Going with the recent “Long or Short” thread I have been thinking for some time that it would be interesting to apply a fish scale to various kayaks to measure the actual loaded drag in a uniform river current. I bought a small digital scale and will give it a try sometime this Spring – but has anyone already tried this? What are the typical results? My thought is to pull the scale out when the group has stopped for lunch at a spot with a decent current that is shallow enough for me to wade out and stand there with a short tow rope attached to the scale.
I’ve towed them behind boats
Back in school we towed some hulls behind boats and measured them in flat water condition you can get pretty reliable results.
Load needed
– Last Updated: Mar-13-10 6:35 PM EST –
That should work OK, as that's essentially what happens in tow tank or water channel testing. The boat should be loaded so that its draft is correct - you just need someone to sit in it while you do it. To be comparable, tests from different boats should have the same load.
Meanwhile, for the results to be useful, you have to measure the current at the same time - this is less straightforward. There are little propeller flow meters that would work, but I'm not sure where you get one or how much they cost. A small one like a pocket anemometer would be ideal.
Eric Nyre has a coasting test routine
that he says gives useful results. Maybe he’ll notice this thread.
looking at those #‘s in greyhawks’ link
… it seems the faster the hauls go through the water the greater the resistence becomes … that looks to be consistent with all the boats listed .
So if one wishes to reduce resisitence (total haul drag in lbs. of opposing force) , one should be trying paddle the boat as slow as possible ??
Even the most inefficient high volumn barge type hauls should be more effecient going slower (resistence wise) , than the sleekest of hauls being paddled to make them go fast .
Yes
Yes, drag is a function of velocity. The issue is how much drag is generated by the various hull shapes at equivalent speeds. Alternatively, you can consider the velocity reached by different hulls with specified input power - hulls with less drag will be faster.
Ounces or pounds
Yes, I’ve seen that chart before, but it gives you some mysterious coefficients rather than hard numbers. Just what are you supposed to do with those coefficients to calculate the actual drag? I think the hard numbers – so many ounces of drag in this particular current with the boat owner seated in the boat – will be much more interesting.
Where did all the racers go?
A few years this topic would be discussed ad nauseam.
Ounces or pounds
– Last Updated: Mar-15-10 9:46 PM EST –
"Posted by: davejjj on Mar-15-10 6:26 PM (EST)
Yes, I've seen that chart before, but it gives you some mysterious coefficients rather than hard numbers."
Drag is a function of speed squared. The coefficient form allows you to translate that to pounds based on the speed of the boat.
The coefficient form is basically a lowest common denominator. It allows you to compare boats on an equal basis. A lower CD, is a lower drag boat. The absolute value doesn't really matter.
To quote from the referenced article:
"The viscous resistance Rv (drag) can be written as
Rv=1/2*rho*U^2*S*Cv
where rho is the water density and S the wetted surface area of the hull."
Cv is the drag coefficient
Kilograms, not Cd
– Last Updated: Mar-15-10 11:41 PM EST –
Just checking the chart at the link again - those resistance values listed are in kilograms, as it says in little tiny print at the bottom of the page. This is not correct, it should be a force unit (not mass). I believe this is done for historical reasons. I have a mechanical force balance from a defunct tow tank in my office - the readout is in grams, which is the mass of the little counterweight that balances the drag force on an attached model.
To get the drag force, multiply resistance in kilograms by g (9.81 m/s2), giving the result in Newtons, which is the SI unit of force. If you want to convert to Imperial/English units, 1 pound-force = 4.448 Newtons.
The description above of how to use drag coefficient is correct, but that's not what's tabulated in the linked chart. Also, the definition referenced only accounts for viscous drag. A full resistance value would also include form drag and wave drag, and not be exactly proportional to velocity squared (but it would be close).
Seriously, no ordinary person measures .
– Last Updated: Mar-16-10 12:21 AM EST –
... force in Newtons. Sure, it's correct, but as long as all of us are living on the same planet, experiencing exactly the same strength of gravity (the differences in gravity around the planet are too small for us to perceive), using units of mass in this situation is perfectly understandable. We know how it feels to portage a 50-pound canoe, and knowing the value in Newtons of the force exerted on our body in the process doesn't make it feel any different than if all we know is that it takes a 50-pound "force" to keep it there. Same goes for the force needed to move a boat through the water. In short, any of us can understand that an object of a certain mass has a certain weight when it is on Planet Earth, and any of us can perceive that weight in terms of the force necessary to resist gravity. Now, if the average person stepped on a bathroom scale after someone had changed the units of measure to Newtons, they wouldn't have any idea if the scale was giving them correct information or not, because they've never seen a force, not even the force of their own weight, expressed in Newtons before. I enjoyed studying physics in college, but not since then have I had any need to use Newtons as a unit of force. Even engineers use units of mass instead of Newtons when describing the forces exerted on building materials, because there's no need to do it another way.
Thanks
You are on the mark. I just quoted a portion of the article for simplicity. It also talks about skin friction as well. It all adds up…
Yep
That’s the beauty of the English system of measurement. For most people Lbs (force) = Lbs (mass) and it really doesn’t change the price of tea in China.
I’ve always enjoyed expressing rocket thrust ratios in lbs (thrust)/lbs (weight). 
So if I paddle infinitly slowly
It will take no effort at all! Brilliant!
What it means …
– Last Updated: Mar-16-10 10:39 AM EST –
As quantified in layman's terms, here is what it may mean to you:
http://www.paddling.net/message/showThread.html?fid=advice&tid=1257849
And check a couple of post down from the top for the smiple math: http://www.paddling.net/message/showThread.html?fid=advice&tid=1257849#1257893 (copy the entire link including the #1257893 at the end to jump to the right post with the math)
It is deceiving to just look at the resistance numbers. Some will say that if 2 boats differ by 3 onces in resistance (or pull), multiply that by the number of paddle strokes and you get the "real" difference. Afterall, you "pull" with the paddle at each stroke, right? Right, but that's not all. One needs to use the correct math and the # of strokes times pull (or resistance) is not it.
Check the above links for the simple math that can give a meaningful difference b/w boats with difference resistances *over distance* at a given speed. I'm not going to argue what it means to you, just that the difference can be translated into something we are familiar with - car-top loading a 50lb kayak 30 times in a row, for instance -:)
True
I agree with you on the non-intuitive nature of Newtons as a unit of force, I’m more comfortable with them because I teach dynamics. The only real problem comes up when incorporating mass values into computations - as long as they are multiplied by ‘g’ to give a force, I guess it’s all good. For a consumer-oriented chart, I suppose resistance in mass feels better. For use by eggheads, resistance values should be in the correct unit.
I actually prefer the English system for the reason mentioned above, even though there are some conversion issues that come up between pounds-mass and pounds-force (lbm and lbf).
Average power
I believe these comparison are very interesting and much more non-trivial than they appear. The speed of a boat varies during a paddle stroke, so the total resistance varies in time, as does the force applied by the paddler via the paddle. The paddle stroke rate, time of paddle in the water, glide between strokes, etc. all matter. Power input will therefore vary, but the average power can be calculated by summing over a cycle (or many cycles) at constant average velocity. This is the comparison that I think I want to look into - how does a given hull perform (i.e. its average velocity) with specified average power input.