ME 576 Homework #8

1. Bollinger 8.7

2

2

∆

∆

θ

θ

The error, E, is

E=R 1 cos2

−

θ

Since R is 5” and max. error should be 0.001”

1

-1

E2cos 1R

0.001 =2cos 1- 0.04rad 2.295

− ∆ = − = = °

θ

The length of the straight line segment is

L 2R sin 2 5 sin(0.02) 0.200"2∆

= = × × =θ

Time required to traverse is

L 0.200" 60(s) 0.040sec

300" (min)t∆ = = =

υ

R

error (E)

B.C. for the first three segments

( ) cos( ) ( ) sin( )

0 0.00 0.000 2

i i i i i ii t x t R y t R= =θ θ θπ 5.000

1 0.04 0.04 0.200 4.99722 0.08 0.08 0.400 2

−

−

π

π 4.985

3 0.12 0.12 0.599 4.9652 −π

Equations for linear interpolation

( )

( )

1i 1 -1

1

1i 1 -1

1

( ) ( )( ) (t ) --

( ) ( )( ) (t ) --

i ii

i i

i ii

i i

x t x tx t x t tt t

y t y ty t y t tt t

−−

−

−−

−

−= + ⋅

−= + ⋅

2. Bollinger 8.10

2 22 2 2 2

0 4

1

1.5 1.17

( ) ( ) ( ) ( )

(t) (t)

t t

i i f f

dy dxx ydx dt

dy dydx dx

t t t t

= =

+ = ⇒ + =

= =

=

= ⋅

R r r r r

r A b

υ υ

i) Use a single spline

3 2

2

1 2.77 10 3.25

1 4.16 4.5 3.8

0 0 (4) 3(4)0 0 (4) 2(4) 0 1 4 11 0 1 0

= = =

R

B

1

0 0 64 480 0 16 80 1 4 11 0 1 0

0 0 64 481 2.77 10 3.25 0

1 4.16 4.5 3.8

−

= ⋅ =

A R B

1

0 16 80 1 4 11 0 1 0

0.0951 -0.5114 2.7735 1.0000 =

0.3882 -2.3741 4.1603 1.0000

−

1 2 3 4 5 6 7 8 9 101

1.5

2

2.5

3

3.5

4

4.5

ii) Use three splines as in problem 3.

1

3 2

2

1

Segment 1.

1 2.77 4 5

1 4.16 4 0

0 0 (1.33) 3(1.33)0 0 (1.33) 2(1.33) 0 1 1.33 11 0 1

=

=

R

B

11

0 0 2.37 5.330 0 1.78 2.670 1 1.33 11 0 1 0 0

1 2.77 4 5

1 4.16 −

=

= ⋅ =A R B

10 0 2.37 5.330 0 1.78 2.67

4 0 0 1 1.33 11 0 1 0

1.8413 -2.8478 2.7735 1 =

−

-0.1911 -1.1779 4.1603 1

2

3 2 3 2

2 2

2

Segment 2.

4 5 7 5

4 0 2 0

(1.33) 3(1.33) (2.66) 3(2.66)(1.33) 2(1.33) (2.66) 2(2.66) 1.33 1

=

=

R

B 2.66 1

1 0 1 0

2.37 5.33 18.96 21.331.78 2.67 7.11 5.33

1.33 1 2.67

=

2

1 1 0 1 0

3.0938 -18.5625 38 -21

1.6875 -10.1250 18 -6

=

A

3

3 2

2

3

Segment 3

7 5 10 3.25 R

2 0 4.5 3.8

18.96 21.33 4 3(4) 7.11 5.33 4 2(4) B 2.67 1 4 1 1 0

=

=

3

18.96 21.33 64 48 7.11 5.33 16 8 2.67 1 4 1 1 0 1 0 1 0

2.1086 A

=

=-21.7427 75.9776 -80.9776

0.0286 1.1392 -6.6872 11.1862

1 2 3 4 5 6 7 8 9 10 111

1.5

2

2.5

3

3.5

4

4.5

3. Bollinger 8.10

[ ]

2 22 2 2 2

0 4

1

1.5 1.17

* *(0) *(0) * (1) * (1)

*( ) * *( )

t t

i i

dy dxx ydx dt

dy dydx dx

u u

= =

+ = ⇒ + =

= =

=

= ⋅

R r r r r

r A b

υ υ

*1

* 11

Segment 1.

