Tg(γ/2), Γ = π – arccos[(δx, Π – arccos – ElmoMC Multi-Axis Motion Controller-Maestro Motion Control User Manual

Page 24: Γ/2)

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Motion Library Tutorial

Switch Radius Calculation

MAN-MLT (Ver 2.0)

2-3

r

_max

= d

max

*

tg(γ/2) =

50000*

tg

(0.5*

0.1974) = 4951

This value is limiting and produces singular trajectory with the switch arc that

replaces

L

2

.

So we should take some value less than limiting value 4951 for instance 0.5*

r_

max

=

0.5*4951 = 2475

.

Now we can recalculate maximum end velocity (vse) that satisfies this

value:

vse = [r*vac*vae]

1/2

= [2475*500000*0.9]

1/2

= 33376

(any value less equal than 33376 can be used with the switch radius r = 2475).

Example 2.1b

Line 1 is defined by init point (50000, 70000) and end point (60000,20000). Line 2 is
defined by the init point (60000,20000) and its end point (60000,70000). Switch from Line
1 to Line 2 must be executed with the pre-defined switch radius (vsc = 2). We define
cruise velocity vsp = 50000 and end velocity vse = 50000. Vector
acceleration/deceleration vac = vdc = 500000 and switch radius vsr = 6000

1. We calculated minimal switch radius that satisfies the kinematics constraint

r_min = (vse)

2

/(vac*vae) = (50000)

2

/(500000*0.9) = 5555.6

Pre-defined switch radius is greater than r_min so it satisfies kinematics constraints

2. We have to check if switch radius r_switch = 6000 satisfies geometric constraints

Δ

X

1

= 60000 - 50000 = 10000, dY1 = 20000 – 70000 = -50000,

Δ

X

2

= 60000 – 60000 = 0, dY2= 70000 – 20000 = 50000

ΔL

1

= [dX1

2

+ dY1

2

]

1/2

= [(

10000

)

2

+ (

-50000

)

2

]

1/2

= 50990

ΔL

2

= [dX2

2

+ dY2

2

]

1/2

= [0 + (

50000

)

2

]

1/2

=

50000

γ = π – arccos[(ΔX

1

ΔX

2

+ ΔY

1

ΔY

2

)/(ΔL

1

ΔL

2

) =

=

π – arccos

{[(-50000)*(0) + (-50000)*(50000)]/(

50990*

50000)} =

0.1974

The distance from the intersection point that corresponds to the r_switch = 6000

d

= r_switch/tg

(

γ/2) =

6000/

tg

(0.5*

0.1974) = 60593

The calculated value exceeds

ΔL

1

and

ΔL

2

meaning that the chosen switch radius does

not fit geometric constraints and must be decreased:

r

_max

=

min(

ΔL

1,

ΔL

2

)tg(γ/2)

= 50000*

tg(0.5*0.1974) = 4951.

It is possible to choose any value r_switch that satisfies r_min < r_switch and r_switch <
r_max
but in this case r_min > r_max. So as r_switch, use 0.9min(r_min , r_max) =
4455.9.

The chosen value exceeds

r_min so it does not fit the kinematics constraints and the end

velocity must be decreased:

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