Vector navigation
5/16/14
Objectives
Vector navigation
Assessment
B = (0, +8) m
Add two-dimensional vectors graphically using the head-to-tail method.
•
Add two-dimensional vectors algebraically using the component method.
Physics terms
1. Three displacement vectors A, B, and C are given below. Add these vectors graphically using the head-to-tail method, and draw the resultant.
A = (+4.0, 0) m
•
•
Cartesian coordinates
•
polar coordinates
C = (-8, -4) m
2. Add these vectors algebraically by adding the components. Report the resultant vector in component form.
When will this happen?
How could a car drive itself?
1
5/16/14
Displacement vectors A single straight-line movement in any direction is described with a displacement vector.
Displacement vectors 100 m 20° E of N
A robotic car starts at the origin and takes a drive consisting of three straight legs . It drives . . . • 10 meters east • 5 meters north • 5 meters west Where does it end up?
Displacement vectors
Graphical vector addition
Where does it end up?
The car travels . . .
There are two ways to answer this question:
• 10 meters east
• graphical vector addition • adding vectors algebraically using components
Graphical vector addition
Sketch the first vector d1 starting at the origin.
Graphical vector addition
The car travels . . .
The car travels . . .
• 10 meters east
• 10 meters east
• 5 meters north
• 5 meters north • 5 meters west
Sketch the second vector d2 starting from the head of the first vector.
Sketch the third vector d3 starting from the head of the second vector.
Do you see why this is called the head-to-tail” method?
2
5/16/14
Graphical vector addition
Graphical vector addition
Now draw the resultant vector. The resultant vector is the single vector that creates the same change in position.
d = d1 + d2 + d3
The resultant vector shows the result of this trip. It points from where the car started to where it ended.
BE CAREFUL! Which way should the resultant vector point?
Graphical vector addition Does this graphical method work for computers?
Graphical vector addition Does this graphical method work for computers? NO! Computers don’t have hands. They can’t draw. Computers don’t have eyes. They can’t read a graph. Computers can only add, subtract, and multiply numbers.
Another method Computers need a numerical method for adding vectors. Vectors can be described numerically in Cartesian or polar coordinates.
Adding vectors using components The Cartesian method uses the vector components along the x, y, and z axes. This allows you to add vectors algebraically without having to draw a vector diagram.
3
5/16/14
Adding vectors using components Look at the car problem again. This time we solve it without drawing a vector diagram.
Adding vectors using components +y
Assign directions to the positive and negative x and y axes.
A robotic car starts at the origin and takes a drive consisting of three straight legs .
Two numbers, called components, determine the magnitude and direction of any vector.
d1 = 10 meters east
N
-x W
E +x
d2 = 5 meters north d3 = 5 meters west S
Where does it end up?
-y
Adding vectors using components
Adding vectors using components
+y Write each vector in component form.
d1 = 10 meters east = (+10, 0 ) m
+y Write each vector in component form.
N
-x W
E +x
d1 = 10 meters east = (+10, 0 ) m
N
-x W
E +x
d2 = 5 meters north = ( 0, +5 ) m
S
S
-y
-y
Adding vectors using components +y Write each vector in component form.
Adding vectors using components +y
Add the x components to find the total x displacement.
N
N
Add the y components to find the total y displacement. d1 = 10 meters east = (+10, 0 ) m
-x W
E +x
d2 = 5 meters north = ( 0, +5 ) m
d1 = 10 meters east = (+10, 0 ) m
D -x W
E +x
d2 = 5 meters north = ( 0, +5 ) m
d3 = 5 meters west = ( -5, 0 ) m
d3 = 5 meters west = ( -5, 0 ) m S
-y
D = (+5, +5) m
S
-y
4
5/16/14
Adding vectors using components Adding vectors is just adding components separately in the x and y directions.
