Punter vs. Paddler

[leong]

I just read this article about punting in Cambridge, England. It begged for the following brainteaser:

Assume a punter and a kayaker practice racing in still water and it always results in a tie. Then they enter a down-river race. Will one of them now win and why?

-Leon

Note: Punting refers to boating in a punt. The punter propels the punt by pushing against the riverbed with a pole. A punt should not be confused with a gondola that is propelled by an oar rather than a pole.

[Inverseyourself]

The Punter's pushing off the ground is rendered less effective by the increased speed over ground. The ground that the punter pushes off of essentially "disappears" under the pole. This results in less forward speed relative to the moving water surface compared to the still water race. The kayaker, on the other hand, maintains the same speed relative to the water surface of the moving river; it was enough for a tie on still water but now is enough for a win on moving water.

Edited November 27, 2015 by Inverseyourself

[Pintail]

I don't understand or follow your reasoning at all, Inverseyourself...apologies! In any case, the River Cam is <very> languid and slow-moving, due to the flatness of that part of the country -- I know it, I assure you. Does it have any bearing on the question? I don't know, honestly...it will take a Leon to explain it to me! (And you always know when he has time on his hands: his academic questions and hypotheses abound -- in winter-time!)

;^)

Edited November 28, 2015 by Pintail

[billvoss]

The punter and the kayaker always tie on still water because they are friends, and neither wants to risk their friendship by beating the other. That will not change in current.

However, if they have a falling out, and physics controls the outcome...

In still water both achieve speed S relative to the ground, and relative to the water.

In a down river race we also have current speed C.

Current speed will actually vary being slower near the river bed.

Neglecting minor affects such as air resistance, the kayaker should achieve speed S + C in a down river race.

The punter previously pushed at speed S against the ground not the current. As current moves the punt down river the ground moves the wrong direction like a treadmill. So to tie the kayaker the punter must now push at speed S + C against the ground. As current speed C increases, this becomes more and more challenging and eventually for high values of C becomes impossible for human muscles. The water resistance against the punt required to tie the kayaker has not changed. However, the water resistance and inertial effects on the pole itself will change during different poling stages as the pole is forced to move faster than was required in still water, and different parts of the pole move through water moving at slightly different speeds relative to the ground.

Pretending to be a physicist, my bottom line is that though both will make better speed over ground as C increases, the kayaker will benefit more as C increases. For small values of C the race will be very close.

As a white water kayaker my answer is that as C increases, the river becomes more turbulent eventually capsizing both craft. The punter swims; while the kayaker rolls up and wins!!!!

[Inverseyourself]

Exactly, almost to the word, what I said, Bill

[billvoss]

Exactly, almost to the word, what I said, Bill

Of course, I looked at your answer, then jazzed it up to make it look like you had copied off my answer.

[leong]

Okay, both Andy and Bill are quite right.

A thought experiment for Sir Christopher to help him get it. Suppose you were climbing up a ladder to the sky at, say, 2 feet per second. Then the ladder began to slip down into the earth at, say, 2 feet per second. So now you’ll have to climb the rungs at 2 feet per second just to stay at the same position relative to the ground. To continue moving “up” at 2 feet per second you’ll have to climb the rungs at 4 feet per second. That will require a lot more power from you, right.

Here’s a slightly more physics-like explanation using power of why the kayaker will do better in the downriver race. (Note that power = force * speed). For simplicity, let’s use these numbers:

4 knots – Still-water racing speed for both the kayaker and punter
5 pounds - Drag force of water to go 4 knots faster than the water (assume it’s the same for the kayak and punt boat)
2 knots – speed of current


20 pound-knots will be the power required by both the paddler and punter to go at 4 knots in the still water (5 pounds * 4 knots). Going downriver, the kayaker will be moving at 6 knots (2 free knots due to the current + 4 knots provided by his paddling). So the kayaker’s required power (force * speed) is still 20 pound-knots (5 pounds * 4 knots relative to the river). But for the punter to move at 6 knots downriver, his power requirement would be 30 pound-knots (5 pounds [to increase his ground speed from the free 2 knots to 6 knots] * 6 knots (the speed of his pole against the non-moving river bed). So, in this example the punter would require 50% more power (30 vs 20) to keep up with the kayaker. So, obviously the kayaker will win the downriver race.

