ee 570: location and navigation - navigation mathematics...

EE 570: Location and NavigationNavigation Mathematics: Translation

Kevin Wedeward Aly El-Osery

Electrical Engineering Department, New Mexico TechSocorro, New Mexico, USA

In Collaboration withStephen Bruder

Electrical and Computer Engineering DepartmentEmbry-Riddle Aeronautical Univesity

Prescott, Arizona, USA

February 4, 2016

Vector Notation for Translation Translation Between More Than Two Coordinate Frames ExampleKevin Wedeward, Aly El-Osery (NMT) EE 570: Location and Navigation February 4, 2016 1 / 15

Lecture Topics

1 Vector Notation for Translation

2 Translation Between More Than Two Coordinate Frames

3 Example


Translation Between Frames

Define the vector ~rαβ from the origin of {α} to the origin of {β}.specifies translation between frames

xαyα

zα

xβ

yβ

zβ31

1.5~rαβ

~r ααβ =

xααβyααβzααβ

=

−1.00

31.50

Cαβ =

0.785 −0.366 0.5000.242 0.924 0.296−0.571 −0.111 0.813

Now have means to describe rotation and translation betweencoordinate frames.



Resolve, i.e., coordinatize, ~rαβ wrt frame {β}.

xαyα

zα

xβ

yβ

zβ~rαβ xβ

yβ

zβ

~r βαβ =

xβαβ

yβαβ

zβαβ

=

−0.914

2.97

1.61

= Cβα~r ααβ

Same vector, so same “direction” and length.



Reverse vector ~r , i.e., now from origin of {β} to origin of {α}.

notation:

~rβα = −~rαβ

xαyα

zα

xβ

yβ

zβ31

1.5~rβα

~r αβα =

xαβαyαβαzαβα

= −~r ααβ =

−(−1.00)−(3)−(1.50)




notation: ~rβα =

−~rαβ

xαyα

zα

xβ

yβ

zβ31

1.5~rβα

~r αβα =


= −~r ααβ =

−(−1.00)−(3)−(1.50)




notation: ~rβα = −~rαβ

xαyα

zα

xβ

yβ

zβ31

1.5~rβα

~r αβα =


= −~r ααβ =

−(−1.00)−(3)−(1.50)


Translation (more than two coordinate frames)

Consider three coordinate systems {a}, {b}, {c} that have translationand rotation relative to each other.

Knowing relationships between frames {a}, {b}, and {c}, i.e., ~rab ,~rbc , ~rac , C a

b , Cbc , and C a

c , location of point p can be described inany frame, i.e., ~p a or ~p b or ~p c .

xb

yb

zb

xa

ya

za

xc

yc

zc

~rab

~rbc

~rac

p



Determine the location of the point p relative to {a} given location ofpoint p is known relative to {b}.

xb

yb

zb

xa

ya

za

xc

yc

zc

~rab

p~pbp

~pap

~pap =

~rab + ~pbpIn what frame?~p aap = ~r a

ab + ~p abp

or~p bap = ~r b

ab + ~p bbp

or~p cap = ~r c

ab + ~p cbp

Shorthand notation: ~p a ≡ ~p aap




xb

yb

zb

xa

ya

za

xc

yc

zc

~rab

p~pbp

~pap

~pap = ~rab + ~pbp

In what frame?~p aap = ~r a

ab + ~p abp

or~p bap = ~r b

ab + ~p bbp

or~p cap = ~r c

ab + ~p cbp





xb

yb

zb

xa

ya

za

xc

yc

zc

~rab

p~pbp

~pap

~pap = ~rab + ~pbpIn what frame?

~p aap = ~r a

ab + ~p abp

or~p bap = ~r b

ab + ~p bbp

or~p cap = ~r c

ab + ~p cbp





xb

yb

zb

xa

ya

za

xc

yc

zc

~rab

p~pbp

~pap

~pap = ~rab + ~pbpIn what frame?~p aap = ~r a

ab + ~p abp

or~p bap = ~r b

ab + ~p bbp

or~p cap = ~r c

ab + ~p cbp




Given ~p aap = ~r a

ab + ~p abp and/or the diagram, how would one find ~p b

bp?

xbyb

zb

xa

yaza

xc

yc

zc

~rab

p~pbp

~pap

use given relationshipor vector addition⇒ ~p a

bp = ~p aap − ~r a

ab

now need to referenceto {b}Cba ~p

abp =

Cba

(~p aap − ~r a

ab

)⇒ ~p b

bp = ~p bap − ~r b

ab



Given ~p aap = ~r a


bp?

