Elmomc Multi-Axis Motion Controller-Maestro Motion Contro Uživatelský manuál stáhnout pdf (Strana 23)

Motion Library Tutorial Switch Radius Calculation

MAN-MLT (Ver 2.0)

2-2

vsr ≤ min(0.5ΔL

, 0.5ΔL

)*tg(γ/2)

(2.1-4)

and kinematics constraint

end

≤ [r*vae*AC

)]

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)

/(vac*vae) = (50000)

/(500000*0.9) = 5555.6

2. The calculated distance from the intersection point that corresponds to r_min = 5555.6

ΔX

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

ΔX

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

ΔL

= [dX1

+ dY1

]

1/2

= [(10000)

+ (-50000)

]

1/2

= 50990

ΔL

= [dX2

+ dY2

]

1/2

= [0 + (50000)

]

1/2

= 50000

γ = π – arcos[(ΔX

ΔX

+ ΔY

ΔY

)/(ΔL

ΔL

) =

= 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

and d > ΔL

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.

max

= min(ΔL

,ΔL

) = 50000

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