Time Power: A Proven System for Getting More Done in Less Time Than You Ever Thought Possible

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displaystyle P={\frac {dW}{dt}}={\frac {d}{dt}}\int _{\Delta t}\mathbf {F} \cdot \mathbf {v} \,dt=\mathbf {F} \cdot \mathbf {v} . Peak power and duty cycle [ edit ] In a train of identical pulses, the instantaneous power is a periodic function of time. Likewise, the power dissipated in an electrical element of a circuit is the product of the current flowing through the element and of the voltage across the element. In the case of a periodic signal s ( t ) {\displaystyle s(t)} of period T {\displaystyle T} , like a train of identical pulses, the instantaneous power p ( t ) = | s ( t ) | 2 {\textstyle p(t)=|s(t)|

Other common and traditional measures are horsepower (hp), comparing to the power of a horse; one mechanical horsepower equals about 745. Other units of power include ergs per second (erg/s), foot-pounds per minute, dBm, a logarithmic measure relative to a reference of 1 milliwatt, calories per hour, BTU per hour (BTU/h), and tons of refrigeration. Mechanical power [ edit ] One metric horsepower is needed to lift 75 kilograms by 1 metre in 1 second. These relations are important because they define the maximum performance of a device in terms of velocity ratios determined by its physical dimensions. displaystyle W=\int _{C}\mathbf {F} \cdot d\mathbf {r} =\int _{\Delta t}\mathbf {F} \cdot {\frac {d\mathbf {r} }{dt}}\ dt=\int _{\Delta t}\mathbf {F} \cdot \mathbf {v} \,dt.Let the input power to a device be a force F A acting on a point that moves with velocity v A and the output power be a force F B acts on a point that moves with velocity v B. Specifying power in particular systems may require attention to other quantities; for example, the power involved in moving a ground vehicle is the product of the aerodynamic drag plus traction force on the wheels, and the velocity of the vehicle.

displaystyle \mathrm {MA} ={\frac {T_{\text{B}}}{T_{\text{A}}}}={\frac {\omega _{\text{A}}}{\omega _{\text{B}}}}. The ratio of the pulse duration to the period is equal to the ratio of the average power to the peak power. displaystyle P=\lim _{\Delta t\to 0}P_{\mathrm {avg} }=\lim _{\Delta t\to 0}{\frac {\Delta W}{\Delta t}}={\frac {dW}{dt}}. displaystyle P={\frac {dW}{dt}}={\frac {d}{dt}}\left(\mathbf {F} \cdot \mathbf {x} \right)=\mathbf {F} \cdot {\frac {d\mathbf {x} }{dt}}=\mathbf {F} \cdot \mathbf {v} .displaystyle \mathrm {MA} ={\frac {F_{\text{B}}}{F_{\text{A}}}}={\frac {v_{\text{A}}}{v_{\text{B}}}}. As a simple example, burning one kilogram of coal releases more energy than detonating a kilogram of TNT, [6] but because the TNT reaction releases energy more quickly, it delivers more power than the coal.