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asguard101 Lieutenant SG
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Joined: 20 Jul 2008 Posts: 170 Location: Florida
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Posted: Wed Jul 29, 2009 8:43 pm Post subject: Hydran Hellbore damage resolution |
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I have a couple of questions for hellbores, I think I saw them answered before, but I couldn't find it.
1. I know hellbores are a seperate volley from the firing ship, does each ship fire have a seperate Volley or all hellbore total for one volley? i.e. 2 DD firing hellbores and PH-1. does each ship have one PH-1 volley and 1 Hellbore volley, or is it 1 PH-1 volley from each ship and a combine Hellbore volley from both ships?
2. In the above example, is there or who decided which volley goes fire?
3. When determining weakest shield, is it based off the shields strength at the time of firing, or the weakest shield at the time the hellbore's volley is reconcilled? |
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terryoc Captain
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Joined: 07 Oct 2006 Posts: 1386
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Posted: Wed Jul 29, 2009 8:57 pm Post subject: |
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Asguard,
All fire from one ship in one impulse is one volley. Even Hellbores. There are some partial exceptions to this:
* Hellbores create six "volleys", one per shield, because of the wrap-around damage they do.
* Any damage done by phasers will be combined with one of the Hellbores' six volleys. So if the Hydran is firing his phasers at the enemy ship's #1 shield, and hellbores too, you'll have six volleys: Shield #1 gets a volley of all the damage from the phasers plus hellbore damage; shields #2-#6 will each get a volley of hellbore damage.
* Hellbores can't use directed targeting, so if the phasers use directed targeting, the phasers are a different volley from the hellbores. If the hellbores are overloaded, NONE of the weapons fired by that ship in this impulse can use directed targeting.
2. The firing player decides in what order each volley is resolved, same as always.
3. Weakest shield at the time the volley is resolved. _________________ "Captain" Terry O'Carroll, fourteen papers published including six best of issue
"Man, Terry, you are like a loophole seeking missle!" - Mike West
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asguard101 Lieutenant SG
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Joined: 20 Jul 2008 Posts: 170 Location: Florida
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Posted: Wed Jul 29, 2009 9:06 pm Post subject: |
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terryoc wrote: | Asguard,
All fire from one ship in one impulse is one volley. Even Hellbores. There are some partial exceptions to this:
* Hellbores create six "volleys", one per shield, because of the wrap-around damage they do.
* Any damage done by phasers will be combined with one of the Hellbores' six volleys. So if the Hydran is firing his phasers at the enemy ship's #1 shield, and hellbores too, you'll have six volleys: Shield #1 gets a volley of all the damage from the phasers plus hellbore damage; shields #2-#6 will each get a volley of hellbore damage.
* Hellbores can't use directed targeting, so if the phasers use directed targeting, the phasers are a different volley from the hellbores. If the hellbores are overloaded, NONE of the weapons fired by that ship in this impulse can use directed targeting.
2. The firing player decides in what order each volley is resolved, same as always.
3. Weakest shield at the time the volley is resolved. |
I understand almost every thing, execpt one thing. The hellbore damage, from what I understand is issues out this way, please correct if I'm wrong.
Range of 1 a hellbore would do 10 points to the weakest shield, and then 2 points to the other 5 shields. I don't see how you example of shield #1 would work, unless it happened to be the weakest shield. |
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mjwest Commodore
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Joined: 08 Oct 2006 Posts: 4075 Location: Dallas, Texas
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Posted: Wed Jul 29, 2009 9:38 pm Post subject: |
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Let's do some examples. In all examples, a single hellbore and a single Ph-1 are fired. Range is 1. The hellbore hits and the phaser does 5 points of damage. Note that in all of the examples below, the volley order given is not required, but simply what I chose for the examples.
Ex #1: The #1 shield of the target ship is facing the firing ship, but the #5 shield is weakest. Hellbore is fire in normal mode, but directed damage is not used. In this case, we apply the first number in the hellbore damage (10) to the weakest shield (#5), and divide the second number (also 10) to the other shields as evenly as possible. Conveniently in this case, that works out to 2 points of damage each. Damage is divide thus:
Volley 1: Shield #5 takes 10 points of damage from the main hellbore element.
Volley 2: Shield #2 takes 2 points of damage from the hellbore.
Volley 3-5: Same as volley 2 on shields #3, #4, and #6.
Volley 6: Shield #1 takes 2 points of damage from the hellbore, but also takes the 5 points of phaser damage, for a total of 7 points in the volley.
Ex #2: As above, but directed damage is being used with the phaser. In this case the volleys resolve like this:
Volley 1: Shield #5 takes 10 points of damage from the main hellbore element.
