Archive for the Rules Category

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


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.


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.



Han Solo aggressive docking

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

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.


Posted in Design system, Intercept, Rules, Traveller on January 8, 2019 by Anders Backman

Falling off the hull

I recently learned that Dropbox have changed how direct download links works so it became really inconvenient for you to download stuff from me (they tried to get you to register at DropBox and yadda yadda). Anyhow, I have fixed all the download links so you just click on the link and save, just the way it was intended, sorry about the inconvenience.

So, just go to the Downloads page and click click on what you want.

PS I have updated the rulebook with a brief combat example on page 13, and some other small updates here and there DS

Intercept maps

Posted in Intercept, Rules on January 8, 2019 by Anders Backman


What is this Marre-red maneuver Sir?

It was invented by a rather colourful pirate named Mauricio Redondo way back and is used when you come out of the sun and assume your enemy is hiding in the planetary shadow. You approach the planet building up quite some speed, say 30 to 50 clicks a second, and drift as you pass above or below the planet with your powerplant off. Keep drifting ’til you think you passed the enemy then power up the reactor and start braking hard. You are now ‘south’ of your opponent facing him, in the planet shadow with your enemy to sun-ward yet no risk of any sun-blinding. Then you take him out son.


Intercept comes with a bunch of ready-made maps, empty, with a small or large planet or our beloved Earth and Luna system. Aside from the grid squares and boxes there are some features that might not be that obvious. This little post will teach you what they are and how to use them. Oh, and if anyone want a custom map with planets and asteroids give me a message here on the blog and I’ll add them.

Map coordinates

The Intercept map consist of individual squares and larger boxes, each box holding five by five squares, each box has a column letter and a row number to help identify which box you mean, individual squares within a box are numbered left to right and then top to bottom. The top left square of the top left box is A1-1-1 for example and the top right square of the third box on the second row of boxes is C3-5-1.

map coordinates

The top left box of each mapsheet has the columns and rows numbered for you.

Note that the rightmost column of boxes is labeled G/A and the lowest two rows are labeled 10/1 and 11/2. If you wish to play on an area larger than a single mapsheet the rightmost column form the first column of the next sheet and the lower two rows for m the first and second of the sheet below. This overlap is there to simplify maneuvering straddling the two sheets. Note that playing on more than one sheet complicates the game quite a bit so don’t do it unless you feel like it is really needed. Simply stating that ships flying outside the sheet are lost works fine too and is much simpler.

The box G/A 10/1 also has the columns and rows numbered for you, this is to help you remember that this box might be the top left box of the next sheet to the right and down.


The sun is always shining from the top of the mapsheet as indicated by an arrow. To the left of the arrow there’s a box where you write the actual Sunfactor used in the scenario. The Sunfactor is 6 for Earth & Luna or any planet in the biozone. Each orbit outside of the biozone and the Sunfactor goes down by 1, down to 0 and for each orbit inside the biozone increase the Sunfactor by 1, Mars is 5, Venus 7 and Mercury 8 for example. Sunfactor 9 or above will be hot enough for the ships to require special rules outside of the scope of this little blog post, stay tuned.

If you have detailed information about the star system you play in, specifically the relative luminosity of the central star (the Sun has 1.0) and the orbit radius in AU (Earth has radius 1.0) use this formula (Round to the nearest whole number but never below 0):

Sunfactor = 6 – 4 * Log10( L / R )

L = Luminosity, R = Orbit radius in AU

Turn number and A/B order

Sensor scans are done in A/B order and the final Initiative tie breaker is also done inA/B.  A/B order means that side A scans first and wins initiative ties on odd turns and B on even turns. There is a row of boxes at the bottom of the mapsheet, cross them off as each turn is performed. The leftmost uncrossed box tell you if it is an A or B turn.

mapsheet legend lower

Fractional thrust etc)

Ships with factional thrust (1.33, 0.85 etc) can always thrust the integer part on any turn but the fractional part may give them +1 G to use on certain turns. This is shown on the ship DataCard but the information is also available in the turn number boxes (the 0.25+, 0.75+, 0.5+ values). Let’s say your ship has a thrust of 1.33, it will thrust 1 G at every turn but as 0.33 >= 0.25 it will thrust 2 G during the first turn of every four. The 0.85 G ship has less than 1 G of thrust so all of its thrust is fractional; 1G at the first, second and third turn but no thrust on the fourth turn of every block of four.

Map out your future – but do it in pencil. – Jon Bon Jovi

That’s all, stay cool, and in the shadows people.