I posted a previous version of this thing elsewhere, back when Arioch was still using Digital Webbing to host his forum and some of us had to seek other venues if we wanted to discuss Outsider. A lot of things have changed since then.
After putting this off for forever and a half, I thought I'd try to clean it up and adjust it to newer information. There are a lot of reasons why I've been reluctant to try to fix this up, and not all of them have to do with issues of time or effort. Most of the ideas based on the previous calculations I'd come up with have already been proven wrong, and the revised math doesn't help; for instance, I like the new estimates for the length of a solon much less than the old ones. I'm pretty bad at math, there are likely to be many important facts that I missed, and a lot of my calculations are based on ambiguous statements, guesswork, and extrapolation from incomplete data. Almost every time I've run through the calculations I've come up with different results, which is always a bad sign. I expect to be wrong about even the most basic things here. And all conditions are subject to change, since Outsider is a living project rather than some ancient artifact carved in stone, so even things which might be correct right now will doubtless become wrong in time.
I've done a lot of revision on this, but I'm so terrible at math that I could easily have "fixed" a correct number to an incorrect one, and really I'm the last sort of person who should attempt this sort of thing in the first place.
There are several items open to interpretation and debate (or author fiat, should Arioch wish to strike me down), and some vital areas where there are no concrete numbers at all. Every time I reached such an item, I chose an interpretation that I felt was most likely to yield concrete results, even when said interpretation wasn't most likely to be true. I usually tried to verify the values I used, or at least to approach them somewhat rationally, but keep in mind, everything I did, I did in order to proceed forward as far as possible, rather than just until I hit a reasonable stopping point. I used to think I did well enough, but time has proven me wrong.
You probably shouldn't try to read this all at once; I sure as hell couldn't have written it all at once.
What can I say, I needed a weekly Outsider fix, and couldn't get one on the internet. It was going to be either this or learning how to draw Loroi myself. Naked Loroi. Wrestling. I may have made the wrong choice.
One more note here. I would never invest this much of my free time into any other webcomic, especially not with math involved. Not even if it was porn. Not even if it was really good porn. So, if you can take anything away from this post at all, I suppose it should be that Outsider is better than porn. Or that I'm really really crazy and kinda dumb. Or both, I guess.
1: Of Dice And Men
SpoilerShow
Since this is not specifically a gaming forum, it would be improper for me to assume everyone who will read this post is familiar with dice-based tabletop games. Somewhere around 99% of you can probably skip this section anyways.
I'm going to be talking about dice for a while.
Dice-based probability is pretty important to a lot of things here, so I feel obligated to go over it.
Arioch used a d6 system, meaning it runs off of standard 6-sided dice, the sort that you probably have a few dozen sets of in your house and are already familiar with if you ever played such classic games as Monopoly, Risk, Yahtzee, or Throw Small But Hard Objects At Your Brother.
Rolling a fair d6 gives you a 1-in-6 chance of getting any number between 1 and 6. This seems self-explanatory.
However, there are times in the Combat Sim where you would be required to roll 2 dice at once and use the sum, making possible sums any integer from 2 to 12. Assuming each d6 is fair, each d6 is an independent variable. That means each one has a 1/6 chance of any number from 1 to 6 coming up, regardless of circumstances or the result of the other die roll. So the total number of possible arrangements for 2d6 won't be 12, nor 11, but 36.
Imagine that one die is blue and that one die is red. The blue die has a 1/6 chance of any result, the red die has the same, and there are 11 possible sums,. But there are arrangements of each die that give certain individual sums an increased or decreased chance of occurring; for instance, there is only one arrangement that provides a sum of 2 (snakeyes), but you can get a sum of 3 with either a blue 2 and a red 1 or a blue 1 and a red 2. Since there are six possible states for the red die regardless of which of its six possible states the blue die has landed in, each possible state has a probability of 1/6 * 1/6, even though many of those states result in the same sum. The result of this is that a 2d6 roll's results follow a curve, rather than an even distribution.
Laid out a bit more explicitly:
This gets even worse if something involves a 3d6 roll, wherein you have 16 possible sums but 216 possible states. Fortunately/unfortunately, there are no 3d6 rolls in the Combat Sim (that I can recall), but there are instances of dependent variables. Smallcraft dodge rolls are the most pertinent example of this.
Dodging is an optional rule, and it's legitimacy is questioned even in the .pdf, so this probably won't seem important to anyone but me.
Let's say a given craft has a 6+ Dodge Bonus, meaning it only dodges if you roll a 6 on 1d6. So it's got a 1/6th chance of dodging a successful attack. So the odds of a successful hit against it are a multiple of 1/6*(beam chance-to-hit). So if you've got to make a roll of 8+ and then face a 6+ dodge, the result is a 34.72% chance to damage; the dodge roll subtracted about 7% from the chance to hit, whereas the dodge roll would subtract 16.67% from the chance to damage against a 2+ to-hit roll. The result is a more even spread of chance-to-damage than the chance-to-hit table would give, though it does lead to some... unlikely results, especially at point-blank range. A more complicated procedure might be more realistic.
That said, I'm going to proceed under the assumption that dodge rolls are a valid simulation of an important aspect of Outsider space combat.
So here's how a 6+ dodge affects the entire to-hit chart.
A 5+ dodge would double the Dodge Prob. and Effect of Dodge columns, 4+ would triple them, and so on. The effect on Damage Prob. is pretty self-explanatory there.
Another thing about the Beam To-Hit Table. One single table covers the chance-to-hit of every beam weapon in the game, regardless of the beam's nature. Lasers are as likely to miss as charged particle beams, even though only one moves at the speed of light. Although the to-hit table's preamble says it is modified by size and acceleration, it doesn't have anywhere to separate craft by their accelerative abilities, just size class; for the beam table to work as-is, the Bell would have to be rated a few sizes larger than the Tempest, but by the rules laid out in the size class table, the Bell would be around size 1, whereas the Tempest would be size 5. This is something which could present problems.
However.
The Bell example is something which would pretty much never become relevant in the use of this game as a combat sim. Both the Loroi Union and the Umiak Hierarchy have yet to declare war upon humanity, human ships have yet to open fire upon their vessels, and Loroi and Umiak vessels have yet to engage human ships at ranges where misses are plausible even by the combat sim as-is. While the potential for an exploit exists, it is irrelevant to any practical consideration. The important considerations are not every possible battle arrangement, but rather arrangements for battles between Loroi and Umiak forces. For now, at least.
There's also the issue of Umiak versus Loroi acceleration, and Umiak versus Loroi beam weaponry (and Loroi beam weapons vs. other kinds of Loroi beam weapons). Umiak ships have less acceleration and thus should be easier to hit than Loroi vessels of the same size, and Loroi and Umiak use different kinds of beam weapons which could have an easier or harder time hitting enemy vessels based upon any number of factors. It would be a bit too convenient if Umiak Plasma Foci exactly compensate for Loroi acceleration, and the idea that Loroi lasers travel at the same speed as Loroi blasters is just ridiculous.
But Umiak and Loroi weapons don't have to have exactly the same chance to hit over all ranges; they just have to be close enough at such ranges as they are actually used. Umiak beam weapons are able to pierce Loroi screens within 15 hexes; Loroi Pulse Cannons are useful all the way out to 30 hexes. The weapons page of Insider mentions that plasma beam weapons can contact a target at light-second ranges, suggesting the key difference isn't the chance to hit, but the distance at which they may penetrate one another's defensive screens. It's possible that the difference between actual Pulse Cannon and Plasma Focus chances to hit would be negligible at the range within which Umiak weapons are actually used. Lasers, having a practical range of around 3 hexes, also find the different qualities shaping their own chance-to-hit no longer relevant in practice.
The Starship Combat Sim is by its nature mere approximation. To reduce frustration many things may have been lost along the way. But in order to be a suitable approximation, it has to be closer to being right than being wrong. So, for instance, if a 2d6 roll were meant to be made against a given table, and a given event requires a roll of 11 or greater, then the probability of said event has to be closer to 8.33% than to either 16.67% or 2.78%. It doesn't have to be exactly at 8.33% for 11 or more to be the right roll to assign, it could be 6% or 12%, but it couldn't be 5% or 13%, as those would require that event be assigned to 12 or 10 respectively to have the best approximation a 2d6 spread can offer. There are certain ramifications to this; the Umiak could be much less likely to hit certain Loroi vessels at 15 hexes than the Loroi are to hit equivalent Umiak vessels at the same distance, and their chances to hit could still potentially be represented by the same diceroll. And even if the Umiak would require a different value than the Loroi at, say, 20 or 25 hexes, since Umiak maximum effective weapons range is within 15 hexes, then the approximation would still work well enough.
I am not Arioch, but were I he, I would have based the to-hit table off of the chances a Pulse Cannon has of hitting a standard Umiak vessel of a given size class at a given distance, since other weapons are probably close enough to the pulse cannon beam velocity that their own unique qualities would be largely irrelevant at the ranges within which they are actually used; the chance of a miss seems to be relevant largely at Pulse Cannon ranges, and much less so within PF or Blaster ranges. The difference between Loroi and Umiak acceleration is also minimized in this manner. Dodge rolls already aid in distinguishing between the different acceleration rates of fighters and other smallcraft. The Historians could badly upset this balance, but that's not pertinent at the moment.
There is one dice-related thing where I feel competent enough to suggest an adjustment. The Critical Engine Hit table does not have any possible "Delayed Reaction" result. I think I've seen three engine hits in the comic; the hit to the Bell's severed aft section, causing what the crit table would call a Catastrophic Secondary Explosion, the hit to the Winter Tide, causing a delay culminating with a Catastrophic Secondary Explosion, and, though I'm much less sure about this one, a hit to an Umiak aft on page 84, causing what looks to me like a secondary explosion, possibly catastrophic in nature. I suggest that a roll of 7 be assigned to "Delayed Reaction, roll again next segment." 7 is the most likely roll to come up on a 2d6, so it makes crunch sense and even becomes likely that Winter Tide should spend so many segments in a delayed reaction, culminating with a catastrophic secondary explosion. It also makes such an engine crit repairable with a Legendary Chief Engineer if you roll well enough before it erupts in a catastrophic secondary explosion, so players can experience for themselves the desperate hope that they can get that starboard reactor under control before it catastrophically explo--
...aw, hell.
The real problems I have are uncertainty dealing with the Wave Loom Device, and issues over the effect of the proplyd on CtH and damage. And also just plain CtH and damage outside of the proplyd too.
Which is unfortunate, because everything that will follow will depend on being able to find concrete values based on those things.
There are a few issues of clarity which I ought to lay out here.