-Leon

PS
Brainteaser part 2: How about an upriver race?

PPS
Hmm, I gotta make a decision fast. Should I take out the kayak or the Sunfish? Retirement sure has its hard-to-make decisions. <Sir Christopher, is this a correct hyphenation?>

[Inverseyourself]

Upriver: Again the Kayaker. His cadence is just higher than the Punter's could ever be. What counts upriver is cadence. If your cadence is very low, you loose ground whenever your paddle blade does not "grip" water. The Punter looses ground whenever his pole does not touch the bottom and he takes more time pulling the pole up, then pushing it down to the river bottom. The kayaker looses a little bit of ground as well between paddle blade entries but the transition left-right or right left is much faster. The higher his cadence, the less ground he looses.

[leong]

Andy,

Good analysis. That’s an interesting take on the problem. I don’t think “cadence” is the proper term, however. Cadence is the number of strokes per unit time. You’re talking about only shortening the time interval between the left blade exiting and the right blade entering. For a given paddle length, size and hand position, I think cadence is proportional to boat speed. That is, you can’t increase cadence without increasing boat speed. But an increase in power is necessary to increase boat speed (I’m disregarding stroke form and ergonomics).

Anyway, I get your point about back drift. You're saying that back drift is less for a paddler because his left to right transition is probably faster than a punter’s pulling the pole up, then pushing it down. That certainly adds to the complexity of the problem. Nevertheless, I believe that the punter has the advantage going upcurrent. Here’s a thought experiment showing his advantage for a “stressing” set of conditions:

Say the maximum still-water speeds are equal for the punter and the paddler. Say 5 knots. Next suppose the current speed is 6 knots, in the opposite direction. For this case, the paddler can make no headway, in fact he drifts backwards at a speed of 1 knot. Now when the punter’s pole is planted at the bottom of the river, he can stop any backward motion. So, at least for this instant in time, the punter is going faster than the paddler (0 knots > -1 knots). Without getting into a complicated power analysis (it would be complicated because of the discontinuity of the drive force for each), I think the punter has the advantage going upcurrent.

An easy power argument could be made using the following simplified thought experiment: Think of a amphibious vehicle with two drive modes: wheels on the bottom (analogous to the punter) and a propeller (analogous to the paddler). You could show that for traveling upcurrent the propeller mode requires more power than the wheel mode for a given speed. Of course, I’ve left out the complications of the rolling friction at the bottom of the river and the efficiency of the propeller.

Although I think the punter has the advantage going upcurrent, I’m not certain that you are wrong. A more definite answer requires a very complicated analysis, more than I’m willing to tackle.

What do you think now?

-Leon

PS
A real-life analogy from today: I was paddling against a strong wind (about 25 knots) and a strong tidal current (about 2 knots). The GPS said that my back drift was over 3 knots when I stopped paddling! After a half-hour of very hard paddling I wished that I had a long enough paddle (or pole) to plant it in the bottom of the bay to hold my position so I could catch my breath. Eventually, I grabbed onto a lobster pot (or whatever it’s called in Florida) to hang on to rest.

Edited November 29, 2015 by leong

[Inverseyourself]

Hi Leon:

The problem I see with the "punter-wheels-on-bottom" analogy is that the punter loses contact with the bottom when pulling the pole up to plant it at the next upriver position. On the other hand, if he changes position really fast by essentially inching or 10-inching up the river, he could possibly reach the same "do whatever advances you"-frequency as the paddler or an even higher one and lose minimal ground in between "strokes". He would have the added benefit of using his whole body weight on the one pole to move forward, whereas the paddler only uses (the muscles on) one side of his body. Hmmm...not so sure who's going to win anymore.

[leong]

Methinks (perhaps) there’s no single answer to who wins the upcurrent race, the punter or the paddler. It might depend on the still-water speed of the paddler (punter), the current speed, the water depth, etc.

