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vo2
September 7th, 2013, 10:18 AM
I overheard one of the local club team coaches prepping the kids before a technique based drill make the statement that the faster you go the more drag you encounter. He was having them ensure they kept their shoulder to cheek on their free sets to narrow the frontal profile. While I am not questioning the coach I just wanted to know if one of you smart Masters swimmers might be able to dumb that down for me.

So to ensure I lay this out to the best of my understanding which is probably wrong: for two physical clones swimming freestyle in lanes next to each other with completely identical technical strokes down to the mm. The drag for the 1:20 pace swimmer will be less than the swimmer peeling off 1:10 splits? I'm a big dummy so wrapping my head around that idea just isn't sinking in.

Thanks for the Saturday morning hydrodynamics lesson!

Allen Stark
September 7th, 2013, 11:16 AM
The last time I posted on this was post # 35 on this thread http://forums.usms.org/showthread.php?22888-Total-Immersion-Does-it-work&p=290479&highlight=#post290479 .This is easy to prove to yourself.Pull your hand slowly through the water.Now pull your hand quickly through the water.Which is harder? Or push off the wall at the pool as hard as you can with your head up and your arms wide,now do the same as streamlined as possible.Which way goes further?

Swimosaur
September 7th, 2013, 11:46 AM
Start here:

Drag (physics) (http://en.wikipedia.org/wiki/Drag_(physics))

Drag depends on the properties of the fluid and on the size, shape, and speed of the object. One way to express this is by means of the drag equation:


http://upload.wikimedia.org/math/9/9/a/99a6015b6a230860c9b1517b238e5de9.png (Note, drag is proportional to velocity squared)

Also,

Power

The power required to overcome the aerodynamic drag is given by:


http://upload.wikimedia.org/math/e/3/1/e31430f0898268091f410282a89503b1.png (Note, power is proportional velocity cubed)

Have fun!

__steve__
September 7th, 2013, 01:18 PM
Proof that paleontologists also make good physicists

foolishfloater
September 7th, 2013, 04:59 PM
Hi Vo2,

Fluid mechanics is very tough subject. The formulas Swimosaur provided are correct but the first is only a simplification of drag.

The simple answer is that the actual drag a swimmer must overcome increases rapidly as their speed increases. There might be an exception to this corresponding to the "transition" between laminar and turbulent flow. That said, the simple answer " Drag rises quickly as you swim faster..." is about 99.9% correct.

FF

Sojerz
September 8th, 2013, 01:03 AM
The amount of drag increases with the square of the velocity (see Swimosaur above). If one swims 100 yards in 60 seconds that's 300 feet/60 sec, the velocity (v) is 5 feet per sec. If you then swim the 100 yards in 120 sec the velocity is 300/120 and the velocity is 2.5 feet per second. The first 100 is 2x the velocity (v) of the second 100.

But the drag force (being proportional to the square of the velocity) has increased by 2x squared, which is 4x. So one encounters 4x the amount of drag when one swims twice as fast, and consequently the force required to overcome that drag (Swimosaurs FD) is also 4x times as much.

As Swimosaur points out above, the power required is proportional to the cube of the velocity, so swimming at 2x the velocity requires 8x as much power (2x cubed). It's easy to see that the amount of force and power required goes up very quickly as one swims faster unless you do something to reduce the drag coefficient.

The above assumed that the drag coefficient (CD) remained the same throughout both 100s. But the amount of force and power required is also directly proportional to the drag coefficient CD. Reduce the drag coefficient by a factor of two and the amount of force and power required will be reduced by 2.

Because the force required is increasing by the square, and the power required is increasing by the cube, the faster you swim the more important reductions in the drag coefficient become thru streamlining.

What does this say about the force and power required and drag of elite (fast) swimmers?

vo2
September 8th, 2013, 06:41 AM
The amount of drag increases with the square of the velocity (see Swimosaur above). If one swims 100 yards in 60 seconds that's 300 feet/60 sec, the velocity (v) is 5 feet per sec. If you then swim the 100 yards in 120 sec the velocity is 300/120 and the velocity is 2.5 feet per second. The first 100 is 2x the velocity (v) of the second 100.

