leong Posted December 13, 2015 Posted December 13, 2015 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 Quote

leong Posted December 13, 2015 Author Posted December 13, 2015 (edited) How did you choose that answer, Katherine? Edited December 13, 2015 by leong Quote

Inverseyourself Posted December 13, 2015 Posted December 13, 2015 Same ground speed on both days...? Quote

JohnHuth Posted December 13, 2015 Posted December 13, 2015 (edited) I'll take a guess - in the 1 kt current. Secondary effect is arms getting tired on the return part of the upper blade pushing against the wind. Edited December 13, 2015 by JohnHuth Quote

leong Posted December 13, 2015 Author Posted December 13, 2015 I'll take a guess - in the 1 kt current. Secondary effect is arms getting tired on the return part of the upper blade pushing against the wind. 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.-LeonPSSpeaking 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. Quote

Lbeale Posted December 13, 2015 Posted December 13, 2015 If I was Marge,I would get behind a bigger and stronger paddler and ride his/hers draft and they can help block the wind -- such a girly thing to do, I know -- Quote

Katherine Posted December 14, 2015 Posted December 14, 2015 Ahh, Leon, the same way I make all decisions :-) Quote

leong Posted December 14, 2015 Author Posted December 14, 2015 Ahh, Leon, the same way I make all decisions :-)If this is a typical guess then your batting average must be pretty high. Quote

leong Posted December 14, 2015 Author Posted December 14, 2015 If I was Marge,I would get behind a bigger and stronger paddler and ride his/hers draft and they can help block the wind -- such a girly thing to do, I know -- 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 Quote

leong Posted December 15, 2015 Author Posted December 15, 2015 I'll take a guess - in the 1 kt current. Secondary effect is arms getting tired on the return part of the upper blade pushing against the wind. 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 poundsDrag 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 poundsD3 (water drag at 3 knots): 2 poundsDW (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 poundsSince power = speed * drag force we have,For 1-knot upcurrent: power = 4-knots * 4-pounds = 16 knot-poundsFor 15-knot headwind: power = 3-knots * 6-pounds = 18 knot-poundsSo, 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. -LeonPSIn 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. Quote

Phil Allen Posted December 18, 2015 Posted December 18, 2015 Morning Leon. Risky saying this, but I think you missed something more obvious than Hills equation. When paddling upwind in calm water, the paddle of your 'good paddler' is fixed in position over the ground with no wind induced movement of the blade. In contrast in moving water the paddle slips over the ground along with the boat. So without increasing the stroke rate or stroke force (bigger blade) there is less effective propulsion in moving water than calm water. bestPhil Quote

leong Posted December 18, 2015 Author Posted December 18, 2015 (edited) 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 Edited December 18, 2015 by leong Quote

JohnHuth Posted December 19, 2015 Posted December 19, 2015 knot-pounds?? First time I've seen those units. What's the speed of light in furlongs per fortnight?What is "today drag'?OK, so waves, yup, that'll change things. Quote

leong Posted December 20, 2015 Author Posted December 20, 2015 (edited) knot-pounds?? First time I've seen those units. What's the speed of light in furlongs per fortnight?What is "today drag'?OK, so waves, yup, that'll change things. 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.-LeonPS knot-pounds are my favorite power units when working with kayaks and drag. You want hp, then just multiple by ~ 0.00307186,000mps = 1,799,885,057,678.61 Furlongs Per Fortnight Edited December 20, 2015 by leong Quote

Inverseyourself Posted December 20, 2015 Posted December 20, 2015 What's wrong with FPF ? I've been using this to express speed all my life. Glad there's finally someone else who does. Quote

leong Posted December 24, 2015 Author Posted December 24, 2015 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 Quote

JohnHuth Posted December 25, 2015 Posted December 25, 2015 I haven't had a chance to analyze it. One question is whether you're double counting things that go into Burch's estimate, or, put a different way, what goes into Burch's estimate? For example, if resistance due to encountering waves is a factor, might that not already be present in what Burch has? I *will* get around to checking it out, but the Holiday Season has me hopping. Quote

leong Posted December 26, 2015 Author Posted December 26, 2015 I haven't had a chance to analyze it. One question is whether you're double counting things that go into Burch's estimate, or, put a different way, what goes into Burch's estimate? For example, if resistance due to encountering waves is a factor, might that not already be present in what Burch has? I *will* get around to checking it out, but the Holiday Season has me hopping.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_cadence4 = k * upcurrent-cadence.This results in:upwind-cadence = ¾ * upcurrent_cadenceBut 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.-LeonPSWay 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. Quote

leong Posted December 26, 2015 Author Posted December 26, 2015 (edited) 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_cadence4 = k * upcurrent-cadence.This results in:upwind-cadence = ¾ * upcurrent_cadenceBut 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.-LeonPSWay 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.PPSBelow 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 Edited December 26, 2015 by leong Quote

JohnHuth Posted December 29, 2015 Posted December 29, 2015 I have to understand Hill's equation better - I see you have some links earlier in the posting. You'll have to give me some time to digest it. I *am* a biophysics fan, but I'm not as familiar with his work. On general grounds, I switched to a shorter paddle because I 'test drove' one and for some reason the increased cadence and ease of draw through the water felt like I could maintain a certain speed longer. Not very scientific, I realize, but it seemed to work for me. Quote

leong Posted December 30, 2015 Author Posted December 30, 2015 On general grounds, I switched to a shorter paddle because I 'test drove' one and for some reason the increased cadence and ease of draw through the water felt like I could maintain a certain speed longer. Not very scientific, I realize, but it seemed to work for me. 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 Quote

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.