Lightweight kayaks do matter for the better

[leong]

To help quantify how increasing weight slows down a kayak use the fact from physics that the momentum of a system must be conserved. During a stroke, the momentum of a kayak and paddler must equal the momentum of the water being pushed back by the paddle.

Momentum equals mass (m) times velocity (v).  Let m be the total of mass of the paddler and boat, v be the velocity of the kayak, M be the mass of the water being pushed back by the paddle and V be the velocity of the volume of water being pushed. Thus the conservation of momentum equation is

m*v = M*V

Solving for v

v = (M*V)/m

Now consider a fixed paddle size and a fixed speed of the paddle moving backward (say at the middle of the stroke). For this case M*V is constant. So increasing m and keeping everything else the same, v decreases; i.e. your kayak doesn’t move as fast. This inverse relationship is more significant than you might think. Even a little extra weight will slow you down.

Example: Assume that the combined weight of the paddler and kayak is 175 pounds. Now add 20 pounds more, so the new total weight, W, is 195 pounds. Note that weight is just mass times acceleration due to gravity (W = m* g). If you do the math this heavier kayak (increased displacement) will go about 10% slower if you paddle exactly the same way. Of course you can use a heavier kayak and/or add ballast and/or go off of your diet to get the increased weight and the resulting speed decrease. Ugh!
 
Note 1. Only when the blade is in the water is the kayak being pulled forward. In between, the kayak is slowing down during a coast. So the momentum argument overestimates the total slowdown during a total stroke cycle.

Note 2. Various articles like this one https://web.archive.org/web/20170313200355/http://roguepaddler.com/weight.htm make two claims arguing against lightweight kayaks:

1. “Although you expend more effort to propel a heavier load, some of that effort is gained back between strokes when the weight may actually help to keep the kayak moving.” Actually this is only partly true because the increased water drag/displacement of a heavier kayak also slows down the coasting between powered strokes. The increased speed of a lighter kayak during the powered phase of the stroke cycle easily outweighs this glide argument.
2.  Heavier kayaks track better. Although this is true (disregarding skegs and rudders) the author’s numerical example of a constant 1-degree tracking penalty of 8 minutes after 8 hours of paddling assumes a constant 1-degree bias. In reality, a kayak that doesn’t track well would tend to zigzag (alternating between 1-degree to the left followed by 1-degree to the right). I computed the actual zigzag penalty in my Mathcad worksheet “Heading Error Penalty”. The correct answer is only 4 seconds, not eight minutes as the author computed for a constant 1-degree bias.

For additional information about how a paddle works see Guillemot Kayaks at,
http://www.guillemot-kayaks.com/guillemot/information/kayak_design/how_a_paddle_works

 

[josko]

Hmmm....

[jason]

When you have 400lbs of kayaker and gear 4LB's of boat one way or another shouldn't make that much of a difference.  

[leong]

2 minutes ago, jason said:

When you have 400lbs of kayaker and gear 4LB's of boat one way or another shouldn't make that much of a difference.  

Yes, when the ratio of the delta weight to the starting weight is small the speed percentage change will be correspondingly small. But surprisingly, going from 400 lbs to 404 lbs slows you down by 1% (at least according to my formula). Nevertheless, I think that 1% seems too much. Also, there is less of a glide slowdown with a real big displacement.

This website http://www.paddlinglight.com/articles/the-lightweight-secret/ estimates the drag increase for increasing displacement weight is 0114% per ounce. That equates to 0.73% per 4 lbs. Note that my formula doesn't account for drag increase. It's just for inertia due to changes in weight.

[Inverseyourself]

On a slightly different note, apart from the lightness of carbon-kevlar boats or hulls, the increased stiffness is (I'll just say probably now) conducive to increased speed as well.

[leong]

On 9/14/2017 at 2:36 PM, Inverseyourself said:

On a slightly different note, apart from the lightness of carbon-kevlar boats or hulls, the increased stiffness is (I'll just say probably now) conducive to increased speed as well.

Stiff boats are faster for pretty much the same reason a well inflated tire is faster. Flex costs energy. But I don't think you'd notice the energy lost on a standard fiberglass boat compared to a stiffer carbon boat. Nevertheless, racers want to save every second.