D = d1 + d2 + d3
d1 = 10 meters east = (+10, 0 ) m
Add these three force vectors to find F4. F4 = F1 + F2 + F3 where
F1 = (2, 2) N F2 = (4, -4) N
D
d3
d2 = 5 meters north = ( 0, +5 ) m d3 = 5 meters west = ( -5, 0 ) m
Test your knowledge
F3 = (4, 8) N
d1 d2
D = (+5, +5) m
Test your knowledge
Investigation
Add these three force vectors to find F4. F4 = F1 + F2 + F3 where
In Investigation 6A, you will add vectors to solve a series of mazes.
F1 = (2, 2) N F2 = (4, -4) N F3 = (4, 8) N
F4 = (10, 6) N
Force vectors add the same way displacement vectors do!
Part 1: Navigating a maze Find a series of vectors that can solve the maze without crossing a wall. What is the shortest path through the maze? You may make up to 20 displacements.
Click on this simulation on page 180.
Part 1: Navigating a maze Enter x and y coordinates for vectors here. The Simulate button runs your course on the screen. Reset erases all vectors
5
5/16/14
Part 1: Navigating a maze
How does this relate to self-driving cars? 1. How could a computer store a route?
There are four preset mazes to solve. The fifth button opens a free-form area where you can program a route of your own.
Performance evaluation – a STEM skill
2. What kind of driving logic might be used by a computer? 3. What additional information would be useful to the driving computer? 4. What might cause the computer to make a mistake?
The ErgoBot
This knob engages or disengages the drive motors on both wheels.
Test your vector addition skills on a real maze! Download your maze solution into an ErgoBot – a robot vehicle that accepts commands in vectors.
Program your ErgoBot
Set the starting position The Program button programs your vectors in polar coordinates to the ErgoBot.
The ErgoBot must be at the start and pointing in the x-direction.
y
It will turn 90° as its first move for this maze.
The Run button executes your program.
x
6
5/16/14
What did you learn?
Reprogram your ErgoBot Set the diameter of the red circle that represents the ErgoBot.
How would you change this path to miss the cones?
Find a new course that takes the ErgoBot’s size into account.
What did you learn?
Accounting for uncertainty A real vehicle has some uncertainty in its position. It must move as if it were larger to allow for errors and still avoid known obstacles.
This is better! Does the ErgoBot know exactly where it is? Discuss precision and motion.
What did you learn? What is the effective size of the ErgoBot if you want to miss all the cones 9 out of 10 times?
Investigation Click on the second simulation on page 180 to experiment with force vectors.
7
5/16/14
Investigation
Investigation
Part 2: Force, inertia, and a race course
Part 2: Force, inertia, and a race course
1. In this interactive simulation, two players race through a course to the finish line.
3. Using the setup button, you may choose a different race course or switch to oneplayer mode.
2. During their turns, players may apply a force of 1 N in any direction (or choose zero force instead).
Investigation
Assessment
Questions
1. Three displacement vectors A, B, and C are given below. Add these vectors graphically using the head-to-tail method, and draw the resultant.
a. How does an upwards applied force change the velocity of the cart?
A = (+4.0, 0) m
b. Are there negative forces applied in this activity? Explain.
B = (0, +8) m C = (-8, -4) m
c. Explain the role of inertia in this activity.
Assessment
Assessment
1. Three displacement vectors A, B, and C are given below. Add these vectors graphically using the head-to-tail method, and draw the resultant.
A = (+4.0, 0) m
1. Three displacement vectors A, B, and C are given below. Add these vectors graphically using the head-to-tail method, and draw the resultant.
A = (+4.0, 0) m C
C
B = (0, +8) m
B = (0, +8) m B
C = (-8, -4) m Now draw the resultant. Which way does it point?
B C = (-8, -4) m
A
D=A+B+C
D A
8
5/16/14
Assessment
Assessment
2. Add these vectors algebraically by adding the components. Report the resultant vector in component form.
A = (+4.0, 0) m
2. Add these vectors algebraically by adding the components. Report the resultant vector in component form.
A = (+4.0, 0) m C
B = (0, +8) m
B = (0, +8) m B
C = (-8, -4) m
D=A+B+C
C = (-8, -4) m
D A
D
D = (-4, +4) m
9
Objectives
Vector navigation
Assessment
B = (0, +8) m
Add two-dimensional vectors graphically using the head-to-tail method.
