Question

One axis of an RRL robot is a linear slide with a total range of 750 mm. The robot's control memory has a 10-bit capacity. It is assumed that the mechanical errors associated with the arm are normally distributed with a mean at the given taught point and a standard deviation of 0.10 mm. Determine (a) the control resolution for the axis under consideration, (b) accuracy, and (c) repeatability.

Answer 1

a) To find the solution of this point we need to calculate the relation of the control resolution, that is,

$CR=\frac{R}{2^B-1}$

Where

R= Range of the joint

B= Storage capacity

Making the substitution of the previous values we have,

$CR=\frac{750}{2^{10}-1}=0.733mm$

B) For the now we need to calculate the accuracy, that is,

$A_c = \frac{CR}{2}+3\sigma$

Where

$\sigma$ = Standard deviation

$A_c=$The accuracy of the robot

Making the substitution,

$A_c = \frac{0.733}{2}+3(0.1)\\A_c = 0.667mm$

c) At end we can to calculate the repeatability, that is,

$Re=6\sigma\\Re=6(0.1)\\Re=0.6mm$