Difference between revisions of "Warp"

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===The formulae===
 
===The formulae===
CCP provided formulae for both distance traveled and velocity reached after ''t'' seconds of acceleration.  
+
CCP provided formulae for both distance traveled and velocity reached after ''t'' seconds of acceleration. If ''d'' is distance in meters, ''v'' is speed in meters per second, ''k'' is a (sort of) constant defined as the warp speed (in AU/s) and a = 149,597,870,700 meters (1 AU).
  
 
<math>\pagecolor{Black}\color{White}
 
<math>\pagecolor{Black}\color{White}
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d & = e^{kt} \\
 
d & = e^{kt} \\
 
v & = k*e^{kt}\\
 
v & = k*e^{kt}\\
 +
v_{max} & = k * a\\
 
\end{align}
 
\end{align}
 
</math>
 
</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 = 149,597,870,700 meters (1 AU).
+
===Distance===
 +
To calculate distance traveled while accelerating
  
 +
<math>\pagecolor{Black}\color{White}
 +
\begin{align}
 +
d & = e^{kt} \\
 +
v & = k*e^{kt}\\
 +
& = k*d\\
 +
\therefore d & = \frac{v}{k}
 +
\end{align}
 +
</math>
 +
 +
The distance covered while accelerating to v<sub>max</sub> is
 +
 +
<math>\pagecolor{Black}\color{White}
 +
\begin{align}
 +
d & = \frac{v_{max}}{k}
 +
& = \frac{k*a}{k}
 +
& = a
 +
\end{align}
 +
</math>
 +
 +
This means that every ship covers exactly 1 AU while accelerating to its maximum warp speed.
 +
 +
 +
===Time===
 
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)
  
Line 41: Line 66:
 
v & = k * a\\
 
v & = k * a\\
 
\therefore t & = \frac{\ln{(\frac{v}{k})}}{k}\\
 
\therefore t & = \frac{\ln{(\frac{v}{k})}}{k}\\
t & = \frac{\ln{(a)}}{k}\\
+
& = \frac{\ln{(a)}}{k}\\
 
\end{align}
 
\end{align}
 
</math>
 
</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==
 
==Deceleration==

Revision as of 00:37, 24 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. If d is distance in meters, v is speed in meters per second, k is a (sort of) constant defined as the warp speed (in AU/s) and a = 149,597,870,700 meters (1 AU).

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

Distance

To calculate distance traveled while accelerating

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

The distance covered while accelerating to vmax is

[math]\pagecolor{Black}\color{White} \begin{align} d & = \frac{v_{max}}{k} & = \frac{k*a}{k} & = a \end{align} [/math]

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


Time

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}\\ & = \frac{\ln{(a)}}{k}\\ \end{align} [/math]

Deceleration

Deceleration is calculated slightly differently. Instead of using 'k' to calculate distance and velocity, it uses 'j', which is defined as [math]\pagecolor{Black}\color{White}\min(\frac{k}{3},2)[/math]. A different rate of deceleration is used to prevent ships suddenly transitioning from "many, many AU away" to "on grid and out of warp" more rapidly than other pilots (or the server / client) can keep up with.

This changes the formulae used slightly. Remember that distance travelled is the integral of velocity.

[math]\pagecolor{Black}\color{White} \begin{align} v & = k * e^{jt}\\ d_{decel} & = \int_{0}^{\infty}k*e^{jt}\,dt = \frac{k*e^{jt}}{j}\\ \therefore d_{decel} & = \frac{v}{j} \end{align} [/math]