leong

I've used the dual mode as well. But, on weekends, it's disturbing to hear all the traffic on 16. So, I'd rather hear the quiet afforded by the squelch. I'm on the water to relax with friends, not to help the CG. Okay, turn me in. I'm friends with the guys guarding Mar-a-Lago.

Do any of you sea kayakers obey the following rule? I don’t. Vessels not required to carry a marine radio – for example, recreational vessels less than 65 .6 feet (20 meters) in length, but which voluntarily carry a radio – must maintain a watch on Channel 16 (156 .800 MHz) or VHF Channel 9 (156 .450 MHz), the boater-calling channel, whenever the radio is operating and not being used to communicate. Sometimes when paddling in a group we all monitor a specified channel (usually 69 or 72) to be ready to help each other, if the need arises. Let the large powered vessels fellow the above rule.

>> Maybe the takeaway from all of this is that if we're going to take advantage of the key lock so we can stuff the radio into a PFD pocket, we need to commit the unlock sequence to muscle memory so it becomes automatic.  I have no argument with anything you said, Dan. But my one and only point (or takeaway) is that the channel 16 key should (as described in the Icom M73 manual) select Channel 16 when pushed. But it doesn't if the key pad lock is on. I don't care that it doesn't say it. I care that it shouldn't be that way for safety. If the VHF was designed according to fail safe criteria, the channel 16 key should always take you to channel 16, no matter what. Otherwise, it's like an aircraft autopilot with a switch to disable it that doesn't work under certain conditions. No way around it, it's a stupid design! And it would doubly stupid if your DSC distress button wouldn't work while in key lock!! I bet it does. Someone will eventually sue if it doesn't. BTW: There aren't volume up/down keys on the Icom M73. The volume control is a knob that also turns the unit on and off.

I think you started a new thread, Robert. One for those kayakers that help the CG monitor channel 16 for distress calls. I don’t even have my radio on, unless I’m paddling with a group. My almost encounter with a mega-yacht was just an anecdote that served as a reminder of the poor design of most handheld VHFs. Nothing more, nothing less. Clearly there are circumstance where not being able to get to 16 using the button would be a problem. PS When you lock the channel on the Icom M73 it stays locked even after turning the radio off and back on again. Also note that the channel lock button is labeled H/L and is a dual purpose button;, it selects high, middle or low power when pushed, unless you hold it down and then it toggles between lock on and lock off. Try and remember all that when you're in the water in an emergency situation, especially if it's a borrowed VHF radio.

Jason, with all due respect, you're wrong. Yes, a second or two doesn't matter. But what if you don't remember which is the unlock button? Or what if you've capsized in a storm (ask Keith) and are bouncing around in the water and you want to put out a Mayday call ASAP and can't find the damn unlock button or don't know the procedure for the radio you borrowed that day. I’ve never heard of anyone locking their radio to, say, channel 72 only because they're afraid that it might accidentally switch it to channel 16. That’s not the purpose of the locking a radio to channel 72. It’s to prevent it from accidentally switching to any other channel including 16. But if you need to go to 16 then the emergency channel 16 button should work whether the radio is locked or not! It’s not a catastrophe if you accidentally use channel 16. At worst, the CG will shoo you off. But it could be a catastrophe if you can’t get to channel 16. That’s what fail safe designs are for; i.e., to prevent catastrophes. If the radio is in your hatch or in a duffel bag it wouldn't be on. When the radio isn't on pressing the emergency channel 16 button will do nothing at all. -Leon

Almost every day I padddle out to the ocean through the Palm Beach inlet. To avoid powerboats, I stay close to the south side of the inlet, far from the channel. Here are a couple of pictures of a dock that I paddle past on the way out. Little did I know that I also need to be careful of flying cars while paddling past the docks. A few week ago, a Ferrari went flying off this dock. Luckily, it happened a few hours before I paddled there. Below are the Ferrari adventure stories: https://www.wpbf.com/article/dash-cam-video-shows-driver-of-ferrari-plunging-into-water/25703591 https://www.palmbeachpost.com/news/20190118/submerged-ferrari-in-palm-beach-oh-god-thats-good Perhaps I should be careful of powerboats when I go pedaling on roads near the ocean. ?

You've got it backwards, Jason. I'm not against the channel lock. I like it. I'm against the design that when the channel lock is set the channel 16 emergency button is disabled. In an emergency you should be able to get to channel 16 ASAP; i.e., you shouldn't need to take the time to unlock (and you may not remember how to do it) to make an emergency call. As I said, it's a very stupid and unsafe design. PS It's freezing today in Southeast FL (high of about 62 and very windy). Went pedaling instead of paddling.