But the drag force (being proportional to the square of the velocity) has increased by 2x squared, which is 4x. So one encounters 4x the amount of drag when one swims twice as fast, and consequently the force required to overcome that drag (Swimosaurs FD) is also 4x times as much.

As Swimosaur points out above, the power required is proportional to the cube of the velocity, so swimming at 2x the velocity requires 8x as much power (2x cubed). It's easy to see that the amount of force and power required goes up very quickly as one swims faster unless you do something to reduce the drag coefficient.

The above assumed that the drag coefficient (CD) remained the same throughout both 100s. But the amount of force and power required is also directly proportional to the drag coefficient CD. Reduce the drag coefficient by a factor of two and the amount of force and power required will be reduced by 2.

Because the force required is increasing by the square, and the power required is increasing by the cube, the faster you swim the more important reductions in the drag coefficient become thru streamlining.

What does this say about the force and power required and drag of elite (fast) swimmers?

Two things. First off thank you for that explanation. Secondly. I'm shocked that I *think* I actually understand what you are laying out there. THAT is shocking because NASA isn't tracking me down to fill in at Mission Control on weekends;)

What does it say about elites? Well, my highly uninformed positions says it tells us they absolutely OWN low drag positions and have probably been taught the smartest way to go faster if you don't have drag issues worked out is to fix those issues b/f trying to over power them.

Thanks again that is good stuff!

So to gain a 10% increase in speed a swimmer would need a 33% increase in power if the CD remains unchanged. But if that same swimmer reduces drag by 10% it would net an 11% increase in speed. Is my mAthS koRrecT?

ande
September 9th, 2013, 10:06 AM
"the faster you go the more drag you encounter."

Probably.
Here's a couple thoughts to consider:

1) when a swimmer is swimming freestyle, what if, the faster a swimmer moves through the water, the less of his body is actually in the water. Like a boat.

2) what if, the faster a swimmer SDKs underwater, the more the water around their body compresses their body making their profile slightly thinner. Also the deeper a swimmer SDKs, the more the water compresses them.

It would be interesting to see the complex physics calculations to really figure this out.
One that took into account body mass shape, skin, hair, suit coverage, compression, floatation . . .
Also if there's any truth to the notion that the faster a swimmer swims the less of their body is actually in the water.

__steve__
September 9th, 2013, 06:47 PM
I never realized the compression at depth thing, interesting

I think the only reason we can swim as fast as we do on the surface is because the air provides us with a medium of low resistance to recover our paddles, and as we travel faster at the boundary it also pushes us up to the lower pressure air on a plane.

I wonder if a near maximum velocity exist for surface swimming, where going any faster wouldn't be possible, any remaining room for improvement is in starts, reaction, and underwater streamline, and how close current records are to it.

sunruh
September 10th, 2013, 01:18 PM
2) what if, the faster a swimmer SDKs underwater, the more the water around their body compresses their body making their profile slightly thinner. Also the deeper a swimmer SDKs, the more the water compresses them.

It would be interesting to see the complex physics calculations to really figure this out.


this has been known for many years.....33 feet down equals 1 atmosphere of pressure.

using this tool -> http://www.calctool.org/CALC/other/games/depth_press

you can see that the difference between 1 foot under and 4 feet under is 0.09 atm.
even if you go crazy and SDK down to 8 feet (double dog dare you to ande) you will only get .21atm more sqeeze than just that initial 1 foot under.

engineers rule!
steve

vo2
September 10th, 2013, 02:05 PM
Thanks for all the input folks this is great. I was fiddling around with what I overheard from the club coach and darned if it wasn't immediately noticeable when I kept my shoulder to cheek. I had unknowingly been allowing a big gap between there simply b/c I didn't know it mattered. What I felt was just easier momentum, less surgey, which I took as less resistance allowing me to keep my speed up a bit better. Thanks again for dumbing it down for me:)

knelson
September 10th, 2013, 03:30 PM
1) when a swimmer is swimming freestyle, what if, the faster a swimmer moves through the water, the less of his body is actually in the water. Like a boat

This is important because swimming at the surface creates an additional type of drag: wave drag. Just like you said, you want to ride as high as possible, like a boat does, to minimize this drag. The equation Swimosaur quoted above is for pressure or form drag. The added wave drag at the surface is why underwater kicking can be faster than surface swimming.

