Swim smarts: Engineers? Drag increases as speed increases?
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!
Parents
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?
Reply
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?
