leong Posted October 23, 2005 Posted October 23, 2005 Part 1. I recently bought a new farmer john wet suit with relief zipper (exhibit NRS Ultra John http://www.rei.com/online/store/ProductDis...ory_rn=4501480). Thursday while paddling back to Manchester from Marblehead a nature call required the use of the relief zipper. Note it was quite windy (probably 20+ knots) with high frequency whitecaps. I removed the front of the spray skirt and proceeded to use my plastic pee bottle. From a seated position the lower zipper was too high to perform the operation so I had to lift myself up off the seat. Doing this caused a capsize. I rolled up, pumped out, replaced the spray skirt and was soon on my way without further incidents, but with a full bladder. The new wetsuit was purchased for the express purpose of having a relief zipper. Since it doesn't work well while in the boat I'm now considering exchanging it for the female version (the relief zipper goes much lower). Question 1. Are there any major problems I'll have using a Farmer Jane? I assume fit is the only problem -- I can handle the hazing. I'm 5 ft 8 inches with chest-waist-hip/butt measurements of approximately 44-36-41. Any recommendations? Part 2. You're practicing hand rolls in a circular pool surrounded by a circular patio (i.e. concentric circles). A genie appears and offers to grant you any wish. But you have to answer one question first: Allowing just one linear measurement of some length, what should you measure to find the area of the circular patio in terms of pi? You don't know anything about the radius of the two circles (the pool radius or the radius of the outer circle formed by the outer edge of the patio). Question 2. What length should you measure? Question 3. If the measurement is n, what is the area in terms of n and pi? Thanks in advance for answers to question 1. A 12 pack of beer and/or a kayak fishing lesson to the first person correctly answering question's 2 and 3. Quote
markstephens Posted October 23, 2005 Posted October 23, 2005 1. I think you should have relieved yourself after the capsize, THEN rolled up.2. Assuming that you can't actually measure the radius, measure the diameter. 3. where d = the diameter, .5d x .5d x pi = nThis is just good ol' "pi r-squared", but in fact everybody knows that pie are round, not square. Quote
jeffcasey Posted October 23, 2005 Posted October 23, 2005 Hi Leon - Pick a point on the edge of the pool. Draw a radius from that point to the center of the pool. Draw another line from the center of the pool to outside edge of the patio, but pick a point on the edge of the patio so that the line from there to the first point is tangent to the pool (i.e. 90 degrees from the radius you drew of the pool). You now have a triangle. Measure the distance from your first point (on the edge of the pool) to your point on the edge of the patio. If this is "n", the area of the patio is pi*n^2. I drink Tremont Ale, and I've seen you fish (and even taken pictures)...so I'll take the beer. - cheers, Jeff Quote
leong Posted October 24, 2005 Author Posted October 24, 2005 Mark,There are two diameters involved ... the outer diameter (a chord through the center of the pool connecting two points on the circumference of the patio) and the inner diameter (a chord through the center of the pool connecting two points on the circumference of the pool. Your formula gives the area of the pool or the sum of the areas of the pool and patio, depending on what diameter you mean to use. But the question relates to the area of the patio alone. Sorry no beer. Quote
Bill Gwynn Posted October 24, 2005 Posted October 24, 2005 The area of the annulus/patio can be determined by measuring the chord (n). The chord is a straight line measured from one side of the patio to the other side with the line touching the edge of the pool. So... the formula for the area of the annulus is Pi times 1/2 of n squared.pi*n/2^2I wish I could draw a picture to explain it better.Oh...don't know the answer to the farmer jane question, but i'll be glad to tease you about wearing it. Quote
BOB L Posted October 24, 2005 Posted October 24, 2005 The main thing is the units you use to measure if you measure the radius of the patio in units of the radius of the pool you can then express the area of the patio as pi((n^2)-1)in Units of Squared Pool radius. I'd like to know if there is another way to do it.Bob Quote
leong Posted October 24, 2005 Author Posted October 24, 2005 Hi Bob L,There is a way to do it, but not your way. You made it a different and trivial problem. You can only take one measurement! If you measure the radius of the pool you can't then measure the radius of the patio (center of pool to outer circumference of patio). See response to Mark. Quote
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