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Old 10-04-2018, 07:39 AM   #1
hal
 
Join Date: Aug 2004
Location: Buffalo, New York
Default CT rules, or what am I missing here? Help???

Hello Folks,
As I continue on with a project of sorts, I ran into this question and had a "Doh!" moment. I'm seriously hoping that I'm the one who is being stupid here and hope someone can help either confirm that I'm right, or point out where I'm wrong (and boy do I hope I'm wrong!!!!)

Quoted from the classic book 2 rules...

"2. Space: A playing surface is required, representing space as a two-dimensional surface at the scale of 1 :100,000,000; one millimeter equals 100 kilometers. Three meters equal one light-second."

So that is the scale.

Elsewhere it states that each "turn" is 1,000 seconds.

So this is where I'm having the "Doh!" moment. Velocity = Acceleration * Time right? So in theory, assuming that the authors rounded the acceleration of 1 G to 10 meters per second, then 10 meters/second * 1000 seconds should equal to 10 Meters * 1000 (units after factoring out seconds would be meters) should equal 10,000 meters right? The authors of the rules specified the following as step 3 on page 26 of Book 2: Starships. It says this...

"3. Thrust: Maneuver drive thrust is measured in Gs (gravities) expressed as a vector of both length and direction. While direction is variable, the length of the arrow is represented at the scale 100 mm equals 1 G (1,000 seconds acceleration at 1 G will produce a velocity change of 10,000 km, or 100 mm in scale, per turn)."

Per the math involved, shouldn't this be 10 Km per turn instead of 10,000 km? If so, shouldn't a velocity change only be 0.1 mm per G of accleration instead of 100mm?

Is this a case where I've always accepted the author's dictum and never bothered to check it?

I HAVE to be wrong - don't I?

What I should have realized is this:

Distance traveled during the acceleration is: 1/2 * A * T^2

This means that in 1,000 seconds, a ship will travel a total of 10/2 x 1000^2 meters, or 5,000,000 meters. Turning this to Km, I get 5,000 km. Dividing by 100 to get mm movement on the table, it becomes 50mm

Last edited by hal; 10-04-2018 at 08:15 AM. Reason: Addenda
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Old 10-04-2018, 11:27 AM   #2
ak_aramis
 
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Default Re: CT rules, or what am I missing here? Help???

Quote:
Originally Posted by hal View Post
Hello Folks,
As I continue on with a project of sorts, I ran into this question and had a "Doh!" moment. I'm seriously hoping that I'm the one who is being stupid here and hope someone can help either confirm that I'm right, or point out where I'm wrong (and boy do I hope I'm wrong!!!!)

Quoted from the classic book 2 rules...

"2. Space: A playing surface is required, representing space as a two-dimensional surface at the scale of 1 :100,000,000; one millimeter equals 100 kilometers. Three meters equal one light-second."

So that is the scale.

Elsewhere it states that each "turn" is 1,000 seconds.

So this is where I'm having the "Doh!" moment. Velocity = Acceleration * Time right? So in theory, assuming that the authors rounded the acceleration of 1 G to 10 meters per second, then 10 meters/second * 1000 seconds should equal to 10 Meters * 1000 (units after factoring out seconds would be meters) should equal 10,000 meters right? The authors of the rules specified the following as step 3 on page 26 of Book 2: Starships. It says this...

"3. Thrust: Maneuver drive thrust is measured in Gs (gravities) expressed as a vector of both length and direction. While direction is variable, the length of the arrow is represented at the scale 100 mm equals 1 G (1,000 seconds acceleration at 1 G will produce a velocity change of 10,000 km, or 100 mm in scale, per turn)."

Per the math involved, shouldn't this be 10 Km per turn instead of 10,000 km? If so, shouldn't a velocity change only be 0.1 mm per G of accleration instead of 100mm?

Is this a case where I've always accepted the author's dictum and never bothered to check it?

I HAVE to be wrong - don't I?

What I should have realized is this:

Distance traveled during the acceleration is: 1/2 * A * T^2

This means that in 1,000 seconds, a ship will travel a total of 10/2 x 1000^2 meters, or 5,000,000 meters. Turning this to Km, I get 5,000 km. Dividing by 100 to get mm movement on the table, it becomes 50mm
You are the one missing something. Two key somethings.