•
Add two-dimensional vectors algebraically using the component method.
Physics terms
1. Three displacement vectors A, B, and C are given below. Add these vectors graphically using the head-to-tail method, and draw the resultant.
A = (+4.0, 0) m
•
•
Cartesian coordinates
•
polar coordinates
C = (-8, -4) m
2. Add these vectors algebraically by adding the components. Report the resultant vector in component form.
When will this happen?
How could a car drive itself?
1
5/16/14
Displacement vectors A single straight-line movement in any direction is described with a displacement vector.
Displacement vectors 100 m 20° E of N
A robotic car starts at the origin and takes a drive consisting of three straight legs . It drives . . . • 10 meters east • 5 meters north • 5 meters west Where does it end up?
Displacement vectors
Graphical vector addition
Where does it end up?
The car travels . . .
There are two ways to answer this question:
• 10 meters east
• graphical vector addition • adding vectors algebraically using components
Graphical vector addition
Sketch the first vector d1 starting at the origin.
Graphical vector addition
The car travels . . .
The car travels . . .
• 10 meters east
• 10 meters east
• 5 meters north
• 5 meters north • 5 meters west
Sketch the second vector d2 starting from the head of the first vector.
Sketch the third vector d3 starting from the head of the second vector.
Do you see why this is called the head-to-tail” method?
2
5/16/14
Graphical vector addition
Graphical vector addition
Now draw the resultant vector. The resultant vector is the single vector that creates the same change in position.
d = d1 + d2 + d3
The resultant vector shows the result of this trip. It points from where the car started to where it ended.
BE CAREFUL! Which way should the resultant vector point?
Graphical vector addition Does this graphical method work for computers?
Graphical vector addition Does this graphical method work for computers? NO! Computers don’t have hands. They can’t draw. Computers don’t have eyes. They can’t read a graph. Computers can only add, subtract, and multiply numbers.
Another method Computers need a numerical method for adding vectors. Vectors can be described numerically in Cartesian or polar coordinates.
Adding vectors using components The Cartesian method uses the vector components along the x, y, and z axes. This allows you to add vectors algebraically without having to draw a vector diagram.
3
5/16/14
Adding vectors using components Look at the car problem again. This time we solve it without drawing a vector diagram.
Adding vectors using components +y
Assign directions to the positive and negative x and y axes.
A robotic car starts at the origin and takes a drive consisting of three straight legs .
Two numbers, called components, determine the magnitude and direction of any vector.
d1 = 10 meters east
N
-x W
E +x
d2 = 5 meters north d3 = 5 meters west S
Where does it end up?
-y
Adding vectors using components
Adding vectors using components
+y Write each vector in component form.
d1 = 10 meters east = (+10, 0 ) m
+y Write each vector in component form.
N
-x W
E +x
d1 = 10 meters east = (+10, 0 ) m
N
-x W
E +x
d2 = 5 meters north = ( 0, +5 ) m
S
S
-y
-y
Adding vectors using components +y Write each vector in component form.
Adding vectors using components +y
Add the x components to find the total x displacement.
N
N
Add the y components to find the total y displacement. d1 = 10 meters east = (+10, 0 ) m
-x W
E +x
d2 = 5 meters north = ( 0, +5 ) m
d1 = 10 meters east = (+10, 0 ) m
D -x W
E +x
d2 = 5 meters north = ( 0, +5 ) m
d3 = 5 meters west = ( -5, 0 ) m
d3 = 5 meters west = ( -5, 0 ) m S
-y
D = (+5, +5) m
S
-y
4
5/16/14
Adding vectors using components Adding vectors is just adding components separately in the x and y directions.
