Find the maximum length of an offset so that the displacement on paper from both sources of error should not exceed 0.025 cm. Given that the offset is measured with an accuracy of 1 in 25, and the scale is 1 cm = 30 m.

Explanation

This calculation determines the longest permissible offset length based on accuracy constraints and plotting limitations.

Combined Effect of Errors: Surveying offsets are susceptible to errors from both angle setting and distance measurement. The formula used, √(2) * l / n, estimates the maximum expected resultant displacement *on the ground*. It typically arises from a model where the maximum displacements from angular error and linear error (`l/n`) are assumed to be approximately equal in magnitude and are combined vectorially (often treated as orthogonal components) to find the overall maximum likely displacement.

Scale Conversion: Field measurements and their associated ground displacements must be translated to their representation on the map or plan. This is done using the scale factor (meters on the ground per centimeter on paper). Dividing the estimated ground displacement by the scale factor gives the corresponding displacement on the paper.

Plotting Accuracy Limit: Technical drawings have a limit to how precisely points can be plotted and distinguished by the human eye. A common standard is that errors smaller than about 0.025 cm (or 0.25 mm) are considered negligible on the paper. This problem uses this limit as the maximum acceptable displacement on the paper arising from the combined field errors.

Determining Maximum Field Measurement: By setting the calculated paper displacement (expressed in terms of the unknown length ‘l’, the known accuracy ‘n’, and the scale) equal to the maximum allowable paper displacement (0.025 cm), we can algebraically solve for ‘l’. This ‘l’ represents the maximum offset length that can be laid out under the given accuracy and scale constraints without exceeding the plotting error limit on the paper.