leong

Andy, I agree with everything said above by Sir Christopher. He’s our esteemed grammatician, airplane pilot and paddler, Although a hand bearing compass is a cool device, and it's fun to learn how and to use any type of compass for navigational purposes, a low cost GPS does a lot more (as long as it works, that is). Allow me to plug the one thing that a GPS can do that can't be reliably done any other way (even with all the tricks that John Huth has developed). As we all know one cannot simply point their kayak at a waypoint and expect to get there along a straight direct-line course-over-ground. To stay on this straight line we have the concept of ferry angles (you offset the kayak’s heading by the ferry angle). But, unless you know the kayaks side drift (due to wind, current, waves, etc.) accurately, you can’t compute an accurate ferry angle. Furthermore, conditions vary so the necessary ferry angle varies in accordance. With a GPS, the whole concept of computing ferry angles becomes moot. Here’s how a GPS works to get you along a straight line to a waypoint: The GPS knows the kayak’s location and the waypoint’s location in some fixed earth coordinate system. It doesn’t know it’s own orientation (what direction it’s facing or even if it’s pointing backwards or straight down). It also doesn’t know the heading angle of the kayak. But it does know the Course-Over-Ground (COG) that the kayak is moving along as you paddle. If the COG coincides with the direct straight line from your present location to the waypoint, the GPS’s arrow (at least it’s an arrow with my GPS) points to the top of the screen. If the kayak’s COG for this latest update is to the left of the present direct straight line, then the GPS’s arrow points to the right (indicating that you should change your heading to the right), and vice versa. In effect, the GPS knows that the COG is the vector sum of the kayak’s forward speed and whatever is the cross-velocity vector. So, the GPS doesn’t really have any idea of current velocity (I mean water speed and direction) or wind or anything else that’s pushing the kayak. All it really knows is the COG and the latest direct straight line to the waypoint. As long as you keep the arrow pointing to the top of the screen (via heading changes as necessary) you will be following a direct straight line to the waypoint. From many conversations with NSPN members, I’ve found that most don’t realize that ferry angles don’t come into play when you use a GPS. That is, they think a GPS always keeps your heading pointing to the target. If that were the case, the course over ground to the waypoint would be a curved path whenever there is any side drift. (For constant side drift that curved path is called a pursuit curve.) As shown above, you do follow a direct straight line to the waypoint. So this feature of how a GPS works makes the GPS a really cool gadget. Obviously, a GPS has many other cool features. -Leon

Gary, I just got back from a play and began to think about why you think that my answer is wrong. Perhaps it’s because you are not familiar with the jargon I used when moving reference frames are needed. So I’m going to try and convince you that you’re wrong another way. This time by using a different version of your ferry mind experiment. So here goes: A small ferry is on the Piscataqua River drifting down current at 3 knots. The ferry captain wants to increase his speed to 4 knots. Down stream in front of the ferry is the Piscataqua River Bridge with a supertanker under the bridge. The supertanker’s engines are off; i.e. it’s just drifting down stream at the speed of the current. The ferry captain desires to increase his speed by 1 knot to reach 4 knots. Two ropes are available for him to pull in 1. A rope attached to the supertanker. 2. A rope attached to the Piscataqua River Bridge. Assume it will take of force of 50 pounds of tension on the rope (the drag force at 1 knot) to gain the additional 1-knot. Case 1. The ferry boat captain (FBC) pulls in the rope attached to the supertanker until it’s just taut (0 tension neglecting the weight of the rope). Since the supertanker and ferry are moving with the current at 3 knots, the FBC will have to pull the rope closer to him at a speed of 1 knot to bring the ferry’s speed up to 4 knots. When he does this the tension on the rope will increase to 50 pounds (exactly the drag force of moving against the river at 1 knot). So the power requirement is 50 pound-knots (50 pounds times 1 knot). Case 2. Instead, FBC pulls in the rope attached to the bridge until it’s momentarily taut (0 tension if you neglect the weight of the rope). Of course he has to continue pulling in the rope at a speed of 3 knots to just keep it taut, but still 0 tension. But since the bridge is stationary, and the FBC wants the ferry to move at 4 knots, he will have to pull the rope in even faster, to at rope speed 4 knots. As in case 1, the tension of the rope will increase to 50 pounds (exactly the drag force of moving against the river at 1 knot). So, in case 2, the power requirement is 200 pound-knots (50 pounds times 4 knots). I hope this will convince you. Respectfully, -Leon

