Swim smarts: Engineers? Drag increases as speed increases?

Former Member
Former Member
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!
  • Proof that paleontologists also make good physicists
  • Former Member
    Former Member
    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
  • The last time I posted on this was post # 35 on this thread forums.usms.org/showthread.php .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?
  • Start here: 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: upload.wikimedia.org/.../99a6015b6a230860c9b1517b238e5de9.png (Note, drag is proportional to velocity squared) Also, Power The power required to overcome the aerodynamic drag is given by: upload.wikimedia.org/.../e31430f0898268091f410282a89503b1.png (Note, power is proportional velocity cubed) Have fun!
  • 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?
  • Former Member
    Former Member
    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?
  • 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.
  • "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.
  • 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 -> www.calctool.org/.../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
  • 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.
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