It's actually 6 power per hex of web to generate. 3 PCs can *just* make a (completed) radius one globe on a single turn. You'd have to get creative with speed changes to make sure the web is always anchored and one of them might have to burn some battery to get some extra movement in.

By Michael Grafton (Mike_Grafton) on Saturday, December 28, 2013 - 05:16 am: Edit

I have always loved my submission "The Web of Lloth" back in a term paper long ago.

Because it ISN'T just the same old thing...

And the very best way to "dig a foxhole" fast is to use an entire squadron of PFs equipped with web devices. IIRC I submitted a term paper about that too...

By Michael Kenyon (Mikek) on Monday, December 30, 2013 - 11:35 am: Edit

Matt,

First, I was including the power to move the hex, which admittedly was 6 1/2, so I do agree that I shouldn't have rounded.

Second, how do you intend to get 3 PCs to make a web? You start at the same point and have to proceed from there with the path drawn between you. You can't have any more than 2 do it unless the 3rd is acting as anchor for the other two, but at the point, he's not contributing to the laying of web at all ...

Mike

By Matthew Potter (Neonpico) on Monday, December 30, 2013 - 01:34 pm: Edit

Mike: I was thinking they would start (roughly) in position to lay their first hex of web immediately. If two of them start adjacent to eachother, you don't have to worry about their section of the web being anchored (those two PCs would immediately be at opposite ends of a 2-hex web). The third PC, admittedly, would usually need to start two hexes away from one of the PCs. However, if you are willing to burn batteries in EA, they all could start in the same hex.

Here's the math for EA (assuming they are starting in the same hex and facing the same way):
Generate 12 warp, 2 impulse, and have 2 full batteries.
Each PC spends 1.5 on HK (no need for AFC)
Each PC spends 12 to generate 2 web.
This leaves 1/2 point of power.

One PC spends 1/3 power for one movement. It will lay the web in the hex that all three PCs start in, and will lay a hex adjacent to them.

Another PC burns 1/6th a battery (spent on something non-moving to free up the warp power) to spend 2/3 warp for 2 hexes of movement.

The third PC burns 1/2 a battery to spend 1 warp power for 3 movement. The first PC would need (through speed-changes from a fast-ish speed to stopped. Speed 8 for a movement, then speed 4 for the second movement, would work) to have laid it's second web by the time the third PC moved twice and laid it's first web (thus providing an anchor for the third-PC's web).

If you're willing to spend more batteries to free up warp power on this turn, you could even perform a few tacs (and maybe even the movement to inside the "fort"). Any power to reinforce the web would have to wait until the following turn. The PCs just don't have the battery power available to pump up the web strong enough to be a firebreak, after all that movement.

By Josh Driscol (Gfb) on Monday, December 30, 2013 - 10:38 pm: Edit

I'm just not sure what the tree fort or pinwheel would do for Mark in any of our campaign battles that require you to go on the offensive, ie start at the edge of the map.

In any defensive type battle he can use campaign fortification web to harden the defenses and will have bases to make most attacks costly.

In the game we played I feel that the big mistake he made was not overloading disruptors when the range was narrowing. There is no reason the Tholian cant dip into range 4-8 against a Seltorian, the ships may not be as sturdy but they are better shielded.

I think the Seltorian could be one of the tougher matchups for a Tholian in these open space battles because they have such excellent phaser suites, but the particle cannon seems to me to be inferior to the disruptor in almost every way.

But when you don't overload the disruptors to keep speed up and then allow the range to get to 4 or less things will even out with OL PC's hitting as frequently and for slightly more damage than disruptors and you have to worry about a follow up shot in 12 impulses if the engagement is early in the turn.

He could have used more EW to ECM and less to ECCM but would have probably just forced an adjustment from me from battery but that's at least some resources used.

By Michael Kenyon (Mikek) on Monday, December 30, 2013 - 10:51 pm: Edit

Ahhh, I see where you're going and there is a single problem I'm seeing and it comes down to an option.

EITHER you are generating Linear web (G10.11) or Globular Web (G10.12).

If it's the former, you can construct it as you describe, the end points cannot leave or move. You've locked two of your ships into the outside of the fort. That seems like a horrible idea.