Volley 2: Shield #2 takes 2 points of damage from the hellbore.
Volley 3-5: Same as volley 2 on shields #3, #4, and #6.
Volley 6: Shield #1 takes 2 points of damage from the hellbore. Since the phaser is using directed damage, it is not combined into a single volley with the hellbore damage on the same shield.
Volley 7: Shield #1 takes 5 points of damage from the phaser using directed damage.
Ex #3: Same as the first example, but this time the hellbore is fired overloaded. Because of this, the phaser may not use directed damage. The first number (15) is applied to the weakest shield (#5), and the second number (also 15) is applied as evenly as possible to the other shields, which works out to 3 points to each. The damage ends up being:
Volley 1: Shield #5 takes 15 points of damage from the main hellbore element.
Volley 2: Shield #2 takes 3 points of damage from the hellbore.
Volley 3-5: Same as volley 3 on shields #3, #4, and #6.
Volley 6: Shield #1 takes 3 points of damage from the hellbore, but also takes the 5 points of phaser damage, for a total of 8 points in the volley.
I hope this makes things clearer ... _________________
Federation Commander Answer Guy |
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asguard101 Lieutenant SG
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Joined: 20 Jul 2008 Posts: 170 Location: Florida
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Posted: Wed Jul 29, 2009 10:02 pm Post subject: |
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Thank you for the info, that did clear up my thinking on it. |
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Steve Robinson Ensign
Joined: 22 Apr 2009 Posts: 9
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Posted: Thu Jul 30, 2009 5:41 pm Post subject: |
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In example #2, since the damage to the #1 shield was from 2 volleys, could the firing player decide to have the volley from the phaser resolved first in an attempt to reduce the #1 shield to the "weakest"? |
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mjwest Commodore
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Joined: 08 Oct 2006 Posts: 4075 Location: Dallas, Texas
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Posted: Thu Jul 30, 2009 6:49 pm Post subject: |
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Steve Robinson wrote: | In example #2, since the damage to the #1 shield was from 2 volleys, could the firing player decide to have the volley from the phaser resolved first in an attempt to reduce the #1 shield to the "weakest"? |
No. The weakest shield determination is made before applying any damage from the volleys from a single ship. _________________
Federation Commander Answer Guy |
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m1a1dat Lieutenant JG
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Joined: 17 Dec 2008 Posts: 99 Location: 91320
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Posted: Wed Aug 05, 2009 6:13 am Post subject: |
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mjwest wrote: | Steve Robinson wrote: | In example #2, since the damage to the #1 shield was from 2 volleys, could the firing player decide to have the volley from the phaser resolved first in an attempt to reduce the #1 shield to the "weakest"? |
No. The weakest shield determination is made before applying any damage from the volleys from a single ship. |
Really? That's not the way we have understood it and played it. Looking through the rules now, it doesn't specifically say it that way. This is how we interpreted it. Now it seems we may have been wrong.
Damage Allocation rule ( 3d2 ) says one volley must be completely resolved before the next and the attacker gets to choose the order of the volleys. So we played it directed damage from the phasers first then determine weakest shield for the hellbore. And the reverse for ESG's, do the ESG damage first then the directed damage for maximum effect with those weapons.
The rule for firing Hellbores ( 4K3d ) basically just says that if the weapon hits the main damage is done to the weakest shield.
So is it now that the weakest shield is determined before any dice are rolled? For each hellbore ship firing? Or maybe at the same time as weapons fire is declared and before dice are rolled as in split shield boundaries cases? That would mean that multiple helbore ships firing on a target with more than one weak shield, can have their fire spread all over, such as with an old style Romulan with its all around equal shields.
Ok, rant off. Just wanting to know how to get what you said from what is in the rules. Maybe this is another case where the rules need to be reworded. Thanks |
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pmiller13 Lieutenant JG
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Joined: 12 May 2009 Posts: 64
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Posted: Wed Aug 05, 2009 6:14 pm Post subject: |
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Keep in mind that in FC the hellbore damage will not be spread evenly across all shields when they are all equal (or if there are more than 1 that are the smallest and equal).
