Help with a swmming math problem

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.

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.