PS

Lest the reader thinks this brainteaser has no practical value consider canoe poling. Canoe poling may be a lost art, but there are certain advantages to having the skill (and the pole). As they say, “You can easily push your boat upstream and then steer down.”

Here are a few articles about canoe poling (hmm, the video below begs for some new stupid kayak tricks at Walden Pond ... are you listening Les?):

http://nwwoodsman.com/Articles/CanoePoling.html
http://www.outdoors.org/publications/outdoors/2008/how-to-canoe-pole.cfm

Respectfully,

-Leon

Edited November 29, 2015 by leong

[billvoss]

Okay, both Andy and Bill are quite right.

5 pounds - Drag force of water to go 4 knots faster than the water (assume it’s the same for the kayak and punt boat)

But for the punter to move at 6 knots downriver, his power requirement would be 30 pound-knots (5 pounds [to increase his ground speed from the free 2 knots to 6 knots] * 6 knots (the speed of his pole against the non-moving river bed). So, in this example the punter would require 50% more power (30 vs 20) to keep up with the kayaker.

PS

Brainteaser part 2: How about an upriver race?

Perhaps you should have stopped with "Andy and Bill are quite right" Leong. I don't think your formula applies.

By the way, my first read of your description was that 5 pounds of drag existed when traveling at 4 knots. Not that 5 pounds * 4 knots was the appropriate formula, since we all know that drag increases exponentially, not linearly as speed increases.

However, back to my main objection. The punt's drag is predominantly a function of the punt against the water, not a function of the punter's pole. Just because the punter is pushing on the ground does not negate the benefit of the current. So you still need to credit the punter with 5 pounds times 2 knots in your (power = force * speed) formula example. If the punter requires exactly the same amount of power to match the kayaker on still water a 4 knots, the punter does not require 50% more power to match the kayaker's speed when both have a 2 knot current assist.

Consider this mind experiment. Old manually powered ferries hauled on a stationary cable to move the water craft. Assume the ferry and the kayak achieve the same speed in races over still water. Now repeat the race going down current. If the ferryman is manually pulling directly on the cable, the situation is almost equivalent to the punt. However, equip the ferry with gears like a bicycle, and assume the friction losses from the gearing are the same in current as in still water. Then I believe the ferryman and the kayak will still be essentially tied going downstream. The difference is that now the gearing can be adjusted so that the ferryman uses the exact same aerobic output at the exact same cadence in any reasonable current. So the ferryman does not suffer from current moving faster than the punter can pole.

Regarding going upstream. The NH AMC White Water group includes some canoe poling enthusiasts. Though I have not poled a canoe, my understanding is that the huge advantage of poling is that it is practical to pole upstream in current too fast to paddle upstream. I believe a combination of poling upstream and paddling downstream was used by fur traders in America before motorized transport was invented. So you are on the right track there Leong. I also agree with your analysis that if the speed of the current exceeds the vessel's still water race speed, the poler will certainly beat the paddler, since the paddler is washed downstream.

[leong]

Perhaps you should have stopped with "Andy and Bill are quite right" Leong. I don't think your formula applies.

By the way, my first read of your description was that 5 pounds of drag existed when traveling at 4 knots. Not that 5 pounds * 4 knots was the appropriate formula, since we all know that drag increases exponentially, not linearly as speed increases.

However, back to my main objection. The punt's drag is predominantly a function of the punt against the water, not a function of the punter's pole. Just because the punter is pushing on the ground does not negate the benefit of the current. So you still need to credit the punter with 5 pounds times 2 knots in your (power = force * speed) formula example. If the punter requires exactly the same amount of power to match the kayaker on still water a 4 knots, the punter does not require 50% more power to match the kayaker's speed when both have a 2 knot current assist.

Consider this mind experiment. Old manually powered ferries hauled on a stationary cable to move the water craft. Assume the ferry and the kayak achieve the same speed in races over still water. Now repeat the race going down current. If the ferryman is manually pulling directly on the cable, the situation is almost equivalent to the punt. However, equip the ferry with gears like a bicycle, and assume the friction losses from the gearing are the same in current as in still water. Then I believe the ferryman and the kayak will still be essentially tied going downstream. The difference is that now the gearing can be adjusted so that the ferryman uses the exact same aerobic output at the exact same cadence in any reasonable current. So the ferryman does not suffer from current moving faster than the punter can pole.

