In architectural drafting and similar disciplines, a detailed scale drawing of a part of a larger construction will often be produced. These drawings are needed so that the builder can proceed with the job having a clear set of instructions and specifications.
For this exercise, we work back from a finished object to recreate something like a scale drawing.
The following photograph is of a section of a railway bridge across the Hawkesbury River in Brooklyn, NSW. We will make a representation of the section within the white rectangle with some of the important dimensions.
The complete bridge has eight steel trusses similar to the one pictured. This particular truss spans a distance of $105.92$105.92 m. This information enables us to deduce the scale of the photograph and the subsequent enlargement.
The photograph on your computer screen is already a scale representation of the object and the actual scale will depend on the computer monitor being used. We assume, for the exercise, that the span in the picture is $126$126 mm or $0.126$0.126 m. That is, a distance of $0.126$0.126 m in the photograph represents $105.92$105.92 m in the object. Thus, the scale is$0.126:105.92$0.126:105.92 or approximately $1:840$1:840.
The white rectangle is $17$17 mm wide on the screen and so covers a width $0.017\times840=14.28$0.017×840=14.28 m on the bridge.
The enlarged rectangle from the photograph is close to the scale drawing that we wish to produce. We can deduce the measurements of the important components in the bridge.
Assume that the enlarged rectangle has width $93$93 mm on the screen. This distance represents $14.28$14.28 m on the bridge. So, we see that the scale in this enlargement is $93:14280$93:14280 or approximately $1:154$1:154. All that remains now is to measure the distances on the enlargement, which is, in effect, our scale drawing, and to use the scale to calculate what the corresponding distances are on the bridge.
The vertical struts are $0.123\times154\approx18.9$0.123×154≈18.9 m and $0.126\times154\approx19.4$0.126×154≈19.4 m. The distance between centers of the vertical struts is $0.07\times154\approx10.8$0.07×154≈10.8 m.