>>By the way, my first read of your description was that 5 pounds of drag existed when traveling at 4 knots. Yes, that’s true and consistent with what I said. >>Not that 5 pounds * 4 knots was the appropriate formula, since we all know that drag increases exponentially, not linearly as speed increases. I assume that you’re talking about the case where the kayak is going at 6 knots downriver. But his speed relative to the river is still 4 knots. So there is no drag increase, it’s still 5 pounds. Non linear drag increase has nothing to do with the problem here, since there’s no increase in “water” speed (although an increase in “speed over ground”). >>The punt's drag is predominantly a function of the punt against the water, not a function of the punter's pole. Yes, of course. And I used that fact. >>Just because the punter is pushing on the ground does not negate the benefit of the current. You don’t understand that the force in the power equation is relative to what’s being pushed against. The paddler’s force is against the moving water so the speed of the push is still 4 knots. But for the punter, he’s pushing against the fixed earth, so the speed of his push is the full 6 knots of “speed over ground”. Consider this thought experiment: You paddle and/or punt at 4 knots in still water requiring a power of 20 pound-knots, assuming the 4 knot water drag is 5 pounds. Now suppose the current is 100,000 knots (neglect turbulence, air drag, etc.)! The paddler will have no trouble paddling at a speed of 100,004 knots (speed-over-ground) going downcurrent, right? But do you really think that the poler can also increase his speed from 100,000 knots to 100,004 knots? The power required to do that would be 5*100,004 pound-knots! Note that in the first case (paddler) the force is with respect to a moving reference frame (the moving water with respect to the earth). In the second case (punt boat) the force is with respect to a fixed reference frame (mother earth). This applies to all physical systems, whether they’re kayaks punt boats, ferries, etc. Sir Isaac Newton was aware of this. So, no, I shouldn’t have quit while you mistakenly thought that my detailed analysis was wrong. It’s quite correct. No biggie, though. It’s a subtle topic. Peace, -Leon Gotta go finish lunch and hit the swells soon. I’ll respond tonight if necessary,

Methinks (perhaps) there’s no single answer to who wins the upcurrent race, the punter or the paddler. It might depend on the still-water speed of the paddler (punter), the current speed, the water depth, etc. PS Lest the reader thinks this brainteaser has no practical value consider canoe poling. Canoe poling may be a lost art, but there are certain advantages to having the skill (and the pole). As they say, “You can easily push your boat upstream and then steer down.” Here are a few articles about canoe poling (hmm, the video below begs for some new stupid kayak tricks at Walden Pond ... are you listening Les?): http://nwwoodsman.com/Articles/CanoePoling.html http://www.outdoors.org/publications/outdoors/2008/how-to-canoe-pole.cfm Respectfully, -Leon

Andy, Good analysis. That’s an interesting take on the problem. I don’t think “cadence” is the proper term, however. Cadence is the number of strokes per unit time. You’re talking about only shortening the time interval between the left blade exiting and the right blade entering. For a given paddle length, size and hand position, I think cadence is proportional to boat speed. That is, you can’t increase cadence without increasing boat speed. But an increase in power is necessary to increase boat speed (I’m disregarding stroke form and ergonomics). Anyway, I get your point about back drift. You're saying that back drift is less for a paddler because his left to right transition is probably faster than a punter’s pulling the pole up, then pushing it down. That certainly adds to the complexity of the problem. Nevertheless, I believe that the punter has the advantage going upcurrent. Here’s a thought experiment showing his advantage for a “stressing” set of conditions: Say the maximum still-water speeds are equal for the punter and the paddler. Say 5 knots. Next suppose the current speed is 6 knots, in the opposite direction. For this case, the paddler can make no headway, in fact he drifts backwards at a speed of 1 knot. Now when the punter’s pole is planted at the bottom of the river, he can stop any backward motion. So, at least for this instant in time, the punter is going faster than the paddler (0 knots > -1 knots). Without getting into a complicated power analysis (it would be complicated because of the discontinuity of the drive force for each), I think the punter has the advantage going upcurrent. An easy power argument could be made using the following simplified thought experiment: Think of a amphibious vehicle with two drive modes: wheels on the bottom (analogous to the punter) and a propeller (analogous to the paddler). You could show that for traveling upcurrent the propeller mode requires more power than the wheel mode for a given speed. Of course, I’ve left out the complications of the rolling friction at the bottom of the river and the efficiency of the propeller. Although I think the punter has the advantage going upcurrent, I’m not certain that you are wrong. A more definite answer requires a very complicated analysis, more than I’m willing to tackle. What do you think now? -Leon PS A real-life analogy from today: I was paddling against a strong wind (about 25 knots) and a strong tidal current (about 2 knots). The GPS said that my back drift was over 3 knots when I stopped paddling! After a half-hour of very hard paddling I wished that I had a long enough paddle (or pole) to plant it in the bottom of the bay to hold my position so I could catch my breath. Eventually, I grabbed onto a lobster pot (or whatever it’s called in Florida) to hang on to rest.

