Water Resistance

RE: Water Resistance

Former Member — Wed, 26 Sep 2012 15:33:10 GMT

And that is why streamlining is so important. Using 0.5 for the drag coefficient was a total guess, by the way. A swimmer in a good streamline probably has a lower drag coefficient than that. I would guess so. Even bad swimmer might. A 150 lbs swimmer pushes off the wall at 15 ft/sec and 25 lbs resistance would stop him in 7.5 yards That's with the silly over simplification that the drag doesn't change with speed. I could probably do the caluculus with the drag coeeficients and all, but I'm WAY too lazy. Where's that 17 year old college junior?

RE: Water Resistance

Former Member — Sat, 22 Sep 2012 07:24:07 GMT

That is crazy. It's exactly what I was looking for. Is there a way to simplify that 25 lbs. of drag in relation to duration or distance for me? a velocity of 5 ft/sec Like in terms of a weight lifter doing 25 lb. reps? I realize we are simplifying this, but it's just for fun.

RE: Water Resistance

__steve__ — Sat, 22 Sep 2012 06:06:32 GMT

Probably the reason tighter splits are faster overall

RE: Water Resistance

Sojerz — Sat, 22 Sep 2012 03:51:37 GMT

Not really. It's a constant 25 lb force you must overcome. You can multiply the force by time to get the power required. Yah, swimming at a constant speed doesn't let you put the weight down.

RE: Water Resistance

knelson — Sat, 22 Sep 2012 03:43:22 GMT

As Kirk mentioned, Form matters. If your cross sectional shape was that of a 4'x4' piece of plywood, the drag coefficent would go way up and it would take way more force. And that is why streamlining is so important. Using 0.5 for the drag coefficient was a total guess, by the way. A swimmer in a good streamline probably has a lower drag coefficient than that.

RE: Water Resistance

Sojerz — Sat, 22 Sep 2012 03:38:56 GMT

That is crazy. It's exactly what I was looking for. Is there a way to simplify that 25 lbs. of drag in relation to duration or distance for me? Like in terms of a weight lifter doing 25 lb. reps? I realize we are simplifying this, but it's just for fun. Yep, it's the same "pounds' unit of force. It takes 25 lbs. of force to lift a 25 lb weight (on the earth's surface) and it takes about the same 25 lbs of force to overcome drag at 5 ft/sec. As Kirk mentioned, Form matters. If your cross sectional shape was that of a 4'x4' piece of plywood, the drag coefficent would go way up and it would take way more force.

RE: Water Resistance

knelson — Sat, 22 Sep 2012 03:33:43 GMT

That is crazy. It's exactly what I was looking for. Is there a way to simplify that 25 lbs. of drag in relation to duration or distance for me? Like in terms of a weight lifter doing 25 lb. reps? I realize we are simplifying this, but it's just for fun. Not really. It's a constant 25 lb force you must overcome. You can multiply the force by time to get the power required.

RE: Water Resistance

Sojerz — Sat, 22 Sep 2012 03:27:38 GMT

Yes, it can be measured in pounds. The drag created by the swimmer on the surface is a little tricky because both form and wave drag must be considered. A streamlined swimmer underwater probably produces somewhere around 25 pounds of drag. You can use the equation drag = 1/2 * density * velocity^2 * drag coefficient * frontal area. I assumed a drag coefficient of 0.5, a velocity of 5 ft/sec (this is 50 yards in 30 seconds) and an area of 2 square feet. There will be more drag at the surface because of the addition of wave drag. Good explanation Kirk - note that the amount of drag increases with the square of the velocity so at a velocity of 5 ft/sec you have 4x the drag of a velocity of 2.5 ft/sec. And therefore 5ft/sec requires about 4x as much effort to overcome drag at 2.5ft/sec. Moral: swim slower, you'll greatly reduce drag, and won't have to worry as much about it. :D I've been working at this.

RE: Water Resistance

jaadams1 — Sat, 22 Sep 2012 02:25:24 GMT

RE: Water Resistance

knelson — Sat, 22 Sep 2012 01:27:36 GMT

Can anyone explain to me how much resistance water creates while swimming? Is there a equivalent in pounds? or is there another way it is measured? Yes, it can be measured in pounds. The drag created by the swimmer on the surface is a little tricky because both form and wave drag must be considered. A streamlined swimmer underwater probably produces somewhere around 25 pounds of drag. You can use the equation drag = 1/2 * density * velocity^2 * drag coefficient * frontal area. I assumed a drag coefficient of 0.5, a velocity of 5 ft/sec (this is 50 yards in 30 seconds) and an area of 2 square feet. There will be more drag at the surface because of the addition of wave drag.