leong

John, You’ve already demonstrated the effect of Hill’s equation when you switched to a shorter paddle. The increased cadence probably brought you closer to your power peak. Note that as you increase your cadence starting at 0 cadence, the maximum power that you can produce increases until you reach the energetically optimum cadence (EOC). Of course, you’re not paddling at maximum effort, except for a short sprint. Say you usually paddle at 30% of your maximum power. The curve of 30% of maximum power vs. cadence will also increase until you reach some optimum cadence. I haven’t seen any data on this for paddling but I’m pretty sure that your optimal cadence (given an effort of 30% of maximum power) will be approximately the EOC cadence, probably a little less. Unfortunately because you can’t change the “gear ratio” for paddling very much, paddling cadence is just about proportional to paddling speed. When I raced a bike I was very interested in this topic. For example, to go at my maximum speed in a race (maximum power), what gear ratio should I use to get my actual cadence as close as possible to the EOC? And for a club rides below maximum speed, what gear ratio should I use to maximize my endurance for the distance of the trip? I never scientifically answer the second question. Nevertheless, my experience was that I should select a gear ratio that provided a cadence that it was a little lower than the EOC. I’ve been paddling (sometimes sailing) almost every day in strong winds in southeast FL. Oh how I wish paddles came with changeable gears like bicycles do! -Leon

PPS Below I found a graph of human power vs. cadence. It shows that as you decrease your cadence below the “Energetically Optimal Cadence” (the peak of the curve) the power that you can produce drops quickly. Although the graph was made for pedaling it is just as relevant for paddling. -Leon

John, Okay, let’s assume that the resistance to encountering waves is already present in Burch’s tables and I’m double counting this drag. I don’t think so, but let’s remove the 0.5 pounds of drag that I added; it was just a guess for completeness anyway. The modified resultant power for going upwind becomes 3* (F + D3) = 3* (3.5 + 2) = 16.5 knot-pounds. This is still greater than the upcurrent power of 16 knot-pounds. But I’ll go one more for you. I think that it’s very unlikely, but suppose that the drag going upwind (water drag + wind drag) and upcurrent drag (just water drag) to go at the same SOG are equal. Call this drag D. So now we compute the required power for equal SOGs as follows for the Marjorie brainteaser: Upwind power = 3 * D (SOG = 3-knots) Upcurrent power = 4 * D (SOG =3-knots) So now we have upwind power < upcurrent power. So you might argue that Marjorie might be able to paddle upwind (SOG) faster than she can paddle upcurrent (SOG). However, Hill’s Equation may be the deciding factor. To use Hill’s Equation we need to have an expression for cadence. A reasonable assumption is that water speed is proportional to cadence. Thus we have, 3 = k * upwind_cadence 4 = k * upcurrent-cadence. This results in: upwind-cadence = ¾ * upcurrent_cadence But if 4 knots is the maximum water-speed that Marjorie can paddle in the absence of wind, it implies that her cadence is no greater than the energetically optimum cadence (EOC). And since we just demonstrated that upwind-cadence < upcurrent_cadence, Hill’s Equation says that the maximum power that she can produce to paddle upwind is less than the maximum power she can produce to paddle upcurrent. Depending on her particular “Power vs. Cadence” curve (it varies from person to person), it’s possible that she can paddle upcurrent (SOG) faster than she can paddle upwind (SOG), even in this case of equal drags where less actual power is required to paddle upwind. The determining factor is whether or not the muscles can generate the required power at the reduced cadence. Merry Xmas to all. -Leon PS Way back it was the fashion to use long paddles. To go at the same speed the required cadence of a longer paddle was less than the required cadence of a shorter paddle. More recently, the advice has been to use shorter paddles. I believe this follows from studies addressing elite kayak sprinters where the goals were to determine the optimal relationship between fatigue in athletes and their performance. A good part of these studies concentrated on the techniques and equipment necessary to drive cadence to the EOC point. Shorter paddles are one result of the studies. Luckily this has filtered down to the sea kayaking folks.

