After my September mention of the so-called Brian’s Law, Thom (many of you may remember Thom from HubBub a few years ago) wrote asking what it was. I was a bit surprised we had never discussed it during his 2 1/2 year tenure here, but Thom has a pretty good head for these kinds of things so I’m sure he’s right that it never came up. The drawing of my cable guide in my last post brings up the subject again; as for some it may raise a question of determining how to illustrate the profile of the mating surface where the down-tube (a cylinder) intersects the guide.
If you cannot tell by now, this post has nothing to do with bicycles, but only orthographic projection of intersecting cylinders and a related technique used in engineering drawing. It specifically refers to a technically incorrect (however practically useful) shortcut I discovered in college. If this fails to interest you, consider not reading any further, ‘cause it ain’t gonna get any better from here. I promise my next post will be more bicycle related.
First off… no, I did not name it. Some classmates in college did. (Incidentally, I didn’t come up with the title Shimergo either. I applied it only after others began to universally refer to my article that way when passing the hyperlink around, nearly 10 years ago.)
Secondly, it’s not a law. It would be more like a rule-of-thumb shortcut, but in fact it is incorrect. As far as I can tell however, it remains practically useful under specific conditions. For illustrative purposes only, it works extremely well.
Third… well, if you’re still reading, you probably want to know what the #&@% I’m talking about:
Orthographic projection is a drafting technique commonly used to indicate the shapes and dimensions of a complex 3-dimensional object, including architectural plans. It consists of six principal views of course: top, front, right, left, back, and bottom. As a general rule the fewest number of views necessary to effectively communicate all relevant information should be used. Three is probably most common, and two is often adequate for simple items. More than three is perfectly acceptable if required, and additional views, at oblique angles, can even be defined to ease communication – although that would no longer be orthogonal, but instead auxiliary projection.
It is important to remember that all perspective is removed, as depth is defined entirely by an adjacent view. That is, each individual view is a parallel projection of an object’s defining features, as perpendicularly viewed on its respective side, onto a 2-dimensional plane. This is useful of course because paper and computer screens are 2-dimensional mediums.
What is handy about orthographic projection is that objects, such as endpoints of lines, centers of circles, intersections, radii, etc. can simply be transferred (projected) from one viewing plane to another. If done correctly, all information can be gleaned from what appears to be very little at first. A fine example of this can be found in these two fun exercises,
where you are asked to complete the illustrations. In case you wanted to try your hand (after printing the images and using a pencil) without knowing the answers, here is the first solution, and here is the second.
So, how does one orthographically project the intersection of two cylindrical objects? What is taught, can be universally applied, and is technically correct, uses your own points created along the curve by dividing the curve into known segments, and projecting them into the plane you require. This incredibly tedious (especially by hand) technique in fact can be applied to any object, and is especially useful for complex curves, or illustration of cylindrical and elliptical objects intersecting with any surface, particularly at an oblique angle.
But what if it’s just a simple situation of real cylinders, of differing diameters, assembled in a true perpendicular intersection? Maybe these constraints sound excessive, but I have found it to occur rather commonly. It is very rare on a typical bicycle frame – maybe the top-tube intersecting with either the head-tube or seat-tube, and of course the seat-tube or down-tube joining the bottom-bracket shell, but coping bicycle frame members generally doesn’t require accurate representation of what the copes will look like projected orthographically.
Given a problem like this in class once, and being lazy when it comes to overly tedious tasks, I opted to spend my class time instead trying to develop a shortcut. What occurred to me was the projected curve should always be the same radius as the larger of the two cylinders. Now, before you say, “that’s overly simplistic,” remember that I already admitted this is incorrect. Then, go try it. It does work. It may be incorrect, but it works to a tolerance well within any reasonable expectations for practical use. Also, remember there are rules to this, and it will only work in a specific situation:- The cylinders must be different in diameter. Cylinders of the same diameter project an “X” onto the view plan – one indicator that the “law” must change as the cylinders approach each other in size (thereby rendering it a non-law).
- The cylinders must be perpendicular to each other relative to the view plan. Any variance from perpendicularity immediately illustrates why this shortcut idea is technically incorrect.
- The cylinders must be true, and not elliptical, in cross-section.
It was a customer of mine, a professor here in Cleveland, also named Brian, who finally figured out why the projection is actually not an arc with a radius equal to that of the larger cylinder. The projection is hyperbolic, and without constant radius at all. This seemed obvious only after he eventually pointed it out. The curve does however occur so near the apex, well inside the hyperbola’s focal point, that it indeed appears to have a constant radius, which effectively approximates that of the larger cylinder.
Finally, as this projection is really only used for illustrative purposes, there seems to be no reason not to apply this non-“law” to accurately portray the visual appearance of such an intersection. One might argue using the point I tried to make regarding the physics of pendulums. I cannot imagine discouraging teaching the truth first, but use of effective techniques, however technically incorrect, seems appropriate for sake of efficiency in practical application, especially for illustrative purposes only.
If this wasn’t enough fun, here’s one to think about for a little while, provided by the owner of the Univega (and cable guide) featured in the previous post:
Given three circles of different diameters, all co-planar but arranged any way you wish, draw adjoining tangents among them, and continue the tangents to their respective points of convergence. Note that the three points of convergence are co-linear; that is, while only two points are required to define a line, all three of these rest on the same line. This is true. Prove it.
A hint was provided: although all objects (circles and lines) exist in the same plane, the only known solution required reference to the third dimension. Again, note that the circles can be any diameter (as long as they’re different) and can have any possible arrangement.
If you figure it out, please do not send me your solution. I have not solved it, although telling me you did might provide some motivation to get cracking.