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 November 9th, 2009, 12:59 AM #1 Newbie   Joined: Nov 2009 Posts: 2 Thanks: 0 Drag force, analytical solution Hi It's been a very long time since i did any advanced math. I am currently studying bodies under water. The forces acting on them are * Gravity F=m*g * Buoyancy F=roh*Vol*g * Drag F=0.5*roh*vel^2 * Area * Cd Simplifying I get a = accelleration v = velocity k1 = constant 1 k2 = constant 2 a = k1 - k2 *v^2 So is there a analytical solution for displacement for this equation? I have solved it numerically but would like to have an analytical solution if possible. All the problems with drag (parachuting, golf balls, baseballs, free fall etc etc) either solve this numerically or simplify drag to be F=k*v Any ideas?
 November 13th, 2009, 01:29 AM #2 Member   Joined: Nov 2009 Posts: 72 Thanks: 0 Re: Drag force, analytical solution There is. Let $x(t)$ be the displacement. Then your equation is $\ddot{x}(t)=k_1-k_2\dot{x}^2$ which, by defining $v(t):=\dot{x}(t)$, can be written as $\dot{v}(t)=k_1-k_2v^2.$ Integrate by parts: \begin{align*} \int_{v_0}^v \frac{dv}{k_1-k_2v^2}&=\int_{t_0}^t\ dt\\ \frac{1}{\sqrt{k_1k_2}}\left(\tanh^{-1}\ v\sqrt{\frac{k_2}{k_1}}-\tanh^{-1}\ v_0\sqrt{\frac{k_2}{k_1}}\right)&=t-t_0\\ \tanh^{-1}\ v\sqrt{\frac{k_2}{k_1}}&=(t-t_0)\sqrt{k_1k_2}+\tanh^{-1}\ v_0\sqrt{\frac{k_2}{k_1}}\\ v&=\sqrt{\frac{k_1}{k_2}}\tanh{\left((t-t_0)\sqrt{k_1k_2}+\tanh^{-1}\ v_0\sqrt{\frac{k_2}{k_1}}\right)} \end{align*} To simplify the equations define $\tau=\frac{1}{\sqrt{k_1k_2}}$ and $v_t=\sqrt{\frac{k_1}{k_2}}$, so that the above can be written as $v(t)=v_t\tanh{\left(\frac{t-t_0}{\tau}+\tanh^{-1}\frac{v_0}{v_t}\right)}$ And so then $x(t)=v_t\int_{t_0}^t \tanh{\left(\frac{t-t_0}{\tau}+\tanh^{-1}\frac{v_0}{v_t}\right)}\ dt$ If you want a nicer formula than that, you can do a u-substitution, but that'd be pretty messy. But if you can take initial velocity to be zero and $t_0=0$, then $v(t)=v_t\tanh{\frac{t}{\tau}}$ so $x(t)=v_t\int_0^t\tanh\ {\frac{t}{\tau}}\ dt=x_0+v_t\tau\ln{\ \cosh{\ \frac{t}{\tau}}}$
 November 13th, 2009, 02:08 AM #3 Newbie   Joined: Nov 2009 Posts: 2 Thanks: 0 Re: Drag force, analytical solution amazing! Thanks so very much

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