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!

 

Improving Bumping with Mass Matching: Part II - The Data!

Its the most wonderful time of the year, and the holiday season is a good time for contemplation on meaning and methods.  When I had first looked at the mass-match bumping data that I promised you in November, I found some interesting things.  This post took a little bit longer than I had anticipated because I wanted to present a plausible hypothesis for what is happening.  Now that this is done, I hope it will motivate you to explore the bump behavior in game further and hurl your adversaries across the grid with greater energy than ever before.  Enjoy the post, and please contact me with questions or corrections. 

In my last post, I proposed a way to improve how far you can bump objects of much different size by thinking a bit about impedance matching, or mass matching, to couple more energy to the bumped ship.  Once again, the on-grid motion mechanics of EVE are based on real physics models.  Understanding this physics deeply is a way to learn how to do things that were previously believed impossible.  I continued in this line of reasoning by applying mass matching to the analogy of a Newton's cradle device and applied it to ship collisions. 

Recall in the previous post that I showed that if we use an intermediate ship to match the masses between the bumping ship and target ship, that we can deliver more of the bump energy to the target.  That is, one ship bumping another with no intermediate will bump it with velocity, 
$\Large v_{2-ship}(t=0^+) = v_1 \frac{2 m_1}{m_1 + m_3}$  
Whereas, if an intermediate mass, m₂, is present, the target ship bump velocity is,
$\Large v_{3-ship}(t=0^+) = v_1 \frac{4 m_1 m_2}{(m_1 + m_2)(m_2 + m_3)}$ 

To maximize the distance that we bump our target, I find the best mass for the intermediate ship is, $m_{2,optimal} = \sqrt{m_1 m_3}$ . The additional distance that you will bump the target in this case can be written as the ratio, 
$\LARGE \frac{x_{3-ship}(t \rightarrow \infty)}{x_{2-ship}(t \rightarrow \infty)} = \frac{2 (K + 1)}{(\sqrt{K} + 1)^2} $.
Readers will recall that K is the ratio of the bump target ship to the initial bumping ship, $K = m_3 / m_1$.  In the case where the square-root of K is much larger than 1, we can simply approximate this ratio as a factor of 2.  In most cases, however, the additional bump distance for capitals is 40% to 60%.  Based on this analysis, I setup an experiment where I bump a Thanatos with either a 500MN MWD Vagabond, or with an intermediate Orca bumped by the 500MN MWD Vagabond. On to the experimental results...

Data Doesn't Lie

I have shown how to calculate the bump distance of objects in EVE.  Lets see if bumping with an intermediate ship helped to improve the bump distance, and if so, by how much?


Figure 1 shows our comparison point with a traditional bump setup -- a 500MN MWD Vagabond hull bumping a stationary Thanatos.  Keep in mind that for each of these data points, I waited around 8 x τ_Thanatos = 318 seconds, for the velocity of the bumped ship to drop below 0.1 m/s.  Of course, if I could accurately sample the ship velocity immediately after the bump, none of that would be necessary because I could compute the settle distance from τV_BUMP, but I'm too lazy to setup OBS.  As you can see in Figure 1, there is good agreement between data and theory.  

Figure 1: Data for the two-ship Carrier bumping is shown.  Time constant for Thanatos in this situation is, τ_Thanatos = 39.7s.  Note that I make a small correction to the bump distance of the carrier by 500m (See Part I for notes on zero-meter radius), and I show quantized bump distance greater than 10,000 meters. 

Next, I positioned an unfitted Orca between 200m and 400m of the target Thanatos.  This distance should not be critical to the results as the only requirement is that the Orca does not have time to slow down after being bumped.  If the distance between the stationary Orca and the Thanatos is much smaller than τ_OrcaV_MAX,Orca, which is approximately 3000m, you should get similar results.   I then bump the Orca into the Thanatos using the 500MN Vagabond, with varying velocities. 