Per (4K3d) The first number is the number of damage points applied to the weakest shield (if there is more than one 'weakest shield' the ship owner picks one of them);
This means that if a hellbore hits a ship w/6 equal shield the owner of the ship picks which one takes the first number on the hellbore damage chart and the other shields get the second number spread across them. In other words don't fire hellbores at a ship w/6 equal shields till you have dented one of them. |
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asguard101 Lieutenant SG
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Joined: 20 Jul 2008 Posts: 170 Location: Florida
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Posted: Wed Aug 05, 2009 11:16 pm Post subject: |
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I have a concern about the PBV for the Hydrans, let’s have a comparison of a few of the ships without fighters:
Hydran Federation
FF 61 78
CA 148 147
CC 150 151
HBC 180 189
LDN 180 235
HDN 212 276
BB 390 333
For some reason the Hydran ship start lower that their equals, then around the CC Balance up, then go back to a lesser BPV, until the BB shows up then it leaps ahead of everyone else. I don’t understand why the BB got such a huge BPV leap. Can anyone help? |
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junior Captain
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Joined: 08 May 2007 Posts: 803
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Posted: Thu Aug 06, 2009 12:44 am Post subject: |
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When talking about Hydran ships of a certain size class (and smaller than a BCH), you need to mention whether you're referring to a Hellbore ship or a Fusion ship. In this case, it appears that you're referring to Hellbore ships (which makes more sense, since Fusion-only ships are always fewer points than their opponents).
The Cuirassier is most likely as low as it is points-wise because it's such a small ship. For instance, whereas every other frigate in the game mounts at least two heavy weapons, the Cuirassier only mounts a single Hellbore. The Knight Destroyer, at 90 points, weighs in at slightly more than its two primary opponents - the Lyran DD and the Klingon F5 (which is essentially a weak destroyer). The Tartar Medium Cruiser weighs in at 125, which puts it pretty much on par with the Klingon D5.
As for the Battleship...
I haven't looked at the sheet for it in a while, but my recollection is that it's a massive overlybeweaponed monstrosity (even for a battleship) that also happens to be the only ship in the game that mounts more than four Hellbores (six in front, two in back iirc).
That's probably why it's so expensive.
Also, the Fed Battleship is the second least expensive battleship in the game. Most of the others are actually within 20 points or so of the Monarch (and then there's the Romulan Battleship, which weighs in at 415 points...). |
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terryoc Captain
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Joined: 07 Oct 2006 Posts: 1386
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Posted: Thu Aug 06, 2009 3:43 am Post subject: |
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The Hydran BB weighs in at over 500 points if it has a full load of fighters (12 IIRC).
It's a monster. _________________ "Captain" Terry O'Carroll, fourteen papers published including six best of issue
"Man, Terry, you are like a loophole seeking missle!" - Mike West
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pinecone Fleet Captain
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Joined: 03 May 2008 Posts: 1862 Location: Earth
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Posted: Thu Aug 06, 2009 2:12 pm Post subject: |
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Six HB's is enough to turn just about anything besides a BB into Molten Metal. You can't beat it at close range, you can't beat it at long range. (FUS and HB's) |
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mjwest Commodore
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Joined: 08 Oct 2006 Posts: 4075 Location: Dallas, Texas
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Posted: Sat Aug 08, 2009 6:03 pm Post subject: |
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m1a1dat wrote: | mjwest wrote: | Steve Robinson wrote: | In example #2, since the damage to the #1 shield was from 2 volleys, could the firing player decide to have the volley from the phaser resolved first in an attempt to reduce the #1 shield to the "weakest"? |
No. The weakest shield determination is made before applying any damage from the volleys from a single ship. |
Really? That's not the way we have understood it and played it. Looking through the rules now, it doesn't specifically say it that way. This is how we interpreted it. Now it seems we may have been wrong. |
First, if everyone is in agreement with how a rule is supposed to work, don't ever let me spoil the fun. After all, that is the primary purpose of Federation Commander: to have fun.
The issue is that having the phasers and hellbores be divided into separate volleys is an artificial mechanism introduced to not penalize the ability to use directed damage just because hellbores are fired normally. Adding in this exception does not mean that it should be used to give even more advantage to this situation.
Therefore, regardless of whether the phasers are fired in directed mode or not, the weakest shield is determined prior to that "double-volley" exception. Doing otherwise will instead advantage hellbores even more than they already are. (And believe me, hellbores do not need any more advantages.)
Quote: | So is it now that the weakest shield is determined before any dice are rolled? For each hellbore ship firing? Or maybe at the same time as weapons fire is declared and before dice are rolled as in split shield boundaries cases? That would mean that multiple helbore ships firing on a target with more than one weak shield, can have their fire spread all over, such as with an old style Romulan with its all around equal shields. |
Just to be clear, this only applies in one case: firing phasers and normal load hellbores from one ship at another ship. In any other case, this doesn't apply. (OK, you could have normal load fusions in there. And on an Orion, who knows what could be used. But, let's just go with this.) In that case, what would normally be a single volley is divided into two separate volleys in order to allow the phasers to use directed targetting. In this one specific case, you need to determine weakest shield prior to both of the volleys.