If it's globular web, you cannot have the third ship join per (G10.121) [first two words of the rule indicate that globular web can only be laid by two ships].

By Matthew Potter (Neonpico) on Tuesday, December 31, 2013 - 03:44 am: Edit

Josh: The reason to make this sort of fort (usually in some distant area of the battle and before things get hot) is to make a place where cripples can go to repair and not be shot at, yet also not be forced to disengage from the field. Additionally, this might be an area where the Tholians might cause the opponent to chase them to. Once there, the Tholians might convince the opponent that they must web-dive in order to root out the Tholians from the area (with the usual consequences for doing so: the Tholians get a free range-2 shot on the unit of their choice or a range-1 shot with bricked shields and overloads.)

Mike: I had always intended the "fort" (as I described earlier) to be a globular web.

I see what you mean referenced with the first line of (G10.121). But I have always taken that to describe the most common situation and the easiest to visualize (to the reader of the rules). I had never really considered that to be an absolute.

As I described in my previous post (where PC #1 lays the web in the hex they all start in, and then on adjacent. PC #3 continues that "arm" of the globe), only two units are ever laying the globe. (first #1 lays some web, opposite #2. Then #1 stops and anchors until #3 lays it's first web. at that time, #2 and #3 are making the globe.)

For those who are having a hard time visualizing the situation I'm describing, here's a blow-by-blow:

PC #1, #2, and #3 are all starting in the same hex, facing the same direction. Turn modes are satisfied and they are free to plot any speed under 10. All three (as per my previous post) plot to lay 2 hexes of web each. PC #1 plots a speed of 3 to impulse 12 (1 move), and then speed 0 for the rest of the turn. PC #2 plots a speed of 2 for the turn. PC #3 plots a speed of 3. There are other ways to plot their speeds to allow for tacs or to put this globe up in half a turn, but this is simple.

Impulse 1: PC #1 lays the first web hex (G10.211). This is where all three of the PCs are located.
Impulse 11: PC #1 and PC #3 turn left. PC #1 lays it's second hex of web, thereby extending the globular web (G10.1181), with PC #2 anchoring the other side (G10.121).
Impulse 12: PC #1 stops. #1 and #2 are currently anchoring the globular web end-points and are in adjacent hexes.
Impulse 16: PC #2 turns right and lays it's first hex of web. This extends the web. The web is 3 hexes long. PC #1 and #2 are still the anchor points.
Impulse 22: PC #3 turns right and lays it's first hex of web. The web is now 4 hexes long. PC #2 and PC #3 are now the anchor points of the web. It begins to look like a 1-hex radius globe around the hex that was in front of the PCs at the beginning of the turn.
Impulse 32: PC #2 turns left and PC #3 turns right. They end the turn in adjacent hexes. Both lay their second hex of web, extending the web from 4 hexes to 6. This completes the globe as a 1-hex radius "fort" of strength 0. If not reinforced, the globe will deteriorate exactly 7 turns after the first web hex was laid (G10.43).

By Gregory S Flusche (Vandor) on Tuesday, December 31, 2013 - 02:17 pm: Edit

No need for that i have done just what You describe. It is hard to do under fire. Then you have to get the other player to attack it. It let you snipe from outside overload range then duck into the web when the other player goes into overload range.

By Mark Lorenc (Scrappinak) on Wednesday, January 01, 2014 - 10:12 pm: Edit

I really appreciate all the ideas. I am definitely going to spend a decent amount of time reviewing web rules so I can add to the conversation. Mike, I'd love to have copies of your term papers.

By Alan Trevor (Thyrm) on Thursday, January 02, 2014 - 12:38 pm: Edit

I guess I'll throw in my .02 quatloos worth. One problem that hasn't been addressed yet is that, even if the the Tholians manage to get their globular web formed, if the web is at low strength it will provide very little defense against seeking weapons. So if the Tholians are considering forming a globular web during the battle, they need to assess the enemy's seeking weapon capability and decide whether they can not only form the web, but reinforce it to sufficient strength before they are overwhelmed by incoming drones or plasma. If they have doubts about their ability to do this in the face of an aggressive enemy, they may be better off not even trying, and fighting a maneuvering battle instead. The other problem with the globular web tactic, against a "phaser-2 empire" (early Klingons) or "phaser-3 empire" (early Kzinti) is that it means the fight will take place at a range that minimizes the advantage fo your phaser-1s.

