A point ‘P’ on a CNC controlled XY-stage is moved to another point ‘Q’ using the coordinate system shown in the figure below and rapid positioning command (G00).

A pair of stepping motors with a maximum speed of 800 rpm, controlling both the X and Y motion of the stage, are directly coupled to a pair of lead screws, each with a uniform pitch of 0.5 mm. The time needed to position the point ‘P’ to the point ‘Q’ is ______ minutes. (round off to 2 decimal places).

The correct answer is 1.5.

Based on the coordinate system shown in the figure, we can identify the coordinates of the two points:
- The starting point P is located at (200, 300).
- The destination point Q is located at (800, 600).

First, we find the required travel distances in the X and Y directions:

$\Delta x=800-200=600\text{mm}$

$\Delta y=600-300=300\text{mm}$

The motors drive lead screws directly with a pitch of 0.5 mm at a maximum speed of 800 rpm. The maximum linear velocity (feed rate) of the stage along either axis is:

$V=\text{Rotational Speed}\left(N\right)\times \text{Pitch}\left(p\right)$

$V=800\text{rpm}\times 0.5\text{mm/rev}=400\text{mm/min}$

With G00 (rapid positioning), the X and Y motors control the stage axes independently and simultaneously at their maximum speed. Therefore, we calculate the time taken to travel along each direction separately:

Time taken to traverse in the X-direction:

$T_x=\frac{\Delta x}{V}=\frac{600}{400}=1.5\text{minutes}$

Time taken to traverse in the Y-direction:

$T_y=\frac{\Delta y}{V}=\frac{300}{400}=0.75\text{minutes}$

Since both axes move simultaneously, the total positioning time required to move from point P to point Q is determined by the axis that takes the maximum time to reach its coordinate:

$T_{\text{total}}=max(T_x,T_y)=1.5\text{minutes}$

Thus, the motor for the Y-direction finishes its travel and stops after 0.75 minutes, while the motor for the X-direction continues to run until the total target position is reached at 1.5 minutes.