\nYesterday was an incredible day here, especially given the forecast. It was bright and sunny, with a temp that reached over ten degrees Celsius. Naturally I went out into the yard to get some sun, of which none is predicted before Friday.<\/p>\n
\nThis time of year, when one goes out into the yard, one might notice cones and branches fallen to the ground from the trees. We’ve had some storms, after all. Said cones and branches need to be picked up, if only to allow lawn mowing, although that won’t likely happen again for months.<\/p>\n
\nThe yard clean-up, which I tend to tackle on days like yesterday, is one of those fascinating jobs that might never get completed, but requires work anyway. Establishing the time line for such a job can be interesting to contemplate.<\/p>\n
\nThe way I tackle the cleanup is to pick up the obvious debris first: I move around the yard, picking up what I notice. After the first pass, less-obvious debris become more noticeable, and so on, so the system of picking up the most obvious debris continues. The question becomes how long one will spend on a given occasion.<\/p>\n
\nLet’s imagine, perhaps, that at any given time, 20% of the debris is obvious. Therefore, on the first pass, 20% of the original debris get removed, while 80% remain. On the second pass, 20% of the remaining 80%, or 16% of the orginal debris, get removed, so 64% remain. On the third pass, 20% of the remaining 64%, or 12.8% of the original debris, get removed, leaving 51.2%, and so on.<\/p>\n
\nIn terms of passes, one can establish a half-life for the debris. Converting the percents to decimals, one arrives at 0.5=0.8^n, where 0.8=80% is the amount of debris that remain after a given pass relative to before it (rather than to the original total), with n being the number of passes. Next, taking log of both sides, then pulling the exponent in front as a multiple, ln0.5 = nln0.8, so n=(ln0.5)\/(ln0.8) = 3.1 passes. Note that, above, with arithmetic, it was found that after three passes, 51.2% of the debris would remain, so a half-life of 3.1 passes seems to agree.<\/p>\n
\nThat said, if a pass of the yard takes ten minutes, then half the debris would be removed after 31 minutes. After another 31 minutes, half of the half remaining, or another 25%, would be removed, leaving a quarter of the original amount.<\/p>\n
\nHowever, to someone looking at the yard from the door, the remaining 25% of debris is likely much less noticeable than the first 25%, so even with the last 25% still on the lawn, it likely looks much, much better than before the cleanup was started.<\/p>\n
\nDoes that make one excited to get out and continue yard cleanup? That’s best known to the reader:)<\/p>\n
Source:<\/p>\n
Travers, Kenneth J. et al. Using Advanced Algebra.<\/em> Toronto: Doubleday Canada Limited, 1977.<\/p>\nJack of Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.\n","protected":false},"excerpt":{"rendered":" Self-tutoring about yard work: the tutor relates it to exponential decay. Yesterday was an incredible day here, especially given the forecast. It was bright and sunny, with a temp that reached over ten degrees Celsius. Naturally I went out into …<\/p>\n