I'm going to be citing both the pages of the comic and portions of the sim quite a bit here. To reduce confusion, when I cite the comic, I will do so by page number, and when I cite the combat sim, I will do so by section. While both the comic and the .pdf have a "Page 30," only the game has a "Section F1.1.2," and so any references to any "Page 30" will indicate page 30 of the comic, rather than the combat sim.
I use terms and abbreviations frequently in this document, to help stay within the character limit. Most of them are pretty straightforward, and come from either the comic or the .pdf themselves. They are:
51st: The 51st Strike Group of the Loroi Imperial Fleet. Stillstorm's group.
WLD: Wave Loom Device; the biggest gun on the Tempest. Abbreviated as Wld in the .pdf, capitalized because I think it scans better in written text this way.
PC: Pulse Cannon, the main long-range killing tools of the Loroi fleet.
PF: Plasma Focus, the main beam weapon of the Umiak fleet.
Hex: One hexagon on the tabletop. 10,000 kilometers. I don't know whether the hexagon is inscribed or circumscribed upon said 10 kkm diameter, but that's irrelevant here, as I use the term to signify 10,000 km of distance exactly.
_d: Position displacement value on the tabletop game; 1d for one turn means moving one hex in that turn, 2d moves 2 hexes, etc. The sim rounds 1d up to 16 km/s, but the exact figure is 15,625 m/s. I will be using 15,625 m/s because those 375 m/s can add up quite a bit.
Segment: One period of a turn, during which the game player assigns various actions. 80 seconds.
Turn: Eight segments. The easiest way of bookkeeping a ship's movement in the game. Mostly ignorable.
G: For my purposes here, one gravity is exactly 9.8 meters per second per second.
Now that that's all out of the way, I can finally get started.
I'm going to be talking about dice for a while.
Dice-based probability is pretty important to a lot of things here, so I feel obligated to go over it.
Arioch used a d6 system, meaning it runs off of standard 6-sided dice, the sort that you probably have a few dozen sets of in your house and are already familiar with if you ever played such classic games as Monopoly, Risk, Yahtzee, or Throw Small But Hard Objects At Your Brother.
Rolling a fair d6 gives you a 1-in-6 chance of getting any number between 1 and 6. This seems self-explanatory.
However, there are times in the Combat Sim where you would be required to roll 2 dice at once and use the sum, making possible sums any integer from 2 to 12. Assuming each d6 is fair, each d6 is an independent variable. That means each one has a 1/6 chance of any number from 1 to 6 coming up, regardless of circumstances or the result of the other die roll. So the total number of possible arrangements for 2d6 won't be 12, nor 11, but 36.
Imagine that one die is blue and that one die is red. The blue die has a 1/6 chance of any result, the red die has the same, and there are 11 possible sums,. But there are arrangements of each die that give certain individual sums an increased or decreased chance of occurring; for instance, there is only one arrangement that provides a sum of 2 (snakeyes), but you can get a sum of 3 with either a blue 2 and a red 1 or a blue 1 and a red 2. Since there are six possible states for the red die regardless of which of its six possible states the blue die has landed in, each possible state has a probability of 1/6 * 1/6, even though many of those states result in the same sum. The result of this is that a 2d6 roll's results follow a curve, rather than an even distribution.
Laid out a bit more explicitly:
Code: Select all
Chance of rolling:
2: 2.78%
3: 5.56%
4: 8.33%
5: 11.11%
6: 13.89%
7: 16.67%
8: 13.89%
9: 11.11%
10: 8.33%
11: 5.56%
12: 2.78%
Chance of rolling equal to or greater than:
2: 100%
3: 97.22%
4: 91.67%
5: 83.33%
6: 72.22%
7: 58.33%
8: 41.67%
9: 27.78%
10: 16.67%
11: 8.33%
12: 2.78%Dodging is an optional rule, and it's legitimacy is questioned even in the .pdf, so this probably won't seem important to anyone but me.
Let's say a given craft has a 6+ Dodge Bonus, meaning it only dodges if you roll a 6 on 1d6. So it's got a 1/6th chance of dodging a successful attack. So the odds of a successful hit against it are a multiple of 1/6*(beam chance-to-hit). So if you've got to make a roll of 8+ and then face a 6+ dodge, the result is a 34.72% chance to damage; the dodge roll subtracted about 7% from the chance to hit, whereas the dodge roll would subtract 16.67% from the chance to damage against a 2+ to-hit roll. The result is a more even spread of chance-to-damage than the chance-to-hit table would give, though it does lead to some... unlikely results, especially at point-blank range. A more complicated procedure might be more realistic.
That said, I'm going to proceed under the assumption that dodge rolls are a valid simulation of an important aspect of Outsider space combat.
So here's how a 6+ dodge affects the entire to-hit chart.
Code: Select all
Roll >/= Initial CtH Dodge Prob. Damage Prob. Effect of Dodge
2 100% 16.67% 83.33% 16.67%
3 97.22% 16.67% 81.02% 16.20%
4 91.67% 16.67% 76.39% 15.28%
5 83.33% 16.67% 69.44% 13.89%
6 72.22% 16.67% 60.19% 12.04%
7 58.33% 16.67% 48.61% 9.72%
8 41.67% 16.67% 34.72% 6.94%
9 27.78% 16.67% 23.15% 4.63%
10 16.67% 16.67% 13.89% 2.77%
11 8.33% 16.67% 6.94% 1.39%
12 2.78% 16.67% 2.31% 0.46%
Another thing about the Beam To-Hit Table. One single table covers the chance-to-hit of every beam weapon in the game, regardless of the beam's nature. Lasers are as likely to miss as charged particle beams, even though only one moves at the speed of light. Although the to-hit table's preamble says it is modified by size and acceleration, it doesn't have anywhere to separate craft by their accelerative abilities, just size class; for the beam table to work as-is, the Bell would have to be rated a few sizes larger than the Tempest, but by the rules laid out in the size class table, the Bell would be around size 1, whereas the Tempest would be size 5. This is something which could present problems.
However.
The Bell example is something which would pretty much never become relevant in the use of this game as a combat sim. Both the Loroi Union and the Umiak Hierarchy have yet to declare war upon humanity, human ships have yet to open fire upon their vessels, and Loroi and Umiak vessels have yet to engage human ships at ranges where misses are plausible even by the combat sim as-is. While the potential for an exploit exists, it is irrelevant to any practical consideration. The important considerations are not every possible battle arrangement, but rather arrangements for battles between Loroi and Umiak forces. For now, at least.
There's also the issue of Umiak versus Loroi acceleration, and Umiak versus Loroi beam weaponry (and Loroi beam weapons vs. other kinds of Loroi beam weapons). Umiak ships have less acceleration and thus should be easier to hit than Loroi vessels of the same size, and Loroi and Umiak use different kinds of beam weapons which could have an easier or harder time hitting enemy vessels based upon any number of factors. It would be a bit too convenient if Umiak Plasma Foci exactly compensate for Loroi acceleration, and the idea that Loroi lasers travel at the same speed as Loroi blasters is just ridiculous.
But Umiak and Loroi weapons don't have to have exactly the same chance to hit over all ranges; they just have to be close enough at such ranges as they are actually used. Umiak beam weapons are able to pierce Loroi screens within 15 hexes; Loroi Pulse Cannons are useful all the way out to 30 hexes. The weapons page of Insider mentions that plasma beam weapons can contact a target at light-second ranges, suggesting the key difference isn't the chance to hit, but the distance at which they may penetrate one another's defensive screens. It's possible that the difference between actual Pulse Cannon and Plasma Focus chances to hit would be negligible at the range within which Umiak weapons are actually used. Lasers, having a practical range of around 3 hexes, also find the different qualities shaping their own chance-to-hit no longer relevant in practice.
The Starship Combat Sim is by its nature mere approximation. To reduce frustration many things may have been lost along the way. But in order to be a suitable approximation, it has to be closer to being right than being wrong. So, for instance, if a 2d6 roll were meant to be made against a given table, and a given event requires a roll of 11 or greater, then the probability of said event has to be closer to 8.33% than to either 16.67% or 2.78%. It doesn't have to be exactly at 8.33% for 11 or more to be the right roll to assign, it could be 6% or 12%, but it couldn't be 5% or 13%, as those would require that event be assigned to 12 or 10 respectively to have the best approximation a 2d6 spread can offer. There are certain ramifications to this; the Umiak could be much less likely to hit certain Loroi vessels at 15 hexes than the Loroi are to hit equivalent Umiak vessels at the same distance, and their chances to hit could still potentially be represented by the same diceroll. And even if the Umiak would require a different value than the Loroi at, say, 20 or 25 hexes, since Umiak maximum effective weapons range is within 15 hexes, then the approximation would still work well enough.
I am not Arioch, but were I he, I would have based the to-hit table off of the chances a Pulse Cannon has of hitting a standard Umiak vessel of a given size class at a given distance, since other weapons are probably close enough to the pulse cannon beam velocity that their own unique qualities would be largely irrelevant at the ranges within which they are actually used; the chance of a miss seems to be relevant largely at Pulse Cannon ranges, and much less so within PF or Blaster ranges. The difference between Loroi and Umiak acceleration is also minimized in this manner. Dodge rolls already aid in distinguishing between the different acceleration rates of fighters and other smallcraft. The Historians could badly upset this balance, but that's not pertinent at the moment.
There is one dice-related thing where I feel competent enough to suggest an adjustment. The Critical Engine Hit table does not have any possible "Delayed Reaction" result. I think I've seen three engine hits in the comic; the hit to the Bell's severed aft section, causing what the crit table would call a Catastrophic Secondary Explosion, the hit to the Winter Tide, causing a delay culminating with a Catastrophic Secondary Explosion, and, though I'm much less sure about this one, a hit to an Umiak aft on page 84, causing what looks to me like a secondary explosion, possibly catastrophic in nature. I suggest that a roll of 7 be assigned to "Delayed Reaction, roll again next segment." 7 is the most likely roll to come up on a 2d6, so it makes crunch sense and even becomes likely that Winter Tide should spend so many segments in a delayed reaction, culminating with a catastrophic secondary explosion. It also makes such an engine crit repairable with a Legendary Chief Engineer if you roll well enough before it erupts in a catastrophic secondary explosion, so players can experience for themselves the desperate hope that they can get that starboard reactor under control before it catastrophically explo--
...aw, hell.
The real problems I have are uncertainty dealing with the Wave Loom Device, and issues over the effect of the proplyd on CtH and damage. And also just plain CtH and damage outside of the proplyd too.
Which is unfortunate, because everything that will follow will depend on being able to find concrete values based on those things.
There are a few issues of clarity which I ought to lay out here.