Assuming equal BPV and a Klingon opponent, the Tholian will have better shields and be more maneuverable, but will have less power and be nominally outgunned. But they are only actually outgunned at certain ranges. Since Mark Lorenc's original question specified "earlier ships", I am going to use a Tholian DD+PC against a Klingon F5+E4 as an example. All ships are turn mode A but the Tholians are nimble. The Tholians have 8 phaser-1s, 2 phaser-3s, and 2 disruptors. The Klingons have 9 phaser-2s and 4 disruptors and two drone racks. The Klingons have nominally more firepower but that really depends on the range at which the fight is conducted. The E4's disruptors have a maximum range of 10 so if the fight is conducted at ranges 11-15 the Tholians have slightly better firepower due to their phaser-1s, and much better shields. In the long run, the Klingons would lose if the fight is conducted at this range. So the Klingons have to force the fight closer. The Tholians should try to force an oblique pass at range 5, using ECM and erratic maneuvers to keep the Klingons from hurting them in the 6-10 range bracket and timing their erratic maneuvers to end as they cross from range 6 to range 5. (Nimble status helps here.) If the Tholians can do this, their phaser-1s (and superior flank shielding) will give them a big advantage over the Klingon's phaser-2s. An exchange that takes place at range 4 is also good for the Tholians but I prefer trying for range 5 because there is less chance of a miscalculation (an unanticipated Klingon speed change, for example) allowing the Klingons to reach range 3.

Obviously, different forces (to include different/ non-historical opponents) would call for different tactics. But in the example given - a Tholian DD+PC versus a Klingon F5+E4 - wouldn't even try to build a web. I would much rather take my chances in a maneuvering fight, and fly the Tholians a "phaser boats". Those disruptors on the DD? Those are something to arm if I still have power left after spending as much power as the situation dictates on movement, phasers, and EW.

None of the above is intended to disparage the notion of the Tholians forming a small globular web. Rather, it is intended to encourage the Tholians to evaluate whether that tactic is appropriate in a specific situation,

By Michael Kenyon (Mikek) on Monday, January 06, 2014 - 12:04 pm: Edit

Alan,

I like your thinking and agree heartily.

Mike

By Michael Kenyon (Mikek) on Monday, January 06, 2014 - 12:14 pm: Edit

Mark & Matt,

Just saw that my question from a while back got answered by SPP. Looks like it's just a 2-ship ordeal. Minimum of 3 turns for 2 PCs to generate a web, and (from memory) that leaves the batteries toast.

For strength 0 web, I don't see it as a useful scenario IN THE GENERAL CASE. It'll take another 210 points of power (in the middle years) to get to the point where you've got the web topped off. That's probably not happening in a battle, the best you're hoping for is something to stop direct fire and maybe slow down a lot of slow-medium drones.

Places where I can see it being useful ...
* Trying to defend ... something .. cripple, planet, etc.
* Facing early drone bombardment, have one guy power the web and everyone else draw drones into it. From GW on, the drones are moving fast enough that you're not going to be able to stop them from getting through.
* Where you're severely outgunned in the mid-range
* Where, for whatever reason your opponent is slow enough that you've got the time to do it up right.

By Matthew Potter (Neonpico) on Monday, January 06, 2014 - 12:59 pm: Edit

Actually, with the "only two ships can make a globe" (which is realized by the limitation on (G10.1161) to apply only to linear webs), 2 PCs can make a minimum-radius web over two turns and be able to pump enough power into it to be a fire-break. I'd shown above that 2 PCs can lay 4 web in a single turn. On the second turn, they each lay a hex of web and can put 6 power (each) into the web (a little more if they use the batteries). With speed changes, this thing could be completed and reinforced by impulse 5 of turn 2.

By Michael Kenyon (Mikek) on Monday, January 06, 2014 - 01:51 pm: Edit

Okay, I'm not sure that I'm following. A PC has 14+2 power.