[leong]

A reader questioned my analysis saying the conservation of momentum argument applies to accelerating a boat from a standstill, not to steady-state boat speed.  

Did you ever notice while kayaking when you start the catch you have to pull hard to accelerate the kayak (the blade sort of stands still in the water and you lever past it)? Remember, between each paddle stroke a kayak slows down (it must slow down because of the water drag). Assume it slows down by 0.1mph (my guess, but use another value if you want to). Assume a paddle is in the water 0.5 seconds (my guess, but use another value if you want to). So the stroke has to increase the kayak’s velocity by 0.1 mph in 0.5 seconds. So the average acceleration during the stroke is

(0.1/0.5)/3600 = 5.6 * 10exp(-5) [mph/sec squared]

Now from a standing start say it takes 15 seconds (my guess, but use another value if you want to) to accelerate the kayak to 5 mph. So the average acceleration at startup is

(5/15)/3600 = 9.3 * 10exp(-5)  [mph/sec squared]

So, using my estimated values, the acceleration while the paddle is in the water is comparable to the startup acceleration.

Although the momentum argument is not the whole story, it demonstrates its own energy requirement due to mass. Obviously, additionally drag from increasing weight also slows a kayak down for different reasons.

-Leon

 

[rfolster]

On 9/13/2017 at 3:56 PM, leong said:

To help quantify how increasing weight slows down a kayak use the fact from physics that the momentum of a system must be conserved. During a stroke, the momentum of a kayak and paddler must equal the momentum of the water being pushed back by the paddle.

Momentum equals mass (m) times velocity (v).  Let m be the total of mass of the paddler and boat, v be the velocity of the kayak, M be the mass of the water being pushed back by the paddle and V be the velocity of the volume of water being pushed. Thus the conservation of momentum equation is

m*v = M*V

Solving for v

v = (M*V)/m

Now consider a fixed paddle size and a fixed speed of the paddle moving backward (say at the middle of the stroke). For this case M*V is constant. So increasing m and keeping everything else the same, v decreases; i.e. your kayak doesn’t move as fast. This inverse relationship is more significant than you might think. Even a little extra weight will slow you down.

I decided to use actual numbers and compare:

Werner Cyprus blade = 610 sq cm = 41 lbs of water (lots of math calculations done, so feel free to double check this)
Blade Velocity = 4 knots (based on average paddling speed instead of using some incomprehensible time vs distance in nanoseconds or something like that)
Total Weight = 245 lbs (200 for paddler, 45 for boat)
v(velocity) = (41*4)/245 = .669388 knots

If we decrease the boat weight by 5 lbs, the results would be:  v(velocity) = (41*4)/240 = .683333 knots.  This difference of 5 pounds in boat weight provided .01395 knots of increased momentum which is only 0.3486% of the overall speed of 4 knots.  If we take Jason's boat on a month-long expedition, we have the following results: v(velocity) = (41*4)/400 = .41 knots, so a 67% increase in total weight yields a 10% reduction of speed from the original 4 knots.  This makes sense to me as I might estimate my speed has been reduced by half a knot between an empty boat on a day trip and a full boat for camping.

 

9 hours ago, leong said:

A reader questioned my analysis saying the conservation of momentum argument applies to accelerating a boat from a standstill, not to steady-state boat speed.  

Did you ever notice while kayaking when you start the catch you have to pull hard to accelerate the kayak (the blade sort of stands still in the water and you lever past it)? Remember, between each paddle stroke a kayak slows down (it must slow down because of the water drag). Assume it slows down by 0.1mph (my guess, but use another value if you want to). Assume a paddle is in the water 0.5 seconds (my guess, but use another value if you want to). So the stroke has to increase the kayak’s velocity by 0.1 mph in 0.5 seconds. So the average acceleration during the stroke is

(0.1/0.5)/3600 = 5.6 * 10exp(-5) [mph/sec squared]

Now from a standing start say it takes 15 seconds (my guess, but use another value if you want to) to accelerate the kayak to 5 mph. So the average acceleration at startup is