To answer your question, yes. You must make the weakest shield determination prior to resolving damage from one ship to another ship. Note that this is only for one ship at another ship. When the second ship fires its weapons, those decisions must take place using the new situation. So, your theoretical Romulan will still get hammered on one shield, as the first hellbore ship's damage will influence the choices that have to be made on the next hellbore ship's fire.
Let's take an example to show what I mean: We have two Hydran Knights firing on a Klingon D7. Our unfortunate D7 is, up to this point, completely undamaged. Each of the Knights is firing both hellbores as normal loads, both Ph-2s, and its Ph-1. (They are saving the Ph-Gs for drone defense and close-in work.) Range is four hexes. The D7 is, for whatever reason, not firing (let's assume it already fired). Also, just for fun, let's assume the Knights are in the same hex, and are exactly on a hex spine from the D7. Just to have numbers, the D7 is in hex 0606 facing A, and the Knights are in 1006 facing E.
So, Knight #1 is chosen to fire first. Now, even though the Knight is almost guaranteed to do multiple volleys of damage, at this point this is treated as a single volley (one ship firing at a single target). So, all pre-volley decisions must be made at this point. Since he is unlikely to penetrate with his phasers, Knight #1 decides to not use directed targetting. He is at the shield boundary for the D7's #2 and #3 shields. The D7 chooses shield #2 as the "facing" shield. Since hellbores are being fired, he must also choose which shield is "weakest". At this moment, he has 5 equal shields, so he can choose any shield but shield #1 as his weakest. He chooses #5 (since it is on the opposite side of his ship from the Knights). Since the Knight did not use directed damage for the phasers, and, displaying moderately bad luck, only hits with one hellbore. This results in 6 points of damage against the facing #2 shield from the phasers, and 8 points of damage against the "weakest" #5 shield from the main hellbore element. For the remaining hellbore damage, the D7 gets to choose the order, so he applies 1 point of damage to shields #1 through #4, and #6 to resolve 5 of the 7 points. The 6th point goes to the #1 shield (since it is the strongest). Finally, with the last point of damage, he intentionally takes it on the #2 shield. As a result, the D7's shields (from #1-#6) are 28-14-21-21-14-21.
Now, it is Knight #2's turn. The D7 (obviously) decides this Knight is facing the #3 shield. For the weakest shield, the D7 has an interesting choice. Because he took the last point of hellbore damage on the #2 shield, he has two "weakest" shields from which to choose. (Had he taken the point of damage anywhere else, then the weakest shield would have unequivocably been the #5 shield.) For a variety of reasons, the D7 chooses to make the #2 shield weakest in this case. Knight #2 decides to use directed targetting with the phasers, so he must decide what to fire first. He choses to fire the hellbores first, and (with better luck than his companion) hits with both. This means 16 points of damage are applied to shield #2, which, after the D7 reinforces, results in one point of non-directional damage of burn-through. The other hellbore damage is spread across the other shields. Since it is 14 points, four of the shields will get three points of damage, while one will get only two. So, the D7 takes two points of damage on the #5 shield, and three points on the #1, #3, #4, and #6. Now, the phasers are fired, causing 10 points of damage (he rolled very well with the Ph-2s). Since the #3 shield is facing, it takes the damage. It started at 21 points (after Knight #1's fire). It lost three points to the hellbore, and is down to 18. The 10 points of damage are then applied, meaning it is reduced to 9, with one point of directional burn through.
Now, it is highly unlikely that two Knights would end up on the hex spine like this. Had this not happened, the D7 would have take some serious internals (rather than just two points of burn through). But, I hope this shows the idea. _________________
Federation Commander Answer Guy |
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Jiraiya1969 Lieutenant JG
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Joined: 17 May 2007 Posts: 90
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Posted: Tue May 27, 2014 5:13 am Post subject: |
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mjwest wrote: | Ex #2: As above, but directed damage is being used with the phaser. In this case the volleys resolve like this:
Volley 1: Shield #5 takes 10 points of damage from the main hellbore element.
Volley 2: Shield #2 takes 2 points of damage from the hellbore.
Volley 3-5: Same as volley 2 on shields #3, #4, and #6.
Volley 6: Shield #1 takes 2 points of damage from the hellbore. Since the phaser is using directed damage, it is not combined into a single volley with the hellbore damage on the same shield.
Volley 7: Shield #1 takes 5 points of damage from the phaser using directed damage. |
In this example, what would the point have been in firing the Ph-1 in directed-targeting mode? It's only 5 points and would not penetrate the shield.
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