I'm going to be citing both the pages of the comic and portions of the sim quite a bit here. To reduce confusion, when I cite the comic, I will do so by page number, and when I cite the combat sim, I will do so by section. While both the comic and the .pdf have a "Page 30," only the game has a "Section F1.1.2," and so any references to any "Page 30" will indicate page 30 of the comic, rather than the combat sim.
I use terms and abbreviations frequently in this document, to help stay within the character limit. Most of them are pretty straightforward, and come from either the comic or the .pdf themselves. They are:
51st: The 51st Strike Group of the Loroi Imperial Fleet. Stillstorm's group.
WLD: Wave Loom Device; the biggest gun on the Tempest. Abbreviated as Wld in the .pdf, capitalized because I think it scans better in written text this way.
PC: Pulse Cannon, the main long-range killing tools of the Loroi fleet.
PF: Plasma Focus, the main beam weapon of the Umiak fleet.
Hex: One hexagon on the tabletop. 10,000 kilometers. I don't know whether the hexagon is inscribed or circumscribed upon said 10 kkm diameter, but that's irrelevant here, as I use the term to signify 10,000 km of distance exactly.
_d: Position displacement value on the tabletop game; 1d for one turn means moving one hex in that turn, 2d moves 2 hexes, etc. The sim rounds 1d up to 16 km/s, but the exact figure is 15,625 m/s. I will be using 15,625 m/s because those 375 m/s can add up quite a bit.
Segment: One period of a turn, during which the game player assigns various actions. 80 seconds.
Turn: Eight segments. The easiest way of bookkeeping a ship's movement in the game. Mostly ignorable.
G: For my purposes here, one gravity is exactly 9.8 meters per second per second.
Now that that's all out of the way, I can finally get started.
SpoilerShow
Beam weapon to-hit probabilities are on a table in section D2.10. The weapons tech page on Insider site says that beam weapons range is 300,000 km to hit, and that Loroi PCs can penetrate shields at that range but the Umiak PFs needs to be within 150,000 km to pierce screens. The Beam To-Hit and weapon damage tables bear this out, conditionally; you'd need a volley of two shots to manage to pierce a typical Umiak destroyers' shields at 30 hexes, which can take 7-8 damage, and PCs do a max of 8 damage at 30 hexes if you're lucky. You'd need an immense number of PCs to be able to bring to bear the number of volleys required to both hit and do damage to an enemy vessel at any further range than 30 hexes. However, the Tempest has another beam weapon, the WLD, which could be used for extreme range kills but is supposedly nearly impossible to use properly for a variety of reasons. It makes sense to me that the WLD have more killing power while still working on the same to-hit table as the PCs, because my understanding is that the WLD operates on essentially similar principles to the PCs but with vastly more energy. The WLD table shows that it does a small amount of damage up to 60 hexes away, and becomes more effective at closer ranges and based on how charged it is [section E4.2.7]. One charge accumulates per dedicated segment, a maximum of 8 charges can be accumulated, and at less than maximum safe charge the WLD would ream an Umiak Command Cruiser from 45 hexes away according to the armor and shield values laid out in Section G1.2 (the second G1.2; instead of going to I after H the letter goes back to G, which had been previously used for "Optional Rules"). That sort of extreme range killing power might be irrelevant in this battle, since you're pretty unlikely to hit a size 2 or 3 vessel with a beam weapon at that sort of range. So for my purposes I'll arbitrarily restrict the WLD's maximum useful firing range to 40 hexes.
Page 69 says there are eight Heavy and sixty-four Medium vessels, and I'll assume that's an accurate number even though there are a few reasons to doubt it. So killing a Heavy would take out a good portion of the Umiak force's offensive ability. I'd done a bit of math to find out how significant that portion might be, and eventually (after many errors and loose approximations) ended up with the seemingly-trivial figure of 2% to 3%. You don't have a great chance to hit what you're targeting with a WLD at 31-40 hexes, not when they're size-2 or 3, but it's still a chance. A Heavy vessel gives you a 28% chance to hit from 36-40 hexes, and about 42% at 31-35. But even with just three charges at 8x damage per charge, the WLD could be enough to take a Heavy out of the game if it hit. I'd take that shot unless there was something specific preventing me from doing so; sure, a 28% chance to get rid of 2% of the enemy's force seems trivial, but as long as it constitutes a trivial advantage rather than a net disadvantage, anyone would go for it. Stillstorm seems the aggressive type, too. But she didn't try to take that shot. So we ought to look to reasons not to fire.
I see four reasons. The first is that the game .pdf could be so wildly inaccurate as to render any and all judgement based upon it invalid. The second is that the WLD could be impossible to use for one reason or another; this includes the Tempest's heat saturation level being high enough that using it would damage the ship, or any preexisting damage to the ship at all (I'm willing to believe the device is so delicate that it is destroyed by one point of damage anywhere, not just the WLD itself). The third is that the charge time could take long enough that the enemy would already be in PC range while you're wasting time with the WLD; the Tempest has eight PCs, so with PCs instead of the WLD you could kill three-ish destroyers at ~30 hexes, instead of just one cruiser (getting about 4% rather than about 2%); the pre-fire effects of the WLD making it a poor choice. The fourth and most important is that the post-firing effects of the WLD would be extreme enough and take long enough to make using the WLD undesirable; specifically, loss of main batteries due to the WLD during time the Umiak would spend in PC range.
The first option is the most likely to be correct, followed by the second, but I'm going to dismiss them anyways.
The most important argument against accepting the first reason is that it does not allow me to get results. I found no concrete facts or figures about the WLD available anywhere outside of those few pages. The most strongly suggested reason to discard the .pdf's figures is the statement that weapons are overpowered. However, I am going through these calculations to find circumstances which could be easily achieved by the Umiak under which the WLD would be less than the best option in order to explain its disuse, and to do that does not require it's strength to be balanced with respect to enemy defenses, merely with respect to other weapons. To the extent that I understand the issue the first reason does seem the most reasonable explanation to me, but because it produces no results I will pursue alternatives.
I dislike the second possibility because it also yields no results, and because it strikes me as poor form. We didn't get to see the previous battle, so if Arioch wanted Tempest's WLD to be disabled I think he should have shown us some battle damage on the ship during the exterior shot at the end of the prologue, or had some screens on the bridge indicating that the ship is damaged in a fairly obvious way (like, a section being a livid red on an otherwise all-green layout of the Tempest or something). As it is, all that we have that might indicate such a thing is the knowledge that the Tempest was in an intense battle immediately prior to this one, that the WLD is considered destroyed after taking a single point of damage [section E4.2.7] and that the 51st has already taken heavy casualties and expended most of its heavy ordinance [page 59]. These could hint at damage, but, they don't really inform the audience that the Tempest is damaged, and the Tempest looked undamaged to me back on page 17. I have similar issues with things like the heat saturation being too high, especially since the only comment we've had regarding any sort of temperature on the Tempest is that the air is cold. And with pretty much every other way of disabling the WLD that I can think of. So I'll be looking for answers from the third and fourth reasons; charge time, and post-discharge effects.
Charge time issues require the minimum time to charge and discharge the WLD to exceed the Umiak main fleet's travel time between pages 68 and 76. To get an accurate minimum value for the Umiak vessels' travel time between pages 68 and 76, I will look to the gunboats on pages 69, 77, and 78.
On page 77 we see gunboats being shot at with blasters and on page 78 we see a gunboat shooting at the Loroi. That particular configuration of gunboat looked like a Quad Gunboat, which, in the sample ships [section G1.2] carry four Type 2 SR PFs, giving them an effective range of about 3 hexes. There are probably many variants with different weapons loads, and the gunboat on page 78 only fires two of its PFs when we see it, but 3 hexes is a useful figure and it seems a decent fit.
As of page 78, we still had yet to see an Umiak medium or heavy vessel fire its main batteries, even though we see a gunboat fire; since the first portion of the main force firing their first volley would be one hell of a big deal, combat-wise. I'd think it would be shown, and in sequence, meaning the gunboat entered firing range before the main fleet did.
We've also seen Winter Tide shot by a Heavy vessel, so it's not like the main force was waiting to shoot at the non-point-defense specialist vessels (something that I had to consider back when I started writing this, when we were all much younger).
It can get a little hard to pin down the exact sequence of events in a comic where you can expect a number of things to be happening at the same time or fairly close to one another. Even with a clearly laid out sequence, even if the author of the comic planned out every single thing down to the last second. But in order to explain the gunboats being shown firing before the KTKh we saw, either the gunboats hit firing range first, or the gunboats and the KTKh would fire at about the same time. The latter makes it seem less likely to me that Stills would have a segment or two to "reserve nothing" on point-defense, so instead of the torpedo barrage having bought time for the Umiak fleet to close in to weapons range it would have distracted the Loroi while the Umiak were firing. Which isn't what Alex said he observed.
I'd bet on Van Squad being out of main Umiak PF range when the gunboats open fire. However, I'm not ruling out the idea that the Umiak are in intercept range simultaneously with their gunboats' arrival. So my figure for the minimum travel time will be based on the idea that the KTKh is right at 15 hexes when the gunboat fires; the real travel time might be higher or lower, but it might not be much higher, and couldn't be significantly lower.
So, this means that the gunboat was 3 hexes or less from Van Squad when the KTKh is 15 hexes away or more; a difference of at least twelve hexes. The gunboats were not kicked loose by page 69, but rather by page 71, and thus had an initial velocity and initial position matching the main fleet up to that time. Gunboats have a thrust of 30-36Gs while a KTKh has a maximum of 28Gs. However, the KTKh isn't necessarily going to have attacked at maximum thrust; it's possible that they decided, in this specific situation, to keep it close by the rest of the attack force, and accelerate at around 24G to stay with the Type KK or 25G to stay around the Type K cruisers. A reason to do this is that getting too separate from the rest of the fleet would give the Loroi one segment in which to shoot the leading force without that segment having any possibility of shooting anyone else; enough space and time of separation between small enough forces (again, I'm assuming just one KTKh) turns a rush into defeat in detail. Allowing the enemy to separate and prioritize targets is a bad thing. The whole force WILL pass through Loroi weapons range, and it WILL take a given amount of time for the slowest part of the force to do so, and the 51st is presumably unable to kill the entire force in that time, but a portion of the force arriving ahead of the rest actually extends the force's total time within range without doing anything to extend the stationary 51st's time within each Umiak weapon's range.
However, that reason ceases to have much meaning as soon as the slower Umiak vessels are within 30 hexes of the Loroi PCs, so even if that were the tactic adopted, the KTKh would have probably been accelerating at full for at least the last 12 hexes of the distance to the Rapier. And it's honestly not that great of a reason to start with, at the very least because the purpose of this attack was to see if the Loroi would stay put because that's out of character,. Planning around the expectation of them doing something totally unreasonable and out of character is just... bad. It doesn't matter even if it could have helped here, it'd still be poor strategy.
