The Power of Passwords: POS Shield Energy

Brief Notes On A High Energy Event 

I have been interested in analyzing the Drebuchet technique in the context of EVE's motion model for some time.  I was pleasantly surprised to find that the Rooks & Kings YouTube channel had a helpful clue along these lines. In this video, RnK explains how this technique works, and show some applications.  The essential idea is to take advantage of a ship being ejected from a POS shield to bump ships with tremendous energy.

 My interest in analyzing the motion of ships in these conditions is that POS shield ejection represents a new physical interaction that is not yet modeled in my notes.  I've modeled the effect of ships bumping ships, as well as ships bumping immovable objects, but what is the equivalent bump of changing a POS-shield password? 

Normally, I would try to measure the energy of the POS shield by seeing how far ships are flung.  What is their final resting distance from the starting position?  While this might be easy to do conceptually, it is much easier if someone takes the data for me.  Furthermore, the bump data from this video will provide a way to double-check any other measurements I make in the future.  

On with the data -- The video below has a clue at the 5:28 mark that can help me to determine how much energy is in a POS shield ejection. 




Is there anything more satisfying than a titan moving at 20km/s?  A couple things to note about this.  First, the narrator explains that the dreadnought is put in siege mode prior to ejection with the POS password change.  This means that during the event, the mass of the bumping ship is +900% or 10x the nominal mass.  Second, even though there is no explicit image showing the alignment of the bump dread and the titan target, I am going to assume alignment is perfect. 

Figure 1: Thanks to the velocity overlay, the Avatar velocity can be read.  It appears that the maximum is near 20km/s for this bump event!  Furthermore, note they have used a Naglfar Dreadnought to bump this titan. 

 You can see that the moment when the bump event occurs, where the Titan accelerates from stationary to 20km/s instantaneously.  We know the mass of both objects and we know the velocity of the Titan immediately after the collision. 

As a matter of perspective, the energy change required to accelerate a titan to 20,000m/s is approximately 2 x 10¹⁷ joules, which is a remarkable energy.  Consider that the energy released by a one megaton nuclear weapon is approximately 4 x 10¹⁵ joules.  So the POS shield energy delivers at least the energy of a 50 megaton warhead all to one object.  That would be a lot of energy, but keep in mind that we are looking at the energy of only one of the two ships, and unless the mass of the bumping ship matches the mass of the target ship, there will be residual energy reflected in the bump ship as well.  

Fortunately, we know the masses of the objects as well as one of the final velocities, leaving the velocity of the bumping ship before and after the bump event as unknowns.  All collisions in EVE are elastic, so we also have two equations -- one for conservation of momentum and one for conservation of energy.   I'm not including the equations here but you can review them in Part III of my notes or in the posts on bumping with mass-matching techniques.

Two Equations, Two Unknowns -- You Know What To Do

I solve for both the before bump and after bump velocities of the siege-mode dreadnought.  I find the velocities before and after the bump, respectively, 

$\Large v_1(0^-) = v_2 \frac{m_2 + m_1}{2m_1}$
$\Large v_1(0^+) = v_2 \frac{-m_2 + m_1}{2m_1}$

I'll assume that the Dreadnought is in siege resulting in the masses as follows:

• m₁ = m_SiegeNaglfar = 1.1 x 10¹⁰ kg
• m₂ = m_Avatar = 2.2 x 10⁹ kg

The velocity of the bump ship before and after the bump are, 

• v₁(0-) = 12000 m/s 
• v₁(0+) = 8000 m/s 

Prior to striking the titan, all of the energy is in the Naglfar, meaning that we can revise our calculation of the total POS shield energy from the motion of the one ship at that time, prior to the collision.  I find the total energy is 8 x 10¹⁷ joules, equivalent to a blast from 200 megaton warhead. 

Putting this in the context of being bumped by other ships, it would take all of the energy from over 700 thousand 500MN stabbers to deliver this much energy in a single bump, which isn't even realistic.  

If you left the titan to drift to a stop, it would come to a rest at τV_MAX = 81.4s x 20,000m/s = 1,600km.  Because a siege-mode Dreadnought has 10x the time-constant as non-siege-mode, the bump distance should be almost 6000km.  Indeed, the POS password is mightier than the sword.

Always More Questions

 You can have a lot of ships inside a POS shield.  Do they all get the same energy?  Or is there an equivalent bump mass that the POS shield delivers to the ships.  Like characterizing matter with high energy particle collisions, the POS shield could be characterized by looking at the resting distance of launching ships with a spectrum of masses. 
 

Acknowledgement

Thank you to Rooks and Kings for posting their technique with quantitative information, making it possible for EVE physics to move forward. 

The CSM XI Quiz - Candidate Questions (and answers)

[Apr 2017 - All equations should now be fixed and free of codecogs-related HTTPS bugs.  Thank you for your patience. 

Thank you,
S. Santorine]

But dreadful is the mysterious power of fate —

there is no deliverance from it by wealth or by war,

by towered city, or dark, sea-beaten ships.