Okay, both Andy and Bill are quite right. A thought experiment for Sir Christopher to help him get it. Suppose you were climbing up a ladder to the sky at, say, 2 feet per second. Then the ladder began to slip down into the earth at, say, 2 feet per second. So now you’ll have to climb the rungs at 2 feet per second just to stay at the same position relative to the ground. To continue moving “up” at 2 feet per second you’ll have to climb the rungs at 4 feet per second. That will require a lot more power from you, right. Here’s a slightly more physics-like explanation using power of why the kayaker will do better in the downriver race. (Note that power = force * speed). For simplicity, let’s use these numbers: 4 knots – Still-water racing speed for both the kayaker and punter 5 pounds - Drag force of water to go 4 knots faster than the water (assume it’s the same for the kayak and punt boat) 2 knots – speed of current 20 pound-knots will be the power required by both the paddler and punter to go at 4 knots in the still water (5 pounds * 4 knots). Going downriver, the kayaker will be moving at 6 knots (2 free knots due to the current + 4 knots provided by his paddling). So the kayaker’s required power (force * speed) is still 20 pound-knots (5 pounds * 4 knots relative to the river). But for the punter to move at 6 knots downriver, his power requirement would be 30 pound-knots (5 pounds [to increase his ground speed from the free 2 knots to 6 knots] * 6 knots (the speed of his pole against the non-moving river bed). So, in this example the punter would require 50% more power (30 vs 20) to keep up with the kayaker. So, obviously the kayaker will win the downriver race. -Leon PS Brainteaser part 2: How about an upriver race? PPS Hmm, I gotta make a decision fast. Should I take out the kayak or the Sunfish? Retirement sure has its hard-to-make decisions. <Sir Christopher, is this a correct hyphenation?>

I just read this article about punting in Cambridge, England. It begged for the following brainteaser: Assume a punter and a kayaker practice racing in still water and it always results in a tie. Then they enter a down-river race. Will one of them now win and why? -Leon Note: Punting refers to boating in a punt. The punter propels the punt by pushing against the riverbed with a pole. A punt should not be confused with a gondola that is propelled by an oar rather than a pole.

On the other hand my favorite type of response to a trip I propose is when Igor (who I don’t know) says, That’s a great trip proposal. Sounds like a fine way to spend a day. I would love to join. How far is it and when would we get off of the water? Will you be rowing too fast for me? How cold is the water there? Sorry I can’t make it. I need to buy a kayak and oars first. Igor

The following kayaker usually shows up early: You can bet on it that with a measure of certainty I will most likely join you; that is, if my spouse's family hopefully leaves on time and my loaned-out kayak is soon returned and I don’t forget to remember again. -Leon (probably)

Be careful Sir Christopher. You can use too many hyphens and be suspended from this bulletin board. I wouldn’t want that to happen to you. Something similar happened to this author. Anyway, I’m turning a blind eye to what low and middle hyphenator NSPNers have to say on this thread. Or should it be, “Anyway, I’m turning a blind eye to what low- and middle-hyphenator NSPN-ers have to say on this thread”? -Leon

The top ten foghorn sounds are here. Vote for your favorite. -Leon PS I heard that the next release of MRASS will allow you to select your favorite foghorn.

All VHF radios have channel 83A. Except to activate MRASS, channel 83A is only for USCG business. Any VHF radio (whether from a kayak, surfski, battleship, swimmer, etc.) can be used to activate the foghorn as long as there's a line-of-sight path to the MRASS receiver. -Leon

Pru, Thanks for the heads up. I just watched the 20-20 episode on-line. Of course I agree with Todd Wright’s assessment that the missing paddle barrel connector and drain plug would be of minor consequence. Nevertheless, even though removing these items wouldn’t be sufficient to murder someone, if, indeed, the girlfriend did tamper with these items it sure seems like her intent was murder. Intent to murder is a serious crime even if the specific preparations didn’t work. Given that her intent was murder then the weather gods were her accomplices; i.e. the weather and river conditions were the proximate causes of the boyfriend’s death. -Leon PS A recurring scream of mine is this: I wish that there were some standard nomenclature to distinguish sea kayaks and sea kayakers from the novices that paddle floating vessels that look like plastic bathtubs We don’t call people who walk up hills mountaineers, do we? PPS Does anyone have a personal email address for my old friend Dr. Sanjay Gulati? I’m reluctant to call his medical office.