This particular force has more Medium vessels than could form into formations of 5 with the 8 heavies, which to me makes it seem more likely that the KTKh was part of a first wave of faster attack vessels with accelerations of 28Gs; several destroyer types are capable of 28Gs, some are capable of significantly more. If I were the Umiak the chance that an initial, somewhat faster attack wave could hit a few more Loroi vessels even if the Loroi decided to run would be worth the potential additional loss of vessels through separation, as a principle. What's more, the figures I end up with don't give much of a disadvantage through separation anyways, the times to PC range just aren't different enough (see Calculation 7).
Minimum time of travel is the minimum time by which gunboats could have gained a 12 hex lead on the KTKh. That can be minimized by using the highest figure for acceleration on the part of the gunboats and the lowest possible figure for acceleration on the part of the KTKh; I will assume them to have the same initial velocity and initial distance because I'd assume the fleet to have started their attack run fairly unified and I'm not going to run through every single possible scenario, that way lies madness.
The lowest figure for acceleration on the part of the KTKh is to assume they decided to keep pace with the KK for the entire time. 24Gs.
It seems more reasonable to assume that the KTKh was moving at full thrust the whole time. 28Gs.
I'll assume the quad gunboat to have been moving at 36Gs the entire time, because this is hard enough as it is.
So that's a difference in acceleration of between 8 to 12 Gs, to produce a distance of 120,000,000 meters.
This minimum travel time means that even at the closest starting point possible, the Loroi had more than enough time to charge and fire the WLD at a range greater than 30 hexes.
Minimum distance at a given time and acceleration is arrived at by assuming an initial velocity of zero (since an initial negative velocity would mean that the Umiak exited the cloud while heading away from the Loroi, and doing something like that would require some sort of hyperspace thing, I think).
Minimum Umiak starting position would be 39 hexes away from Van Squad, except for a few problems.
Remember that it takes only 3 segments to build up a charge that can do 24 damage even past 40 hexes (with, frustratingly, no info given on 31-39 hexes); enough to blow past the shields and armor of an Umiak Command Cruiser and still have plenty left over. If charge time is to prevent the WLD from working the Umiak over, the Umiak fleet has to move from that starting position at 39 hexes to 30 hexes during the charge and fire segments. But the KK has a thrust of slightly less than 5; it takes one full turn to move 5 hexes from an initial velocity of zero, and minimum distance required an initial velocity of zero. I cannot reconcile a travel time of 1428.57 seconds with any figure that makes charging the WLD take too long, and the same is true of 1749.64 seconds. It would take the KK almost 11 segments to go from minimum starting distance to Tempest's PC range, which would give Tempest enough time to charge the WLD up 3 times and fire it, and two more segments cooling, and then repeat the entire process, before the enemy crossed into PC range; the Beam To-Hit table gives such a situation a ~28% chance to hit with the first WLD blast, and a ~42% chance to hit the second time, giving overall odds of killing at least one Heavy vessel at a little less than 70%, with the odds of killing two a little under 12%, before the PCs even become relevant. So there's just no way the Umiak start out that close. The time of travel is long enough that it always permits Tempest to charge up the WLD in time to fire it at an Umiak heavy vessel at extreme range at least once before the vessel would have entered PC range. The KTKh experiences this problem whether it's moving at 24 Gs or 28; there is always, always time to charge up enough for at least one shot. I understand that the weapons are overpowered according to every statement on the matter, but I was only talking about 3 charges there, you can cut the damage from weapons across the board in half and just compensate by charging the WLD six times, which would still be below the point where it automatically saturates the ship's heat systems or damages the ship. And if you halved weapons across the board, then PCs would no longer work all the way out to 30 hexes, but the WLD would still work at 40, actually serving to make it a better choice rather than a less desirable one. The WLD would have to bear no resemblance to the form it had in the Combat Sim whatsoever to change this. And then I would be sad.
Thus, of the reasons to not fire the WLD I mentioned before, the only remaining involves the aftereffects of firing, rather than the time spent charging or firing.
So post-fire cooldown is the most important figure for the rest of this exercise. Too bad I don't have a concrete value for that.
Moving ahead anyways, can't let a little thing like total ignorance stop me now.
Section E4.2.7.1 gives us a series of excellent reasons not to fire the WLD when you expect you'll need to use any of the ship's other systems or weapons any time soon. However, those reasons only apply after a ship's heat saturation surpasses its heat rating or if more than 6 charges are released at once, and it seems unlikely that one initial volley with the WLD at 6 or fewer charges would push them over the top. In fact, if their heat saturation is at zero, they could charge it to 7 and have no real negative effect, and charge it to 8 with the only negative effect being that the WLD is rendered unavailable for a maximum of 480 seconds, seconds during which I doubt the WLD would have been used anyways since the Umiak fleet might be within main battery distance by the time the WLD could have charged again.
So I tried to come up with a reasonable figure.
Section E4.2.7 tells us that charging the WLD uses the same rules as charging the Jump Drive. E5.2.1 tells us that charging the jump drive sucks out all main power so things like main batteries and main engines don't work, but secondary batteries, shields, maneuvering thrusters, and other auxiliary systems still work. I don't think firing the WLD would knock out any systems that aren't affected by its charging, so Tempest's point defense lasers would still work fine, but her engines and main guns present the real issue.
Loroi main engines probably need to be active in order to avoid being hit by beams, but this also only becomes relevant if those beams could get past the shield in the first place since the screens are notably not knocked out by main power failure (section A3.2.1; screens are run on auxiliary power), and according to Insider and the damage tables the Umiak can't really do that past 15 hexes. So the WLD knocking out main engines only becomes an issue if it's still the case when the Umiak are within 15 hexes; the Tempest could easily set up a movement vector away from the Umiak axis of approach prior to firing the WLD if maximizing distance from their approach lane is a concern, after all. So the Umiak would have to move from my arbitrarily set maximum WLD range of 40 hexes to the PF's maximum range of 15 hexes in the time it takes the engines to recover. According to the to-hit table, the only chance of missing the Tempest with a beam at 15 hexes or less--hell, the only chance of missing her at 27 hexes or less since she's a size 5 GCS--is a dodge roll supplied by either a Legendary Helmsman or Legendary Captain, but I'm guessing that even a living legend would still need the engines to actually be turned on in order to dodge. But such a dodge would still only be necessary within 15 hexes of distance; the PCs are relevant at 30 hexes.
So it is the weapons that present the most obvious issue. I will assume, for the sake of finding a minimum, that any impingement upon a target's time within effective PC range will disqualify the WLD from use.
So now comes the question of the time between WLD discharge and the return of Main Power.
There are four rolls on the Gremlin Table [section E4.2.7.1] that give us down-times if a ship's heat systems are saturated, which seems like the best place to look for insight on how the device affects main power.
2 of those 4 keep the main batteries out of commission for D6 segments; the others knocks the WLD out for D6 or 2D6 segments.
So 6 segments seems like the upper limit for the WLD disabling main batteries, and that only gets brought up if the ship exceeds its temperature threshold.
Since the average of a D6 roll is 3.5, and 2 out of 6 outcomes on the Gremlin Table disable Main Batteries, it seems to me that a roll on the Gremlin Table produces a theoretical average of 1 1/6 segments cooldown for the Main Batteries.
The Gremlin Table says it's intended for use when a ship has gone past it's heat rating, and I have no reason to believe Tempest has done so. But I needed a usable figure real bad, and this seemed a better way than just making something up.
Since PCs spend a full second charging between firing segments anyway, I'll multiply in the 1/6th part instead of rounding down. Since one idle segment translates to a wait of 160 seconds between firings, and since I got 1 1/6th idle segments,160 * 7/6 = ~187. So I'll assume a delay of 187 seconds after firing the WLD before PCs can fire.
...I might as well have pulled a number out of a hat, and had the same chance of being right.
So the KTKh has to move a full 10 hexes distance in 187 seconds with an acceleration of 28Gs. We can find the minimum velocity at 40 hexes required to do that.
Supposing the WLD's effects on the PCs are as I have interpreted them, then in order for the WLD to not be a good opener for the Loroi player, the Umiak player's fleet would have a minimum displacement value of slightly under 33 hexes/turn when 40 hexes from the Loroi player (rounding up, and assuming the KTKh applies full acceleration at 40 hexes).
I assume the Umiak to be capable of detecting the energy surge involved in the WLD charging, because it must be a whole lot larger than the energy surge that the Bell detected prior to being torn in half.
So if the Umiak know the WLD's capabilities, and it is indeed capable of destroying a Heavy 40 hexes away, and they see it charging, then even if they'd decided to clump all of their ships together, there is still no way the KTKh isn't going to rush.
The KTKh has the longest effective ranged PFs of any relevant Umiak ship, but it is, I think, overall, less valuable in this particular situation than the other heavies, should their weaponry be brought to bear, and by the time they've crossed to the ballpark of the the 40 hex range in this particular situation, it would have become obvious that those weapons WILL be able to get within their proper range.
So by rushing, the KTKh potentially spares another, more raw-damage capable Heavy a WLD to the face. And it becomes, however slightly, less likely to be hit by the WLD in the first place. And also if the Loroi choose to target a slower cruiser, then the KTKh significantly increased its own chances of surviving the non-WLD Loroi beam attacks without doing anything to significantly diminish the survivability of any other Umiak ship. And the reasoning I came up with for the KTKh to go at 24 or 25 or 26 Gs in the first place were pretty hollow, and was not anything I would have acted on in the situation.
But this is gonna bug me forever if I don't.
There are numbers for acceleration, which have given time of travel to a given distance from Van Squad, and now the WLD's figures (which I kinda made up) have yielded minimum values for velocity at a position relative to the Tempest.
Time to put those numbers together.
Page 69 says there are eight Heavy and sixty-four Medium vessels, and I'll assume that's an accurate number even though there are a few reasons to doubt it. So killing a Heavy would take out a good portion of the Umiak force's offensive ability. I'd done a bit of math to find out how significant that portion might be, and eventually (after many errors and loose approximations) ended up with the seemingly-trivial figure of 2% to 3%. You don't have a great chance to hit what you're targeting with a WLD at 31-40 hexes, not when they're size-2 or 3, but it's still a chance. A Heavy vessel gives you a 28% chance to hit from 36-40 hexes, and about 42% at 31-35. But even with just three charges at 8x damage per charge, the WLD could be enough to take a Heavy out of the game if it hit. I'd take that shot unless there was something specific preventing me from doing so; sure, a 28% chance to get rid of 2% of the enemy's force seems trivial, but as long as it constitutes a trivial advantage rather than a net disadvantage, anyone would go for it. Stillstorm seems the aggressive type, too. But she didn't try to take that shot. So we ought to look to reasons not to fire.