Okay, it looks like no one is following my simple proof (or admitting to it). So forget about math and forget about Hill’s Equation. The proof is in the pudding. Think about the following real life experience below: Yesterday I was paddling continuously at 4 knots in the lee of an island. When I reached the windward side I turned directly into a ~ 25-knot headwind. When I paddling as hard as I could I only made about 1.5-knots (SOG). (Note: that the extra drag of the wind reduced my cadence from ~ 60 rpm to ~ 25 rpm, even with shorter strokes.) When I stopped paddling I checked my drift speed. It was about ~ 1.5-knots if I was parallel to the wind, but as expected, faster when I was broadside to the wind. Results: Paddling at 4-knots (water speed) against 1.5-knot current in still air would obviously result in a SOG of 2.5 knots (4 – 1.5). Yesterday, paddling against a 25-knot wind (“equivalent” to a 1.5-knot current) resulted in a final SOG of 1.5 knots. Result: I could paddle a faster (SOG) into a current than against an “equivalent” wind. And like riding a bike uphill in too high a gear, I couldn’t increase my power when paddling into the wind because my cadence was too low. That’s Hill’s Equation in simple terms. Read about Nobel Prize winner, Archibald Hill, here https://en.wikipedia.org/wiki/Archibald_Hill. -Leon

Probably enough so that Tompkins would have lived to die of old age. -Leon (who even keeps a paddlefloat in his boat)

The following quote from one of the articles implies that Tompkins wasn’t wearing a drysuit, right? River and sea kayakers often wear wet or dry suits, which help preserve body temperature in the event they capsize in cold water. Canoe and Kayak Magazine reported that in a photo taken at the beginning of the trip, Tompkins is pictured wearing a Patagonia dry top. Since Boyles was reportedly holding Tompkins by his clothes, it’s possible that the garment was pulled up around his neck, exposing his skin to the cold water. [Editor’s note: sea kayaks are longer, wider and more stable than river kayaks and generally have rudders operated with internal foot pedals.] I thought that sea kayaks were usually narrower than river kayaks, right? A five-day paddling trip in cold water sans a dry suit is more hubris than I could imagine.

John, “today drag” was a typo, I meant to type “total drag” Note that applying Hill's Equation to the cadence slowdown for upwind paddling is sufficient to demonstrate that it's easier to paddle upcurrent than against an "equivalent" wind. I just included an estimate of the drag from wind-generated waves for completeness. I didn’t need it to prove the point. -Leon PS knot-pounds are my favorite power units when working with kayaks and drag. You want hp, then just multiple by ~ 0.00307 186,000mps = 1,799,885,057,678.61 Furlongs Per Fortnight

Phil, You’re talking about a blade that doesn’t slip (is locked) with respect to the water, right? So, yes, if the river isn’t moving the blade is locked with respect to both the river and ground. But if the river is moving the paddle is locked with respect to river, but is moving with respect to the ground (at the speed of the river). >>So without increasing the stroke rate or stroke force (bigger blade) there is less effective propulsion in moving water than calm water. I’m not sure I understand what you think this implies. The “effective” propulsion with respect to the water is the same whether the water is moving or not. How about this thought experiment: Say (with closed eyes) you’re pushing with a constant force and speed on a box on a walkway (the speed is with respect to the walkway). If the walkway is one of those constant speed moving walkways used at airports, the force of you push with will be no different no matter the speed of the walkway, right? In fact, if the walkway’s movement is vibration-less you won’t even know whether the walkway is moving or not, right. The power to move the box (force * speed of the box relative to the walkway) is the same whether the walkway is moving or not, right? Of course, the distance along the fixed ground that the box moves is different since it depends on the speed of the moving walkway. But the power is the same. -Leon

