ElmoMC Multi-Axis Motion Controller-Maestro Motion Control - Manual (Page 23)

Motion Library Tutorial Switch Radius Calculation
MAN-MLT (Ver 2.0)
2-2
vsr ≤ min(0.5ΔL
1
, 0.5ΔL
2
)*tg(γ/2)
(2.1-4)
and kinematics constraint
V
end
≤ [r*vae*AC
v
)] 1/2
(2.1-5)
where vae – acceleration error (default value – 0.9).
So the user defined parameters must satisfy
vse ≤ [vsr*vae*vac] 1/2
(2.1-6)
vse –
segment end velocity, vsr – switch radius, vac – vector acceleration.
Example 2.1a
Line 1 is defined by its 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). Switching from Line
1 to Line 2 must be executed with a minimal switch radius (vsc = 1). The cruise velocity is
defined as vsp = 50000 and the end velocity vse = 50000. Vector acceleration/deceleration
vac = vdc = 500000
1. The calculated minimal switch radius that satisfies kinematics constraint is
r_min = (vse) 2 /(vac*vae) = (50000) 2 /(500000*0.9) = 5555.6
2. The calculated distance from the intersection point that corresponds to r_min = 5555.6
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 – arcos[(ΔX
1
ΔX
2
+ ΔY
1
ΔY
2
)/(ΔL
1
ΔL
2
) =
= π – arcos{[( -50000 )*( 0 ) + (- 50000 )*( 50000 )]/( 50990* 50000
)} =
0.1974
The distance from the intersection point corresponding to the minimal switch radius
d = r_min/tg (
γ/2) =
5555.6/ tg (0.5* 0.1974) = 56105
d >
ΔL
1
and d >
ΔL
2
which means that the minimal switch radius does not fit the
geometric constraints. Possible solutions: to decrease the end velocity vse or increase
vector acceleration vac. Suppose that the vector acceleration is pre-defined by the
mechanical parameters of the system and decrease end velocity.
d
max
= min(ΔL
1
,ΔL
2
) = 50000