- Sophocles, Antigone - Chorus

  We've all had to make hard choices about recruiting new corporation members.  As in life, choosing who to trust in EVE is critical.  This is also true when we are choosing CSM members to represent us.  Character and experience are important, but they aren't everything.  You also want to choose pilots who understand how EVE's physics model works.  These skills will come in handy when they are analyzing proposed ship changes and balance tweaks.  In anticipation of this CSM voting season, I present a few quick questions that are suitable for pilot interviews as well as help narrow down who really represents the math constituency.

   I present a few quick questions that should be easy to answer in an interview or chat-room context.  No worthy pilot should have any trouble answering these from the top of their head if they understand the intuitive physics principles that govern EVE.  Indeed, most Origin. pilots circa 2012 were able to answers questions of this type with little difficulty.  I am including links to selected CSM candidates answers although I will refrain from expressing any endorsements.  I hope all of you will take the time to inform yourself on all of the candidates and vote before March 25^th.

  As a final note, I want to mention that there are also some questions that prospective pilots should not be able to answer.   I have included a couple of these at the bottom of my post.  This may seem strange, but if you admit pilots who can answer these questions to your corp, its only a matter of time before they are recruiting on your voice communications channels for World of Warcrap — and I have yet to see any interesting physics in other combat games. 



  Good Luck to all the CSM candidates and the CSM XI team!
 
  S. Santorine, Ph.D.

Motion-Related Questions:

   Ship time-constant is something that every pilot should keep in mind to help them determine maneuvering rates.  I didn't actually post this question, but its quick one that you can use on prospective pilots:  

You have a ship mass of 10,000,000 kg and an inertia factor of 0.4.  You give your ship a new motion command, i.e. double-clicking in space.  Assuming no bumping, hostile webbing or active modules, what is the minimum amount of time you have to wait until you have reached over 98% of your final velocity?

  Solving this question can be a simple matter of knowing that e^-4 is less than 0.02 or 2%.  If you don't have that quantity memorized, you can just look at the equation for motion that CCP has shared with the community, 
  $\large v(t) = V_{MAX} \left ( 1 - e^{-t/(M \cdot I)} \right )$

   We just need to calculate when the the (1-e^-t/τ) term will be greater than 0.98.  The time-constant of an EVE ship is just the product of the mass and the 'inertia factor' (which is really the reciprocal of drag constant in EVE, and has units of s/10⁶kg).  So, the ship in this problem statement, τ = 4s, therefore you need to wait 16 seconds to be at 98% of your maximum velocity in any direction.

   The time-constant of your ship is a useful quantity to remember because it is the characteristic time that governs the rate at which an object can change its motion.  This applies to all ship motion including orbits, bumping, tackling and maneuvering.


  Another important constant to keep in mind about a ship is the characteristic maneuvering distance.  This is the distance needed to stop your ship.  It is the distance that you drift in direction x, when you change to moving in direction y.  It is also the orbit distance at which your orbits transition from being ship-velocity-limited to maximum-acceleration-limited. With that motivational preface in mind, I asked Apothne the following friendly question:

You are moving at a constant 1000m/s.  Your ship has a mass of 1,000,000 kg, and has an inertia factor of 3.  If you command the ship to stop (Ctrl-Space) and apply no other motion commands to the ship, what distance does it take for your ship to come to a complete stop, i.e. if you waited forever?

  You can determine this parameter by just writing the homogeneous solution to the one-dimensional motion equations, and integrating the total distance. 

  $\large v(t) = V_{MAX}e^{-t/\tau}$

  $\large x(t) = x(0) + V_{MAX} \int_0^t e^{-t/\tau}$

  $\large x(\infty) - x(0) = \tau V_{MAX}$

   So, final resting position of the ship will be 3000 meters from the position where the stop command was issued.  This ignores server tick alignment, which considering the short time constant, can be a significant contributor.   



   After asking the following question of Joffy, I realized that you can't really do this problem in your head, however, I have written extensively on this subject, so you can always refer to Part III or my more recent posts. 

You decide to bump a lazy miner off their rock.  The mining barge you decide to annoy has a mass of 20,000,000 kg.  Which of the following ship/velocity choices would bump them the farthest, creating the most rage?
a) 10,000,000 kg with a velocity of 1300m/s
b) 20,000,000 kg with a velocity of 1000m/s
c) 100,000,000 kg with a velocity of 600m/s

  Answering this is a simple matter of writing the final bump distances in each case, but we don't need to know the inertia of the target because all of the bump distances will scale with that value.  All you really need to know is the velocity of the mining barge immediately after the bump event, i.e,

  $\large v_2(t = 0^+) = \frac{2 v_1 m_1}{m_1 + m_2}$

   The post-bump velocity in each of the cases above is,
a) 866m/s
b) 1000m/s
c) 1000m/s

   The other purpose of this question is to remind the participant that you have to have five times the mass to make up for a decrease in velocity of 40% for the same bump distance! 