ditto

John, No new posts so here goes my answer to the brainteaser. Let me know if you agree. Based on empirical evidence from my 25 plus years of paddling against wind and current, I’ve come to the conclusion that paddling at the same SOG (speed over ground) against an “equivalent headwind” is much harder than paddling against a current. So using the brainteaser example and data from Burch’s book here’s (I hope) a quantitative proof. Paddling upcurrent, Marjorie’s speed relative to the river is still 4-knots. But her SOG is 3-knots (4-knots paddling speed – 1-knot current speed). Note that Marjorie is an excellent paddler. Because of that she has chosen her paddle and hand position to get her cadence as close as possible to her energetically optimum cadence (EOC). The EOC is that cadence that gives one the maximal aerobic power. Above or below the EOC the power that a muscle can provide monotonically decreases. More on this later. Paddling against the headwind at an equivalent SOG, Marjorie has to paddle at 3-knots (both SOG and speed relative to the river). Drag Budget for 1-knot upcurrent (using Burch’s Fig. 5-3): D2 (water drag at 4-knots): 4 pounds Drag Budget for 15-knot headwind (using Burch’s Figs 5-2 – 5-5): F (Air drag at 18-knot apparent wind [15-knot wind + 3-knot paddling speed]): 3.5 pounds D3 (water drag at 3 knots): 2 pounds DW (additional wind-induced small-wave drag): 0.5 pounds (included for completeness, but it’s not needed to prove the point) TD (today drag into a 15-knot headwind) = (F+D3+DW): 6 pounds Since power = speed * drag force we have, For 1-knot upcurrent: power = 4-knots * 4-pounds = 16 knot-pounds For 15-knot headwind: power = 3-knots * 6-pounds = 18 knot-pounds So, based on power for the same SOG, it requires more power to paddle into a headwind. But that’s not the end of the story. Even if the required power was the same for paddling upwind or upcurrent it’s easier to paddle upcurrent. That's because, when paddling upwind, the kayak slows down and the blade's movement backwards with respect to the water decreases.This reduces your cadence (actually the speed that you can pull your paddle back through the water). Hill's Equation shows that the power that a muscle can produce decreases as the speed of contraction is decreased below some optimal speed (see this and this and that for additional information). So when the cadence is reduced from the EOC the power you are able to produce is also reduced. -Leon PS In case you don’t realize it the reason bicycles have gears is because of Hill's Equation (not to be confused with the little mountains sometimes called hill's). When you try to determine how fast a bike can go, what you do is you match the power available against the power required, at a given speed. This energy budget indicates whether you can go faster, or whether you can even hold your current speed. Power produced is the product of torque times cadence (rpm). Starting from 0 cadence where power is 0, power monotonically increases as cadence increases until the EOC point is reached. The EOC cadence is where power is maximized. Beyond the EOC cadence power decreases. The beauty of bicycles is that they have multiple gear ratios. So for a particular drag force (gravity on hills, air drag, rolling friction, etc.) one can choose an appropriate gear ratio that allows one to pedal at (or close to) the EOC cadence. Too bad paddlers don’t have variable gears like pedalers do (perhaps peddlers do). About the only things that we can do to “change gears” are: 1. Change paddle length. 2. Change distance from your hands to the paddle blades. 3. Change stroke length. 4. Change blade size. One I go on an upwind run I don’t have the luxury of changing paddle length or blade size. But I can “choke” up more on the paddle and take shorter strokes. The shorter stroke not only increase cadence but also (partially) counteracts the reduced glide when paddling into a wind. But these changes are not nearly as effective for changing a kayak’s “gears”, as they are for a bike where you can just change gears to almost perfectly match the EOC for any total drag.

Oh, knowing that I wouldn't have been so negative. I certainly appreciate all the work done by those who run the site. -Leon

Good try anyway, Dear Les, The problem is that Marge is the best paddler out there. Whenever she's around I draft behind her. Oops, I hope I’m not doing such a girly thing. -Leon

If this is a typical guess then your batting average must be pretty high.

Rob/Katherine I belong to several kayaking Meetup groups in Florida. Although, they might have potential (I don’t know) none function nearly as good as NSPN. But, let’s assume that a meetup option will serve our needs just as good as current NSPN. But why change over unless there’s an advantage? That’s what I meant in my earlier post where I stated, “I was negative about the switch-over because I couldn't see any advantages. Perhaps you could tell us what are the advantages for the user and/or maintainers?” As we engineers used to say about modifying a system: " Sometimes the enemy of good is better" Respectfully, -Leon, who’s been a member of NSPN from day one!

