• 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.
  • We, the systems administration staff, apologize for this unexpected outage of the boards. We have resolved the root cause of the problem and there should be no further disruptions.

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
383
trb456
T
Spartan159
Atmospheric pressure effects on transition to and from orbit and other tidbits
2
Replies
27
Views
869
Spinward Flow
Spinward Flow
D
ISO Ship Vector Movement Software
Replies
5
Views
3K
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