(5/15)/3600 = 9.3 * 10exp(-5)  [mph/sec squared]

So, using my estimated values, the acceleration while the paddle is in the water is comparable to the startup acceleration

I think an issue I have with your time assessment is that all other things are to remain equal.  I am specifically referring to your love of power and work.  In order for your formula comparison to be accurate, the same amount of work needs to be performed at each calculation.  If you apply the same amount of work for the steady-state boat to increase the stationary boat by the same 0.1 mph per stroke, it will never increase its speed since you will loose the .1 mph between strokes.  However, even with that aside, if we were to say that the boat could accelerate to 5mph using your formulas, the time it would take to accelerate the stationary boat to 5mph would be more like 50 seconds based on 5mph (final speed) /.1mph (per stroke).  This would change your average acceleration per stroke to 2.78*10exp(-5) [mph/sec squared] to which I contend is not comparable to the acceleration while paddling since it is half the speed.

Then again, I know nothing about physics, so I can only do the actual math.  I will concede that a lighter boat will go some percentage faster than a heavier boat of the same exact size, shape, and materials.  However, once you start adding in other factors such as wind, waves, skegs/rudders, and what you ate for breakfast, I think that the returns diminish and the weight difference must increase.  That is why you see surf skis weighing in around 25lbs and expedition kayaks around 50lbs.  Different tools for different tasks, but 5 lbs either way will not make much difference in the real world.
 

 

[leong]

On 9/13/2017 at 3:56 PM, leong said:

didn't mean this quote. How do you remove it?

Rob,

Your calculation for M (mass of water) and V (blade velocity) is unnecessary if all you want to do is compute the ratio of velocities for the conservation of momentum approach.

To compute the percentage change in velocity for two different boat weights you perform this divide
(MV/245)/(MV/240)

Because MV cancels out the result is

240/245 ~ .98 (a 2% decrease in speed)

But perhaps you’re onto something; i.e. your values of M and V result in a very small kayak velocity (~ 0.67 knots as you calculated). That would mean that the contribution of momentum to accelerate the kayak is trivial compared to water drag.

However, I think the problem is that your value for M (which you express as weight = 41 pounds) is wrong. With each stroke your paddle is imparting momentum to some theoretical volume of water. And if you know what this theoretical volume of water is you can calculate M.  But how do you what is the theoretical volume of water in the conservation of momentum equation? It’s a complex problem in fluid dynamics. But if your estimate of M is reasonable then you’ve shown that drag is the more significant part of the slowdown with increasing weight.

I am aware that the answer I gave is an overestimate of the total percentage of slowdown just due to the momentum argument; that’s because it doesn’t account for the slowdown due to just drag. But the Guillemot Kayaks link I gave does use a conservation of momentum argument, just like I did.

-Leon

 

[leong]

6 hours ago, rfolster said:

 

I think an issue I have with your time assessment is that all other things are to remain equal.  I am specifically referring to your love of power and work.  In order for your formula comparison to be accurate, the same amount of work needs to be performed at each calculation.  If you apply the same amount of work for the steady-state boat to increase the stationary boat by the same 0.1 mph per stroke, it will never increase its speed since you will loose the .1 mph between strokes.  However, even with that aside, if we were to say that the boat could accelerate to 5mph using your formulas, the time it would take to accelerate the stationary boat to 5mph would be more like 50 seconds based on 5mph (final speed) /.1mph (per stroke).  This would change your average acceleration per stroke to 2.78*10exp(-5) [mph/sec squared] to which I contend is not comparable to the acceleration while paddling since it is half the speed.

 

I made up some example numbers for this. My point was to show that while you are paddling, each stroke must accelerate the boat. I included the comment  because someone (outside of this thread) said that, in steady state, you're not accelerating.  I'm not trying to equate the work of startup vs.steady state or even the average acceleration. When I paddle fast it's quite noticeable that I pull very hard at the start of the catch. That results is a short burst of acceleration. But I just (based on my guess of 0.1 mph and 0.5 secs) computed the average acceleration. Nothing more, nothing less.

Edited September 16, 2017 by leong

[leong]

On 9/16/2017 at 6:57 AM, rfolster said:

Blade Velocity = 4 knots (based on average paddling speed instead of using some incomprehensible time vs distance in nanoseconds or something like that)

Rob,

Note that the blade velocity should be with respect to the water, not with respect to the moving boat. If it's with respect to the boat you have to subtract the boat velocity to get the velocity with respect to the water.

[rfolster]

17 hours ago, leong said:

Rob,

Note that the blade velocity should be with respect to the water, not with respect to the moving boat. If it's with respect to the boat you have to subtract the boat velocity to get the velocity with respect to the water.

That was a confusing point for me since I have always understood that there should not be any velocity of the blade.  Isn't the blade to remain stationary with respect to the water and the boat pulled passed it?

[Jim Snyder]

1 hour ago, rfolster said:

That was a confusing point for me since I have always understood that there should not be any velocity of the blade.  Isn't the blade to remain stationary with respect to the water and the boat pulled passed it?

I think that's how we want it to feel but in reality unless the water was in a sense a solid, this is not possible.

[leong]

3 hours ago, Jim Snyder said:

I think that's how we want it to feel but in reality unless the water was in a sense a solid, this is not possible.

Exactly. But a larger blade will move a larger mass of water and thus does not move through the water as fast. From a mechanical efficiency standpoint that is a good thing, because the kinetic energy of the water is lost. However, there is a trade-off. It also reduces your cadence and that may be a bad thing. That’s because, according to Hill’s Equation, there is an optimal cadence where human muscles can put out their maximum power. So that’s one reason people use smaller blade paddles and, perhaps, GPs.

Edited September 18, 2017 by leong

[leong]

6 hours ago, rfolster said:

That was a confusing point for me since I have always understood that there should not be any velocity of the blade.  Isn't the blade to remain stationary with respect to the water and the boat pulled passed it?

That's called locking the blade. It's theoretically not possible but this how-to video shows how to almost do it.

 

[Jim Snyder]

I cant' believe I've gotten sucked into this.

These guys can think they have their paddle "locked in" but it's a fact that the only way the paddle doesn't move with respect to the ground under the water is if the boat is moving forward at precisely the same speed as the paddle is moving backwards relative to the boat. These excellent paddlers approach that as the boat nears it's maximum hull speed but the paddle still has to lose position to the ground slightly to compensate for the very minimal resistance on the boat. Without this compensation the boat would slowly grind to a halt like a pendulum if the paddle never went back faster than the boat went forward.

[mhabich]

In the video the end of shaft/top of blade may be stationary wrt the water, but the blade below the water is moving through the water, since the paddle pivots about the top of blade. 

[leong]

6 hours ago, Jim Snyder said:

Without this compensation the boat would slowly grind to a halt like a pendulum if the paddle never went back faster than the boat went forward.

 

Correct. Obviously the paddle is moving, but not much. A wing paddle locks better than a flat Euro paddle which locks better that a GP paddle.

>> I cant' believe I've gotten sucked into this.

Resistance is futile.

1 hour ago, mhabich said:

In the video the end of shaft/top of blade may be stationary wrt the water, but the blade below the water is moving through the water, since the paddle pivots about the top of blade. 

I think the center of rotation (pivot point) is close to the center of the paddle. Say half of paddle length is 100 cm. Thus the top of the blade is ~ 90 cm from the center of rotation. Thus the bottom of the blade moves ~11% faster than the top of the blade (100/90).

Edited September 20, 2017 by leong

[leong]

1 hour ago, leong said:

Correct. Obviously the paddle is moving, but not much. A wing paddle locks better than a flat Euro paddle which locks better that a GP paddle.

>> I cant' believe I've gotten sucked into this.

Resistance is futile.

I think the center of rotation (pivot point) is close to the center of the paddle. Say half of paddle length is 100 cm. Thus the top of the blade is ~ 90 cm from the center of rotation. Thus the bottom of the blade moves ~11% faster than the top of the blade (100/90).

Correction: I think the center of rotation (pivot point) is close to the center of the paddle. Say half of paddle length is 100 cm. Thus the top of the blade is ~ 80 cm from the center of rotation. Thus the bottom of the blade moves ~25% faster than the top of the blade (100/80).