Archive for the Rules Category

Deterministic Pilot task

Posted in Intercept, Rules on March 5, 2020 by Anders Backman

John Boyd was dubbed “Forty Second Boyd” for his standing bet as an instructor pilot that beginning from a position of disadvantage, he could defeat any opposing pilot in air combat maneuvering in less than 40 seconds.

John Boyd said that a pilot going into aerial combat must know two things: the position of the enemy and the velocity of the enemy. Given the velocity of an enemy, a pilot is able to decide what the enemy is capable of doing. When a pilot knows what maneuvers the enemy can perform, he can then decide how to counter any of the other pilot’s actions. 

Deterministic rules

In order to capture the feel of space combat being a contest of wits between two veteran commanders, like a game of chess, I have added variants of the rules where no luck is involved at all. Movement, initiative, sensors, attack, defense and damage are all governed by deterministic rules. Pages 44-45 of the rulebook hold the various deterministic rules but this blog post will focus on one particular section; the Deterministic Pilot task and initiative determination.

The Pilot default table was discussed here and these rules are somewhat similar. You look up your ships Size on a table after having modified the row used by computer pool use, Pilot skill, damage etc. The table gives you the steps of turning allowed this turn if you didn’t turn at all the previous turn. The Usable steps of turning is the value from the table minus half the number of steps you used the last turn.

 

Usable steps of turning

Look up modified Size, subtract used steps from last turn / 2 (round up)
Highest number of Usable steps of turning win Initiative
A ship has a certain number of steps each turn, based on ship Size, Pilot skill
and Crew damage. Subtract the used number of steps / 2 from the previous turn
(rounded up). A good way to handle this is to write down the turn number / used steps of turn for each ship, so you don’t have to remember how many steps was used the last turn.

Deterministic Pilot task 1

Initiative

Untracked ships win Initiative over tracked ships as usual but higher Useable
steps of turning is used instead of the Pilot task or Pilot default.
Initiative is determined as follows (in order of priority):
1 Untracked ships have higher Initiative
2 Higher Useable steps of turning have higher Initiative
3 Higher Ship tactics have higher Initiative
4 Higher crew station have higher Initiative (Bridge > Full > Limited)
5 Break ties with player A wins on even, B on odd turns, note on DataCard.

Deterministic Pilot task 2

Pilot task result

Whenever the rules call for the Pilot task result (Very Good, Good, Fair, Miss, Bad or Very Bad) determine the degree of success or failure from the above chart. Pilot task results are used when Aerobraking, landing, ramming and docking. The Pilot task result is not used to determine initiative; for example two ships, one with 4 steps of turning to use and the other with 5  both have a Fair Pilot task result but the ship with 5 steps win over the one with 4.

The Deterministic usable steps of turning go very well together with the Deterministic hitlocation rules as these make maneuvering even more critical. Feel free to mix and match whatever Deterministic rules you like, as long as both players agree upon their use before the game starts. Although I have taken care to make them as balanced as possible there’s no guarantee that the balance will remain the same, you have been warned!

Pilot duel example

The next post will cover a mock duel between two trader captains using the deterministic pilot task presented above.

There’s only one course of action
Left for me to take
I’ve tried every switch selection
That might control this state
I think for my protection
I better make it straight
Into Ejection
Better tell Base
Ejection
That I think it’s a case
For Ejection
Explode into Space
Ejection
Protect my Face
Ejection

Ejection – Hawkwind

 

 

Small October update

Posted in Intercept, Rules on October 14, 2019 by Anders Backman

October update

I have made some small updates and rearrangements to the rulebook over and now they are available to downloads. That is all folks, see you next time.

Vector movement game units

Posted in Intercept, Other vector movemet systems, Rules, Science, Traveller, Vector movement on July 16, 2019 by Anders Backman

When you create your own vector movement system, as I am sure everyone does, you need to determine what map scale, turn length and acceleration units are. There is an obvious formula from school that most seem to use and but will argue for why this is wrong and why one should instead use another formula, also from school.

