Relative Velocity and Vectors
Content Materials - Vectors
If an object is in motion, descriptions of that motion typically include both magnitude and
direction information. Any measure that combines magnitude and direction is usually
described as a vector. In the context of this lesson, we need to consider not only the
magnitude of the aircraft and wind velocities, we need to consider the direction.
By the way, physics has specific definitions for English words that sometimes differ
some from common usage. In English, speed and velocity are synonyms. In physics,
speed refers to the magnitude of how fast something is going, while velocity also
includes the direction. 60 mph is a speed, while 60 mph South is a vector.
Vectors can be described graphically or numerically. On a graph, vectors are represented
as an arrow where the length of the arrow represents the magnitude and the direction is
represented by, well, the direction:
Vector B has a magnitude roughly
50% more than Vector A and they
are obviously in different
directions.
A
Whenever you use graphical
representations of vectors, you
should use a protractor and ruler.
Numerical vector representations
come in two typical forms:
A=5
53º
B
B = -3i + -9j
Notice that the vectors are bold and the magnitudes are not. In the first case, vector A has
a magnitude of 5 in the direction 45º measured in a counterclockwise direction from the
positive x axis. Vector B is shown in component form. In other words, B has a
magnitude of 3 in the negative x-direction and a magnitude of 9 in the negative ydirection.
Trigonometry is needed to convert between the different numerical representations. If we
wanted to represent vector A in component form, think of the vector as a triangle:
Trig teaches that the sine is associated with the opposite over the hypotenuse,
the cosine with adjacent over the hypotenuse, and the tangent is associated
with the opposite over adjacent (soh cah toa).
A
sin  
Ay
Ay
Θ = 53º
A
A
cos   x
A
Ay
tan  
Ax
Ax
Ay  A  sin   5  sin( 53 )  4
Ax  A  cos   5  cos(53 )  3
A = 4i + 3j
Let’s convert vector B from component form to the magnitude/angle format.
B  B x2  B y2  (3) 2  (9) 2  9  81  90  9.5
 Bx 
  tan 1   3   18 

9
 By 
  tan 1 
Strictly speaking, we would call this angle 252º or -108º. Trig helps us figure out the
angle within the triangle, we need to interpret what this means
Vector Addition
Assume that an aircraft has air speed of 300 knots in a NE direction and a strong wind of
60 knots blowing due East. The aircraft’s ground speed is the vector sum of these two
numbers. Since they are vectors, we can’t just add 300 and 60. We need to use either
graphical or numerical vector techniques.
The blue arrow represents the
airspeed, the green arrow is the wind
speed and the red is the ground
speed. To add two vectors, you
place the tail of the second arrow to
the tip of the first. Then you draw
the resultant arrow from the tail of
the first arrow to the tip of the
second. Got it?
If you accurately use a ruler and
protractor to draw these arrows, you
can read the ground speed and
direction directly from the graph.
Numerical addition of vectors requires the use of the component form of vectors:
Vair  300  cos( 45  )i  300  sin( 45  ) j  212i  212 j
Vwind  60  cos(0  )i  60  sin( 0  ) j  60i  0 j
V ground  Vair  Vwind  (212  60)i  (212  0) j  272i  212 j
After addition, it is a good idea to convert the resultant vector back in to
magnitude/direction format:
V ground  272 2  212 2  345knots
 212 

  38
272


  tan 1 
Vector Subtraction
We can get real time aircraft ground speed and wind data using the internet. If we want
to determine the air speed, we need to subtract the wind speed from the ground speed. To
subtract vectors graphically, you flip the arrow you want to subtract by 180º and add that
to the vector you want to subtract it from. Using the example above, you would draw the
ground speed vector (345 knots @ 38º), then you would draw the wind speed vector of 60
knots, but due West. When you draw the resultant vector to get air speed, you would end
up with 300 knots @ 45º!
Numerical subtraction would again use component form:
Vair  V ground  Vwind  (272  60)i  (212  0) j  212i  212 j
Vair  212 2  212 2  300knots
 212 

  45
212


  tan 1 