   I asked Steve Ronuken the following teaser: 

Describe a situation where a ship, without the help of another ship, active modules, or going into or coming out of warp, can achieve acceleration that is twice the magnitude that can be achieved with double-click piloting?

  While this is not a deep conceptual question, it does make you think about the definition of acceleration. The answer is, of course, bumping into an immovable object, such as a asteroid or a station will cause you to change direction instantly in the opposite direction.  Collisions with objects are shown to be elastic so an immovable object just reflects all of the incident motion energy into the colliding ship. All my questions are trick questions and this is no exception.  When you bump into an immovable object, EVE changes your direction instantaneously meaning acceleration was infinite for a short time. 


   Niko Lorenzio wanted a quick question from me, so I came up with this entertaining word problem.  I've been in similar situations, although I tend to eat hummus when I play EVE: 

You are stationary in a ship with mass 100,000,000 kg on a station undock in 0.0-space.  You do not have any aggression timer.  You are about to get bumped by a hostile ship.  Unfortunately, you are using your right hand to eat a burrito and you can not use the mouse to enter any piloting directional commands!  The only decision you can make is to turn on your 100MN afterburner by hitting a function key with your left hand, or leave it off.  If your goal is to prevent being bumped out of docking range by minimizing bump distance, should you activate the 100MN AB?

  To solve this, you just need to write the ratio of the ship bump distance with both cases for mass.  I could have, in fact, left out the mass of your ship because it doesn't matter. In this case, however, you need to write the final bump distance that includes the time constant of your ship, 
  $\large x(t \rightarrow \infty) = \frac{2 v_1 m_1}{m_1 + m_2} m_2 I_2$


   In this case, you have to determine whether the quantity is larger or smaller when you add 50Mkg to m₂.  You can see that this quantity is always less than 1, 

  $\large \frac{x(t \rightarrow \infty)_{no AB}}{x(t \rightarrow \infty)_{AB}} = \frac{(m_1 + m_2 + 50)(m_2)}{(m_1 + m_2)(m_2+50)} \leq 1$

   Therefore, if you activate the Afterburner, you will always be bumped a greater distance unless you are piloting actively, which is impossible while eating a burrito. 

Tracking-Related Questions:

  This question is always a good one to ask pilots for turret tracking intuition.  We found that most recruits were able to answer this without much issue.  I thought maybe Xenuria would enjoy it too:

Assume for a moment that you have infinite optimal range and you have two targets to choose from, both the same distance from your turret-fitted ship:  The first target has a transverse velocity of 400m/s and a signature radius of 60m.  A second target has a transverse velocity of 800m/s and a signature radius of 120m.  Which target would you have a greater chance to hit? 

  This is also a bit of a trick question.  To answer this question, all you need to know is that in the exponent for tracking mechanics, target angular velocity and target signature radius always appear as a fraction.  Now, the way I asked the question did not originally indicate the distance to the targets, so I posted a correction highlighted in blue above.  You can answer the question assuming they are at the same distance, or not.

   If they are at the same distance, the ratio of these terms is the same and they are both will have the same hit probability, regardless of what type of turret you use. So, the correct answer for same-distanced targets is that neither of these has a greater chance to hit. Recall the tracking-dependent factor in the hit probability equation, 

$\large p_{hit} = 2^{-\left ( {\frac{s_{res} \omega_{target}}{s_{rad}\omega_{turret}}} \right )^2} $

   Along with the definition for angular velocity, $\omega = v_{transverse} / R $, you have everything you need to answer this for differently-distanced targets as well, although it is easiest if you apply log-2 arithmetic. 


   Mr. Hyde received the following little question: 

Assume for another moment that you have infinite optimal range. You notice that you have exactly a 50% chance to hit a particular target.   You also know that your turrets are medium sized (i.e. signature resolution of 125m) and you have an angular tracking rate of 0.1 rad/sec.  If the target has a signature radius of 125m and is at a distance of 1000 meters, what is the transverse velocity of the target? 

  The solution is quite direct once you have broken down the tracking equation into its parts. 

$\large p_{hit} = 2^{-\left ( {\frac{s_{res} \omega_{target}}{s_{rad}\omega_{turret}}} \right )^2}$


  If the hit probability is exactly 0.5 then you know that,

$\large \frac{s_{res} \omega_{target}}{s_{rad} \omega_{turret}} = 1$


  With the turret resolution and target signature radius the same, it means that the angular rates for the target and the turret are also equal.  We know that the target has an angular velocity of 0.1rad/sec, or 100m/s at 1000meters distance. 




  There are also many good quiz questions in the subject areas of locking time, tackling distance, aggression timing, and spawning area enclosed by of an interdictor sphere.  The issue with these questions is that they usually require the aid of written diagrams or a computer.  Tackle distance is described by a translinear equation for which there is no closed form and iteration must be used to get a precise answer.  Asking candidates for answers along these lines is not really practical, even if it is fun.

Questions that EVE pilot recruits should not know the answer to:

1 - Who is this champion and how would you build them?

 

2 - Who is this person, and what do they do? 

 

Musical Interlude

  Enjoy!