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=(R)/(2^B-1)`

Where

R= Range of the joint

B= Storage capacity

Making the substitution of the previous values we have,

`CR=(750)/(2^(10)-1)=0.733mm`

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

`A_c = (CR)/(2)+3\sigma`

Where

`\sigma` = Standard deviation

`A_c=`The accuracy of the robot

Making the substitution,

`A_c = (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`