John, Yes, that's a secondary effect that’s not nearly as significant as the primary effect that I have in mind for paddling into a strong wind. So let's assume it away for the brainteaser; i.e., assume no wind resistance against your paddle. Then same question. -Leon PS Speaking of good wind, today it's too windy (gusts into the 30's) in SE Florida for me to sail or paddle. I gave it the ole college try in the Sunfish and capsized twice on the downwind legs (the GPS indicated 15 knots while I was on a beam reach). So back to my reading and computer-ing.

How did you choose that answer, Katherine?

Kayak Navigation by David Burch, has some good information about paddling upwind (or downwind). Especially Figure 5-4 which quantifies upwind slowdown. For example, suppose you can paddle at a sustainable 4 knots in still air. A 15-knot headwind will slow you down to approximately 3-knots (a drag program I wrote gets about the same approximate answer). So here’s an easy brainteaser (but it has some interesting ramifications that I’ll get to later): Assume that Marjorie is an expert forward-stroke paddler. She can paddle continuously at 4 knots. On Monday morning Marjorie paddled against a 1-knot current in still air. On Friday she paddled against a 15-knot headwind in still water. More assumptions: 1. She used the same boat and paddle both days (and everything else is the same including her breakfast choice). 2. The water was smooth both days. 3. There's a one-knot slowdown for paddling against a 15-knot headwind (from Burch’s book). Question: On which day was Marjorie’s average ground speed faster? -Leon

I made the same point above in post # 7 and I'm still waiting for an answer!

Yes, Pintail, that's an excellent point. In fact, even the regular floating bathtubs called recreational (kayaks?) that use paddles are not sea-worthy. I found this out when I borrowed a wide 12-foot recreational sit-in kayak and took it out for an ocean spin. The initial stability was much greater than needed; however, I had to continually brace with each broadside wave. And, sans a sprayskirt, I had to eventually start pumping out water. -Leon

A recycled topic from the past: A neighbor just brought down a Hobie Kayak that uses the MirageDrive. I’ll probably try it soon. It got me thinking of this thread. I did some googling and found this well written review about the MirageDrive. The author (I used to be a member of his kayak fishing club in Fort Lauderdale) is very knowledgeable about kayak hydrodynamics. Read the article if you’re interested in lots of details about the MarageDrive and how manufacturers hype their products based on erroneous tests or no data at all. The review clearly explains why I lost the short distance race and easily won the long distance race (and the Barton/Chalupsky team lost the tug-of-war) to the Hobie with MirageDrive. The main reason for our losses is “With the MirageDrive there is very little pause between what amounts to very short strokes. Compare to the paddler(s) whose strokes are necessarily longer, and with a greater pause between strokes.” That's why for a static race (tug of war) or very short distance race the MirageDrive has a big advantage over a paddled kayak. Several of us said essentially the same thing in this thread. -Leon

I already filled out the questionnaire. I was negative about the switch-over because I couldn't see any advantages. Perhaps you could tell us what are the advantages for the user and/or maintainers? -Leon

Hi again, Bill. [Note: our posts crossed so my previous post was answering your penultimate post, not your last one. This time I know how Gary appeared instead of Bill … because I copied an earlier message into my editor for reference and forgot to erase the heading when I erased everything else.] When the power levels are close I think we both agree that current speed, water conditions, ergonomics, biomechanics and paddle/pole choice might be the determining factors. Because of that there may be too many variables to predict the winner. Given such fuzziness, perhaps my brainteaser is ill posed; i.e. because it cannot have a single definite answer. But these types of brainteaser serve the purpose of thinking about many controlling variables. At least they do for me. Nevertheless, I think the approach of choosing the winner based on minimum power has some good theoretical merit. Of course, as someone smarter than me once said, “In theory, practice and theory are the same, in practice, they are not.” But I do wish that you could agree on how to quantify power when someone in a fixed frame is pushing something attached to a moving frame. Study my “box on a moving table” example. It’s much simpler to analyze than the punting example. If you (or anyone else) have comments to any of my previous posts in this thread have at it. Respectfully, -Leon

