User:DLangrish

$$mg\sin\theta=\mu mg\cos\theta+ \frac{\rho AC_dv^2}{2}$$

Where:
 * m = mass
 * g = gravitational acceleration constant for earth
 * sin(theta) = factor of weight acting down the slope
 * mu = co-efficient of friction
 * cos(theta) = factor of weight acting perpendicular to the slope
 * rho = air density
 * A = cross-sectional area
 * Cd = coefficient of drag
 * v = velocity

Re-arranged in terms of v: $$v=\sqrt{\frac{2(mg\sin\theta-\mu mg\cos\theta)}{\rho AC_d}}$$ Example to estimate terminal velocity for me in a speed tuck on a 10% slope. $$v=\sqrt{\frac{2(83*9.81\sin(5.71)-0.022*83*9.81\cos(5.71))}{1.2*0.37*1}}=16.87ms^-1$$

That's 37.4mph. Not unrealistic I would hope.

$$mg\sin\theta=\mu mg\cos\theta+ \frac{\rho AC_dv^2}{2}$$

$$k= \frac{2(\rho_s-\rho_m)g}{9\eta}$$

$$\eta= \frac{2(\rho_s-\rho_m)g}{9k}$$ $$v_t=kr^2$$

$$F=6\pi r\eta v$$

$$C=\frac{2F}{\rho v^2A}$$

$$m = m_{added}.\frac{1}{(\frac{f_{1}}{f_{2}})^{2}-1}$$

$$ m\ddot{x} + c\dot{x} + kx = Fe^{j\omega t} $$

$$ Z = \frac{F}{\dot{x}} $$

$$ \dot{x} = \dot{x}_{0}e^{j\omega t} $$

$$ Z = \frac{F}{\dot{x}} \frac{e^{j\omega t}}{e^{j\omega t}} = j\omega m + c - \frac{jk}{\omega} $$

$$ Re(Z) = c $$

$$ Im( \tilde{Z}) = \omega m - \frac{k}{\omega} $$

$$ \delta = \frac{1}{n}ln \frac{x_1}{x_n+1} $$

$$ \zeta = \frac{ \delta}{ \sqrt{(2 \pi)^2 + \delta^2}} $$

$$ \eta = 2 \zeta $$

$$ \delta = ln \frac{x_1}{x_2} $$

$$ \zeta = \frac{ \omega_2 - \omega_1}{2 \omega_n} $$

$$ \tilde{H} $$

$$ \tilde{R} $$

$$ u = \sqrt{ \frac{2(p_{T} - p)}{ \rho}} $$

$$ \tilde{Z} $$

$$ f_{c} = \frac{1.84c}{ \pi d} $$

$$ q = Au $$

$$ Power = \frac{1}{2} \rho u^2q $$

$$ \Delta \dot{E}_{mech} \approx (p_9 - p_8)q $$

$$ ( \frac{P}{P_0} )^ \frac{1}{ \gamma} = \frac{ \rho}{ \rho_0} = ( \frac{T}{T_0} )^ \frac{1}{( \gamma-1)} $$

$$ P = e^ \frac{20.386-5132}{T} $$

$$ \omega = \sqrt{ \frac{2k}{m_2}} $$

$$ P_{total} $$

$$ p_i $$

$$ P_{total} = p_{air} + p_{w} $$

$$ \frac{P_{total}}{P_{w}} = \frac{p_{air} + p_{w}}{p_{w}} = \frac{p_{air}}{p_{w}} + 1 $$

$$ pV = nRT, p = \frac{nRT}{V} $$

$$ \frac{p_{air}}{p_w} $$

$$ \frac{p_{air}}{p_w} = \frac{( \frac{n_{air}RT}{V})}{( \frac{n_{w}RT}{V})} = \frac{n_{air}}{n_w} $$

$$ h = u + p \frac{1}{ \rho} $$

$$ dh = du + dp \frac{1}{ \rho} - p \frac{dp}{ \rho^2} $$

$$ \frac{d \rho}{ \rho^2} $$

$$ dh = \frac{dp}{ \rho} $$

$$ \frac{dh}{dz} = \frac{1}{ \rho} \frac{dp}{dz} $$

$$ p(z) = p_0 - z \rho g $$

$$ \frac{dp}{dz} = - \rho g $$

$$ \frac{dh}{dz} = -g $$

$$ c_p \frac{dT}{dz} = -g $$

$$ \frac{dT}{dz} = \frac{-g}{c_p} $$

$$ - \int \frac{dT}{dz}dz = \int \frac{g}{c_p}dz $$

$$ -T(z) = \frac{g}{c_p}z - constant $$

$$ \frac{constant}{c_p} = konstant = T(z)c_p + gz $$

$$ konstant = h(z) + gz $$

$$ h(z) + gz = h_0 $$

$$ \int \frac{dh}{dz}dz = - \int gdz $$

$$ h(z) = -gz + constant $$

$$ [M]\{\ddot{v}\} + [K]\{v\} = \{F\} $$

$$ \{v\} = [A]\{F\} $$

$$ \omega^2 = 13.424 \frac{EI}{ \rho Al^4} $$

$$

[A] = \begin{bmatrix} 1/3 & 5/6 & 4/3 \\ 5/6 & 8/3 & 14/3 \\ 4/3 & 14/3 & 9 \end{bmatrix} \frac{1}{EI} \left ( \frac{l}{3} \right )^3

$$

$$ \omega_n = k^{2}_{i} \sqrt{ \frac{EI}{ \rho A} } $$

$$ cos(k_i l)cosh(k_i l) = -1 $$

$$ h = C_{p}T $$

$$ c_p \frac{dT}{dz} = -g $$

$$ \frac{dT}{dz} = \frac{-g}{c_p} $$

$$ - \int \frac{dT}{dz}dz = \int \frac{g}{c_p}dz $$

$$ -T(z) = \frac{g}{c_p}z - constant $$

$$ c_{p}constant = konstant = T(z)c_p + gz $$

$$ konstant = h(z) + gz $$

$$ - \frac{dT}{dz} = \frac{g}{c_p} $$

$$

T(z) = \psi z + T_{0} $$

$$ \frac{dp}{\rho^2} $$

$$ z_{dew} = \frac{1}{ \psi} \left ( \frac{p_{SAT}(z_{dew})}{ \beta} - T_0 \right ) $$