ElmoMC Multi-Axis Motion Controller-Maestro Motion Control - Manual (Page 24)

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: