Kayak Resistance and Speed

[leong]

Say that in an all-out sprint you can get your slowpoke kayak up to 5 knots. According to Sea Kayaker Magazine’s resistance curves, the total drag for your kayak at that speed is 4.213 pounds. You’re thinking of buying a fast new kayak, so you look up the speed vs. drag data for that faster kayak. Here it is (speed in knots -> total drag in pounds):

Spd -> drag

5.0 –> 3.312

5.5 –> 3.830

5.7 –> 3.914

6.0 –> 4.213

Notice that the drag at 6 knots for the faster kayak is the same as the drag for your current kayak at 5 knots.

Question: How fast will you be able to sprint with the faster kayak?

Be careful, think about your answer.

[cfolster]

Leon,

Since I'm bored and you put this out there for all to answer, I'm going to attempt it simply using logic.

You will be able to sprint at 6 knots in the new kayak, because if you were able to overcome the drag in the old kayak at get it up to 5 knots, you will be able to overcome the same amount of drag in the new kayak and get it to it's max speed of 6 knots.

Now I'll let those who are better at math and have actually done velocity calculations answer correctly.

Cathy

[rfolster]

This type of question always frustrates me since there is not enough information to solve the problem. What you need to know is how much force is being asserted to make the boat move through the water. The amount of force should vary in direct proportion to the speed of the boat. More force = faster speed. You should be able to find a correlation between the amount of force applied and the speed of a given boat. However, in this instance you can not solve for a direct correlation between force, speed, and drag since the numbers are not consistent across the data range you provided. If we divide the drag by the speed, we get the amount of force needed to achieve each speed. Here is the math:

Old Boat:

4.213(pounds) / 5.5(knots) = .8425(force)

New Boat:

3.312(pounds) / 5.0(knots) = .6624(force)

3.830(pounds) / 5.5(knots) = .6964(force)

3.914(pounds) / 5.7(knots) = .6867(force)

4.213(pounds) / 6.0(knots) = .7022(force)

Now I would have said that you just need to find out the ratio between the amount of force applied to the amount of resistance of the new boat to find the corresponding speed. If you then applied the old boat force, you would get the new boat speed. However, this is not possible since the new boat specifications say it takes less force to travel 5.7 knots than it did at 5.5 knots, which should not be possible. The lack of consistent information makes this mathematically impossible to solve.

Therefore, Cathy's calculation is correct based on the real-world expectation that approximately the same amount of force is being applied to the same resistance. Since the same unknown amount of force is applied to the same drag (4.213), then the only variable is the speed, which would end up being the calculated 6 knots.

[leong]

Leon,

Since I'm bored and you put this out there for all to answer, I'm going to attempt it simply using logic.

You will be able to sprint at 6 knots in the new kayak, because if you were able to overcome the drag in the old kayak at get it up to 5 knots, you will be able to overcome the same amount of drag in the new kayak and get it to it's max speed of 6 knots.

Now I'll let those who are better at math and have actually done velocity calculations answer correctly.

Cathy

Oh, Cathy, you got that right (the part that I'm quite boring). For the time being I won’t reveal the right answer to the question, though.

[leong]

This type of question always frustrates me since there is not enough information to solve the problem. What you need to know is how much force is being asserted to make the boat move through the water.

The drag is the force and you have enough information to solve the problem. I'll ignore your analysis for the time being to allow you the time to reconsider.

[djlewis]

Hey, Leon, I'll bet you a hundred dollars...

[leong]

Hey, Leon, I'll bet you a hundred dollars...

Hard to believe you're willing to lose again!

[dan_f]

Hint - assume that paddler output power is constant. Power = Force*velocity. Leon told you force=drag.

[cfolster]

Oh, Cathy, you got that right (the part that I'm quite boring). For the time being I won’t reveal the right answer to the question, though.

Never boring Leon! You do the most to keep this board interesting. I'm waiting on pins and needles to hear the answer, but I also wonder why more folks are not attempting to answer . . . this is fun!

[dan_f]

Another hint - the amount of energy that the paddler supplies is: Energy=Force*distance. The amount of energy that the paddler supplies to go a certain distance

against a constant drag force is independent of the amount of time that the paddler takes to go that distance, However, we all know that it matters if we supply that energy

over 10 minutes or 20 minutes, so just matching the drag forces on the two boats and reading off the corresponding speeds is not the right answer. We must consider power,

the rate of energy expenditure per unit time. If you were confused by my equation above, speed and velocity are equivalent here.

[leong]

Everything that Dan_f said is exactly right and is more than enough to answer the question. However, I’ll give one more hint. In an all-out sprint with the slowpoke kayak you are using all of your available power; i.e. it’s your maximum effort. Your maximum effort (power) with the fast boat will be the same, right? Nuff said. Now the answer should be obvious.

[djlewis]

Hard to believe you're willing to lose again!

Oh, so you accept the bet, even before you know what we're betting on?! In that case, make it $10,000 -- like a recent losing presidential candidate. You too can be a loser!

Shake?

[leong]

Never boring Leon! You do the most to keep this board interesting. I'm waiting on pins and needles to hear the answer, but I also wonder why more folks are not attempting to answer . . . this is fun!

Thanks, Cathy. One last hint:

Suppose you and your identical twin were having a race, you in the slow kayak and your twin in the fast kayak. According to the stated original question, you’d be going at 5 knots (your all-out speed). Do you think your twin could paddle 6 knots in the fast kayak? But at 6 knots, she would be going 20% faster than you were going. Given that you both would be overcoming the exact same drag, wouldn’t that mean that she was working 20% harder than you were?

With all the hints you might know the right answer now. So what is it? Like you, I’m waiting on pins and needles to hear your answer.

[dan_f]

Well Leon, I'm afraid that we've both (once again?) overestimated general interest in physics problems. Your paddler can paddle at 5 knots with 4.213 lbs (force) of drag, producing a power output of 21.065 lbs-knots. With the same power in the new boat the paddler can go 5.5 knots against 3.830 lbs of drag. It might be fun to convert this strange unit of power to one that is more familiar: 21.065 lbs-knots =48.7 watts if I haven't goofed. This isn't too far off what I might expect since on a bicyle we can generate about 125 W. In reality a strong paddler can probably produce double or triple the power output of the hypothetical paddler in Leon's example.

[BillyD]

Dear Dan.....do not underestimate the interest (over 300 reads at this posting).....some, like myself are reluctant to post, not being fluent in the physics parlance.

But, I'm thinking I can follow this string. and concur with the apparent conclusion. Thanks to Leon for the brain teaser and all resonders for joining in the fun.

One request, however......Dan, is it possible to convert that "strange unit" into Henways instead of Watts?

[leong]

is it possible to convert that "strange unit" into Henways instead of Watts?

Only these dudes http://kenhardley.com/henways/ would know how to convert to Henways. But in horsepower units the power is roughly 0.065 horsepower.

By the way, my post was meant more as a practical application of kayak speed/resistance curves than as a physics lesson or brainteaser. I’ve noticed that some fairly sophisticated kayaking articles erroneously equate drag to problems like the one that I posted nearby.

The most obvious use of equating power for kayaks is when (everything else being equal) you want to choose the kayak that, for your particular power level, is the fastest for you. If they’d listen to me, Sea Kayaker Magazine would provide speed/power curves instead of speed/drag curves, then everyone would get it “right”.

Leon

PS

I think Dan_f probably meant about 48.5 watts (not 48.7 watts), not that it matters.

[BillyD]

Thanks again Leon...Seriously.....my attempt to inject humor is in no way meant to diminish the practical value of your post.... I suspect I speak for many when I say that my interest in kayaking has been greatly enriched by the conversations you get started on the NSPN board. I can hardley wait for the next one

BD

[dan_f]

Hi Leon, I should have rounded to the nearest watt since my conversion factors were rounded off. My question for you is how much sustained power a strong kayak racer actually produces?

[rfolster]

My question for you is how much sustained power a strong kayak racer actually produces?

1.21 gigawatts!

[leong]

Hi Leon, I should have rounded to the nearest watt since my conversion factors were rounded off. My question for you is how much sustained power a strong kayak racer actually produces?

Interesting question. I looked up the Blackburn Challenge (16 nautical mile, mostly offshore race) times for the two fastest finishers ever using the Epic 18X kayak: 1) Olympian Greg Barton and 2) the best-ever “normal” paddler. I entered their respective average speeds into the John Winters performance prediction spreadsheet KAPER (maintained by Lisa Huntington). Here are the results:

Barton finishing in ~2.5 hours (avg. speed = 6.4 knots): 0.201 HP (~ 150 watts)

Best-ever “normal” paddler finishing in ~3.0 hours (avg. speed = 5.33 knots): 0.101 HP (~ 75 watts)

It’s interesting to note that just increasing the speed from 5.33 knots to 6.4 knots (about 1.07 knots) requires a doubling of power! It’s quite amazing that Barton can maintain double the power for 2.5 hours, especially since this “normal” paddler usually wins every elite race that he enters, unless, of course, if Barton is competing in the same division.

Obviously, the power output for short sprints is significantly higher (perhaps double?).

[rfolster]

The drag is the force and you have enough information to solve the problem. I'll ignore your analysis for the time being to allow you the time to reconsider.

OK Leon - so I will admit some stupid mistakes on my part in that I should have multiplied instead of divided (I don't know if that would have helped me) and I used the word force in stead of power. But now let's consider the analysis of the actual data. Were they just made up for this exercise or did you find them for a particular boat?

[leong]

OK Leon - so I will admit some stupid mistakes on my part in that I should have multiplied instead of divided (I don't know if that would have helped me) and I used the word force in stead of power. But now let's consider the analysis of the actual data. Were they just made up for this exercise or did you find them for a particular boat?

The values were made up for the exercise. I used realistic drag values for 5 knots. Then I chose the other drag values so that the drag of the fast kayak at 6 knots would match the drag of the slow kayak at 5 knots. Other than at 5 knots, the drag values are unrealistic.

[dan_f]

This short article plots a number of kayak drag force curves and shows that many boats have nearly identical drag forces at speeds less than 4 knots:

http://www.seakayakermag.com/2007/07e-newsletters/December/fastkayak.htm

[EEL]

shows that many boats have nearly identical drag forces at speeds less than 4 knots:

http://www.seakayake...r/fastkayak.htm

Boats which are close in surface area will also be very close in frictional resistance up to 4 knots. Surface area is a major determinant of frictional resistance. Wave resistance, which is a major factor above 5 knots, is a minor contributor to total drag at 4 knots and below which in turn means hull shape is not much of a factor at 4 knots and below. Boats with comparatively low total drag at speed above 5 knots can have higher total drag numbers at 4 Knots and below than boats which have high total drag numbers above 5 knots.

Ed Lawson

[leong]

Boats which are close in surface area will also be very close in frictional resistance up to 4 knots. Surface area is a major determinant of frictional resistance. Wave resistance, which is a major factor above 5 knots, is a minor contributor to total drag at 4 knots and below which in turn means hull shape is not much of a factor at 4 knots and below. Boats with comparatively low total drag at speed above 5 knots can have higher total drag numbers at 4 Knots and below than boats which have high total drag numbers above 5 knots.

Ed Lawson

Yes, you’re right. Also, one tradeoff in kayak design usually not discussed is the “effective waterline length”. Traditional kayaks tend to have sharp vee bows that reduce the effective waterline length. Given two kayaks of equal waterline lengths (and most everything else the same) the one with the higher effective waterline length will have less wave resistance, but at a cost of higher frictional resistance. That’s why racing kayaks usually have higher volume bows below the waterline; i.e. the extra volume increases the effective waterline length and thus reduces wave resistance. Racers don’t care about the slightly increased frictional resistance of full bows since they’re paddling at speeds way above 4 knots.