Plotting example

Mapscale

Typically one decide on the map scale first (how large will the hexes, squares, inches or centimeters be?). Deciding on scale is mainly about what you want in your maps, do you want planets to be one unit or less in scale? Do want to show Earth and the moon on the same map? Do you want to fit the inner solar system on your map, like Triplanetary? And so on.

Some examples:

  • Intercept 10 000 km per square, 15 minute turns, 1G per square.
  • Intercept large scale 100 000 km per square, 60 minute turns, 1G per square.
  • Traveller LBB 1000 km per inch, 5 minute turns, 1G per inch.
  • Mayday by GDW 300 000 km per hex, 100 minute turns, 1G per hex.
  • Triplanetary ~10 million km per hex, 1 day per turn, ? G per hex.

Intercept let you play using two different scales and switch back and forth as you like, there’s even a smaller scale in the works if I can iron out the problems with planets taking up large parts of the map, that scale will be 1 000 km, 4 minute turns and 1G per square as usual.

We will later calculate what acceleration Triplanetary is likely using based on the distance and time scales and formulas learned.

Formulas

High school math teaches us two formulas for determining distance traveled, one for when applying constant acceleration from a standstill and another when having constant speed.

The two formulas are:

Formula one and two

Notice how formula 2b is the same as formula 1 but without the 1/2 multiplier. Distances traveled become twice as far in formula 2 so one of them must be wrong, right?

Not so fast! The formula from high school actually looks like this:

Formuila one with prior speed

Formula (3) also take the velocity from the previous turn into account (the v0 term). As v equals a multiplied by t we get our beloved formula (2b) as the first term, or something similar at least.Why is the first term twice as big as the second term? Well, the the first term assumes the speed is constant through the time segment t while the second term treat is as increasing, the distance traveled can be seen as areas in graphs with speed plotted versus time, like this:

Formula and area

If we use formula (3) to determine total distance traveled while t keeping t as the turn length and v as v (n-1) where n is the number of turns we’ll see that as the number of turns increase the grey area will more and more resemble a rectangle (the triangle of the last turn will contribute less and less of a fraction of distance traveled).

Formula graph

The grey area is the distance traveled. If we call one rectangle as 1.0 and one triangle as 0.5 we get the following distances:

  • Turn 1: 1 triangle plus 0 rectangles = 0.5
  • Turn 2: 2 triangles plus 1 rectangle = 2.0
  • Turn 3: 3 triangles plus 3 rectangles = 4.5
  • Turn 4: 4 triangles plus 6 rectangles = 8.0

and so forth…

You see that as the number of turns increases the number of rectangles increases faster than the number of triangles. So, as the number of turns increase the the number of rectangles will outstrip the number of triangles.

In the vector movement systems of Triplanetary, Mayday, Traveller, Intercept etc we use a vector that both represent velocity and acceleration however. So we either decide that one unit length should be correct for acceleration from a standstill but wrong for drifting or accelerating with a prior velocity (1) or we decide that one unit length should be correct for drifting and approach correct when handling prior velocity (2).

Too much theoretical bullshit you say? OK, let’s do a practical example.

Let’s travel from a standstill to the moon as see which of formula (1) or (2) most closely fit (3). We ignore braking at the moon just go to the moon as fast as possible. The average distance between the earth and the moon is 380 000 km so let’s go with that.

Units for formula (1)

  • A = 10 m/s^2
  • T = 1000 s
  • S = 5000 km

Distance earth – moon will be 76 squares.

Traveling 76 squares with 1 unit acceleration will be

1+2+3+4+5+6+7+8+9+10+11+12 = 78 units after 12 turns (12 000 s) = 3 h 20 m
(We overshot the moon by 2 squares but this is the closest we could get)

Units for formula (2)

  • A = 10 m/s^2
  • T = 1000 s
  • S = 5000 km

Distance earth – moon will be 38 squares.

1+2+3+4+5+6+7+8+9+10 = 46 units after 10 turns (10 000 s) = 2 h 45 m
(We overshot the moon by 8 squares but this is the closest we could get)

The correct value using formula (3) and setting v0 to zero is (8 718 s) = 2 h 25 m

 

Sorry about the long winded explanation but for some reason most vector systems get this wrong. Doesn’t matter when you play of course but say you want to travel from earth to the moon using actual mapboard movement you’d find that the travel time would not match the calculated value.