Turn 1 - 12.667 for two hexes of movement with laying, + 2.5 for housekeeping = 15.167, there's most of your battery.

Turn 2 - 14+.833 power, you need to spend 2.5 on housekeeping, 6.333 for the hex, that's 8.833, giving you 6 power left over. That needs to be distributed after you've laid the web and takes two impulses to put 6 power in. I'll presume that you're talking max to put in in a single impulse. That means you plot speed 7 and put 4 in. Then, as you say, you'll be done on impulse 2.5. The web will have a strength of 8 (between the two PCs), or 1.333 per hex. After 2.32, it'll become 0.333 per hex and be considered a 0-strength web.

In turn 3, assuming that you are still within a hex of the web, you can put all power you haven't used on phasers, let's be generous and presume 0 back into the web to bring it back up. Assuming two PCs again, that would be 9 power to add each:
* 3.1: 2 + 8 = 10 = 1 point web
* 3.2: 10 + 8 = 18 = 3 point web
* 3.3: 18 + 2 = 20 = 3 point web

At that rate, as you're losing 6 points of power and adding 18 points, every turn, your web gets 2 points stronger every turn. So, that means...
Turn 4: 5 point web
Turn 5: 7 point web
Turn 6: 9 point web - we can stop a slow drone
Turn 7: 11 point web
Turn 8: 13 point web - we can stop mod drones, cause breakdowns
Turn 9: 15 point web
Turn 10: 17 point web
Turn 11: 19 point web
Turn 12: 21 point web - stopping med drones

With 3 PCs, it's a considerably better:
Turn 3: 4 point web
Turn 4: 8 point web - stopping slow drones
Turn 5: 11 point web
Turn 6: 14 point web - stopping mod drones, causing breakdowns
Turn 7: 17 point web
Turn 8: 21 point web - stopping med drones

In fairness, if you've got 4-5 PCs you can make a solid web in 3-4 turns.

This all presumes that you've got several turns where a number of your PCs are completely unmolested by the enemy. I don't see that happening. I see one of three situations happening:
a) you have a vastly smaller force than your opponent, he gets there and blows up one of the anchors before you complete - game over.
b) you have a vastly larger force than your opponent, you keep him away and you create you fort which will be useful to have the post-game celebration in but probably won't factor.
c) you have a similar size force to your opponent, you're dedicating 120 BPV of which to creating the fort meaning while you might be able to play keep away with the rest of your force they're both not helping to fortify your force nor are they likely to be fully intact when the fort is built.

I'd say that there's some case for (c) being a reasonable tradeoff, but it's really hard.

The equation changes a little based off of:
a) range - if you've got the time to spare, build the fort
b) terrain - if you have a "corner" you can limit what you need to do
c) ships - a CC + DD can complete a strength 1 web in one turn. Other ships can then fortify the snot out of it.

If you are starting at 40 hexes range, that's a possible option. At 30, I'm still going to get an overload shot from me fleet at one of your ships before you manage to close the loop.

To be clear, I'm not saying it's not doable nor am I saying it's not wise in some circumstances. I'm just saying that the list of those circumstances would be seem to be fairly limited and it's an option that you're very well betting the farm on. If you don't HAVE to do it, you may well be better of not.

By Mark Lorenc (Scrappinak) on Monday, January 06, 2014 - 01:55 pm: Edit

This analysis really helps me out and generally agrees with what I was figuring. The fight which raised the original question forced a raced confrontation in the middle of the map to fight over an abandoned hulk. The phaser boat approach seems like the only reasonable one in such a case. I did not consider leveraging EM more on the DDs and need to study up on tactics regarding it.