I see four reasons. The first is that the game .pdf could be so wildly inaccurate as to render any and all judgement based upon it invalid. The second is that the WLD could be impossible to use for one reason or another; this includes the Tempest's heat saturation level being high enough that using it would damage the ship, or any preexisting damage to the ship at all (I'm willing to believe the device is so delicate that it is destroyed by one point of damage anywhere, not just the WLD itself). The third is that the charge time could take long enough that the enemy would already be in PC range while you're wasting time with the WLD; the Tempest has eight PCs, so with PCs instead of the WLD you could kill three-ish destroyers at ~30 hexes, instead of just one cruiser (getting about 4% rather than about 2%); the pre-fire effects of the WLD making it a poor choice. The fourth and most important is that the post-firing effects of the WLD would be extreme enough and take long enough to make using the WLD undesirable; specifically, loss of main batteries due to the WLD during time the Umiak would spend in PC range.
The first option is the most likely to be correct, followed by the second, but I'm going to dismiss them anyways.
The most important argument against accepting the first reason is that it does not allow me to get results. I found no concrete facts or figures about the WLD available anywhere outside of those few pages. The most strongly suggested reason to discard the .pdf's figures is the statement that weapons are overpowered. However, I am going through these calculations to find circumstances which could be easily achieved by the Umiak under which the WLD would be less than the best option in order to explain its disuse, and to do that does not require it's strength to be balanced with respect to enemy defenses, merely with respect to other weapons. To the extent that I understand the issue the first reason does seem the most reasonable explanation to me, but because it produces no results I will pursue alternatives.
I dislike the second possibility because it also yields no results, and because it strikes me as poor form. We didn't get to see the previous battle, so if Arioch wanted Tempest's WLD to be disabled I think he should have shown us some battle damage on the ship during the exterior shot at the end of the prologue, or had some screens on the bridge indicating that the ship is damaged in a fairly obvious way (like, a section being a livid red on an otherwise all-green layout of the Tempest or something). As it is, all that we have that might indicate such a thing is the knowledge that the Tempest was in an intense battle immediately prior to this one, that the WLD is considered destroyed after taking a single point of damage [section E4.2.7] and that the 51st has already taken heavy casualties and expended most of its heavy ordinance [page 59]. These could hint at damage, but, they don't really inform the audience that the Tempest is damaged, and the Tempest looked undamaged to me back on page 17. I have similar issues with things like the heat saturation being too high, especially since the only comment we've had regarding any sort of temperature on the Tempest is that the air is cold. And with pretty much every other way of disabling the WLD that I can think of. So I'll be looking for answers from the third and fourth reasons; charge time, and post-discharge effects.
Charge time issues require the minimum time to charge and discharge the WLD to exceed the Umiak main fleet's travel time between pages 68 and 76. To get an accurate minimum value for the Umiak vessels' travel time between pages 68 and 76, I will look to the gunboats on pages 69, 77, and 78.
On page 77 we see gunboats being shot at with blasters and on page 78 we see a gunboat shooting at the Loroi. That particular configuration of gunboat looked like a Quad Gunboat, which, in the sample ships [section G1.2] carry four Type 2 SR PFs, giving them an effective range of about 3 hexes. There are probably many variants with different weapons loads, and the gunboat on page 78 only fires two of its PFs when we see it, but 3 hexes is a useful figure and it seems a decent fit.
As of page 78, we still had yet to see an Umiak medium or heavy vessel fire its main batteries, even though we see a gunboat fire; since the first portion of the main force firing their first volley would be one hell of a big deal, combat-wise. I'd think it would be shown, and in sequence, meaning the gunboat entered firing range before the main fleet did.
We've also seen Winter Tide shot by a Heavy vessel, so it's not like the main force was waiting to shoot at the non-point-defense specialist vessels (something that I had to consider back when I started writing this, when we were all much younger).
It can get a little hard to pin down the exact sequence of events in a comic where you can expect a number of things to be happening at the same time or fairly close to one another. Even with a clearly laid out sequence, even if the author of the comic planned out every single thing down to the last second. But in order to explain the gunboats being shown firing before the KTKh we saw, either the gunboats hit firing range first, or the gunboats and the KTKh would fire at about the same time. The latter makes it seem less likely to me that Stills would have a segment or two to "reserve nothing" on point-defense, so instead of the torpedo barrage having bought time for the Umiak fleet to close in to weapons range it would have distracted the Loroi while the Umiak were firing. Which isn't what Alex said he observed.
I'd bet on Van Squad being out of main Umiak PF range when the gunboats open fire. However, I'm not ruling out the idea that the Umiak are in intercept range simultaneously with their gunboats' arrival. So my figure for the minimum travel time will be based on the idea that the KTKh is right at 15 hexes when the gunboat fires; the real travel time might be higher or lower, but it might not be much higher, and couldn't be significantly lower.
So, this means that the gunboat was 3 hexes or less from Van Squad when the KTKh is 15 hexes away or more; a difference of at least twelve hexes. The gunboats were not kicked loose by page 69, but rather by page 71, and thus had an initial velocity and initial position matching the main fleet up to that time. Gunboats have a thrust of 30-36Gs while a KTKh has a maximum of 28Gs. However, the KTKh isn't necessarily going to have attacked at maximum thrust; it's possible that they decided, in this specific situation, to keep it close by the rest of the attack force, and accelerate at around 24G to stay with the Type KK or 25G to stay around the Type K cruisers. A reason to do this is that getting too separate from the rest of the fleet would give the Loroi one segment in which to shoot the leading force without that segment having any possibility of shooting anyone else; enough space and time of separation between small enough forces (again, I'm assuming just one KTKh) turns a rush into defeat in detail. Allowing the enemy to separate and prioritize targets is a bad thing. The whole force WILL pass through Loroi weapons range, and it WILL take a given amount of time for the slowest part of the force to do so, and the 51st is presumably unable to kill the entire force in that time, but a portion of the force arriving ahead of the rest actually extends the force's total time within range without doing anything to extend the stationary 51st's time within each Umiak weapon's range.
However, that reason ceases to have much meaning as soon as the slower Umiak vessels are within 30 hexes of the Loroi PCs, so even if that were the tactic adopted, the KTKh would have probably been accelerating at full for at least the last 12 hexes of the distance to the Rapier. And it's honestly not that great of a reason to start with, at the very least because the purpose of this attack was to see if the Loroi would stay put because that's out of character,. Planning around the expectation of them doing something totally unreasonable and out of character is just... bad. It doesn't matter even if it could have helped here, it'd still be poor strategy.
This particular force has more Medium vessels than could form into formations of 5 with the 8 heavies, which to me makes it seem more likely that the KTKh was part of a first wave of faster attack vessels with accelerations of 28Gs; several destroyer types are capable of 28Gs, some are capable of significantly more. If I were the Umiak the chance that an initial, somewhat faster attack wave could hit a few more Loroi vessels even if the Loroi decided to run would be worth the potential additional loss of vessels through separation, as a principle. What's more, the figures I end up with don't give much of a disadvantage through separation anyways, the times to PC range just aren't different enough (see Calculation 7).
Minimum time of travel is the minimum time by which gunboats could have gained a 12 hex lead on the KTKh. That can be minimized by using the highest figure for acceleration on the part of the gunboats and the lowest possible figure for acceleration on the part of the KTKh; I will assume them to have the same initial velocity and initial distance because I'd assume the fleet to have started their attack run fairly unified and I'm not going to run through every single possible scenario, that way lies madness.
The lowest figure for acceleration on the part of the KTKh is to assume they decided to keep pace with the KK for the entire time. 24Gs.
It seems more reasonable to assume that the KTKh was moving at full thrust the whole time. 28Gs.
I'll assume the quad gunboat to have been moving at 36Gs the entire time, because this is hard enough as it is.
So that's a difference in acceleration of between 8 to 12 Gs, to produce a distance of 120,000,000 meters.
Code: Select all
Calculation 1
f''(t) = 12(9.8 m/s^2) = 117.6 m/s^2
f'(t) = 117.6t + 0 m/s
f(t) = 58.8(t^2) + 0t + 0 m = 120,000,000 m
t^2 = ~2,040,816.33
t = ~1,428.57 seconds
With a KTKh acceleration of 24 Gs, the minimum time of travel from page 71 to page 78 is 1,428.57 seconds.
f''(t) = 8(9.8) m/s^2
f'(t) = 78.4t + 0 m/s
f(t) = 39.2(t^2) + 0t + 0 m = 120,000,000 m
t^2 = 3,061,224.49
t = ~1,749.64 seconds
With a KTKh acceleration of 28 Gs, the minimum travel time expands to 1,749.64 seconds.
Minimum distance at a given time and acceleration is arrived at by assuming an initial velocity of zero (since an initial negative velocity would mean that the Umiak exited the cloud while heading away from the Loroi, and doing something like that would require some sort of hyperspace thing, I think).
Code: Select all
Calculation 2
Time = 1,428.57 seconds
Acceleration = 24 Gs
f''(t) = 235.2 m/s^2
f'(t) = 235.2(t) + C_1 m/s
f(t) = 117.6(t^2) + C_1(t) + C_2 m
Assuming null initial velocity means assuming C_1 = 0.
f(1,428.57) = 117.6(1,428.57^2) + C_2 m
At the end of time t, the KTKh is within PF range of van squad; 15 hexes away.
The distance traveled at 24 Gs starting from null velocity over the course of time t takes them to 15 hexes at most from Van Squad.
Therefore C_2 is less than or equal to 150,000,000 m, but it seems unlikely to be too much less.
240,000,000 m + 150,000,000 m = 390,000,000 meters.
