ElmoMC Multi-Axis Motion Controller-Maestro Motion Control - Manual (Page 25)

Motion Library Tutorial Switch Radius Calculation
MAN-MLT (Ver 2.0)
2-4
vse = [r_switch*vac*vae]
1/2
= [4455.9*500000*0.9]
1/2
= 44778.9
Example 2.1c
(Motion Mathematic Lib Samples \Line To Line\LineLineMinRad_Ex_2c -
www.elmomc.com)
Line 1 is defined by its init point (300000, 900000) and end point (700000,200000). Line 2 is
defined by the init point (700000,200000) and its end point (1100000,700000). Switch from
Line 1 to Line 2 must be executed with the minimal switch radius (vsc = 1). We define the
cruise velocity vsp = 50000 and the end velocity vse = 50000. Vector
acceleration/deceleration is vac = vdc = 28000000.
1. We calculate the minimal switch radius that satisfies kinematics constraint by
r_min = (vse) 2 /(vac*vae) = (50000) 2 /(28000000*0.9) = 99.2
2. Geometric description of the two lines intersecting:
ΔX
1
= X
12
– X
11
= 700000 - 300000 = 400000
ΔY
1
= Y
12
– Y
11
= 200000 – 900000 = -700000
ΔX
2
= X
22
– X
21
= 1100000 – 700000 = 400000
ΔY
2
= Y
22
– Y
21
= 700000 – 200000 = 500000
The length of the first line segment
ΔL
1
= [ΔX
12
+ ΔY
1 2
] 1/2 = [400000 2 + (-700000)
2
] 1/2 = 806225.8
The length of the second line segment
ΔL
2
= [ΔX
22
+ ΔY
12
] 1/2 = [400000 2 + 500000 2 ] 1/2 = 640312.4
An angle between two lines can be calculated as
γ = π – arccos[(ΔX
1
ΔX
2
+ ΔY
1
ΔY
2
)/(ΔL
1
ΔL
2
) =
= π – arccos{[ 400000 * 400000 + ( -700000 )( 500000 )]/( 806225.8*640312.4 )} =
1.193887 = 68.4 o
The distance from the intersection point corresponding to the minimal switch radius
d
= r_min/tg (
γ/2) =
99.2*tg(0.5*1.193887) = 145.95
We see that d < ΔL
1
and
d < ΔL
1
so the calculated switch radius satisfies the geometric
constraints.
Example 2.1d
(Motion Mathematic Lib Samples\Line To Line\LineLineFixedDist_Ex_2d –
www.elmomc.com)