Also, note that these results should not depend on the Orca pilot's skills.  This is because the mass is not affected by skills and the 'inertia' (or drag) doesn't matter because the distance that the intermediate ship moves during the collision is small. As you can see in the data below, at low bumping ship velocities, the data and theory agree again, increasing the capital bump distance by 40%.  Something is happening, however, at higher bump velocities that is not described in the intermediate bump mechanics above!   

Figure 2: Bump distance data using closest optimal intermediate ship mass (optimal = 266Mkg, m_2,Orca = 250Mkg) is shown.  Theory predicts a straight line, however, above 2500m/s, a much larger bump distance is achieved.

What is happening here?!  


At 3000m/s and higher, the carrier is bumped an even greater distance!  Can a simple explanation capture what is going on here? 

Without understanding what the bump model is doing internally all we can do is test hypotheses.  Lets start with a  simple hypothesis based on notions of limitations of the tick rate in the game, breaking up a complex bump interaction into individual events.  

The Double-Bump Hypothesis

Notice that the transition to this longer bump distance happens only at high velocity, above 2500m/s.  What if the bumping ship is hitting both the intermediate ship and the bump target before it is being slowed down by either of the collisions.  Then the intermediate bump ship would then hit the target ship, contributing energy from the bumping ship twice!  In this arrangement, the input energy is being double counted.*

If you derive the motion of the target ship after both bumps, you have to account for the fact that the target is already moving.  I derive the following mess for the target velocity after all of the bump events are complete.  Note that I call this time $t = o^{++}$ because it is after the target has been struck twice,  

$\Large v_3(t = 0^{++}) = v_1 \frac{2 m_1(m_1m_3 + m_1m_2 + 3m_2m_3 - m_2^2)}{(m_2 + m_3)(m_1 + m_3)(m_1+m_2)}$


Plotting this against our data, I see qualitative agreement in Figure 3, compared to this larger bump distance in the 3-ship case.  As with many models, it explains almost all of the data, but this does not mean that it is the only explanation, or that it is correct for all circumstances. 
 

Figure 3: The intermediate ship data, for higher strike velocities, shows first-order agreement with the double-bump hypothesis.

Although it is impossible to get definitive proof, it is plausible that EVE works in this way.  Also, consider this 'hack' from the developer's perspective -- If you needed to write a collision algorithm that worked for any number of ships, with synchronization between any number of clients, and limited to a finite time-step, you'd compartmentalize the ship interactions, too.  So, if my hypothesis is correct, CCP's engineering of this solution meets the challenging needs of the game, even if it does not replicate a perfectly adiabatic process in all situations.  

This presents an obvious and intriguing possibility.  If there is a way to double-count energy, is there a way to triple count it?  What about multiply the input energy by any arbitrary amount?  Can we deliver more energy to the target ship than we put into the collision event? 

Looking at the energy for each of the bump cases in Figure 4, you can see that the double-bump phenomenon is now quite efficient, delivering almost 80% of the energy to the target in spite of a 20x difference in the masses of these ships.  Under normal circumstances, less than 20% of the energy is delivered to the target from a 500MN bump stabber, so an almost 4X improvement is huge! 

Figure 4: At high velocities, the double-bump hypothesis appears to explain most of the energy transfer to the target capital ship.  A 4X improvement in bump distance energy is achieved over using the bumping ship alone. 

I have presented enough on this for one post, and given the community something to ponder and experiment with.  In my view, the next step is to study the double-bump regimen in greater detail, reoptimizing the intermediate bump ship mass as well as experiment with ways to further increase the number of double-bump events.  Is the order of the strikes from the intermediate ship and the bumping ship important?

Can we make triple-bump events?  More?  

The more we study the system, the more questions arise. 