Remember that it takes only 3 segments to build up a charge that can do 24 damage even past 40 hexes (with, frustratingly, no info given on 31-39 hexes); enough to blow past the shields and armor of an Umiak Command Cruiser and still have plenty left over. If charge time is to prevent the WLD from working the Umiak over, the Umiak fleet has to move from that starting position at 39 hexes to 30 hexes during the charge and fire segments. But the KK has a thrust of slightly less than 5; it takes one full turn to move 5 hexes from an initial velocity of zero, and minimum distance required an initial velocity of zero. I cannot reconcile a travel time of 1428.57 seconds with any figure that makes charging the WLD take too long, and the same is true of 1749.64 seconds. It would take the KK almost 11 segments to go from minimum starting distance to Tempest's PC range, which would give Tempest enough time to charge the WLD up 3 times and fire it, and two more segments cooling, and then repeat the entire process, before the enemy crossed into PC range; the Beam To-Hit table gives such a situation a ~28% chance to hit with the first WLD blast, and a ~42% chance to hit the second time, giving overall odds of killing at least one Heavy vessel at a little less than 70%, with the odds of killing two a little under 12%, before the PCs even become relevant. So there's just no way the Umiak start out that close. The time of travel is long enough that it always permits Tempest to charge up the WLD in time to fire it at an Umiak heavy vessel at extreme range at least once before the vessel would have entered PC range. The KTKh experiences this problem whether it's moving at 24 Gs or 28; there is always, always time to charge up enough for at least one shot. I understand that the weapons are overpowered according to every statement on the matter, but I was only talking about 3 charges there, you can cut the damage from weapons across the board in half and just compensate by charging the WLD six times, which would still be below the point where it automatically saturates the ship's heat systems or damages the ship. And if you halved weapons across the board, then PCs would no longer work all the way out to 30 hexes, but the WLD would still work at 40, actually serving to make it a better choice rather than a less desirable one. The WLD would have to bear no resemblance to the form it had in the Combat Sim whatsoever to change this. And then I would be sad.
Thus, of the reasons to not fire the WLD I mentioned before, the only remaining involves the aftereffects of firing, rather than the time spent charging or firing.
So post-fire cooldown is the most important figure for the rest of this exercise. Too bad I don't have a concrete value for that.
Moving ahead anyways, can't let a little thing like total ignorance stop me now.
Section E4.2.7.1 gives us a series of excellent reasons not to fire the WLD when you expect you'll need to use any of the ship's other systems or weapons any time soon. However, those reasons only apply after a ship's heat saturation surpasses its heat rating or if more than 6 charges are released at once, and it seems unlikely that one initial volley with the WLD at 6 or fewer charges would push them over the top. In fact, if their heat saturation is at zero, they could charge it to 7 and have no real negative effect, and charge it to 8 with the only negative effect being that the WLD is rendered unavailable for a maximum of 480 seconds, seconds during which I doubt the WLD would have been used anyways since the Umiak fleet might be within main battery distance by the time the WLD could have charged again.
So I tried to come up with a reasonable figure.
Section E4.2.7 tells us that charging the WLD uses the same rules as charging the Jump Drive. E5.2.1 tells us that charging the jump drive sucks out all main power so things like main batteries and main engines don't work, but secondary batteries, shields, maneuvering thrusters, and other auxiliary systems still work. I don't think firing the WLD would knock out any systems that aren't affected by its charging, so Tempest's point defense lasers would still work fine, but her engines and main guns present the real issue.
Loroi main engines probably need to be active in order to avoid being hit by beams, but this also only becomes relevant if those beams could get past the shield in the first place since the screens are notably not knocked out by main power failure (section A3.2.1; screens are run on auxiliary power), and according to Insider and the damage tables the Umiak can't really do that past 15 hexes. So the WLD knocking out main engines only becomes an issue if it's still the case when the Umiak are within 15 hexes; the Tempest could easily set up a movement vector away from the Umiak axis of approach prior to firing the WLD if maximizing distance from their approach lane is a concern, after all. So the Umiak would have to move from my arbitrarily set maximum WLD range of 40 hexes to the PF's maximum range of 15 hexes in the time it takes the engines to recover. According to the to-hit table, the only chance of missing the Tempest with a beam at 15 hexes or less--hell, the only chance of missing her at 27 hexes or less since she's a size 5 GCS--is a dodge roll supplied by either a Legendary Helmsman or Legendary Captain, but I'm guessing that even a living legend would still need the engines to actually be turned on in order to dodge. But such a dodge would still only be necessary within 15 hexes of distance; the PCs are relevant at 30 hexes.
So it is the weapons that present the most obvious issue. I will assume, for the sake of finding a minimum, that any impingement upon a target's time within effective PC range will disqualify the WLD from use.
So now comes the question of the time between WLD discharge and the return of Main Power.
There are four rolls on the Gremlin Table [section E4.2.7.1] that give us down-times if a ship's heat systems are saturated, which seems like the best place to look for insight on how the device affects main power.
2 of those 4 keep the main batteries out of commission for D6 segments; the others knocks the WLD out for D6 or 2D6 segments.
So 6 segments seems like the upper limit for the WLD disabling main batteries, and that only gets brought up if the ship exceeds its temperature threshold.
Since the average of a D6 roll is 3.5, and 2 out of 6 outcomes on the Gremlin Table disable Main Batteries, it seems to me that a roll on the Gremlin Table produces a theoretical average of 1 1/6 segments cooldown for the Main Batteries.
The Gremlin Table says it's intended for use when a ship has gone past it's heat rating, and I have no reason to believe Tempest has done so. But I needed a usable figure real bad, and this seemed a better way than just making something up.
Since PCs spend a full second charging between firing segments anyway, I'll multiply in the 1/6th part instead of rounding down. Since one idle segment translates to a wait of 160 seconds between firings, and since I got 1 1/6th idle segments,160 * 7/6 = ~187. So I'll assume a delay of 187 seconds after firing the WLD before PCs can fire.
...I might as well have pulled a number out of a hat, and had the same chance of being right.
So the KTKh has to move a full 10 hexes distance in 187 seconds with an acceleration of 28Gs. We can find the minimum velocity at 40 hexes required to do that.
Code: Select all
Calculation 3
f''(t) = 28*9.8 m/s^2 = 274.4 m/s^2
f'(t) = 274.4(t) m/s + C_1
f(t) = 137.2(t^2) m + (C_1)t + C_2
f(0) = 0 m
0 = 0 + 0 + C_2, therefore C_2 = 0
f(187) = 100,000,000 m
100,000,000 = 137.2(187^2) + C_1(187)
100000000/187 = 137.2(187) + C_1
C_1 = 509,102.96
C_1 = f'(0) = velocity at 40 hexes in meters per second.
509,102.96 m/s ÷ 15,625 m/s/_d = ~32.58dI assume the Umiak to be capable of detecting the energy surge involved in the WLD charging, because it must be a whole lot larger than the energy surge that the Bell detected prior to being torn in half.
So if the Umiak know the WLD's capabilities, and it is indeed capable of destroying a Heavy 40 hexes away, and they see it charging, then even if they'd decided to clump all of their ships together, there is still no way the KTKh isn't going to rush.
The KTKh has the longest effective ranged PFs of any relevant Umiak ship, but it is, I think, overall, less valuable in this particular situation than the other heavies, should their weaponry be brought to bear, and by the time they've crossed to the ballpark of the the 40 hex range in this particular situation, it would have become obvious that those weapons WILL be able to get within their proper range.
So by rushing, the KTKh potentially spares another, more raw-damage capable Heavy a WLD to the face. And it becomes, however slightly, less likely to be hit by the WLD in the first place. And also if the Loroi choose to target a slower cruiser, then the KTKh significantly increased its own chances of surviving the non-WLD Loroi beam attacks without doing anything to significantly diminish the survivability of any other Umiak ship. And the reasoning I came up with for the KTKh to go at 24 or 25 or 26 Gs in the first place were pretty hollow, and was not anything I would have acted on in the situation.
But this is gonna bug me forever if I don't.
Code: Select all
Calculation 4
f''(t) = 24*9.8 m/s^2 = 235.2 m/s^2
f'(t) = 235.2(t) m/s + C_1
f(t) = 117.6(t^2) m + (C_1)t + C_2
f(0) = 0 m
0 = 0 + 0 + C_2, therefore C_2 = 0
f(187) = 100,000,000 m
100,000,000 = 117.6(187^2) + C_1(187)
100000000/187 = 117.6(187) + C_1
C_1 = ~512,768.16 m/s
Which is, in game terms... 32.82 hexes per turn.
Rounds up to 33.
Pretty much the same thing.
Time to put those numbers together.
SpoilerShow
In order to put those values together properly, I need a position for Van Squad relative to the Tempest and the approaching Umiak.
There are certain issues that can best be resolved by placing Van Squad's initial position further out than it's final position, but Van Squad's position at the moment it engages the torpedoes ought to be within 3 hexes or less of every other squad, so that each ship can cover every other ship's point defenses. So I put Van Squad's distance to the Tempest as of page 78 at 3 hexes, max. Since it was Winter Tide that first reported contact, I assume them to have been closest to the point of emergence, and since Van Squad's duty is point defense, I will assume them to be posted between the other squads and the enemy. Two hexes up front seems like the safest position for Van Squad, since the others could cover them better, while 3 hexes allows Van Squad to give the most coverage to the others.
Van Squad's job is to shoot torpedoes before they hit friendlies, right? And most turrets in the game are given a 315 degree firing arc, except lasers, which are given a 360 degree arc. So a ship loaded up for point defense can effectively cover an arc with a 315 degree radius. The range a weapon covers is related to how much time it has to fire at its targets, and the maximum range of the Loroi laser looks to be about 3 hexes, even against unarmored torpedoes (average damage output of 1 at 3 hexes). So if a missile is headed for, say, the Tempest, and it needs to get shot down by Van Squad's lasers, then if Van Squad were right next to the Tempest they'd have 3 hexes in which to shoot that missile, but if they were 3 hexes ahead of the Tempest in the direction of the missile, then there would be 6 hexes in which they could shoot that missile. This doesn't apply if the missiles simply head for the nearest available target, in which case positioning Van Squad ahead of the others would increase their vulnerability without significantly increasing anyone else's protection. But to me it looks as though the Umiak torpedoes had targets other than "whoever's closest," so Van Squad's marginally increased risk of being hit by a missile is compensated for by the increased security of the rest of the ships. The problem is that being in between the missiles and the rest of the ships will also put them between the enemies who fired said missiles and the rest of the 51st, diminishes the rest of the 51st's ability to defend Van Squad against said missiles, and that separation as a rule of thumb allows for easier target prioritization and defeat in detail. 2 hexes of distance seems close enough to me for the rest of 51 to be able to pick off enough of the torpedoes that get close to Van Squad without sacrificing much of the coverage Van Squad can provide the others. 3 hexes is pushing it, given the limits of the laser.
I'll get to the problem of Van Squad's initial position soon. For now, final position at 0-3 hexes is what matters.
If the KTKh fired at 15 hexes from the Winter Tide, it would be 15 to 18 hexes from the Tempest.
So now I'll try to put the answers so far together and finally solve the Mystery of the Solon (anyone going to bet it's just Old Man Second wearing a rubber mask and elevator shoes?).
40 hexes - 18 hexes = 22 hexes
There are three ways that I mentioned I could see the KTKh's acceleration going; slowed down to keep pace with the slowest portion of the fleet the whole time, full thrust the whole time, or keeping pace until hitting PC range, at which point full thrust is applied. I'll look at lowest and highest possible thrust, in order to provide the fullest range of values (applying full acceleration after 30 hexes would land somewhere in between).
