Vector movement game units

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

 

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