Gary, Okay, I had some time so here’s part 2 of my response to your last post: >>Yet the same poler who supposedly can not generate 5*100,004 pound-knots would easily generate 5*200,000 pound-knots in your example if the speed of the current was doubled! No, wrong analysis. If the current speed was doubled to 200,000 knots than the poler could go to sleep (perhaps use his pole as a rudder) and he’d move at 200,000 knots. So the force he would need to push with would be 0 pounds. So his power would be 0 pound-knots (0*200,000). On the other hand, if he wanted to go at 200,000.1 knots, then his power requirement would be f * 200,000.1 pound-knots, where f is the force necessary to punt at 0.1 knots in still water. >>As a thought experiment, suppose my young nephew throws a toy, and applies enough force to throw the toy 10 feet in 6 seconds across a waiting area. Later, while riding in a jet plane going 600 miles an hour, the nephew applies the same force causing the same toy to travel approximately 1 mile and 10 feet down the aisle in 6 seconds. I can tell you that my nephew contributed the same amount of power in both cases, but the jet definitely contributed way more, allowing the second throw to easily win a "Distance in 6 seconds race!" This is irrelevant to the situation that we’re discussing. If instead, your nephew had extremely long legs and could apply the force against the stationary ground (even though he was moving at 600 MPH) then he couldn’t throw the ball much more than 10 feet (perhaps 10.001 feet). You didn’t comment on my Piscataqua River example as I stated it. I was hoping that that example would clarify things for you. You used that example as the jumping off point to form a different example. I chose not to analyze it for the time being because it’s more complicated than necessary. I’m not clear on what our disagreement is about. Apparently, you think that my calculation for power (for the punter) is wrong and is right for the paddler. So tell me what you think the punter’s power is for the original problem. In fact to make things easier and clearer, let’s use another very simple example that’s designed to see if we really have fundamental differences in the punter vs. paddler brainteaser. Say there’s a block of wood on a flat table. [Note, since kinetic friction is independent of speed, the frictional force to slide the block at any speed is the same.] Say this force is F. So the power to slide the block at speed S is F*S. Now suppose the flat table is moving horizontally at the speed St (in the same direction as the block is sliding). For a guy standing on the table, the power to slide the block at speed S relative to the table is still F * S, right? Now suppose someone’s on the ground following the moving table. He also want’s to push the block at speed S, relative to the table. To do this he has to walk at ground speed (S + St), right? What power is required for this guy on the ground? I say the power is F *(S + St). What say you? If you agree with my answers here then I can’t see how you can disagree with my punter vs. paddler analysis. -Leon