Remember that part where I said an Umiak forward strike force moving at 28 Gs might outpace its 24 G counterpart enough to significantly increase Umiak casualties, but for our minimum-seeking purposes it probably really really really wouldn't?
(And also all the parts where I complained about not being able to use my old, 25 G based figures anymore? I miss 25 Gs. I really do.)
This is the part where I show some work there.
So I'll just assume the KTKh's acceleration is a constant 28Gs, have a minimum total time of travel as of page 78, and have (plausible, I hope) figures for both the initial velocity and position.
Van Squad's initial position depends on a series of unknowns; what exactly maneuver statements like "reforming" and "on the line" really mean on pages 69 and 73, the length of a solon, and the meaning of "intercept range" on page 69. However, if enough rough estimates can fill in enough of those unknowns, I can get a rough projection on the remaining unknown. The inestimable factor is the length of a solon, but filling in the necessary empty spaces around it with sufficiently accurate guesswork should make a practical estimate of a solon possible. So instead of doing that, I've tried in my customary manner; wild, uninformed guessing.
Winter Tide reported contact on page 68. On page 69, they estimated 996 solon to intercept, and Stills called Van Squad back into line. If "intercept" referred to the Umiak fleet being within PF range of the forward squadrons at their final position, then 996 solon would be ~1750 seconds at least. That seems too long. 1.76 seconds per solon gives the typical Loroi a resting heartbeat of about 34 beats per minute, meaning their average heart rate would be comparable to that of a world-class human athlete. Of course, their blood is different than ours, their heart is presumably different as well, their cellular respiration processes are unknowable and so is their oxygen use, and all that, but 34 bpm still feels out of bounds.
If intercept refers to other things, like the Umiak entering PC range (being intercepted by the Loroi, rather than intercepting the Loroi), or the Umiak entering PF range of Van Squad at it's initial position (intercept under current conditions, rather than intercept after reformation for point defense), that changes things.
I already calculated the minimum time of travel to PC range from my other figures, and the Tempest remained, for practical purposes, stationary; if "intercept" refers to getting within Tempest's PC range, then by their measure, 996 solons lasts about 1498 seconds. That gives the Loroi an average resting heartbeat of about 40 bpm. Which is excellent but not quite superhuman. Who knows what that really means when your blood's blue, though, I mean, their circulatory system is obviously already pretty different, what with the alien transport mechanism and all. (Hemocyanin kinda sucks, though. Like, a lot).
Then the figure of 996 solon could refer to the time it would take for the KTKh to get within PF range of the Winter Tide, at its original position on picket duty.
As of page 73, Van Squad was still reforming, and would only be ready in another 37 solon. I don't know how much time passed between page 69 and page 73, but I'm going to guess anyway, because it matters. Events and dialogue between page 69 and page 71 seemed more or less continuous to me, as though those pages took place nearly in real time without much condensing. When I tried reading the pages out loud to myself at what felt a proper tempo, almost a minute and a half had passed. Since the Umiak deployed their light craft on page 71, and since the weapons phase is towards the end of a segment for movement and timing purposes, and since page 71 is halfway between pages 69 and 73 and a segment is 80 seconds, aided by the time it took to read, I would guess that ~2 segments passed between pages 69 and 73. This means that Winter Tide was probably ~3 segments from its proper orientation in the "line." I don't know for sure what that actually means, though; given the persistence of velocity in this game, it could mean that it would take Van Squad another 37 solon at max thrust to get a speed of zero and the proper static position with respect to Center Squad, OR, it could mean that it would take Winter Tide another 37 solon at maximum thrust to reach the proper velocity to drift back to within the rest of the fleet's PD range at the moment of hostile contact, OR it could mean Van Squad was going to spend 37 solon turning about and fiddling around with minor positional/directional corrections, or something else entirely. I am not clever enough to know for sure. If there's any official statement on the matter, I didn't read it.
According to the thrust chart [section C3.0], 3 segments at thrust 6 yields a maximum of 4v but no d value. Even at maximum thrust, if Winter Tide were one or more full hexes removed from their position in "the line," it would take four segments at least to move to the adjacent hex (thrusting all the way), and even more to nullify that velocity and form a static line. Four segments bumps the time between Van Squad being recalled on page 69 and the end of the 37 solon mentioned on page 73 up to five minutes and twenty seconds, which seems like too long, and moving just one hex back to the line doesn't allow me find a minimal value; to that end, I will assume Van Squad's reform time was spent accelerating to establish a vector towards the 51st, rather than reaching a final static position, and I believe that this vector was designed to take them to within 3 hexes in front of Center Squad, with the Wing Squadrons arranged such that each squadron can help cover every other squad's hex with their point defenses. This would be a lot easier if I knew Van Squad's initial velocity and acceleration, or had some way to find that, but I don't. I'm going to assume that they had null acceleration and null velocity, because even though that doesn't quite make sense, it's both easier and more useful than any alternative I can think of. For all I know they were already hauling ass back to the 51st.
If the KTKh were to be moving at 24 Gs the entire time, this figure would go down dramatically.
But that's still a minimum. There are things I could adjust to lower a solon's length, like the number of segments Van Squad spent establishing their vector or the WLD's cooldown time, but the values I ascribed to them were already the best I could do without prescience or Word of God. I'm not going to retroactively change everything just so it all feels okay.
My gut tells me there's a lower value for a solon. At least, lower than 1.5 seconds.
But until word comes down from on high, it's the best I can do.
There are certain issues that can best be resolved by placing Van Squad's initial position further out than it's final position, but Van Squad's position at the moment it engages the torpedoes ought to be within 3 hexes or less of every other squad, so that each ship can cover every other ship's point defenses. So I put Van Squad's distance to the Tempest as of page 78 at 3 hexes, max. Since it was Winter Tide that first reported contact, I assume them to have been closest to the point of emergence, and since Van Squad's duty is point defense, I will assume them to be posted between the other squads and the enemy. Two hexes up front seems like the safest position for Van Squad, since the others could cover them better, while 3 hexes allows Van Squad to give the most coverage to the others.
Van Squad's job is to shoot torpedoes before they hit friendlies, right? And most turrets in the game are given a 315 degree firing arc, except lasers, which are given a 360 degree arc. So a ship loaded up for point defense can effectively cover an arc with a 315 degree radius. The range a weapon covers is related to how much time it has to fire at its targets, and the maximum range of the Loroi laser looks to be about 3 hexes, even against unarmored torpedoes (average damage output of 1 at 3 hexes). So if a missile is headed for, say, the Tempest, and it needs to get shot down by Van Squad's lasers, then if Van Squad were right next to the Tempest they'd have 3 hexes in which to shoot that missile, but if they were 3 hexes ahead of the Tempest in the direction of the missile, then there would be 6 hexes in which they could shoot that missile. This doesn't apply if the missiles simply head for the nearest available target, in which case positioning Van Squad ahead of the others would increase their vulnerability without significantly increasing anyone else's protection. But to me it looks as though the Umiak torpedoes had targets other than "whoever's closest," so Van Squad's marginally increased risk of being hit by a missile is compensated for by the increased security of the rest of the ships. The problem is that being in between the missiles and the rest of the ships will also put them between the enemies who fired said missiles and the rest of the 51st, diminishes the rest of the 51st's ability to defend Van Squad against said missiles, and that separation as a rule of thumb allows for easier target prioritization and defeat in detail. 2 hexes of distance seems close enough to me for the rest of 51 to be able to pick off enough of the torpedoes that get close to Van Squad without sacrificing much of the coverage Van Squad can provide the others. 3 hexes is pushing it, given the limits of the laser.
I'll get to the problem of Van Squad's initial position soon. For now, final position at 0-3 hexes is what matters.
If the KTKh fired at 15 hexes from the Winter Tide, it would be 15 to 18 hexes from the Tempest.
So now I'll try to put the answers so far together and finally solve the Mystery of the Solon (anyone going to bet it's just Old Man Second wearing a rubber mask and elevator shoes?).
40 hexes - 18 hexes = 22 hexes
There are three ways that I mentioned I could see the KTKh's acceleration going; slowed down to keep pace with the slowest portion of the fleet the whole time, full thrust the whole time, or keeping pace until hitting PC range, at which point full thrust is applied. I'll look at lowest and highest possible thrust, in order to provide the fullest range of values (applying full acceleration after 30 hexes would land somewhere in between).
Code: Select all
Calculation 5
At 24 Gs:
If Van Squad is 3 hexes from Tempest in the direction of the Umiak:
f(t) = 220000000 m
f''(t) = 235.2 m/s^2
f'(t) = 235.2t + 512768.16 m/s
220,000,000 = 117.6 t^2 + 512768.16 t + 0
Quadratic Formula: x = (-B +/- sqrt(B^2 - 4AC))/2A
t = (-512768.16 +/- sqrt(512768.16^2 + 4(117.6)(220000000)))/235.2
t = ~393.53 seconds
If Van Squad is 0 hexes from Tempest in the direction of the Umiak:
f(t) = 250000000 m
f''(t) = 235.2 m/s^2
f'(t) = 235.2t + 512768.16 m/s
250000000 = 117.6 t^2 + 512768.16 t + 0
t = (-512768.16 +/- sqrt(512768.16^2 + 4(117.6)(250000000)))/235.2
t = ~442.62 seconds
Minimum total travel time to KTKh firing range by 24Gs of acceleration is ~1428.64 seconds, and at 28Gs it's 1749.64 seconds (see Calculation 1).
Minimum initial distance and speed can be found by subtracting time from 40 hexes to intercept from total time.
Then applying negative acceleration along velocity at 40 hexes along the resulting time.
1428.57 - 393.53 = 1035.04 seconds
-235.2(1035.04) + 512768.16 = ~269326.75 m/s
Initial velocity is minimally about 269326.75 m/s; a little over 17d.
-117.6(1035.04^2) - 269326.75(1035.04) + C_1 = 400000000
117.6(1035.04^2) + 269326.75(1035.04) + 400000000 = C_1
Initial position is minimally about 804749756.79 meters from Tempest.
80 hexes.
1428.57 - 442.62 = 985.95 seconds
-235.2(985.95) + 512768.16 = ~280872.72 m/s
Initial velocity is minimally about 280872.72 m/s; a little under 18d.
Almost two full turns of acceleration.
-117.6(985.95^2) - 280872.72(985.95) + C_1 = 400000000
117.6(985.95^2) + 280872.72(985.95) + 400000000 = C_1
Initial position is minimally about 791245112.818 meters from Tempest.
79 hexes.
