Question:medium
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
Show Solution
Correct Answer: 9
Solution and Explanation
Step 1: A shortcut using the velocity ratio:
Because the commanded path is a straight 45 degree line, the X and Y feed components are equal in the ideal case, both being $ \frac{150}{\sqrt{2}} $ mm/min.
The X component is not disturbed by the error, only Y is reduced by 5 percent.
Step 2: Linking X travel and Y travel through time:
At any instant, the distances covered along X and along Y are both proportional to the same elapsed time $ t $, so:
\[ \frac{Y_{actual}}{X_{actual}} = \frac{V_{y,actual}}{V_{x,actual}} = \frac{0.95 \times V_y}{V_x} \]
Since $ V_x = V_y $ for the 45 degree path, this ratio simplifies directly to 0.95, so the actual mm/min speeds do not even need to be calculated.
Step 3: Applying this ratio at X = 180 mm:
\[ Y_{actual} = 0.95 \times X_{actual} = 0.95 \times 180 = 171 \text{ mm} \]
On the intended 45 degree path, when X = 180 mm, Y should also be 180 mm.
So the error is:
\[ Error = 180 - 171 = 9 \text{ mm} \]
Final Answer:
The Y-axis positional error works out to the same 9 mm with this shortcut, confirming the answer.
\[ \boxed{9 \text{ mm}} \]
Because the commanded path is a straight 45 degree line, the X and Y feed components are equal in the ideal case, both being $ \frac{150}{\sqrt{2}} $ mm/min.
The X component is not disturbed by the error, only Y is reduced by 5 percent.
Step 2: Linking X travel and Y travel through time:
At any instant, the distances covered along X and along Y are both proportional to the same elapsed time $ t $, so:
\[ \frac{Y_{actual}}{X_{actual}} = \frac{V_{y,actual}}{V_{x,actual}} = \frac{0.95 \times V_y}{V_x} \]
Since $ V_x = V_y $ for the 45 degree path, this ratio simplifies directly to 0.95, so the actual mm/min speeds do not even need to be calculated.
Step 3: Applying this ratio at X = 180 mm:
\[ Y_{actual} = 0.95 \times X_{actual} = 0.95 \times 180 = 171 \text{ mm} \]
On the intended 45 degree path, when X = 180 mm, Y should also be 180 mm.
So the error is:
\[ Error = 180 - 171 = 9 \text{ mm} \]
Final Answer:
The Y-axis positional error works out to the same 9 mm with this shortcut, confirming the answer.
\[ \boxed{9 \text{ mm}} \]
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