Regarding going upstream. The NH AMC White Water group includes some canoe poling enthusiasts. Though I have not poled a canoe, my understanding is that the huge advantage of poling is that it is practical to pole upstream in current too fast to paddle upstream. I believe a combination of poling upstream and paddling downstream was used by fur traders in America before motorized transport was invented. So you are on the right track there Leong. I also agree with your analysis that if the speed of the current exceeds the vessel's still water race speed, the poler will certainly beat the paddler, since the paddler is washed downstream.

>>By the way, my first read of your description was that 5 pounds of drag existed when traveling at 4 knots.

Yes, that’s true and consistent with what I said.

>>Not that 5 pounds * 4 knots was the appropriate formula, since we all know that drag increases exponentially, not linearly as speed increases.

I assume that you’re talking about the case where the kayak is going at 6 knots downriver. But his speed relative to the river is still 4 knots. So there is no drag increase, it’s still 5 pounds. Non linear drag increase has nothing to do with the problem here, since there’s no increase in “water” speed (although an increase in “speed over ground”).

>>The punt's drag is predominantly a function of the punt against the water, not a function of the punter's pole.

Yes, of course. And I used that fact.

>>Just because the punter is pushing on the ground does not negate the benefit of the current.

You don’t understand that the force in the power equation is relative to what’s being pushed against. The paddler’s force is against the moving water so the speed of the push is still 4 knots. But for the punter, he’s pushing against the fixed earth, so the speed of his push is the full 6 knots of “speed over ground”.

Consider this thought experiment: You paddle and/or punt at 4 knots in still water requiring a power of 20 pound-knots, assuming the 4 knot water drag is 5 pounds. Now suppose the current is 100,000 knots (neglect turbulence, air drag, etc.)! The paddler will have no trouble paddling at a speed of 100,004 knots (speed-over-ground) going downcurrent, right? But do you really think that the poler can also increase his speed from 100,000 knots to 100,004 knots? The power required to do that would be 5*100,004 pound-knots!

Note that in the first case (paddler) the force is with respect to a moving reference frame (the moving water with respect to the earth). In the second case (punt boat) the force is with respect to a fixed reference frame (mother earth). This applies to all physical systems, whether they’re kayaks punt boats, ferries, etc. Sir Isaac Newton was aware of this.

So, no, I shouldn’t have quit while you mistakenly thought that my detailed analysis was wrong. It’s quite correct.

No biggie, though. It’s a subtle topic.

Peace,

-Leon

Gotta go finish lunch and hit the swells soon. I’ll respond tonight if necessary,

[leong]

Gary,

I just got back from a play and began to think about why you think that my answer is wrong. Perhaps it’s because you are not familiar with the jargon I used when moving reference frames are needed. So I’m going to try and convince you that you’re wrong another way. This time by using a different version of your ferry mind experiment. So here goes:

A small ferry is on the Piscataqua River drifting down current at 3 knots. The ferry captain wants to increase his speed to 4 knots. Down stream in front of the ferry is the Piscataqua River Bridge with a supertanker under the bridge. The supertanker’s engines are off; i.e. it’s just drifting down stream at the speed of the current. The ferry captain desires to increase his speed by 1 knot to reach 4 knots. Two ropes are available for him to pull in 1. A rope attached to the supertanker. 2. A rope attached to the Piscataqua River Bridge. Assume it will take of force of 50 pounds of tension on the rope (the drag force at 1 knot) to gain the additional 1-knot.

Case 1. The ferry boat captain (FBC) pulls in the rope attached to the supertanker until it’s just taut (0 tension neglecting the weight of the rope). Since the supertanker and ferry are moving with the current at 3 knots, the FBC will have to pull the rope closer to him at a speed of 1 knot to bring the ferry’s speed up to 4 knots. When he does this the tension on the rope will increase to 50 pounds (exactly the drag force of moving against the river at 1 knot). So the power requirement is 50 pound-knots (50 pounds times 1 knot).