In the October 19, 2015 issue of “The New Yorker” magazine there’s a mostly negative article about Henry David Thoreau, titled “Pond Scum”. In the November 9, 2015 issue of the same magazine there’s a great rebuttal in “The Mail” section. It’s from, a former NSPN-er and my old kayak race-training partner, Sanjay Gulati. Among other things Sanjay says, “By the way: the only scum floating on Walden Pond is the Walden Pond Scum, a group of kayakers learning Greenland kayaking techniques; I’m a member. We pay tribute to Thoreau’s curiosity about Native Americans, whom he approached as respectfully as he did everyone on the “wrong side” of Route 2 from the citizens of Concord, whose closed-mindedness he disdained.” -Leon

Hey Steve, It's been a while since our Blackburn training run. I've been mostly out of the picture for the last year and a half because of medical problems. But I'm back to training in the cockpit now. I don't think there are any local kayak/surfski races after the Glicker Downwinder that was Oct 10. See this and this and that. On the other hand, if you want to drive south for about 27 hours to Singer Island, FL, we have a weekly informal 8-mile training race. No prizes, just teasing the winner. Best. -Leon PS Thanks for the heads-up about the new Misery Island race. I was unaware of its existence. Up until about 10 years ago Henry Szostek (of Blackburn Challenge fame) used to run a different Misery Island race starting from Misery and out somewhere past Bowditch Ledge and back to Misery. I won it a few times.

Why would you not want to be seen? Navy Seal? Just wondering.

Andy, Right on! It’s probably not obvious to most people but your educated guess is correct; i.e. the two components “cancel” each other out. Accordingly, Jack and Jill will end up sliding the same distance. In fact, at each instant of time, they will have slid the same distance. The elementary physics below demonstrates why your “canceling” guess is correct: F = m*a (from Newton’s second law of motion). The sliding frictional force is equal to -u*m*g (a negative sign was introduced because the sliding frictional force is in the opposite direction of initial sliding velocity). Thus we have m*a = - u*m*g. Dividing through by m we get: a = -u*g. Thus, neglecting air resistance, the deceleration of a sliding object is independent of the mass (or weight) of that object. Therefore, the answer is “c”; i.e. the race will be a tie because Jack and Jill will both decelerate at the same constant rate. Notation used: m is the mass of the object, g is the magnitude of the local gravitational acceleration and u is the coefficient of sliding friction. -Leon PS While reading about cold water paddling I came across the discussion that I linked. That got me to thinking about sliding on ice and the brainteaser came to me. PPS Yes, in a follow-up post you correctly noticed that since Jack and Jill are both in their respective kayaks their clothing has nothing to do with the sliding friction.

Phil, I prefer mint chocolate chip. Doug once said, “It's easier to slow down a VW Bug than a Mac Truck”. That’s correct because the brakes will have to work much harder on the Mac Truck. Nevertheless, in the absence of air resistance, transmission drag and given the same speed before beginning to coast, both vehicles will coast the same distance. Consider that a hint for the Jack and Jill problem. -Leon

Robert, Under your assumption Jill will win because 0.1 ton Jack will break through the ice before Jill does. But, more important, congratulations on your BCU 4*. -Leon

Facing icy waters extends the paddling season to a full year's worth of opportunity to enjoy nature in all of her moods. To this end, Jack (200 lbs.) and Jill (100 lbs.) decide to have an ice race on a frozen river in their identical kayaks. They both run on the ice pushing their kayaks at the same speed and simultaneously jump into their cockpits. They then slide until they come to a stop. The winner will slide longer and farther. Jack says he’ll easily win because he’s heavier and his momentum will carry him farther. Jill says he’s wrong. The possible results are: a. Jack will win? b. Jill will win? c. Neither will win; i.e. it will be a tie? Question: Neglecting air resistance, which is the right answer, a, b or c? Hint: Jack is a professional hockey player and Jill is a mechanical engineer. Ice paddling discussion.

Also GPS's with a super large screen.

Da bear is here.

Sir Christopher et al, You’ve already admitted that you occasionally tie a line to the bow of your kayak on longer trips. Isn’t that an explicit admission that you know that there could be some failure causing your kayak to fly off of the car? In my penultimate post before this one, I provided this link that shows that even a Thule J-Stack kayak carrier can fail. Thus, even with failure-free roof racks and straps, a kayak can still fly off of a cartop because of a J-Stack (or cradle) failure. Secondly, consider how you measure risk. One should be very concerned with high-consequence events even if they’re low probability events. And people underestimate the chances of low probability events because they think about Bell Curve (Normal) distributions. However, the failure statistics of a cartop system for carrying a kayak may, indeed, follow a leptokurtic distribution where extreme outcomes are much more likely than with Normal distributions. Finally, the cost and effort of almost eliminating the risk of a kayak flying off the car is negligible; i.e. always tie the kayak to the front and back bumpers of your car. Respectfully, -Leon

Ya think this setup doesn’t need bow and stern lines?

Yet another reason for bow and stern tie down lines. Stuff breaks! Wasn't me, some bloke in Scotland.