By Matthew Potter (Neonpico) on Monday, January 06, 2014 - 03:34 pm: Edit

At the point this went from doable in a single turn to needing a second turn to create, the fort became less viable. Without going into the reasonability of the tactic, to help those who aren't following how it can be done, I'll post another blow-by-blow:

Turn 1 EA:
PC#1 generates 14 pwr, allocates 1.5 to housekeeping (no need for AFC), 12 to web, and 0.33 for 1 move.
PC#2 generates 14 pwr, allocates 1.5 to housekeeping (no need for AFC), 12 to web, and 0.66 for 2 move (burning 0.166 battery to do so).
1.15: PC#1 lays web in the hex both PCs are starting in.
1.16: PC#2 turns left and lays a hex of web.
1.32: Both PCs turn right and both PCs lay a hex of web. 4 Hexes are laid at this moment, and the open center of the fort is the hex directly in front of where they started the turn.

Turn 2 EA:
Both PCs generate 14 pwr, allocate 1.5 to housekeeping (no need for AFC), 6 to create web, 6 to reinforce web, and 0.33 for 1 move. The second PC can spend the remaining 0.166 power to recharge it's battery. Both plot speed 4 to impulse 9, then 0 for the rest of the turn.
2.8: PC#1 turns left. PC#2 turns right. Both lay a hex of web. The globe is finished and they are in adjacent hexes.
2.9: Both pump 4 power into the web. If this is before Y160, then the strength is 1.333. If this is after Y175, then the strength is 2.666
2.10: Both pump 2 more power into the web. If this is before Y160, then the strength is 2. If this is after Y175, then the strength is 4. If they spent all of their batteries for web this impulse, the strength would be 2.666 or 5.333 respectively.

If they wanted to build the web sooner in turn 2, they could each allocate 2 hexes of move (burning battery to do so) and plot speed 11 to impulse 3, then to 6 until impulse 9, then to speed 2 until impulse 15, then speed 0 for the rest of the turn. That would have the web as reinforced as they can manage, by impulse 5.

Optionally they could spend battery power for AFC on turn 2, if the opponent looked like he would attempt to crash the party. This would give the PCs their phasers. Conversely, they could put the power into TACs and movement, potentially allowing them to end the turn inside the fort or getting a start on acceleration for later turns. But it really boils down to the fact that if they want to do anything but building web, they would need to reach into their batteries.

By Michael Kenyon (Mikek) on Tuesday, January 07, 2014 - 07:42 pm: Edit

Marc,

EM is a great tactic so long as you don't need to either fire or control a seeking weapon, as the Tholians don't really have much in that arena they're more likely to pull that tactic than others. Sadly, as with many tactics, it becomes chicken as to when you're going to drop it and actually attempt to fire.

Matt,

Totally ceded, I'd actually assumed that AFC was a pre-req of web creation, but I can't find that anywhere (asking the question). I'd suggest getting to a situation where you can at least bring up AFC if you've got a chance of an opponent getting within say 10 of you in a turn.

Mike

By Josh Driscol (Gfb) on Tuesday, January 07, 2014 - 08:24 pm: Edit

Have you ever tried not dropping EM before firing?

In a game against Matt I was able to get really close then make an EW adjustment to switch just the right amount of channels to ECCM from ECM at the point of firing.

At very close ranges a 2 shift doesn't really bother a hellbore the same as a tractor-repulsor beam so I figured it was worth a shot.

It worked out ok in our battle for the Hydrans but I doubt the same trick would work on Matt a second time. I can think of several things he could have done differently.

By Michael Kenyon (Mikek) on Thursday, January 09, 2014 - 12:48 pm: Edit

ABSOLUTELY! The problem, as with so many EW games, however, is that you have to catch your opponent a little by surprise to pull it off well. With six channels, limited swap rate and power constraints, the amount that you can toggle back and forth in a non-Andro is fairly limited.

EM is a great tactic for small ships, as it's cheaper than ECM. Also great for Tholians due to the lack of seeking weapons.

However, if all you're doing is doing EM, I can negate that with my ECCM and still have a little ECM up if I'm inclined. If you're putting up EM and ECM you can crank up to a 3 shift and that I've to work hard to avoid. Getting me to react and then swapping over to ECCM to negate your own shift eats a lot of batteries and limits what else you can do, but it's really pretty when it works.

By Steven E. Ehrbar (See) on Sunday, February 22, 2015 - 02:39 am: Edit

I was just re-reading the CL #29 solution to Rebel Reduction, and I noticed something. The Tholians are better off if they can get a couple PCs to replace the higher-BPV NFFs.

