Difference between revisions of "Warp"

From EVE University Wiki
Jump to: navigation, search
Line 102: Line 102:
  
 
===Time===
 
===Time===
 +
As with acceleration, time to decelerate from maximum warp velocity is worked out by rearranging the velocity equation.
 
<math>\pagecolor{Black}\color{White}
 
<math>\pagecolor{Black}\color{White}
 
\begin{align}
 
\begin{align}
d_{decel} & = \frac{v_{warp}}{j}
+
v &= k*e^{jt}\\
= \frac{k*a}{j}
+
\frac{v}{k} & = e ^ {jt}\\
 +
t & = \frac{\ln{(\frac{v}{k})}}{j}
 
\end{align}
 
\end{align}
 +
</math>
 +
 +
While the deceleration from ''s'' to 0 was insignificant in terms of distance, it is significant in terms of time. This means that the time to decelerate is calculated as follows:
 +
 +
<math>\pagecolor{Black}\color{White}
 +
\begin{align}
 +
t_{decel} & = t_{decel\_warp} - t_{decel\_s}\\
 +
& = \frac{\ln{(\frac{v}{k})}}{j} - \frac{\ln{(\frac{s}{k})}}{j}\\
 +
& = \frac{\ln{(\frac{v}{k})} - \ln{(\frac{s}{k})}}{j}\\
 +
& = \frac{\ln{v} - \ln{k} - ln{s} + ln{k}}{j}\\
 +
& = \frac{\ln{v} - \ln{s}}{j}\\
 +
& = \frac{\ln{(\frac{v}{s})}}{j}\\
 
</math>
 
</math>
  

Revision as of 01:26, 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 vwarp is

[math]\pagecolor{Black}\color{White} \begin{align} d_{accel} & = \frac{v_{warp}}{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_{warp} & = k * a\\ t_{accel} & = \frac{\ln{(\frac{v_{warp}}{k})}}{k}\\ & = \frac{\ln{(\frac{k*a}{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.

There is a complication with deceleration calculations. Ships do not drop out of warp at 0 m/s. Instead, they drop out of warp at s m/s, after which normal sub-warp calculations take over.

[math]\pagecolor{Black}\color{White}s = \min(100, v_{subwarp}/2)[/math]

Where vsubwarp is the maximum subwarp velocity of the ship; this varies greatly depending on the ship hull and pilot skills.

Distance

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}\\ & = \frac{v}{j} \end{align} [/math]

The distance covered while decelerating from vwarp is

[math]\pagecolor{Black}\color{White} \begin{align} d_{decel} & = \frac{v_{warp}}{j} = \frac{k*a}{j} \end{align} [/math]

Note that for ships that travel at up to 6 AU/s, k / j = k / (k/3) = 3, so these ships cover 3 AU while decelerating. The complication of not stopping warp at 0 can be safely ignored for distance calculations, because the distance that would be covered while decelerating from 100 m/s is insignificant compared to the ~450 billion meters it takes to decelerate from warp speed to warp drop speed.

Time

As with acceleration, time to decelerate from maximum warp velocity is worked out by rearranging the velocity equation. [math]\pagecolor{Black}\color{White} \begin{align} v &= k*e^{jt}\\ \frac{v}{k} & = e ^ {jt}\\ t & = \frac{\ln{(\frac{v}{k})}}{j} \end{align} [/math]

While the deceleration from s to 0 was insignificant in terms of distance, it is significant in terms of time. This means that the time to decelerate is calculated as follows:

[math]\pagecolor{Black}\color{White} \begin{align} t_{decel} & = t_{decel\_warp} - t_{decel\_s}\\ & = \frac{\ln{(\frac{v}{k})}}{j} - \frac{\ln{(\frac{s}{k})}}{j}\\ & = \frac{\ln{(\frac{v}{k})} - \ln{(\frac{s}{k})}}{j}\\ & = \frac{\ln{v} - \ln{k} - ln{s} + ln{k}}{j}\\ & = \frac{\ln{v} - \ln{s}}{j}\\ & = \frac{\ln{(\frac{v}{s})}}{j}\\ [/math]

Cruising

Distance

The distance covered while cruising is the total warp distance minus any distance covered while accelerating or decelerating.

[math]\pagecolor{Black}\color{White}d_{cruise} = d_{total} - d_{accel} - d_{decel}[/math]

For all but the fastest ships, this will be dtotal - 4 AU.

Time

This one is easy. Time spent cruising is simply

[math]\pagecolor{Black}\color{White}t_{cruise} = \frac{d_{cruise}}{v_{warp}}[/math]