This problem has driven me bats for some time. I think I posted it here several years ago but can't find it, so here it is again: You have a rod with a rope attached to both ends. The length of the rope is greater than the length of the rod but shorter than half PI times the length of the rod. Leaving the rope ends attached, that rope can be shaped into one and only one circular arc. What is the radius of that circle, in terms of the rod length & rope length? i.e. Radius = f(Rod length, Rope length) The best I can do is approach the answer iteratively, but there's gotta be some function, one would think. Rod & rope thickness are as usual assumed to be zero.
Well, if you can't figure it out, I'm not even going to try! I used to be good at practical math (solving real problems as opposed to these odd puzzles) but I've forgotten most of it. I like the problems that at first glance appear to require calculus, but turn out to have a very simple solution if you approach them differently. There's a story about a famous physicist who was asked one of these riddles. (Two trains, 200 miles apart, are heading towards each other each traveling 100 miles an hour. A fly is flying 200 mph from one to the other until it is crushed in the crash. How far does the fly fly?) He gave the answer in an eye-blink. His friend said, "Oh, you found the simple solution." To which the physicist replied, "What simple solution? I summed the infinite series."
The rod in your problem is called the "cord" of a circle. A cord, in this context, consists of two points connected by the arc of a circle. A diamter is a special case of a cord. There are two formulas used for calculating the cord of a circle, depending on your starting information. I will give you both: 1) cord length = 2r * sin(c/2) where r is the radius and c is the angle subtended by the cord at the center of the circle. 2) cord length = 2 * sqrt(r^2 - d^2) where r is the radius and d is the perpendicular distance from the center of the circle to the cord. Tom
Sometimes I can out-funny Tom, but I wouldn't expect to outsmart him on a regular basis. I would have taken the gardener's approach and just laid the stick and the rope on the ground. I've made some surprisingly attractive rock circles and block retaining walls with no math at all.
I'm sorry, DH isn't at home. He could figure this out in a heartbeat. So could our youngest. I could talk you into distraction while we wait for him to get home
Many artists paint without using math. Math is fun and it is useful. But much that is attractive does not require it.
I design and sell paper models. It's all about "flattening" things so the components can be printed on the flat plane of a sheet of cardstock. I keep running into this particular issue when "reverse-engineering" some design in order to fix fitting problems (most models were created with drafting machines, straightedges and french curves decades before the age of CAD and are rife with fitting errors). I'll have a part that's a partial cylinder (a section of fuselage, say) where I know two things: the flattened "width" which in the final model is curled into a semi-circumferential surface, and what the "chord" should be that is the edge to edge distance when the part is in its final shape. To make the necessary corrections when something is awry I need to know what the radius of the finished arc should be - and that is the rod and rope problem: I know the "chord" (the rod) and the semi-circumferential dimension (the rope) but not the radius of the finished arc - and I need to know it so I can design (or redesign) the part that will fit to it. So if it drives me bats it isn't because I'm bats, although opinions have proved varied on that point. I've derived this formula: Semi-Circumferential Dimension = 2 * Radius * Arcsin[Chord / (2*Radius)] But I don't know how to get Radius all by itself over on one side of the equal sign.
"Cord", not "Chord"... unless your model plays music DH looked at it, and said "Not without more sleep, more coffee, and a few minutes to look up some formulas"...
If the rod length is known and the rope can be any length, how many arcs (parts of circles) are possible ?
Sorry, it's "chord" Chord | Define Chord at Dictionary.com I've been using the word for more than four decades so was confident I was spelling it correctly, but scientist that I am I checked Google just to make sure it hadn't changed while I wasn't looking. And it hadn't.
For any given length of rope, only one circular arc is possible. I believe there is also only one possible symmetric parabolic arc (another interesting problem for someone to take on). I don't know if the asymmetric parabolae would all be the same, but that's getting into areas of arcane math someone other than me should tackle.
:hail: I am not worthy. Oh, and DH decided that since he couldn't come up with the formula quickly, he'd go change the sheets for me. I think I came out a winner on that one!
I had no idea. That is so cool. Those 'draw the hidden side' problems in drafting were always fun, and I used to have a great time with graph paper making platonic solids like tetrahedrons with little glue tabs on them. I have some paper TGVs around somewhere, too. Do you have a website? If I take Tom's second equation and solve for r, I get r = 1/2chord +d. But that looks too simple, and I'm quite willing to admit I may be wrong. Algebra classes were a long time ago.
I was referring to Hyo's comment that he didn't need math to create his attractive rock circles: And although there are supposedly some basic proportions often observed by artists, the great paintings did not require serious math to create. In fact, the reason that Mendel's work went unnoticed for such a long time was that most scientists of his day didn't understand the math! If scientists didn't understand math that is simple for us today, how much less must the great artists have possessed of it! Yet they created some really impressive stuff. Modern technology requires a lot of math. The paintings of Rembrandt didn't. Paper airplanes are cool!
Ah, now I see your real question. Just giving this a quick look, I don't believe you will be able to express R as a simple trigonometric function. I believe it will be a converging power series, which means you are back to a numerical solution. I'll give it a little more thought once Christmas gets out of the way. Tom
Or do it the way Hyo suggested: Build a stick-and-string model and measure. Done carefully, you can probably get an answer as near as makes no difference. Or is that what you meant? :blush:
Not quite, if I'm understanding the problem correctly. Now that I know airportkid is making paper models, not gardening, the measurement has to be precise. Otherwise, he'll end up with a wrinkle, or pieces that don't fit properly. The flat piece of paper has to have the correct dimensions on it, or the curve won't come out right when the model is assembled. Maybe he'll have to make one of whatever it is - an airplane hangar? - with separate pieces, get the curve right, and then take it apart to measure the curved piece while it's flat.
