Help with a swmming math problem

We have completely failed to take into account the fluid dynamics of the situation. As effort increases, even the smoothest stroke results in turbulence which would have an effect on the time ratios for the same distance. :D I guess I have always thought of 3/4 pace or 80% pace as an intuitive understanding based on general feeling and experience. 80% is my cruising speed under which I could swim with effort, but could last 5 or 6 K swimming at that pace. 90% is what I would pace myself in the non-sprinting portions of a 1000 or 500. I haven't seen a workout ask for anything less than this nebulous 3/4 or 80% pace.

Thanks for the review. I do not think there is a mistake when I expand out the terms: P_swim = ((C_flow * V^2) * D_pool) * t_interval keeping in mind V = D_pool / t_interval let us now substitute terms for V P_swim = ((C_flow * (D_pool/t_interval)^2) * D_pool) * t_interval one of the t_interval's gets canceled out. Sorry, I don't have a Mech E background so I had to look some things up. Drag Force = (1/2*p*C_D*A)*v^2 Everything in the () can be treated as a constant, which you called C_flow. Drag Force = F_drag F_drag = C_flow*v^2 velocity = distance / time v = d/t work = force * displacement We will assume that displacement and distance are the same W = F*d power = work / time (not work * time) P = W/t To swim at a constant speed, you only need to overcome the force of drag. To convince yourself that this is true, think of iceskating (or rollerskating). Once you get going, you can glide for a long time with no effort, until the friction of the ice slows you down. Acceleration is hard, but once you hit you get going, maintaining speed is easy. W_drag = F_drag * d W_drag = C_flow * v^2 * d W_drag = C_flow * (d/t)^2 * d W_drag = C_flow * d^2/t^2 * d W_drag = C_flow * d^3/t^2 Since C_flow and d are constants, the work needed is inversely related to the time. Thus each second faster you want to go is twice as hard as the previous second. Now if you want to look at the power needed P_drag = W_drag / t P_drag = (C_flow * d^3/t^2)/t P_drag = C_flow * d^3/t^3 Again, ignore the constants (or loose weight and swim shorter distances). A cubic increase in power is needed to drop time. I am confident that the relationship bewtween work (or power) and time is an inverse exponential. Since the original question was about how to calculate a time to swim X, I believe the exponent is 2 and any linear portion of the equation can be ignored because it is dominated.

I am confident that ... any... portion .... can be ignored because it is dominated. Via just a bit of condensing, Q, I have managed to reduce your argument to a more elemental, and elegant, state. You're welcome!

The coach at University of Toronto used to use the term 3/4 Pace. What he wanted was a relaxed almost 100% speed. Do you really think it means to swim 75 yards in the time you swim 100??? I don.t think fluid dynamics even comes into this.

I've used the sharp stress score for quantifying swim training load. Easier in practice. The easiest though, is to use the Sharp score of a swim and multiply by yards instead of minutes. Well technically the easiest is just to use yards, but that is subject to pretty big variations depending on the quality. The method you use to quantify your training load is not nearly as important as what you do with the number though. Phil Skiba's software implements the Banister model and gives great info on training load, effectiveness and taper. So yes, I have looked into it but not found the plusses to outweigh the minuses of entering the data into a spreadsheet for every workout. I use sharp stress score, which is actually implemented into personal swim manager autoamtically.

Has anyone here already tried or is thinking about trying the Swim Score model? Currently I am fair guessing my swim intensity into wko+ but in few weeks when I really start to train seriously, I intend of logging actual power equivalent, calculated using Skiba's Swim Score ref.: www.physfarm.com/swimscore.pdf