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Doing the space-time warp

Carlobrand

SOC-14 1K
Marquis
Okay, my college days are far behind me, maybe I lost a bit of math along the way.

MT referee's manual says a space combat turn is 20 minutes long, a "square" is 25 thousand kilometers, and a ship going 6g can alter their vector by six of those.

Now, as I recall, 6g is 60 m/s^2, using the Traveller "g". If I apply 60 m/s^2 for 20 minutes, I end up with a final velocity of ... v=at, 20 minutes is 1200 seconds, times 60 is 72 kps. That's 86,400 kilometers every 20 minutes if I coast from that point. Distancewise, if I started at speed 0, I ended up covering ... s=1/2at^2, t=1200, t^2=1,440,000, at^2=86,400,000, half of that is 43,200,000 - or 43,200 kilometers.

How in tarnation is that supposed to equal six 25 thousand kilometer squares??? Six 7 thousand kilometer squares, maybe - but then I end up with these 2-hex planets that I have a nasty tendency to put big craters in, to the great dismay of the next of kin of the involved crew, not to mention doing terrible things for my popularity rating with the locals and prompting the enemy to award me medals, which is simply embarrassing. 37 minute turns? 30-minute turns on 16,000 klick squares?? Oh, I'm getting another headache!
 
10*1200*1200/1000=14400km/t per G.

DGP failed their math rolls.

Now, for 25,000km, 1581 sec is the needed amount... 26:21 or so.

Space Combat as a chapter is bad...
 
General formula for movement, where

s = Final displacement
s0 = initial displacement
v0 = initial velocity
a = acceleration
t = time

s = 1/2(a*t^2) + v0*t + s0
 
I got the 1/2 at^2 bit. I noted that in my initial post. I was wondering why we were arbitrarily dropping the 1/2 bit - although I realized belatedly that it seems to be the short-hand method MegaTrav was working under to simplify movement.
 
Why AT^2?

A Turn is T seconds long.
G is your favored unit
in T seconds, you accelerate to GT meters per second. Well call this V. V=GT
For a turn of T seconds, the accumulated vector V is experienced for T seconds. Thus distance D covered in turn is VT. When we unpack V, we get D=VT=(GT)T=GT²

Note that for game purposes, the fact that you don't actually cross the full hex on turns of acceleration is ignored in Mayday and BL. (And Full Thrust. And almost every other ship combat game using vectors.)
 
A Turn is T seconds long.
G is your favored unit
in T seconds, you accelerate to GT meters per second. Well call this V. V=GT
For a turn of T seconds, the accumulated vector V is experienced for T seconds. Thus distance D covered in turn is VT. When we unpack V, we get D=VT=(GT)T=GT²

Note that for game purposes, the fact that you don't actually cross the full hex on turns of acceleration is ignored in Mayday and BL. (And Full Thrust. And almost every other ship combat game using vectors.)

Sigh.

I'm not meaning to get into a physics debate. V is not a constant in this model, so we'd be integrating - if my long-dusty memory of calculus is right. The real-world equation would be x(t)=x(0)+v(0)t+1/2at^2, which is why I was wondering at the missing 1/2. For game purposes, vector sum works adequately to model the battle - I just don't tend to automatically think in those terms.
 
Sigh.

I'm not meaning to get into a physics debate. V is not a constant in this model, so we'd be integrating - if my long-dusty memory of calculus is right. The real-world equation would be x(t)=x(0)+v(0)t+1/2at^2, which is why I was wondering at the missing 1/2. For game purposes, vector sum works adequately to model the battle - I just don't tend to automatically think in those terms.

Wrong.

V is post-turn accumulated vector.

D in this case is how far that vector will carry you each turn, not how far you move during the acceleration. And thus no integration is needed.

The missing 1/2 is only for finding how far you moved DURING the acceleration.
 
Which means your V is the same as my v(0), no?

No. All I'm figuring is change in vector over one turn - Your calculus is irrelevant for scale purposes.

See, If I thrust for 1 turn at 1A, my vector per turn is AT². I will have gone 0.5AT², but each turn thereafter I have that retained vector even if I'm out of fuel. That's thus the most convenient unit for scaling. That on turn 1 I cross only half of the hex is ignored for simplicity, because every other turn, I cross a whole hex on that vector.

A 3-counter system can make it workable to track those lost half-scale units, but it's not worth it long term.

You're trying to figure trip; I'm just figuring scale.

And DGP didn't do the math right for either case.
 
Obscure side note: this is also the "cheat" (or approximation, or simplification, if you prefer) built into Mayday. If you bother to work it out, if you accelerate by 1G for 1 turn, you will have only moved half a hex - due to that elusive 1/2 multiplier you have been discussing - rather than the one hex it moves according to the rules.

(Still, maybe we can just say we round up?) ;)
 
Obscure side note: this is also the "cheat" (or approximation, or simplification, if you prefer) built into Mayday. If you bother to work it out, if you accelerate by 1G for 1 turn, you will have only moved half a hex - due to that elusive 1/2 multiplier you have been discussing - rather than the one hex it moves according to the rules.

(Still, maybe we can just say we round up?) ;)

Not just Mayday... Brilliant Lances, Full Thrust, Babylon 5 Wars, Spacemaster... and more.
 
Ken Burnside addresses this issue in Attack Vector: Tactical (http://www.adastragames.com/index.html) and I have attached a Dropbox link to my copy of a handout from his site that discusses the problem.

http://dl.dropbox.com/u/14464683/vms_vs_dv.pdf

It comes down to the classic gamer's dilemma: how much realism do you want in your simulation? Keep in mind, simplicity and realism are frequently opposite ends of a spectrum when you establish your movement rules.
 
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