D = d1 + d2 + d3
d1 = 10 meters east = (+10, 0 ) m
Add these three force vectors to find F4. F4 = F1 + F2 + F3 where
F1 = (2, 2) N F2 = (4, -4) N
D
d3
d2 = 5 meters north = ( 0, +5 ) m d3 = 5 meters west = ( -5, 0 ) m
Test your knowledge
F3 = (4, 8) N
d1 d2
D = (+5, +5) m
Test your knowledge
Investigation
Add these three force vectors to find F4. F4 = F1 + F2 + F3 where
In Investigation 6A, you will add vectors to solve a series of mazes.
F1 = (2, 2) N F2 = (4, -4) N F3 = (4, 8) N
F4 = (10, 6) N
Force vectors add the same way displacement vectors do!
Part 1: Navigating a maze Find a series of vectors that can solve the maze without crossing a wall. What is the shortest path through the maze? You may make up to 20 displacements.
Click on this simulation on page 180.
Part 1: Navigating a maze Enter x and y coordinates for vectors here. The Simulate button runs your course on the screen. Reset erases all vectors
5
5/16/14
Part 1: Navigating a maze
How does this relate to self-driving cars? 1. How could a computer store a route?
There are four preset mazes to solve. The fifth button opens a free-form area where you can program a route of your own.
Performance evaluation – a STEM skill
2. What kind of driving logic might be used by a computer? 3. What additional information would be useful to the driving computer? 4. What might cause the computer to make a mistake?
The ErgoBot
This knob engages or disengages the drive motors on both wheels.
Test your vector addition skills on a real maze! Download your maze solution into an ErgoBot – a robot vehicle that accepts commands in vectors.
Program your ErgoBot
Set the starting position The Program button programs your vectors in polar coordinates to the ErgoBot.
The ErgoBot must be at the start and pointing in the x-direction.
y
It will turn 90° as its first move for this maze.
The Run button executes your program.
x
6
5/16/14
What did you learn?
Reprogram your ErgoBot Set the diameter of the red circle that represents the ErgoBot.
How would you change this path to miss the cones?
Find a new course that takes the ErgoBot’s size into account.
What did you learn?
Accounting for uncertainty A real vehicle has some uncertainty in its position. It must move as if it were larger to allow for errors and still avoid known obstacles.
This is better! Does the ErgoBot know exactly where it is? Discuss precision and motion.
What did you learn? What is the effective size of the ErgoBot if you want to miss all the cones 9 out of 10 times?
Investigation Click on the second simulation on page 180 to experiment with force vectors.
7
5/16/14
Investigation
Investigation
Part 2: Force, inertia, and a race course
Part 2: Force, inertia, and a race course
1. In this interactive simulation, two players race through a course to the finish line.
3. Using the setup button, you may choose a different race course or switch to oneplayer mode.
2. During their turns, players may apply a force of 1 N in any direction (or choose zero force instead).
Investigation
Assessment
Questions
1. Three displacement vectors A, B, and C are given below. Add these vectors graphically using the head-to-tail method, and draw the resultant.
a. How does an upwards applied force change the velocity of the cart?
A = (+4.0, 0) m
b. Are there negative forces applied in this activity? Explain.
B = (0, +8) m C = (-8, -4) m
c. Explain the role of inertia in this activity.
Assessment
Assessment
1. Three displacement vectors A, B, and C are given below. Add these vectors graphically using the head-to-tail method, and draw the resultant.
A = (+4.0, 0) m
1. Three displacement vectors A, B, and C are given below. Add these vectors graphically using the head-to-tail method, and draw the resultant.
A = (+4.0, 0) m C
C
B = (0, +8) m
B = (0, +8) m B
C = (-8, -4) m Now draw the resultant. Which way does it point?
B C = (-8, -4) m
A
D=A+B+C
D A
8
5/16/14
Assessment
Assessment
2. Add these vectors algebraically by adding the components. Report the resultant vector in component form.
A = (+4.0, 0) m
2. Add these vectors algebraically by adding the components. Report the resultant vector in component form.
A = (+4.0, 0) m C
B = (0, +8) m
B = (0, +8) m B
C = (-8, -4) m
D=A+B+C
C = (-8, -4) m
D A
D
D = (-4, +4) m
9