Code: Select all
Calculation 6
At 28 Gs:
If Van Squad is 3 hexes from Tempest in the direction of the Umiak:
f(t) = 220000000 m
f''(t) = 274.4 m/s^2
f'(t) = 274.4t + 509102.96 m/s
220,000,000 = 137.2 t^2 + 509102.96 t + 0
t = (-509102.96 +/- sqrt(509102.96^2 + 4(137.2)(220000000)))/274.4
t = ~390.94 seconds
If Van Squad is 0 hexes from Tempest in the direction of the Umiak:
f(t) = 250000000 m
f''(t) = 274.4 m/s^2
f'(t) = 274.4t + 509102.96 m/s
250000000 = 137.2t^2 + 509102.96t + 0
t = (-509102.96 +/- sqrt(509102.96^2 + 4(137.2)(250000000)))/274.4
t = ~439.10 seconds
1749.64 seconds total travel time
1749.64 - 390.94 = 1358.70
1749.64 - 439.10 = 1310.54
509102.96 - 1358.70(274.4) = 136275.68 m/s
509102.96 - 1310.54(274.4) = 149490.78 m/s
Initial velocity is minimally between 9d and 10d.
Less than one full turn of acceleration.
(137.2)(1358.70^2) + (136275.68)(1358.70) + 400000000 = 838437979.084 meters
(137.2)(1309.63 ^2) + (149490.78)(1310.54) + 400000000 = 831229583.92 meters
Initial position is minimally around 83-84 hexes from Tempest.
(And also all the parts where I complained about not being able to use my old, 25 G based figures anymore? I miss 25 Gs. I really do.)
This is the part where I show some work there.
Code: Select all
Calculation 7
Initial position:
831229583.92 meters
Initial speed:
149490.78 m/s
Acceleration:
24 and 28 Gs, for the Type K and the KTKh.
Time to 30 Hexes:
831229583.92 - 300000000 = 531229583.92 meters
117.6tt + 149490.78t + 531229583.92 = 0
t = (-149490.78 +/- sqrt(149490.78^2 + 4(117.6)(531229583.92)))/235.2
t = 1582.80 seconds
137.2tt + 149490.78 + 531229583.92 = 0
t = (-149490.78 +/- sqrt(149490.78^2 + 4(137.2)(531229583.92)))/274.2
t = 1498.05 seconds
There are 160 seconds between each shot of a PC.
There are about 85 seconds between the two entry times.
There is less than one PC burst's difference there.
So, let's check exit time then. Entry was at 30 hexes, exit is at -30.
831229583.92 + 300000000 = 1131229583.92
t = (-149490.78 +/- sqrt(149490.78^2 + 4(117.6)(1131229583.92)))/235.2
t = 2530.37
t = (-149490.78 +/- sqrt(149490.78^2 + 4(137.2)(1131229583.92)))/274.2
t = 2379.60
2530.37 - 2379.60 = 150.77
That's closer, but still less than one full PC burst's difference, assuming the Umiak accelerate all the way through rather than slowing. Which we saw them do.
The Loroi would have one more volley between the two groups, but only with respect to blasters and lasers.
That's not sufficient reason for the KTKh to fly slower than full thrust.
The time of travel must be significantly greater than 1749.64 seconds to make clumping the fleet up in a slower-moving mass worthwhile here, and even then.
Thus minimum time from gunboat release to page 78 is 1749.64 seconds, not 1428.57.
Because the idea behind the 1428.57 figure just doesn't make any sense.Van Squad's initial position depends on a series of unknowns; what exactly maneuver statements like "reforming" and "on the line" really mean on pages 69 and 73, the length of a solon, and the meaning of "intercept range" on page 69. However, if enough rough estimates can fill in enough of those unknowns, I can get a rough projection on the remaining unknown. The inestimable factor is the length of a solon, but filling in the necessary empty spaces around it with sufficiently accurate guesswork should make a practical estimate of a solon possible. So instead of doing that, I've tried in my customary manner; wild, uninformed guessing.
Winter Tide reported contact on page 68. On page 69, they estimated 996 solon to intercept, and Stills called Van Squad back into line. If "intercept" referred to the Umiak fleet being within PF range of the forward squadrons at their final position, then 996 solon would be ~1750 seconds at least. That seems too long. 1.76 seconds per solon gives the typical Loroi a resting heartbeat of about 34 beats per minute, meaning their average heart rate would be comparable to that of a world-class human athlete. Of course, their blood is different than ours, their heart is presumably different as well, their cellular respiration processes are unknowable and so is their oxygen use, and all that, but 34 bpm still feels out of bounds.
If intercept refers to other things, like the Umiak entering PC range (being intercepted by the Loroi, rather than intercepting the Loroi), or the Umiak entering PF range of Van Squad at it's initial position (intercept under current conditions, rather than intercept after reformation for point defense), that changes things.
I already calculated the minimum time of travel to PC range from my other figures, and the Tempest remained, for practical purposes, stationary; if "intercept" refers to getting within Tempest's PC range, then by their measure, 996 solons lasts about 1498 seconds. That gives the Loroi an average resting heartbeat of about 40 bpm. Which is excellent but not quite superhuman. Who knows what that really means when your blood's blue, though, I mean, their circulatory system is obviously already pretty different, what with the alien transport mechanism and all. (Hemocyanin kinda sucks, though. Like, a lot).
Then the figure of 996 solon could refer to the time it would take for the KTKh to get within PF range of the Winter Tide, at its original position on picket duty.
As of page 73, Van Squad was still reforming, and would only be ready in another 37 solon. I don't know how much time passed between page 69 and page 73, but I'm going to guess anyway, because it matters. Events and dialogue between page 69 and page 71 seemed more or less continuous to me, as though those pages took place nearly in real time without much condensing. When I tried reading the pages out loud to myself at what felt a proper tempo, almost a minute and a half had passed. Since the Umiak deployed their light craft on page 71, and since the weapons phase is towards the end of a segment for movement and timing purposes, and since page 71 is halfway between pages 69 and 73 and a segment is 80 seconds, aided by the time it took to read, I would guess that ~2 segments passed between pages 69 and 73. This means that Winter Tide was probably ~3 segments from its proper orientation in the "line." I don't know for sure what that actually means, though; given the persistence of velocity in this game, it could mean that it would take Van Squad another 37 solon at max thrust to get a speed of zero and the proper static position with respect to Center Squad, OR, it could mean that it would take Winter Tide another 37 solon at maximum thrust to reach the proper velocity to drift back to within the rest of the fleet's PD range at the moment of hostile contact, OR it could mean Van Squad was going to spend 37 solon turning about and fiddling around with minor positional/directional corrections, or something else entirely. I am not clever enough to know for sure. If there's any official statement on the matter, I didn't read it.
According to the thrust chart [section C3.0], 3 segments at thrust 6 yields a maximum of 4v but no d value. Even at maximum thrust, if Winter Tide were one or more full hexes removed from their position in "the line," it would take four segments at least to move to the adjacent hex (thrusting all the way), and even more to nullify that velocity and form a static line. Four segments bumps the time between Van Squad being recalled on page 69 and the end of the 37 solon mentioned on page 73 up to five minutes and twenty seconds, which seems like too long, and moving just one hex back to the line doesn't allow me find a minimal value; to that end, I will assume Van Squad's reform time was spent accelerating to establish a vector towards the 51st, rather than reaching a final static position, and I believe that this vector was designed to take them to within 3 hexes in front of Center Squad, with the Wing Squadrons arranged such that each squadron can help cover every other squad's hex with their point defenses. This would be a lot easier if I knew Van Squad's initial velocity and acceleration, or had some way to find that, but I don't. I'm going to assume that they had null acceleration and null velocity, because even though that doesn't quite make sense, it's both easier and more useful than any alternative I can think of. For all I know they were already hauling ass back to the 51st.
Code: Select all
Calculation 8
f''(t) = 294 m/s^2
f'(t) = 294t + C_1 m/s
f(t) = 147(t^2) + t(C_1) + C_2 m
initial velocity set to 0, therefore C_1 = 0
initial position set to 0, therefore C_2 = 0
f'(240) = 294(240) = 70560 m/s
f(240) = 147(240)(240) = 8,467,200 m
Van Squad moves at max thrust towards Center Squad across 8,467.2 km and then promptly turns to face the Umiak, moving at a velocity of 70.56 km/s towards the 51st.
The best way I see to minimize the length of a solon is to assume that that velocity persisted all the way up to Page 78.
Recall that the Umiak gunboats were on the move for at least 1749.64 seconds.
Van Squad accelerated for 240 of those seconds, covering an area of 8,467,200 m.
Van Squad now has, at most, zero acceleration, since they had established their vector and I want a maximum figure.
1749.64 - 240 = 1509.64
70560 m/s * 1509.64 s = 106520198.4 m
106,520,198.4 + 8,467,200 = 114,987,398.4 m
Since I am assuming that Van Squad is within 3 hexes of Center Squad as of page 78 for reasons elaborated earlier, we need to add those 3 hexes to that figure to get the maximum.
114987398.4 + 30000000 = 144987398.4 m
So Van Squad's starting position was at most about 14 hexes away from the Tempest in the general direction of the Umiak.
Since effective Umiak PF range has a maximum value of 15 hexes, that means that it was 996 solon until the Umiak were 29 hexes from Center Squad, starting from that minimum 83 hexes away. This produces a higher value for the solon than if "intercept" referred to entering PC range, since that was at 30 hexes, not 29. So this calculation is pointless. Continuing anyway.
f''(t) = 274.4 m/s^2
f'(t) = 274.4t + 136275.68 m/s
f(t) = 137.2t^2 + 136275.68 + 0 = 548437979.084 m
t = (-136275.68 + sqrt(136275.68^2 + 4(137.2)(548437979.084)))/274.4
t = 1563.47 seconds
1563.47/996 = ~1.55 seconds
Comparable with time to PC range in Calculation 7, which gives ~1.5 seconds per solon.Code: Select all
Calculation 9
Too unimportant to bother showing all the work; there's just no way the KTKh was crawling along at the same pace as the KK.
If Van Squad's final position was 0 hexes from Tempest:
((-269326.75 + sqrt((269326.75^2) + 4(117.6)(504749756.79)))/235.6)/996 = ~1.225 seconds per solon.
That translates to a standard Loroi heart rate of 49 beats per minute.
If Van Squad's final position was 3 hexes from Tempest:
((-280872.72 + sqrt((280872.72^2) + 4(117.6)(491245112.818)))/235.6) = ~1.176 seconds per solon
That puts typical Loroi heart rate at 51 bpm.My gut tells me there's a lower value for a solon. At least, lower than 1.5 seconds.
But until word comes down from on high, it's the best I can do.