The advantage of the PC is that it has two bridge boxes and two web boxes, instead of one of each. (It does have two less generated power than the NFF, but has one more battery. We avoid using impulse for movement by moving a speed plot of 16-through-impulse-13, 29-for-14-to-21, 30 after, which is same 25 hexes and even same impulses of movement, but which costs 8.33 instead of 9 since we never go 31 and so never need impulse. 3.67 remaining warp+2 impulse+2 battery = 7.67; 1 for shields, 0.5 for life support, 6 for web =7.5, we're left with 1/6 point in the batteries instead of the NFFs' 1/2, which we don't care about since it never gets used).

In the solution presented in CL29, the rebels do their damage in a single volley to one of the NFFs, hope to do at least 20 points (the average for the phasers, but not guaranteed), and pray for snake-eyes twice to take out the web generator. But with PCs, they can't do this. Since both bridge and flag bridge can only be hit once on the DAC in any volley, two bridge boxes and two web generators mean they have to do two volleys, which means firing on two turns. Except, if the first volley hits and does internals, the PC can just turn a fresh shield. It can still get to where it needs to be (and the heading it needs) for the web anchor/laying by turns and sideslips.

Which means, against a PC instead of an NFF, to eliminate the web generation capability the rebels would need to do 20 damage on each of two volleys with a set of phasers that average 20 damage when fired all at once and a maximum of 48. And on each volley they would need to roll snake-eyes on the DAC twice.

By Alan Trevor (Thyrm) on Tuesday, April 07, 2015 - 02:27 pm: Edit

Continuation of a discussion that started in the "Partical Cannons!" (sic) topic:

Mike, one of the items I wanted to discuss was your assertion in your 12:31 AM post from 31 March, that if I formed 2 PF pinwheels as described, you could destroy them by sacrificing 2 frigates in the web. I don't think you can, for reasons I explained in my 12:06 PM post on 31 March. I think it will cost you a lot more to kill the pinwheels if the Tholian defense is adequately set up. But as I also said in that post, my conclusion is based on exactly one battle in which I tried that against a Seltorian attacker (regettably I don't recall the exact forces but I believe the Tholians had about 1200 points (which included points for the globular web). against a Selt attacker at about 1500. I also know the Tholians had one "Arachnid-P" flotilla stationed at the base and 2 (or it might have been 4) additional casual PFs on mech-links on the supporting ships. I think the base was only a BS but it might have been a BATS. I just don't recall. No X-tech. The Tholians also had some small warships and a fighter squadron (all Spider-IIIs) as well as (I think) two minefield packages. The Seltorians had a PF tender and some casual PFs as well. I don't recall them as having fighters. Nor do I recall how many total web breakers they had.

In any event, after the battle, the Seltorian player and I tossed around some ideas, but neither one of us could think of a way that the Selts could have broken that defense, despite having 300 (give or take) more BPV to play with. It was partly that battle that prompted some of my comments to ADM in the "Partical Cannons!" topic. But a way to break the defense might exist that did not occur to either the Selt or me. So if you (or anyone else) has ideas on that, I would like to know. If there's a weakness in the defense, better to learn about it now.

I should probably say here that I am not generally a fan of using pinwheels. But there are some specific special occasions in which I think they are quite useful and defending a base against Seltorians is one of them. If I understand correctly, a "standard" Tholian flotilla can have only have 2 "W" models. That's plenty for base defense against against most attackers. But against Seltorian web breakers it may not be enough. That's why I like pinwheels in that specific case. They allow me to use (almost) all the energy from 6 PFs to support the web. But the defense then has to be arranged to make the attacker pay an excessively high price to kill those PFs. Given adequate BPV (and in my experience 1200 is easily adequate), the Tholians can arrange the defense so the attackers (other than a few oddballs like the Andros or Ryn) have to pay an enormous price. As I said before, I believe you would have to sacrifice far more than 2 frigates to kill them.

