sobo Posted June 4, 2010 Posted June 4, 2010 me! Lest y'all think this pic was not taken moments ago, please note the envelope on my keyboard. Look familiar?? You may now ask the question... "Hey, Sobo! What time is it??" Quote
hafilax Posted June 4, 2010 Posted June 4, 2010 OK I got one (not multi-variable calculus, but fun nevertheless): show from Newton's laws that the fall factor is length of rope divided by fall distance FF = l/d Be sure to include any assumptions you make (e.g. the rope behaves like an ideal spring where f = kx). Bonus points if you include a free body diagram. The fall factor comes from the fact that the spring coefficient of the rope is inversely proportional to the length of rope that's in use. I'll defer to Richard Goldstone for the rest of the proof: http://jt512.dyndns.org/StandardEqn.pdf Somebody should write an iPhone App version of http://jt512.dyndns.org/impactcalc.html for use on lead. Quote
hafilax Posted June 4, 2010 Posted June 4, 2010 Droppin' science like when Galileo dropped his orange... Quote
112 Posted June 4, 2010 Posted June 4, 2010 Damn meetings.... Use Buckingham's Pie Theory and you don't need no inversly proportional spring relationship shit... http://en.wikipedia.org/wiki/Buckingham_%CF%80_theorem I am waiting for someone to "discover" the "other" significant non-dimensional numbers currently not used in climbing! Quote
G-spotter Posted June 4, 2010 Posted June 4, 2010 Note that keenwash came up with the same answer using the Law of Sines (as one would expect), but in this instance there is no need for such superfluousness in solving the problem. The more elegant solution is to invoke the Pythagorean Theorem. QED Please explain. I was previously led to believe that only the 30-60 and 45-45 triangles can be calculated from knowledge of the length of 1 side. Otherwise 2 sides were necessary and thus the angle irrelevant. Ken, Yes, the 30/60/90 and 45/45/90 triangles make it very easy to calculate the length of any side knowing only the length of one side. But that is not the problem statement here. Trigonometry says that you can solve for the length of any side of a right triangle knowing either 1) the lengths of the other two sides (by using Pythagorean Theorem), or 2) the length of one side and the value of one of the other two angles (those that are not the right angle). Since we are given the (overhanging) angle of the subject pitch of the Leaning Tower as 110 degrees, we know the interior of that angle is 70 degrees. And since the 200-foot side is known, and that side must be perpendicular to the ground line, we have all that is needed to solve by trig (one known side and two known angles). Now that we are on page 3, maybe it's time to point out that he's actually closer than 72.7 feet to the wall at the end of his line. Just not horizontally. Perpendicular distance from the wall = 72.8 * sin 70 = 68.4 feet from the wall at the end of his rope. Quote
sobo Posted June 4, 2010 Posted June 4, 2010 Yes indeed! How perceptive of you, Dru! But I would posit that Plaidman was actually wondering "how far out from the wall would I be" if I were at the end of a 200-foot rope on a 110 degree overhanging wall. While his actual distance from the wall (at his closest approach) would be 68.4 feet as you so aptly demonstrate, he prolly was looking for the horizontal distance from the wall. Only Plaidman knows... What say ye all to a 4th page? I still have more Fat Tires here... Quote
hafilax Posted June 4, 2010 Posted June 4, 2010 I'd want to know how far I would swing out if I released myself from the anchor and how long it to take to be swinging less than a few feet or so. Quote
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