xbyb

zb

xa

yaza

xc

yc

zc

~rab

p~pbp

~pap

use given relationshipor vector addition

⇒ ~p abp = ~p a

ap − ~r aab


abp =

Cba

(~p aap − ~r a

ab

)⇒ ~p b


ab



Given ~p aap = ~r a


bp?

xbyb

zb

xa

yaza

xc

yc

zc

~rab

p~pbp

~pap



ab


abp =

Cba

(~p aap − ~r a

ab

)⇒ ~p b


ab



Given ~p aap = ~r a


bp?

xbyb

zb

xa

yaza

xc

yc

zc

~rab

p~pbp

~pap



ab

now need to referenceto {b}

Cba ~p

abp =

Cba

(~p aap − ~r a

ab

)⇒ ~p b


ab



Given ~p aap = ~r a


bp?

xbyb

zb

xa

yaza

xc

yc

zc

~rab

p~pbp

~pap



ab


abp =

Cba

(~p aap − ~r a

ab

)⇒ ~p b


ab



It is important to remember difference between recoordinatizing avector and finding a location wrt a different frame.

Recoordinatizing: ~p cap = C c

a ~paap

(only frame of reference changes)Location wrt different frame: ~p c

cp = ~r ccb + C c

b~rbba + C c

a ~paap

(vector addition in same frame)6= C c

a ~paap





a ~paap

(only frame of reference changes)

Location wrt different frame: ~p ccp = ~r c

cb + C cb~r

bba + C c

a ~paap


a ~paap





a ~paap

(only frame of reference changes)Location wrt different frame: ~p c

cp = ~r ccb + C c

b~rbba + C c

a ~paap


a ~paap



Determine location of point p from frame {c};⇒ looking for

~pcp

xb

yb

zb

xa

ya

za

xc

yc

zc

~rab

p

~pcp~rbc

~rac

~pbp

~pap

Many approaches givenlabeled vectors/transla-tions.~pcp

= −~rbc + ~pbp

= −~rac + ~rab + ~pbp

= −~rac + ~pap

In what frame?doesn’t matter, solong as sameCan alwaysrecoordinatize givenC ab ,C

bc ,C

ca



Determine location of point p from frame {c};⇒ looking for ~pcp

xb

yb

zb

xa

ya

za

xc

yc

zc

~rab

p

~pcp~rbc

~rac

~pbp

~pap


= −~rbc + ~pbp


= −~rac + ~pap


bc ,C

ca




xb

yb

zb

xa

ya

za

xc

yc

zc

~rab

p

~pcp~rbc

~rac

~pbp

~pap


= −~rbc + ~pbp


= −~rac + ~pap

In what frame?

doesn’t matter, solong as sameCan alwaysrecoordinatize givenC ab ,C

bc ,C

ca




xb

yb

zb

xa

ya

za

xc

yc

zc

~rab

p

~pcp~rbc

~rac

~pbp

~pap


= −~rbc + ~pbp


= −~rac + ~pap


bc ,C

ca


Example - Given

Consider the three coordinate frames {a}, {b}, {c} shown with therotations and translations between some frames given.

xbyb

‖

zb

xa

ya

za

50◦

xc

yc‖

zc

−30◦

~rac

C ab = Rz,50◦

Cbc = Ry ,−30◦

~r aab =

[0 0 2

]T~r bbc =

[3 0 0

]T

findC ac

~r aac

~r cca


Example - Given

Consider the three coordinate frames {a}, {b}, {c} shown with therotations and translations between some frames given.

xbyb

‖

zb

xa

ya

za

50◦

xc

yc‖

zc

−30◦

~rac

C ab = Rz,50◦

Cbc = Ry ,−30◦

~r aab =

[0 0 2

]T~r bbc =

[3 0 0

]TfindC ac

~r aac

~r cca


Example - Find C ac

C ac = C a

bCbc = Rz,50◦Ry ,−30◦

xbyb

‖

zb

xa

ya

za

50◦

xc

yc‖

zc

−30◦

~rac


Example - Find ~r aac

~r aac = ~r a

ab + ~r abc

= ~r aab + C a

b~rbbc

=

002

+ Rz,50◦

300

=

002

+

cos 50◦ − sin 50◦ 0sin 50◦ cos 50◦ 0

0 0 1

300

=

1.932.302.00

xbyb

‖

zb

xa

ya

za

−50◦

xc

yc‖

zc

−30◦

~rac

~rbc

~rab


Example - Find ~r cca

~r cca = −~r c

ac

= −C ca ~r

aac

= − [C ac ]

T ~r aac

= − [Rz,50◦ Ry ,−30◦ ]T

1.932.302.00

=

−3.590

−0.232

xbyb

‖

zb

xa

ya

za

−50◦

xc

yc‖

zc

−30◦

~rac

~rbc

~rab


The End


ee 570: location and navigation - navigation mathematics...

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