Apollo 11 50 years anniversary July 16 1969

Yes, 50 years ago today a couple of Americans started their travel from earth to the moon , certainly not under constant 1 G acceleration and they made damned sure their velocity was as close to zero as possible before they hit the moon. Apollo 11 did the trip in about half a week.

Apollo_11_Flight

 

Han Solo aggressive docking

Posted in Films and TV, Intercept, Rules with tags , , on March 19, 2019 by Anders Backman

Han Solo aggressive docking

C3PO – “Sir, we just lost the main rear deflector shield. One more direct hit on the back quarter and we’re done for!”
Han Solo – “Turn her around!”

And then Han Solo proceed to go head to head against a Star destroyer, barely avoids crashing into the bridge and then miraculously vanishes from the Star destroyers sensors and is gone. The Star destroyer has lost them and captain Needa suffers a reprimand while Millenium Falcon has secretly docked to the enemy ship. Sure, Star Wars is space fantasy and the ships certainly doesn’t follow the laws of physics or even common sense but let’s find out if this particular daring maneuver is at all possible using the Intercept rules? We will only use the standard rules here, no house rules or Star Wars conversions, just plain ol’ Intercept.

Let’s use a Beowulf class Free trader as Millenium Falcon and an Azhanti High lightning class cruiser as the Star destroyer. Let’s break down what is going on in the scene:

Han, Leia and Chewie hastily exit the asteroid to avoid being eaten by a huge improbable worm. A nearby Star destroyer detects them and give chase. Falcon first seem to shrug of the fire from the enemy but then a hit destroys their rear deflector shield. Han reroutes all power to the forward deflector shield, turns around and fly straight at the enemy. Captain Piett order shields up as Falcon barely misses them. The Star destroyer has lost track of the Falcon and assume it has escaped, possibly through some sort of cloaking device albeit a ship that small is unlikely to have one. Falcon has in fact docked on the rear side of the Star destroyers bridge tower.

Top and bottom arcs

Intercept neither have shields nor cloaking devices, the only defense against lasers aside from armor is sandcasters so let’s assume our trader ship has a bunch of these; one laser and two sandcasters in the top turret and one laser and two sandcasters in the bottom turret. The Beowulf won’t ever fire at the Azhanti but we calculate it anyways:

  • Beowulf Pilot task 7+, +4 from Hand Solo being an excellent Pilot
  • Dice pool 1
  • Thrust 1.8 Gs unloaded
  • DAB 20, ARM 19, Surface 13/19, Power 16/19, Thrust 16
  • Effective Range 1 Single 10 MW laser: PEN 24 DAM 24
  • Beams 5+ range 1, 8+ range 2-3, 11+ range 4+, Missiles 8+
  • Two sandcasters defense roll 9+, +2 for dual

The Azhanti has lots of weaponry including the deadly meson spinal mount. Darth Vader wants to capture the ship rather than killing it as he believes Luke is on board. The Azhanti will use its two batteries of 10 MW lasers each and its two batteries of 50 MW each. The Azhanti will also use Spray fire attacks to trade damage for multiple hits hoping to cripple rather than killing, most commanders know what awaits if one displeases Darth Vader.

  • Azhanti Pilot task 12+
  • Dice pool 7
  • Thrust 2.3 Gs loaded
  • DAB 35, ARM 31, Surface 25/31, Power 25/31, Thrust 28
  • Range 1 Triple 10 MW laser: PEN 25 DAM 25
  • Range 3 Single 50 MW laser: PEN 29 DAM 29
  • Beams 0+ range 1, 3+ range 2-3, 6+ range 4+, Missiles 3+
  • No sandcasters

Top and bottom arcs 2

We will simply assume that Han Solo with his excellent piloting skills and a target number of 7 will always be able to turn at least 4 steps (which is the same as saying he will roll 3+ on his Pilot task rolls). We also assume that the cruiser will use many of its computer pool dice to get 2 steps of turning on every turn. Given the fact that when Han Solo initially escaped from Hoth and was chased into the asteroid belt his fancy flying had two cruisers nearly crashing. Surely lord Vader will have none of that so the cruiser captains make damned sure that maneuvering is a top priority.

