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About leong

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    Paddle Upwind

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    Kayak racing, fishing and touring; road bike racing and touring; ski touring; swimming and fitness workouts.

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  1. Ugh, cut a rectangle through the gel-coat/carbon-fiber and push the stern eye out through the bottom. I don't know what the eye is fastened to. Probably some reinforced section. I assume the stern eye looks like this ->http://www.go2marine.com/product/78277F/sea-dog-stern-eye-with-adjustable-finish-nuts.html . I doubt that the rectangular flange on the bottom is just seated up against the bottom of the thin top deck (it's an 18' boat that weighs only 38 lbs.). The repair job for that is doable, but not for me (especially working with carbon cloth). So far the stern eye is just wiggling a little and I try and avoid any force on it. But it anchors the rear deck lines so there's always some horizontal force on it. Anyway, before tackling such a big job I'll wait until it breaks loose itself (hopefully, not while someone is doing a T-rescue on me; note to self: practice and practice rolls
  2. I may have to do that. I'll measure the span.
  3. Excellent, Cathy. They they even have squirrels wearing little pink work clothes. Thanks, -Leon
  4. The hard part will be to seat a socket on an unseen nut over 4 feet away. I think I need something like an endoscope with a remotely controlled wrench. I guess it will be cheaper to buy a new kayak. I'll probably wait until the eye comes out and install something from the outside (a watertight toggle bolt?).
  5. Thanks, David. Good idea. I'd first call QCC (the boat's manufacturer) to get more information about the nut size (not for the squirrel), type and what's down there. So far the stern eye is just wiggling very slightly and I don't use it to carry the boat. -Leon
  6. The stern eye that the stern toggle is attached to is starting to wiggle. It’s almost 5 feet from the back of the stern hatch cover so can’t reach under to tighten it. The only solution I can think of is to train a squirrel to use a ratchet wrench. Any ideas out there? -Leon
  7. The hardest part of landing a big fish is when it's at a right angle beside the boat. I have to lean in the opposite direction to counter the fighting weight. I've capsized a few times and rolled up but ruined a couple of reels. Here are a few pictures. Striper Bluefish Bluefish Barracuda in Florida (I had to tow the fish to the nearest dock to have it gaffed)
  8. Hi, In the spring and summer I fish from a sea kayak usually out of Lanes (bluefish and stripers). I'm now kayak fishing in south Florida (barracuda, tarpon, jacks, etc). I don't bait fish ... I only troll using lures (mostly spoons and sluggos). -Leon
  9. Today I paddled to Mar-a-Lago to meet Japanese Prime Minister Shinzo Abe. I was hoping to demonstrate bracing and rolling to Shinzo and his lovely wife. I usually make the Mar-a-Lago trip alongside the ocean beaches of Palm Beach Barrier Island, but today I took the inside route (between Palm Beach and the mainland) because of the strong winds. This is a pix of Trump Plaza as I passed by it in West Palm Beach. Here are a couple of pix of the stinkpots I passed. Some people are really into conspicuous consumption. Here’s a pix of Mar-a-Lago from about two miles away. As I headed towards shore this Coastie came out to greet me. He was unaware that I was on the way to meet the Japanese Prime Minister. He said I have two choices: 1. Stay far from shore or 2. He’ll try out his bow-mounted machine gun. I chose no. 1. However, there’s a beach on Southern Blvd. that connects the Mainland to Palm Beach and he said it was okay to paddle there. So I did. Here are a few pix of Mar-a-Lago taken near the beach on Southern Blvd. I had a super fast paddle home (wind and tide pushing me) averaged 6 knots over the 8 nautical miles. At the end of the trip I parked my kayak along side of my Sunfish. Oh, did I mention that I never met the Prime Minister.
  10. Pru, A vector is a fancy dancy name for a line segment with a little arrowhead on one end. The rest is just commentary. Look here for a little more information. Luckily you won’t need tensors (they will really blow up your brain). Unless, of course, you want to blow up Kevin's brain. -Leon http://www.dummies.com/education/science/physics/what-is-a-vector/
  11. Excellent report as usual, Pru. BTW, it’s State Stage Fort Park.