1 2.77(1.33) 4 5(1.33)

1 4.16(1.33) 4 0

2 -3 0 11 -2 1 0

-2 3 0 01 -1 0 0

−

= =

R

B

* * * 11 1 1

2 -3 0 11 3.68 4 6.65 1 -2 1 0

1 5.53 4 0 -2 3 0 0

1 -1 0 0

−

= ⋅ =

A R B

4.33 -5.03 3.69 1 =

-0.47 -2.07 5.53 1

*2

* * * 12 2 1

Segment 2.

4 5(1.33) 7 5(1.33)

4 0(1.33) 2 0

7.30 -10.95 6.65 4

4.0 -6.00 0 4−

=

= ⋅ =

R

A R B

*3

* * * 13 3 1

Segment 3

7 5(1.33) 10 3.25(1.33)

2 0(1.33) 4.5 3.8(1.33)

4.97 -8.62 6.65 7

0.05 2.45 0 2−

=

= ⋅ =

R

A R B

3

2*

2

*

(t) (t)

( )

( )

1

3( ) 21 ( ) 1.33 1

0

i

i ii

i i

i

i ii

i

ux t uy t u

ux t uy t

= ⋅

=

=

r A b

A

A

A

4. Use the constant acceleration method.

max

2 2max

2 2max

max

2max

1.697 - 0.693 1.004 45 - (-30) 750.4

1010 min sec6015 0.15sec sec

7202 sec sec14404 sec sec

6 0.1min sec20 0.2 secsec

r

r

r m

z m

m mV

cm mA

revV

revA

m mV

cm mA

Θ

Θ

Ζ

Ζ

∆ = =∆ = = °∆ =

= =

= =

°= =

°= =

= =

= =

θ

First calculate the time intervals for three splines of each axis without the consideration to coordinate them.

r-axis

1

2 1r

3 1

10 1.111 sec60 0.15

x 1.004 60 1.111 4.913 secV 10

1.111 sec

t

t t

t t

∆ = =×

∆ ×∆ = − ∆ = − =

∆ = ∆ =

1

2 2

12

1

axis2 0.5sec4

720 75 3601440

0.208 0.228 sec4

x

x

t

VHowever xA

t

−

∆ = =

∆ = < = =

∴ ∆ = =

θ

Z-axis

1

2

3 1

0.1 0.5sec0.20.4 0.5 3.5sec0.1

0.5sec

t

t

t t

∆ = =

∆ = − =

∆ = ∆ =

Without coordination, the velocity profiles would be like

Modify the velocity profiles according to r-axis

1 2 3t t t t 1.111 4.913 1.111 7.135 seconds∆ = ∆ + ∆ + ∆ = + + =

For r-axis, use max. velocity and accelerations

For θ -axis

1

2 1

2

V 1.111

754.913 1.111 4.913V V

12.45 rad V 0.217sec. s11.206 sec

t

t t

∆ = =Α∆

∆ = − ∆ = ⇒ − =

°∴ = =

°Α =

θ

θ

θ θ

θ

θ

θ

For Z-axis

1

2 1

2

1.111

0.44.913 1.111 4.913

0.0664 sec 0.0598 sec

z z

z

z

Vt

zt tV V

mV

m

Ζ

Ζ

∆ = =Α∆

∆ = − ∆ = ⇒ − =

=

Α =

Boundary conditions

0

1

2

-300.693 0

0 0023.0840.786 0.037

0.167 0.066412.4538.0841.604 0.363

0.167 0.066412.45 451.697 0.4

0 0 0f

r axis axis z axisr z

tr z

r zt

r z

r zt

r z

r zt

r z

− −= − °= =

= === −= =

= ==== =

− === °= =

= ==

θθ

θθ

θθ

θθ

θ

A1 =

-0.0003 0.0757 0 0.6930

-0.0000 5.6031 0 -30.0000

-0.0000 0.0299 0 0

A2 =

0.0000 -0.0003 0.1675 0.6002

-0.0000 0.0000 12.4499 -36.9159

0.0000 -0.0000 0.0664 -0.0369

A3 =

0.0011 -0.0973 1.2170 -2.4412

-0.0015 -5.5736 79.7629 -239.8257

-0.0002 -0.0266 0.4048 -1.0747

0 1 2 3 4 5 6 7 8-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

0 1 2 3 4 5 6 7 8-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

time[s]

Acc

eler

atio

n

Acceleration vs Time

r(m/s2)

z (m/s2)

Theta (rad/s2)