Bill, Oops, sorry I called you Gary for some unknown reason last time >> I still maintain you need to credit the punter with the benefit of the current. He does get the advantage of the current, but I haven’t shown it explicitly. Also, the advantage is not as great as it is for the paddler. Let’s compute the punter’s advantage from the 2-knot current right here: As I said earlier in this thread, with the 2 knot current in his favor the punter requires a power of 30 pound-knots to go at 6 knots. Now let’s compute how much power the punter will need to go 6 knots in still water. First we need the drag force at 6 knots. To compute this drag force we make use of the fact that drag is proportional to the square of speed. So the punt’s drag at 6 knots is (6/4) squared times the drag at 4 knots and that computes to 11.25 pounds. So, without the benefit of current, the punter’s required power is 67.5 pound-knots (11.25 * 6). So a 2-knot current will save the punter 37.5 pound-knots. Or putting it another way, it will require 125% more power to go at 6 knots if the punter doesn’t have the advantage of a 2 knot current in his favor. >> Expressed another way, the current provides some of the power to both vessels. Yes, we agree. >> Expressed another way, you can not use one frame of reference (the ground) for one vessel, and another frame of reference (the water) for the other vessel when comparing the power required to win a race relative to the ground. Actually, I compared the power requirement for each to go at the same ground speed with a 2-knot current in their favor. Since the punter requires more power the paddler would win the race. >> I assert power can be defined as how much force it takes to move something over a distance, divided by the time it takes to move that distance. Actually that’s what I used in an equivalent form. Since speed = distance divided by time we have, Force * distance/time = Force * speed. I think what you’re are missing is that the punters drag force is greater than the paddler’s drag force. That’s because drag force comes from speed relative to the water. In the paddler’s case the this speed is just 4 knots (the extra 2 knots come for free) and the punter’s case this speed is 6 knots (he doesn’t get the 2 knots for free, but as I demonstrated above he does get some advantage). Let’s try the simplest example that I can think of: A flat bed truck has a box on top. Case A: The truck is stationary. In case A if you push the box with force, F, for time, T, at speed, S, the ground distance traveled is S*T and the pushing power is F*S*T. Case B: The truck is moving at a speed of C. Doing the same push as in case A the ground distance traveled is (S+C)*T. But the pushing power isn’t F*(S+C)*T. No, it’s just F*S*T as in case A, right? In Case A, you have an earth-fixed coordinate frame and, in Case B, you have a moving coordinate frame. The constant speed of the truck doesn’t change the power needed to move the box, even though in earth coordinates the box moved farther in Case B. I’m not sure, but I think your objection to my original analysis for the punter vs. paddler corresponds to a wrong answer to Case B. >> As I originally posted, the punter must push at Still Speed + Current Speed (S + C) against the ground, which becomes impossible for a human for large values of C. Yes, it’s impossible and the best way to think of it is that the power grows too large. One more example related to the above and I’ll quit for now: A neat principle of moving friction is that it’s independent of speed. Case C: You push a box at speed S1 and the force is, say, F1. Case D: You push a box at speed S1+100,000 and the force is still F1. Now you’d probably say that Case D is unreasonable because nobody can move at speed S1+100,000. Although that’s true, it doesn’t get to the heart of the matter. The reason Case D is unreasonable is that nobody can generate a power of F*(S1+100,000). But, say a huge trunk has enough power to do both cases. It will burn much much more fuel for Case D than for Case C, even though the pushing force (actually from torque) is the same. I gotta quit now. I’ll try to analyze the rest of your last post tomorrow or later. PS I was so burned out from paddling real fast Sunday that I took out the Sunfish Monday for a leisurely sail. Nevertheless, I still got a paddling workout with Mr. Sunfish (the ¼ mile canal where it lives is too narrow to tack through). I need to take down the sail for entering and exiting the canal. Note: That's a $450 wing paddle I'm using in the pix. I now use an old canoe paddle because it stores more easily.

Nate, Yes of course. I went a little overboard. I've practiced with ranges many times. However, I don’t have good luck unless the range points line up directly with my desired COG, especially in rough seas. My point was that most paddlers mistakenly think that a GPS works by lining up your heading with the waypoint so they still need to calculate a ferry angle if they want a COG directly to the waypoint. In effect, the GPS doesn't explicitly use ferry angles. The way it works is similar to using range points that line up directly with the desired COG. If you mount a GPS on the front deck on a foam block sliced at a 45 degree angle I think it’s just as easy to see as a compass. Of course, if you lose the signal or the GPS fails for any reason, you’re out of luck. So everyone should learn the traditional, time-tested navigation methods for a backup. -Leon PS Here’s a novel use of a GPS: One year I entered the Great Stone Dam Classic kayak race on the Merrimack River. Before the starting gun went off everyone was lined up in the middle of the river. Then at the start everyone immediately paddled over to the left bank, apparently, to be in the slower current (the first half of the race was up river). I stayed in the center for a while and then paddled over towards the left bank, but to the right of the other competitors; i.e., I wanted to be in slightly deeper water. It seems there was some optimal point where the increased speed of the slightly deeper water cancelled out the slowdown due to the opposing current. I used my GPS’s speed indicator to help determine this “sweet” spot. I think it helped; I came in first for true sea kayaks and just behind two training surf skis.