Applications of Mass-matching Bump Technique

In recent history, CCP introduced a new rig called a Higgs anchor.  Whatever CCP's intentions for this module, at first glance it would appear that this rig has made bumping more challenging because it doubles the mass of the target ship.  With mass matching techniques, this effect can be mitigated.  The rig bonuses have quite a profound effect on the ship motion parameters:

Bonuses for the Higgs Anchor Rigs:

• Mass +100%
• Inertia -55%
• Velocity -75%

Looking more closely, the consequences of these parameters is that bumped distance will be reduced significantly for smaller ships when they have this rig fitted.  Figure 5 shows how a Higgs rig will affect the target bump distance for selected bumping ship configurations.  Based on my calculation, the distance ratio is,

$\Large \frac{x(t \rightarrow \infty)_{Higgs}}{x(t \rightarrow \infty)_{No \: Higgs}} = 0.45 \frac{m_1 + 2m_2}{m_1 + m_2}$

When the target mass is much larger than the bumping ship, the bump distance is almost unchanged.  So, capital ship bump distance will not be affected much by Higgs anchors, because the mass is already so large that energy transfer can't be made much worse.  Smaller ships, on the other hand, such as cruiser and BS hulls can try to reduce how far they get bumped. 

Figure 5: Bump distance ratio for ships with a Higgs Anchor rig fitted.  As you can see, capital ship bump distances should not be affected much (10³ Mkg) but smaller ships bump distance will be significantly shorter. 

With the information in my last two blog posts, mass matching techniques give players an option to counter the effects of mass in the game.  Even more significant is the double-bump phenomenon which can significantly extend the bump distance for massive targets.  Stationary mining ships not paying attention to an mass matching ship slipping up along side them deserve to be hurled across space like a pierogi at a pierogi-tossing competition.  

Appendix: More intermediate ships

The case when we have a large number of intermediate ships beckons to us with the possibility of perfectly efficient as an energy transfer between bumper and bumpee.  Of course, in game it would be very difficult to arrange this in an adversarial situation but it is an interesting case to describe formally.  

Start by imagining that you have N total ships.  Let m_i be the mass of i-th ship in the chain.  The first ship, m₁, is the bumping ship.  The final ship, with mass m_N, is the target ship to be bumped.  The N-2 intermediate ships are arranged in a line, and the target ship is placed at the end of this sequence of ships, i.e. m₂, m₃, ..., m_N.  The bumping ship strikes m₂ transferring energy to start the domino motion of Newton's cradle elements which will ultimately strike the target. 

Assuming that the striking ship is not triggering any double-bump effects, as we saw above, the first bump triggers each ship in the sequence to strike the next ship.  I can write the velocity of the next element i+1, based on the velocity of the previous ship and the masses,

$\Large v_{i+1} = \frac{2 m_i v_i}{m_i + m_{i+1}}$

It is easy to generalize this to the solve for the velocity of the final ship, by constructing the product of these terms, as,

$\Large v_{N} = v_1 \left ( \prod\limits_{i=1}^{N-1} \frac{2 m_i }{m_i + m_{i+1}} \right  )$

The obvious question I ask next is, what is the optimal tapering of the masses in the sequence {2, 3, ..., N-1} to achieve maximal energy transfer to the target?  

The equation below fits the bill and contains a trick.  It expresses the mass of the i-th ship as an exponent of the ratio of the target ship to the bumping ship, and then scaling along the array of ships to the α-th root.  You may find this to be of interest if you attempt this derivation yourself, 

$\Large m_i = m_1 K^{(\frac{i-1}{N-1})^\alpha}$

Good luck and let me know what you find.  I hope you have found this post enlightening and good luck with your bumping activities.

Interlude

"If you have two loaves of bread, keep one to feed the body,

sell the other to buy Drum & Bass tracks to feed the soul."

                                                     - Russian Proverb

* Of course, you could also try the assumption that the bumping ship hits the intermediate ship, and then the intermediate ship strikes the target, followed by the bumping ship striking the target.  Order matters in this case, because each step will create different initial conditions for the bumping of the target ship.