First off, I wish to make an announcement both short in length and short in duration: Mechanic's Companion has undergone a minor correction due to a problem with the pdf file. The problem has been corrected and as soon as the printer approves the revision, the presses will run. I expect that by next Monday we'll be back in business as usual. For the meantime, I'm holding orders for Nicholson for a few days.
Second off, Michael Stone, a layout carpenter in a professional Broadway scene shop, was reading our edition of Mechanic's Companion (blatant plug here) and discovered an error in the Geometry chapter. Michael's is the first person to point out this problem, which tells me that todays craftspeople are skipping over the Geometry chapter due to a lack of educational preparation in this most necessary of skills. I would like to state, for the record, that I while I failed High School Algebra (yet was graduated anyhow) I did pass geometry, of which I remember nearly nothing.
Here-with, Msr. Stone's solution:
"I recently bought a copy of the "Mechanics Companion". I'm enjoying it a lot. But, I did spot a small and but important problem in the Practical Geometry section. It concerns the creation of a perpendicular, which is second in importance only to being able to draw a straight line. And, when taught correctly, this is the simplest, most elegant, fastest method I've ever seen for executing this vital process. Small, because to correct the mistake you only have to change one letter of the explanation.
It's in Problem IV, page 19, the accompanying diagram is Figure 4 on plate 3, the previous page. I'll pause while you get your copy . . . .
"To draw a perpendicular from a point at the end of a line.
Let E F be the given straight line, and let F the given point. Take any point a above the line and with the radius a C describe an arc C F B cutting E F at b. Draw b a C; then draw C F and it will be the perpendicular required." underline added for emphasis.
The problem is the underlined C in the second sentence. At this point of the process C is undefined. You can not set your compass, (trammels, nails in a board, what ever) to the radius a C. Frustration follows. Fortunately the fix is easy. Simply change the C to an F.
This method relies on the fact that the length of the hypotenuse of a 30 - 60 - 90 degree triangle is twice the length of the base. Because there are LOTS of ways that 30 - 60 - 90 triangles nest with equilateral triangles, you can use this geometry a number of ways to create a perpendicular. By first creating a equilateral triangle, extending one side and doubling the length of that side, you can find the same point C. George Walker recently showed another method at his Design Matters blog. Just start drawing lines through points, you'll see 30 and sixty degree triangles start popping up all over the place. The equilateral triangle in Nicholson's version is hiding at b a F. Since a F is equal to a C we can substitute the radius a F for the non-existent radius a C.
The nice thing about the method shown in Nicholson is that it has the fewest possible steps. Making it faster and more accurate then other methods. The not nice thing is the eight generations of "mechanics" who read this explanation and thought "UH?" and turned the page without the chance to learn this most elegant of methods.
Enjoy.
P.S. Plate 4, Figure 8 is a Heptagon but it is described in the text as a Octagon, it makes no difference to the methodology.
Michael Stone is a Layout Carpenter at a professional Broadway scene shop and a member of I.A.S.T.E. local One."
Till next, Gary