The trader is leaving the asteroid with a speed of 2 and the Azhanti has the same speed, two squares behind and one to the right. The cruiser immediately give chase and the fight is on.

The cruiser fires its lasers at the trader and the trader defends using the two sandcasters of its belly turret. The sandcasters does not completely stop the laser attack and the cruiser inflicts Severe damage to the Surface location of the trader.

This is the moment C3PO say “Sir, we just lost the main rear deflector shield. One more direct hit on the back quarter and we’re done for!”

The cruiser know the trader is now nearly defenseless and confidently turn and thrust to close the gap.

The trader captain knows he can no longer rely on the sandcasters fending off the cruisers lasers and he cannot outrun the cruiser as the trader’s 1.8 Gs is no match against the 2.3 Gs of the cruiser. He decides to turn the ship around and go for the cruiser’s blind aft centerline. C3PO shouts in protest and Needa flinches as the trader passes the bridge of the cruiser.

The trader disappear from the cruiser’s sensors as it has moved into its blind aft centerline. The next turn the cruiser will no longer learn the trader’s move. It is crucial that the cruiser reacquires the trader or the trader might slip away. Failure is not an option when Darth Vader is your commander.

The captain uses the formidable computer power at hand to turn the cruiser 2 left and thrusts for 2G, turning the cruisers blind aft centerline away from where he thinks the trader will go. The trader decides to capitalize on the cruiser being blind by turning towards the cruiser and thrusts to get on top of it.

What is Han Solo thinking, surely the cruiser will have the common sense of scanning itself when the enemy has lost its track?

The movement phase is over and its time for the Sensor phase. Having lost the track of the trader the cruiser should perhaps go for a wide scan that is guaranteed to contain the trader, a Visual or Infrared 2×2 box Scan maybe?

Every sensor operator and every officer at the naval academy learns that Scans touching the Sunglare column are bad, akin to staring into the sun on guard duty. Sure, Hoth lies outside of the Goldilock zone of its star and the asteroid belt is even further out it will still severely degrade the Scan, what else to do?

The best option might be for the commander to put a small 3×3 square Radar scan on the ship. Radar don’t suffer Sunglare degrading and will make sure the trader cannot hide anywhere near the ship. We who have seen the movie knows this is not what happened – Leia and Han was not captured and brought to Vader’s command ship so why didn’t captain Needa use a radar scan to reacquire the trader?

Captain Needa, after losing track of the trader “I shall assume full responsibility in eluding them apologize to lord Vader – meanwhile, continue to scan the area”

The reason is simple; as Needa has left to ‘apologize’ to Vader the remaining bridge crew on the cruiser does not want to signal to Vader that they have lost track of the prey. Vader’s command ship will surely pick up the radar scan of the cruiser and inform lord Vader of their incompetence.

The cruiser decides on a Visual 1×1 box scan that doesn’t touch the Sunglare and covers the presumed location of the trader.

The cruiser sees nothing and it’s remaining bridge crew learn the fate of captain Needa, failure is not an option in tbe Imperial navy it seems. It is time for the next turn.

Finding nothing from the previous turn’s Scan the cruiser turns left and thrusts for 1G. Han Solo, still untracked, executes the final step in his clever gamble and manuvers to dock with the cruiser (same, position, vector and facing is all that is required).

The trader is now docked with the cruiser and the only way a Scan to detect it is if the scanner explicitly states that they are scanning for docked ships, with their Sense target number up to 7+ from the usual 6+. See page 26 of the rulebook for details on Landed or docked.

C3PO “Captain Solo, this time you have gone too far!”

The trader ship will remain docked until the fleet get ready to jump. Very few people have ever heard about this trick; to dock unseen with an enemy vessel to elude detection, unfortunately for Han Solo, one of these few is Boba Fett who follow them as they undock.

So, there you have it: playing out the escape from the Hoth system using only standard Intercept rules, even finding rationales as to why captain Needa does what he does.

Keep space tidy – dump in jump.