• Welcome to the new COTI server. We've moved the Citizens to a new server. Please let us know in the COTI Website issue forum if you find any problems.

Vector Movement

Book 2 works great on an open floor with cutout planetary models (get a protractor). Many of the worlds are reasonably sized, and slingshot mechanics make it interesting.

Vectors are great fun. I think, based on only a little experience, that the exercise of managing one's vector will probably become *the* centerpoint of the wargame. Transitioning from a vector game back into an RPG feels ... strange. Maybe it makes for a good change of pace, though, in a long-running game. Depends, I guess.
 
Yes, book 2 does work great played as miniatures as suggested. The maneuvering board suggestion just makes it quick and easy to do on a table top. It's sort of a piece of graph paper designed specifically for movement of ships in the way book 2 lays out, only in far greater detail.
 
Magnus von Thornwood said:
That link does not go to a useful PDF. It goes to a PDF of a set of presentation slides. Boo-hiss!

Is there an actual readable version of the "mo board" and some not presentation docs?

I mean it looks cool and neat but so far I can't even make this out at all.
Click to expand...

I found a source for the chart, Weather Graphics, about halfway down, Vector Graphics entry. PDF or JPG download.
 
Classically we did CT movement on graph paper, just like our Harpoon moves.

Nowadays I would probably use Mayday type movement with a double 10-hex mapset, one plotting 10000km and one 100000km. Most CT play will be at 900000km or less.
 
Vectors revisited

Had to work on other things for a while, back now to gaming. Or math as may be.

Given a distance of 489,502,000 meters, a deceleration factor of 30 m/sec and a final speed of relative 0, how do I formulate this to find initial speed?

Wondering if I should just convert to Star Wars motion LOL
 
The deceleration is linear so... it should be 489502000 -30 /30. The initial speed should be 163,732.33
 
Distance is measured in m.
Speed is measured in m/s or how much do my position change every second.
Acceleration is measured in m/s² or how much do my speed change every second.

Decelerating to rest takes the same time as starting from 0 and accelerating.

In the first second we accelerate by 30 m/s², we start at speed 0 and end with speed 30 m/s, the average speed is 15 m/s, so in that second we have travelled 15 m. In the next second we accelerate from 30 m/s to 60 m/s, the average speed is 45 m/s and we travelled another 45 m for a total of 60 m. In the third second we accelerate from 60 m/s to 90 m/s, average speed 75 m/s, travelling another 75 m for a total of 135 m.

With a constant acceleration the speed increases linearly, and the distance travelled increases exponentially. The general formula would be distance = acceleration × time² / 2, so time = sqrt( distance × 2 / acceleration ).
Distance is 489 502 000 m and acceleration is 30 m/s² so time is 5712,57 s.
Speed achieved is 5712,57 s × 30 m/s² = 171 377 m/s.

Decelerating to rest again takes the same time and distance travelled so this is your answer.
 
Enoki said:
The deceleration is linear so... it should be 489502000 -30 /30. The initial speed should be 163,732.33
Click to expand...
No.

(489 502 000 m - 30 m/s²) / 30 m/s² ≈ 16 316 732,33. It is also physically undefined.

You cannot subtract an acceleration from the distance, the result has no defined dimension (unit of measurement). What is 3 apples - 2 oranges?

A distance divided by an acceleration would have the dimension m / ( m/s² ) = s². That is the square of a time, not a speed.
 
AnotherDilbert said:
No.

(489 502 000 m - 30 m/s²) / 30 m/s² ≈ 16 316 732,33. It is also physically undefined.

You cannot subtract an acceleration from the distance, the result has no defined dimension (unit of measurement). What is 3 apples - 2 oranges?

A distance divided by an acceleration would have the dimension m / ( m/s² ) = s². That is the square of a time, not a speed.
Click to expand...

It's a simple TSD problem... Time, Speed, distance. You lose 30 meters in the first second and then 30 additional meters for each additional second until you reach zero. 30 + 30(n). You are starting with a fixed distance to travel so setting equation to 0 gives 163,167,32.33 seconds of deceleration at 30 m/s.
The problem calls for linear deceleration, not a curve. So it isn't m/s^2.
 
Enoki said:
It's a simple TSD problem... Time, Speed, distance.
Click to expand...
Yes.

Enoki said:
You lose 30 meters in the first second and then 30 additional meters for each additional second until you reach zero. 30 + 30(n). You are starting with a fixed distance to travel so setting equation to 0 gives 163,167,32.33 seconds of deceleration at 30 m/s.
The problem calls for linear deceleration, not a curve. So it isn't m/s^2.
Click to expand...
No. You are confusing speed and acceleration, acceleration is a change in speed, not a change in position. You are almost calculating for a constant speed of 30 m/s. Acceleration is always measured in m/s², speed is always measured in m/s, distance is always in metres [m].

Note that 16 316 732,33 s is about six months, a clearly unreasonable result.

See: How to Calculate Time and Distance from Acceleration and Velocity
 
You must log in or register to reply here.

Similar threads

T
MT Only: My MT Errata Odyssey
Replies
1
Views
229
trb456
T
Spartan159
Atmospheric pressure effects on transition to and from orbit and other tidbits
2
Replies
27
Views
496
Spinward Flow
Spinward Flow
D
ISO Ship Vector Movement Software
Replies
5
Views
2K
AndyW
A
O
Range Bands
Replies
7
Views
1K
Supplement Four
S
S
CT Only: TTB Vector Movement
Replies
5
Views
1K
tim_manley
T
Back
Top