Difference between revisions of "Warp"

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(Created page with "==Time taken to warp== It is possible to work out how long it should take for a ship to complete warp (once it enters warp) based on formulae released by CCP. Warp consists ...")
 
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Where ''d'' is distance in meters, ''v'' is speed in meters per second and ''k'' is defined as the warp speed (in AU/s). 1 AU (''a'') is 149,597,870,700 meters.
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Where ''d'' is distance in meters, ''v'' is speed in meters per second and ''k'' is defined as the warp speed (in AU/s). a = 1 AU = 149,597,870,700 meters.
  
 
To calculate the time spent accelerating to warp speed, the equation for ''v'' should be rearranged to be in terms of ''t'', and then solved for the case of ''v'' being equal to the warp speed (in m/s)
 
To calculate the time spent accelerating to warp speed, the equation for ''v'' should be rearranged to be in terms of ''t'', and then solved for the case of ''v'' being equal to the warp speed (in m/s)

Revision as of 23:05, 23 March 2016

Time taken to warp

It is possible to work out how long it should take for a ship to complete warp (once it enters warp) based on formulae released by CCP.

Warp consists of 3 stages:

  1. Acceleration
  2. Cruising
  3. Deceleration

Calculating the time taken to warp is done by calculating the time spent in each of these phases and adding them together. This requires calculating acceleration and deceleration time first, followed by cruise time.

Acceleration

The formulae

CCP provided formulae for both distance traveled and velocity reached after t seconds of acceleration.

[math]\pagecolor{Black}\color{White} \begin{align} d & = e^{kt} \\ v & = k*e^{kt}\\ \end{align} [/math]

Where d is distance in meters, v is speed in meters per second and k is defined as the warp speed (in AU/s). a = 1 AU = 149,597,870,700 meters.

To calculate the time spent accelerating to warp speed, the equation for v should be rearranged to be in terms of t, and then solved for the case of v being equal to the warp speed (in m/s)

[math]\pagecolor{Black}\color{White} \begin{align} v & = k*e^{kt}\\ \frac{v}{k} & = e^{kt}\\ kt & = \ln{(\frac{v}{k})}\\ t & =\frac{\ln{(\frac{v}{k})}}{k}\\ \end{align} [/math]

We want to find the time taken to maximum warp:

[math]\pagecolor{Black}\color{White} \begin{align} v & = k * a\\ \therefore t & = \frac{\ln{(\frac{v}{k})}}{k}\\ t & = \frac{\ln{(a)}}{k}\\ \end{align} [/math]

Substituting the time taken to reach maximum speed in to the formula for distance gives the distance covered while accelerating:

[math]\pagecolor{Black}\color{White} \begin{align} d & = e^{k*t},t = \frac{\ln{(a)}}{k}\\ \therefore d & = e^{k*\frac{\ln{(a)}}{k}}\\ d & = e^{ln{(a)}}\\ d & = a \end{align} [/math]

This means that every ship covers 1 AU while accelerating to maximum speed.

Deceleration