Allen Stark
September 10th, 2013, 04:20 PM
I wonder if a near maximum velocity exist for surface swimming, where going any faster wouldn't be possible, any remaining room for improvement is in starts, reaction, and underwater streamline, and how close current records are to it.
Humans are not very good boats,but apply enough force and the human form will plane(think being towed by a boat or body surfing a relatively large wave.) When planing form drag decreases greatly,but I don't know what happens to wave drag.Anyway,I doubt there is a firm limit as to how fast a person can swim.

Kevin in MD
September 10th, 2013, 04:33 PM
The amount of drag increases with the square of the velocity (see Swimosaur above).

The power required to push a static shape underwater with no wave interaction increases wit the cube of the speed yes.

However, last time I looked at something like this when looking at active drag of actual swimmers, the drag increased with speed raised to the 2.4th power in the study. I have seen one or two others, it varies by measurement method but the drag never goes up with the square of the speed.

As swimmers we are on the surface and have to deal not only with form drag but also wave drag, we are also constantly changing shape and length of our vessel as we swim making the whole thing not amenable to the simple equations we engineers like to use.

smontanaro
September 10th, 2013, 04:39 PM
As long as we are discussing the outer limits, I wonder if it's possible for a human to power some sort of hydrofoil based on swimming. There's this: http://www.human-powered-hydrofoils.com/ but the images all show bike/kayak/jet ski contraptions. I wonder about some sort of tech suit with a hydrofoil...

GregJS
September 10th, 2013, 10:40 PM
Would a tech suit with a hydrofoil be something like a "fish suit?" Has anyone tried creating something like that? I'm picturing a rigid, bullet-shaped suit with a large mono fin out the back and large flippers on the side for the arms. I was watching seals at an aquarium recently and it seemed that with the slightest movement of their side flippers, they'd shoot off with amazing speed.

Maybe I'm in lala land, but it would be cool!

smontanaro
September 10th, 2013, 11:03 PM
Would a tech suit with a hydrofoil be something like a "fish suit?"

Not quite. Think of replacing the human+kayak+hydrofoil bits in this picture:

http://www.human-powered-hydrofoils.com/files/cache/7059a2bd5d79d861d4d666b6154f546d_f105.jpg

with human+weird tech suit+hydrofoil bits. The human effectively becomes the kayak hull. Swim fast enough (I'm really waving my hands here!) and you rise up on the hydrofoil. It seems to me that the tech suit would have to be very stiff (otherwise the hydrofoil would flop around), and once you're up and out of the water, you could really only propel yourself with your arms.

I know, it's almost certainly a bad idea, but, hey, I didn't get much sleep last night and the bike ride home today was in 95+ degree weather... What do you expect?

GregJS
September 11th, 2013, 09:24 AM
Ha ha - I woke up this morning wondering "What the heck was I thinking?" about my idea too, and also came to the conclusion I must've been suffering from sleep deprivation. My fish suit would only work underwater for one thing - hence some kind of "human powered submersible" or maybe just a glorified scuba suit - a kind of mash up of I-Am-The-Walrus-In-A-Yellow-Submarine, I guess.

But if a person could actually get up enough speed to rise up on your hydrofoil, I imagine they'd have such low resistance at that point that they could keep going with much less effort? Maybe they'd just need extra-large paddles on their hands?

Crazy as these ideas sound, you just never know. I had actually pictured this as a kid:

http://en.wikipedia.org/wiki/Zorbing

and now it exists.

Still, better hand these things over to the engineers and let them figure out the details...

Swimosaur
September 11th, 2013, 09:38 AM
As swimmers we are on the surface and have to deal not only with form drag but also wave drag, we are also constantly changing shape and length of our vessel as we swim making the whole thing not amenable to the simple equations we engineers like to use.

This. :agree: