torque is
perpendicular to r and F. The torque vector is the cross product of r with F (r first; the order
matters for cross product):
τ=r×F
Example (torque)
If the force F is perpendicular to the vector r, the magnitude of torque is simply
|τ| = |r| |F| .
A bolt has its head at the origin and points along the z axis. A mechanic pulls with 50 N of force
at right angles at the end of a 0.3 m long wrench mounted on the bolt head. How much torque is
generated?
First we assume that the wrench is mounted perpendicular to the bolt. (This is usually, but not always, true;
sometimes the wrench is not perpendicular to the bolt because something else is in the way.) This
assumption guarantees that the torque generated, which is perpendicular to r and F, will be along the bolt
axis. From the earlier equation,
|τ| = |r| |F| = 0.3 m x 50 N = 15 Nm.
Confirm the above result by computing the cross product with a matrix approach.
We assume the wrench extends along the +x axis and force is applied in the +y direction at the wrench
handle end. If p2=location of handle end=(0.3,0,0), and p1=location of bolt head=(0,0,0), then r = p2-p1=
(0.3,0,0) and force is 50N in +Y direction: F=(0,50,0).
𝒊
𝒋
0.3
0
𝝉 = 𝒓 × 𝑭 = ( 0 ) × (50) = 𝑑𝑒𝑡 (0.3 0
0
0
0 50
𝒌
0)
0
0 0
0.3 0
0.3 0
) − 𝒋 ∙ 𝑑𝑒𝑡 (
) + 𝒌 ∙ 𝑑𝑒𝑡 (
)
50 0
0 0
0 50
= 𝒊 ∙ 𝑑𝑒𝑡 (
= 𝒌 ∙ 0.3 ∙ 50 = 15𝒌
Sometimes we want to compute the torque about a specified axis (rather than about a particular
point). In this case, we can compute torque and its magnitude using the equations just given, but
for r we use the vector from the point of force application to a point on the specified axis. (To
specify an axis, we must specify its direction and one point through which the axis passes.) The
resulting torque vector may have components not along the specified axis. We obtain the torque
along the specified axis by taking the inner product of the resulting torque vector with a unit
vector along the specified axis.
Example (torque)
What is the moment Mz about the z axis, created by a force F=(Fx, Fy, Fz) applied at point
p0=(x0,y0,z0)?
Because the torque axis is specified, we must first find the vector from p0 to a point on the axis. For
convenience, we choose the nearest point on the z axis, i.e. pa=(0,0,z0). (We could choose a different point
along the z axis. It will not affect the final answer for Mz, but it will make the math along the way more
complicated before final simplification.) r is the vector from the axis to p0:
r = p0 - pa = (x0,y0,z0) - (0,0,z0) = (x0,y0,0)
Therefore
𝐹𝑥
𝒊
𝑥0
𝑴 = 𝒓 × 𝑭 = (𝑦0 ) × (𝐹𝑦 ) = 𝑑𝑒𝑡 (𝑥0
𝐹𝑥
0
𝐹𝑧
𝑦0
= 𝒊 ∙ 𝑑𝑒𝑡 (𝐹
𝑦
0
𝑥0
)
−
𝒋
∙
𝑑𝑒𝑡
(
𝐹𝑧
𝐹𝑥
𝒋
𝑦0
𝐹𝑦
𝒌
0)
𝐹𝑧
𝑥0
0
) + 𝒌 ∙ 𝑑𝑒𝑡 ( 𝐹
𝐹𝑧
𝑥
𝑦0
𝐹𝑦 )
= 𝒊 ∙ (𝑦0 𝐹𝑧 ) − 𝒋 ∙ (𝑥0 𝐹𝑧 ) + 𝒌 ∙ (𝑥0 𝐹𝑦 − 𝑦0 𝐹𝑥 )
If the force is purely in the x-y plane (perpendicular to the z axis), then Fz=0 and M is purely in the z
direction. If not, we take the inner product of M with the unit vector along the z axis to get Mz:
𝑦0 𝐹𝑧
0
𝑥
𝐹
𝑀𝑧 = 𝑴 ∙ 𝒌 = (
) ∙ (0) = 𝑥0 𝐹𝑦 − 𝑦0 𝐹𝑥
0 𝑧
𝑥0 𝐹𝑦 − 𝑦0 𝐹𝑥
1
Example: Center of Pressure
The following is taken from a technical reference manual for an AMTI force plate. The outputs
from the force plate are the three components of total force (Fx,Fy,Fz) and the moments (torques,
Mx, My, Mz) about the x, y, and z axes through the origin of the force platform coordinate
system. The actual force is distributed over an area on the plate. The user would like to know
the location of the center of pressure. This is defined as the point on the plate where the total
measured force would have to be applied to get the measured torques. See
diagram(calculusreview3_fig1.doc to edit, .jpg to import to html).
The manual says:
The COP or center of pressure is calculated from the measured forces and moments as follows:
x = (-My + z Fx ) / Fz
y = ( Mx + z Fy ) / Fz
The value z is the distance from the origin of the force platform measurement
system to the top surface of the platform. This value is always a negative value.
Is the manual correct? Find out by computing the expected moments (torques) if the force is
applied at the point (x,y,z). Then invert the equations to solve for (x,y).
Angular velocity
In three dimensions, the angular velocity of a point-like object about a point p is a vector given
by the cross product of r with v:
ω = r × v / |r|2
where r is a vector from point p to the object, and v is the object’s velocity. The direction of
angular velocity is perpendicular to r and v. The magnitude of the angular velocity has units of
radians per second (s-1, since radians are dimensionless) and is given by
|ω| = |v sin θ| / |r|
where θ is the angle between r and v.
The two preceding equations are for the angular velocity of a point-like object about a point. In
biomechanics, one may wish to calculate the angular velocity of a body segment relative to the
lab frame or relative to another body segment. This 3-component angular velocity will be
considered later. Winter, 4th ed., pp.187-188, gives one method, but there are other methods
which have advantages. See my document “Differentiation of biomechanical data.doc.”
In two dimensions, angular velocity is a scalar quantity and is computed as the derivative of
angular position:
ω = dθ/dt
where θ(t) = angle (relative to some reference point ) at time t.
Copyright © 2013 William C. Rose