Icom M73. I think all hand held VHF's have the unsafe design of the lock preventing an immediate switch to channel 16. One of our last paddles (last century) together about 10 of us started from Manchester Harbor. By the time I reached Rockport everyone had dropped out and I continued on to circle Cape Ann counter clockwise and back to Manchester. Bob Burnett drove around the course looking for me. I think he met me back at the put-in. Both of my wrists were sore for a week.

The button to select CG channel 16 is supposed to be for getting to C16 quickly in emergencies. However, it doesn't work when channel lock is set. In an emergency you must first unlock ... and you might not remember how. This happened to me the other day. I wanted to hail a mega yacht heading towards me. I forgot how to cancel the lock (You Push [H/L•LOCK] for 1 second to cancel the lock). I was able to avoid a collision but I had a nice surf ride off of his stern wave. Do the VHF designers take stupid pills?

https://weather.com/science/nature/video/seal-slaps-new-zealand-kayaker-in-face-with-octopus

>>Give the punter a 200 foot pole and it would be planning at 85 knot s but he only has a ten foot pole. Okay, let's consider this with more familiar units (horsepower). Give the punter what ever "gear ratio" he wants, say a long pole that he can handle. Say the punt boat (note: a gondola is not pushed with a pole) is drifting down stream at 80 knots and that the punter can move the boat at 5 knots in still water. Further, assume the water drag at 5 knots relative to the water is 6 pounds. So pushing off the fixed bottom to reach a ground speed of 85 knots (80 knots from the river current plus another 5 knots) requires a power of 510 pound-knots (85 * 6). 1 pound-knot is equivalent to approximately 11 horsepower. So, in HP units, the required power is 5, 610 HP (510 * 11). Obviously, this power is impossible for any punter but superman.

>> Still missing the point the punter can generate 15 gazillion knot pounds of force. Knot-pounds are not force, Ken. Knot-pounds represent the units of power (force times speed). Here's a thought experiment for you. Say at your maximum effort you can push a heavy block of wood on a wood surface at a speed of 1 knot with a pushing force of 100 pounds of force. Do you think you can speed up and push it at a speed of 10 knots with the same force? Don't you realize it would require 10 times the effort (power). The amount of work you can do per unit time is limited by the power you can exert. The force would be the same. However, you couldn't do it unless you were able to increase you power by a factor of 10.. Note: For "dry friction", such as a box on a floor, the sliding friction is independent of speed.

>>You are making the assumption that force is limited and speed is not. No, I'm not. I'm assuming that power is limited. >>So he could push a gondola at five knots or a supertanker at 5 knots but no matter the load, his speed maxes out at five knots because that's how fast his arms can move regardless of load. No. The water's drag on a supertanker is much greater than it is on a kayak or gondola. For example, say the drag on a gondola is 6 pounds at a speed of 5 knots. That means the punter's maximum power is 30 pound-knots. Now assume, for example, the water drag on supertanker is 5000 pounds at 5 knots. So the power to move the supertanker at 5 knots is 25,000 pound-knots. But the punter can only generate 30 pound-knots. So, let's calculate X, how fast the punter can move the supertanker. We have, 30 pound-knots = 5000 * X pound-knots. Solving for X, X = 0.006 knots. That is, at steady state, the punter can push the supertanker at only 0.006 knots. I pulled the numbers out of a hat so the exact answer is probably different. You are correct that the punter's speed of pushing is also limited. But that's not needed to solve the problem. It just introduces an unnecessary constraint that is trivial at the low speeds we're discussing. For example, if the current is 1000 knots, the punter will not be able to push off the ground that is moving away from him at 1000 knots. >>... does the BB leave your hand at 1000 times the speed of the baseball? That's a completely different kind of problem regarding Newton's Second Law of Motion (F = mass * acceleration). We're assuming constant speed in the brain teaser I posted, so acceleration = 0. I can solve that kind of problem, but I'll leave it as an exercise.

It's the rate of doing work that has to be limited; i.e. the power (force * speed). Consider the punter. In still water say his power to go 5 knots is 5*F, where F is the force to punt at 5 knots (5*F is his maximum power). For the punter to go 9 knots down stream he would have to generate a power of 9*F. But he can't do that because his maximum power is 5*F. Of course he could punt at more than 5 knots going downstream, but his force would be less than F. However, the paddler would go at 9 knots down stream in a 5 knot current if he exerted the same power as his still water power.