I noticed your "web of Lloth" defense uses 12 PFWs, with no scouts or leaders. That's efficient for base defense but my objection is that it is much less effective if the PFs have to sortie to a fight away from the base. If they are based from a PF module, they would be effectively without EW support. If they were based off a tender, as in your web of Lloth, the tender could sortie with them but if it was a PC-based tender (based on the BPV in your web of Lloth term paper - it was the PC-based version) then any fight away from the protective webs of the base might require the tender to do substantial maneuvering in order to survive, in which case it wouldn't have the power to provide EW support to the flotilla. But of course, that's not a problem if the flotilla is committed exclsuively to base defense. It just seems less flexible than a "standard" flotilla.

Over to you. Please comment on any of the above that you find interesting or discussion worthy.

By Gregory S Flusche (Vandor) on Tuesday, April 07, 2015 - 05:08 pm: Edit

I am assuming the pinwheel is set up some were along the outer web?. I would not put them in a corner along a edge? I am interested in how you set them up?
I know how i like to place hidden mines in defense of a web and the way to get past them.

By Stewart Frazier (Frazikar2) on Tuesday, April 07, 2015 - 07:33 pm: Edit

Hmmm, IIRC the shielding on a PF Pinwheel is 21 (36 if shield refit) adding 2 for the #1 shield but would only need to hit one generator as the other keeps the pinwheel together [assuming one PFW in the pinwheel)...

By Alan Trevor (Thyrm) on Wednesday, April 08, 2015 - 11:02 am: Edit

Gregory,

The pinwheel is in the ring of empty hexes between the outer and middle web rings. From there it can supply reinforcing energy to the outer ring, to counter the web breakers, but can't itself be shot unless the Selts actually put ships in the web. While it could be placed in a corner of that empty ring, I believe the middle of the side is better. The defense is set up so that no attacking ship can have line-of-fire to the pinwheel without first encoutering at least one large explosive mine plus a lot of "preemptive" phaser fire, delivered when the attacker is right outside the outer ring, or 6 hexes from the base. The intent is that the attackers have most of their weapons destroyed before they have a chance to fire. With only two frigates (as suggested in Mike's comment in the "Partical Cannons!" topic), they won't have enough weapons left to get through the pinwheel shields. Yes, given enough attackers they can get enough ships into position to kill the pinwheel. But it will cost them so much of their starting force that, unless they have a big advantage in BPV, what's left won't be able to deal with the base plus the actual Tholian warships.

I don't recall the exact mine set up. But as a general rule the "bastion" set up to protect the pinwheel would consist of a combination of large mines just outside the outer ring with detection range set at one, and large mines actually in the outer web ring with detection range set at zero. I like to use some alternate set ups to keep an opponent guessing. The mines in the "standard package" have to be spaced more-or-less evenly around the base. But extra mines purchased individually can be set up anywhere. So the "standard package" will be supplemented by individually purchased command-detonated mines that will be concentrated around the bastion. Also, although the explosive mines in "standard package" are automatic, some of the captors are command controlled. Since the requirement to spread the mines evenly relates to gross numbers but not individual types, those command controlled captor mines (phasers, not disruptors) are concentrated around the bastion to supplement the preemptive phaser fire from the PFs, fighters (a squadron of Spider-IIIs will total 24 phaser-3s; shooting at 2 hexes (the fighters in the "empty" ring shooting at attackers just outside the outer ring) they will (assuming no ECM shift) deliver about 72 points of damage all by themselves; Spider-IIPs would be even better) and shuttlecraft, actual warships, and the base. Finally, even the "standard package" contains some chain detonators. Again, these detonators do not have to be spread evenly around the base but can be concentrated around the bastion.

Stewart,

I pretty much always give my PFs shield refits. So the pinwheel shields are 39 off the noses of the 3 PFs, 36 for the other 3 shields. Even standard Arachnids have a 15 point front shield with the refit. The Tholians may have punk fighters (though they are cheap and therefore can be very cost effective in base defense) but their PFs are among the best.

By Matthew Potter (Neonpico) on Wednesday, April 08, 2015 - 02:17 pm: Edit

Actually, a Pinwheel is a term of art. See (C14.0)

Instead you refer to the empty ring of hexes between globular webs. I don't believe there is a term for that, except perhaps to say "the empty portions of a wedding cake".