  12. Rob and all, Just to remind you that the trip between Good Harbor and the end of Dog Bar is usually the roughest section of the circumnavigation around Cape Ann. But more important is that there are no takeouts (except, sometimes, at Braces Cove when there are no breakers). And, Joe, the area along Bass Rocks is very rocky and this portion of the whole circumnav is arguable the most scenic part. -Leon (who’s done the circumnav about a hundred times and is trolling NSPN from the Palm Beaches)
  13. Brian and Phil, I used 2 square feet because I used a large coefficient of drag. The extremely high hull speed (as you say) is not that high. I get my QCC700X to 6 knots (in still air and water) for short sprints. Max hull speed is just a number, not a hard speed limit. The Blackburn Challenge course is almost exactly 18 Nautical miles. The winning Epic 18’s and similar kayaks finish the course in 3 hours, or so. That’s an average speed of 6 knots for a long course. But so-called “max hull speed” of these kayaks is about 5.6 knots (1.34 time the square root of the water-line length of ~ 17.5’) But let all this go. Nothing good on TV so, instead of rerunning the test, I decided to prove my result analytically so there can be no questions on the values that I used (it’s good that I’m retired ). I also gave it a whirl on the water but couldn’t get the proper conditions today. See attached Mathcad worksheet for my analysis. Mathcad - DownCurrent-UpWind and vice a versa.pdf -Leon
  14. Phil, There is a lot of uncertainty in the necessary wind speed because of what to use for the frontal area in the air drag formula. I used a 2 square-foot area for the paddler and kayak. At 9 knots, the wind pressure is ½ pound per square-foot. So the force of the wind is 1 pound and, according to KAPER, that’s the total drag on a kayak going at 2 knots. But let that go. Suppose it does takes about a 20-knot wind to overcome the 2-knot current. Then, for case # 1, the drag increases to about 6 pounds and for case # 2, the drag decreases to about 9.5 pounds. So the qualitative result is the same; i.e. it’s still easier to paddle downcurrent/upwind than vice versa. In my experience in the inlet, it does seem like a 10-knot or so wind is enough to overcome the current. But, I don’t know the exact current speed or wind speed opposite to the current. Also, I don’t account for wind generated waves on the surface that might be moving opposite to the current flow. I think a key point is that a change in current speed costs a paddler more than the same change in relative wind speed. -Leon
  15. Brian, You might want to change your answer based on the following hints: A wind speed of approximately 9 knots against a kayaker has a force of roughly 1 pound and that cancels the 1 pound water drag on a kayak in a 2-knot current in the opposite direction. Case 1: Say you paddle with the current (against the wind) with a speed over ground (SOG) of 4 knots. Then the relative water speed is 2 knots; (4 -2). Also, the relative air speed is 13 knots (9 + 4). Again, at a relative water speed of 2 knots the water drag is 1.00 pounds. Also, the air drag at 13 knots is 2.29 pounds. Thus the total drag you have to overcome is 3.29 pounds (1 + 2.29). Case 2. Instead, say you paddle against the current with the same SOG of 4 knots. Then the relative water speed is 6 knots (4 + 2). Also, the relative air speed pushing you is 5 knots (9 – 4). At a relative water speed of 6 knots the water drag is 11.91 pounds. And the force of the relative air speed of 5 knots pushing you is 0.34 pounds. So the total drag you have to overcome is 11.57 (11.91 – 0.34). Therefore, unless I made a mistake somewhere, I believe the example demonstrates that it’s easier to paddle downcurrent/upwind then vice versa, at any speed (not just as you approach hull speed). My actual paddling experience is consistent with this answer. Also, as you approach hull speed the exponent in the approximate exponential formula for water drag grows larger than 2.0. Also note that it’s less than 2.0 at slow paddling speeds. I validated this using John Winter’s drag program, named Kaper. Note, in a program I wrote I got answers consistent to the two cases at lower speeds. The examples were just easier to present. One mistake I think you made is assuming that both wind resistance and hull drag are the same exponential functions. Yes, to a first approximation, the water drag function is sort of like Dw = W*s^2 and the air drag function is exactly Da = A*s^2 (where s is speed and A and W are constants). But, because water is much denser than air, W >>A so a small change in s changes water drag much more than it changes air drag. The water drag values are from keelhauler.org/khcc/seakayak.htm and the air drag values were calculated from the standard air drag formula like the one published here https://www.grc.nasa.gov/www/k-12/airplane/drageq.html -Leon