The XY table of a NC machine tool is to move from P(1,1) to Q(51,1); all coordinates are in mm. The pitch of the NC drive leadscrew is 1 mm. If the backlash between the leadscrew and the nut is 1.8º, then the total backlash of the table on moving from P to Q is ____________ mm (round off to two decimal places).

The correct answer is 0.25.

Step-by-step Explanation:

First, let us identify the coordinates of the initial point P and the final point Q from the problem statement and the accompanying diagram:

$P=(1,1)$

$Q=(51,1)$

Since the y-coordinate remains constant at 1 mm, the movement of the NC machine table is entirely along the x-axis. The total distance traveled by the table is calculated as:

$\text{Distance}=51-1=50\text{mm}$

The pitch of the NC drive leadscrew is 1 mm. This means one full rotation (360º) of the leadscrew translates to 1 mm of linear table movement. Therefore, the total number of rotations required to move 50 mm is:

$\text{Number of rotations}=\frac{50\text{mm}}{1\text{mm/rotation}}=50\text{rotations}$

We are given that the backlash between the leadscrew and the nut is 1.8º. The linear backlash corresponding to this angle per rotation is:

$\text{Backlash per rotation}=\frac{1.8º}{360º}\times 1\text{mm}=0.005\text{mm}$

For the complete movement of 50 rotations from P to Q, the total linear backlash of the table is:

$\text{Total backlash}=50\times 0.005\text{mm}=0.25\text{mm}$