Case 2. Instead, FBC pulls in the rope attached to the bridge until it’s momentarily taut (0 tension if you neglect the weight of the rope). Of course he has to continue pulling in the rope at a speed of 3 knots to just keep it taut, but still 0 tension. But since the bridge is stationary, and the FBC wants the ferry to move at 4 knots, he will have to pull the rope in even faster, to at rope speed 4 knots. As in case 1, the tension of the rope will increase to 50 pounds (exactly the drag force of moving against the river at 1 knot). So, in case 2, the power requirement is 200 pound-knots (50 pounds times 4 knots).

I hope this will convince you.

Respectfully,

-Leon

[billvoss]

My turn Leon,

I still maintain you need to credit the punter with the benefit of the current.

Expressed another way, the current provides some of the power to both vessels.

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.

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. You can use any frame of reference you want for distance, so long as you use the same frame of reference for both vessels.

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. However, it is impossible because the punter must push the POLE so fast. The punter must only accelerate the punt to S relative to the current for all values of C to tie the paddler.

Now suppose the current is 100,000 knots (neglect turbulence, air drag, etc.)! The paddler will have no trouble paddling at a speed of 100,004 knots (speed-over-ground) going downcurrent, right? But do you really think that the poler can also increase his speed from 100,000 knots to 100,004 knots? The power required to do that would be 5*100,004 pound-knots!

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!

In your Piscataqua River example, put the two ropes into tension so they act like beams. The supertanker's rope is now tied to a second large ship pulling on the rope so it overtakes the supertanker extremely slowly. The bridge is now tied to a stationary anchor. Your ferry attaches an electric motor to the stationary rope and the motor starts spinning and generating electricity. You have in essence a short duration hydro-electric plant! That electricity is because of the power contributed by the current. Disconnect from the stationary rope and connect to the supertanker's rope, the motor does not spin. Now apply a current so that the motor pulls the ferry down the supertanker's rope at 1 knot relative to the rope (4 knots relative to the ground) towards the supertanker. Note how many watts are required to achieve that speed, and the motor's RPMs. Disconnect from the supertanker rope, adjust the motor's gearing so the same motor RPM would pull four times as fast as before. Apply the same wattage as before with the motor connected to the stationary rope. I believe you will find that the motor achieves the same RPM and the ferry achieves the same 4 knot speed relative to the ground as before. The power (watts) required is the same in both cases to accelerate the ferry to 1 knot relative to the current.

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!"

I hope this will convince you.

Respectfully,

-Bill

P.S.

I'm very jealous of your play between posts. Between health issues and contractor hell, I haven't made it out on the water since September 7th, though I'm looking forward to a pool session this Saturday on December 5th.

[leong]

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.

Edited December 1, 2015 by leong

[billvoss]

Hi Leon,

I think we may both remain unconvinced on precisely how to define the Power required for fast current, though we both seem to reach the same conclusion in everything but magnitude.

Last night as I tried to fall asleep, I thought about the results for slow current. Would the paddler always win if the current was a fraction of a knot?

One subtle advantage the paddler has over the poler is the paddler can select from a wide range of paddles. The largest wing paddle money can buy is certainly not always the best choice for all possible race distances and paddlers. Just like the most aggressive gear on a bicycle is not always the most efficient choice.

The poler is normally in the situation of riding a fixed speed bike, with no option to change the gearing, even in the shop. If that fixed gear works well for that poler in that race, great. Otherwise, too bad. However, if the poler could pick the current for the race, the poler would be in the same situation as a fixed speed cyclist who could pick the grade for a race, or chance his cycle's gearing in the shop.

The poler and paddler always tied in a zero current race. I assume the paddler is already always selecting the best paddle for each race. If zero current was a harder than optimal gear for the poler, then by selecting the right current, the poler might be able to improve their bio-mechanics enough to win!

Unfortunately, I don't have enough experience poling to know if zero current poling is ever a harder than optimal gear for a poler.

Enjoy another day in the sun Leon.

-Bill

[leong]

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

[leong]

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