Question:
A linear slot is to be milled in a single pass from the point (0, 0) to (180, 180) on XY plane by a CNC machine. The actual velocity along the Y-axis is 5% less than the intended value. The programmed feed rate along the intended slot is 150 mm/min.
The magnitude of the positional error along the Y-axis when the X-coordinate reaches 180 mm, is ______ mm (in integer).
Note: All coordinates are in mm.
A linear slot is to be milled in a single pass from the point (0, 0) to (180, 180) on XY plane by a CNC machine. The actual velocity along the Y-axis is 5% less than the intended value. The programmed feed rate along the intended slot is 150 mm/min.
The magnitude of the positional error along the Y-axis when the X-coordinate reaches 180 mm, is ______ mm (in integer).
Note: All coordinates are in mm.
The magnitude of the positional error along the Y-axis when the X-coordinate reaches 180 mm, is ______ mm (in integer).
Note: All coordinates are in mm.
Show Hint
Split the feed into X and Y components; X stays unaffected by the error so use it to find time, then find the actual Y distance travelled in that time.
Updated On: Aug 5, 2026
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Verified By Collegedunia
Correct Answer: 9
Solution and Explanation
Step 1: Understanding the setup:
The tool has to move in a straight line from (0,0) to (180,180), so the path makes a 45 degree angle with both axes.
The programmed feed rate of 150 mm/min is the resultant speed along this straight line, and it gets split into X and Y components by the machine controller.
Step 2: Finding intended and actual velocity components:
Since the line is at 45 degrees, both components are equal:
\[ V_x = V_y = 150 \times \cos(45^\circ) = \frac{150}{\sqrt{2}} = 106.07 \text{ mm/min} \]
The actual Y velocity is 5% less than this intended value, while the X velocity stays as programmed:
\[ V_{y,actual} = 0.95 \times 106.07 = 100.76 \text{ mm/min}, \quad V_{x,actual} = 106.07 \text{ mm/min} \]
Step 3: Finding the time to reach X = 180 mm and the actual Y position:
Since X is not affected by the error, the time to cover 180 mm along X is:
\[ t = \frac{180}{106.07} = 1.697 \text{ min} \]
In this same time, the actual Y travel is:
\[ Y_{actual} = V_{y,actual} \times t = 100.76 \times 1.697 = 171.0 \text{ mm} \]
The intended Y position at this instant, on the true 45 degree line, is 180 mm, since X and Y rise together.
Final Answer:
Positional error along Y = 180 - 171 = 9 mm.
\[ \boxed{9 \text{ mm}} \]
The tool has to move in a straight line from (0,0) to (180,180), so the path makes a 45 degree angle with both axes.
The programmed feed rate of 150 mm/min is the resultant speed along this straight line, and it gets split into X and Y components by the machine controller.
Step 2: Finding intended and actual velocity components:
Since the line is at 45 degrees, both components are equal:
\[ V_x = V_y = 150 \times \cos(45^\circ) = \frac{150}{\sqrt{2}} = 106.07 \text{ mm/min} \]
The actual Y velocity is 5% less than this intended value, while the X velocity stays as programmed:
\[ V_{y,actual} = 0.95 \times 106.07 = 100.76 \text{ mm/min}, \quad V_{x,actual} = 106.07 \text{ mm/min} \]
Step 3: Finding the time to reach X = 180 mm and the actual Y position:
Since X is not affected by the error, the time to cover 180 mm along X is:
\[ t = \frac{180}{106.07} = 1.697 \text{ min} \]
In this same time, the actual Y travel is:
\[ Y_{actual} = V_{y,actual} \times t = 100.76 \times 1.697 = 171.0 \text{ mm} \]
The intended Y position at this instant, on the true 45 degree line, is 180 mm, since X and Y rise together.
Final Answer:
Positional error along Y = 180 - 171 = 9 mm.
\[ \boxed{9 \text{ mm}} \]
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