>>One is a safety issue and a performance issue too.

Nice to hear from you, dear doctor Beale. Thanks for the kind words. Whenever I think of you I recall our hilarious circumnavigation of Cape Ann. Do you ever paddle on weekdays? That’s when I paddle until mid-October; then I head south to paddle and/or sail every day in Florida. -Leon

Swearin' ferret.

Yes, in Irish, the punt was the official currency of the Republic of Ireland prior to the coming of the Euro. Ireland has great sea paddling and punting on their canals. However, when I was there the local newspapers were complaining that the bars in Dublin had broken the pound; i.e. they charged more than a pound sterling for a pint.

Who is he and under what situation?

Okay. My model is a simplification of reality, and as such, certain details are excluded from it. The question is always what to include and what to exclude for a given application. For the kayak and punt boat I could model the drag force with the simple function; i.e. drag ~ speed squared. I could apply different duty cycles to the punt pole and the paddle. Or I could go high fidelity and use John Winter's curves of kayak drag vs. speed. But, for the purpose of this exercise, I'm almost certain the results would be the same. I think anymore fidelity would be guilding the lily. The brain teaser doesn't ask by how much will the punter (paddler) will beat the paddler (punter). It just asks which one will win. Think of this. Going up current in a very strong current the paddled boat will go backwards. But the punt boat can lock to the bottom and not lose way. Take a look here. It shows the punter using a smooth and continuous motion.

Okay, say the average power to keep the boat going at 5 knots relative to the water is, say, 100 watts. Aren't you implying that you could put out 100 times that power (10 thousand watts) for1% of a stroke cycle and rest for the remaining 99% of the cycle? I'm not sure how you would model your impulse approach; i.e. Impulse = F delta t = change of momentum.It's hard to believe you could generate a large enough force in a negligible time to keep the boat going against water drag. Anyway, do you think my answers are wrong? Surely you don't think a punter could punt down stream in a 50 knot current. But the paddler would have no trouble. Perhaps I'm missing your points.

Everyone is trying to solve the brain teaser by considering the different mechanics of a paddle vs. a pole, the water and boat hydrodynamics, etc. Obviously, these differences count; however, the solution to the brain teaser comes from the fundamental physics of propelling a “craft” from a fixed point vs. a moving point. So, for simplicity, we’ll demonstrate this by replacing the two boats with two blocks of wood and a rope for the transmission of power. Try this thought experiment: Assume two guys, Puck and Paddy, can both generate the same maximum power. First, pretend the still river is a stationary wooden horizontal platform. The force required to slide a particular block by pulling it with a rope at any velocity with respect to the platform is constant, denote it by F (look up sliding friction). Note that power is the product of force and speed. Therefore, to slide the block to the left at a speed of 5 knots, Paddy, on the platform, uses 5F of power. Say that’s his maximum power. But this is also true for Puck on the ground who also uses his maximum 5F of power. Therefore, everything else being the same, for the stationary platform, the same power is required if you pull the block from the platform or pull it from the ground. That should be obvious. Now pretend the platform is the moving river, say moving to the left at a speed of 4 knots. For Paddy on the platform, if he pulls the block to the left with his maximum power of 5F, it will move the block at 5 knots relative to the platform. However, for Puck on the ground, if he pulls the block to the left with his maximum power of 5F, it will move the block at 1 knot relative to the platform (that’s because he has to move to the left at 4 knots faster to keep up with the moving platform). Therefore, for the same power, Paddy’s block wins the race. If you reverse the example so that both Paddy and Puck are pulling in the direction of the moving platform, Puck’s block will win the race. See for yourself why that is true. Also note if you use actual drag forces for the boats (perhaps drag being approximately proportional to the square of speed) you get the same results for who wins the upriver and downriver races.

No time today. I'll post the way to do it soon. It has nothing to do with paddling or punting, although that is a minor contribution. It has to do with whether you propel off the moving water or from the fixed earth. That's a good hint.

Both the paddler and the punter are limited by the maximum power they can produce. To keep it simple, assume that both small craft have exactly the same drag force versus speed curves. But this isn't necessary to solve the problem.

Andy, On the contrary, the punter is using a long skinny racing punter and the paddler is using this kayak. But, remember, both athletes are able reach 5 knots at maximum effort (racing speed). -Leon