First
not halving the actual distance covered is a standard vector gaming conceit, as it cuts the math involved heavily

Second
YOu're calculating vector in m/sec, not m/turn
Given 1000 sec turns
Vps = G * Time
Vpt = Vps * Time

Vps = 1000 * 10 m/s = 1e5 m/s
Vpt = 1e5 m/s * 1000 = 1e8 m/t = 1e5 km/s

there's your 10,000 km/turn/gee

Yes, Marc explicitly rounds G=10 m*s² and C=3e9 mps
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Old 10-04-2018, 07:04 PM   #3
hal
 
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Location: Buffalo, New York
Default Re: CT rules, or what am I missing here? Help???

Quote:
Originally Posted by ak_aramis View Post
You are the one missing something. Two key somethings.

First
not halving the actual distance covered is a standard vector gaming conceit, as it cuts the math involved heavily

Second
YOu're calculating vector in m/sec, not m/turn
Given 1000 sec turns
Vps = G * Time
Vpt = Vps * Time

Vps = 1000 * 10 m/s = 1e5 m/s
Vpt = 1e5 m/s * 1000 = 1e8 m/t = 1e5 km/s

there's your 10,000 km/turn/gee

Yes, Marc explicitly rounds G=10 m*s² and C=3e9 mps
Math check...

Vpt = VPS x TURNS.

So if the number of turns is 10, then it would be VPS x 10, which would in turn, be 1000*10 * 10 meters, or 1000 x 10 x 10 /1000 km
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Old 10-06-2018, 01:16 AM   #4
Mike Wightman
 
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Default Re: CT rules, or what am I missing here? Help???

In CT 77 edition there was an optional rule right at the end of ship combat to only use half of the displacement from acceleration that the rules suggest earlier - i.e. properly model Newtonian movement.
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Old 10-07-2018, 03:07 AM   #5
ak_aramis
 
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Default Re: CT rules, or what am I missing here? Help???

Quote:
Originally Posted by hal View Post
Math check...

Vpt = VPS x TURNS.

So if the number of turns is 10, then it would be VPS x 10, which would in turn, be 1000*10 * 10 meters, or 1000 x 10 x 10 /1000 km
Not TURNS, but turn length.

Given unit time Time = duration of turn in seconds
Vps = A * Time
Vpt = Vps × Time
Vpt = (A × Time) × Time
Vpt = A × Time²

Note that VPT is in the distance in meters travelled over one 1000 second turn after one turn of acceleration.

Last edited by ak_aramis; 10-07-2018 at 03:13 AM.
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Old 10-07-2018, 10:54 AM   #6
hal
 
Join Date: Aug 2004
Location: Buffalo, New York
Default Re: CT rules, or what am I missing here? Help???

Quote:
Originally Posted by ak_aramis View Post
Not TURNS, but turn length.

Given unit time Time = duration of turn in seconds
Vps = A * Time
Vpt = Vps × Time
Vpt = (A × Time) × Time
Vpt = A × Time²

Note that VPT is in the distance in meters travelled over one 1000 second turn after one turn of acceleration.
Initially, I was of a mind to disagree, until I took the time to set up a spread sheet.

A column was turns

B column: =A2*100 (this gives me mm per turn per CT rules)

C Column: =B2*100*1000/1000 (converts 1mm to 100km to meters divided by 1000 seconds for meters per second)

D Column: =10*1000*A2 (Accleration * time per turn * turns)

E Column: =1/2 * 10 * (1000*A2)^2/1000/100 (real displacement in mm where 1000*A2 is time. divide by 1000 to convert meters to km, divide by 100 to convert to mm scale on table)

F Column: =B3+F2 (this basically tallies total displacement in mm per the rules taking this turn's move plus last turn's built up move.)

End Analysis
After one turn's movement, the game's displacement was higher than the real displacement by 100%

After two turns, it was 50% too high
After three turns, it was 33.33 percent too high
It continues to drop percentage wise as far as being too high, until after 40 turns, it is only 2% too high.

So real displacement formula versus "game rules displacement" works out surprisingly well without the optional rule 50mm per G instead of 100. When I added in an optional rule column for analysis, I found out quickly, that the optional rule would not have come close to the real world formulas. I suspect that is why it was dropped from subsequent publishing.

In the end? The Meters per second value based on time works equally well for the real formula versus a derived one based on vector length providing one DOES take into account that built up velocity really is Distance travelled divided by 1000 seconds to get meters per second value.

The formula =B2*100*1000/1000 was taking the value of B2 (which is 100mm x turns) and converting it first, into Kilometers (hence the 100 value) * 1000 meters per kilometer to convert to meters (hence the 1000 value) dividing by 1000 to get it in terms of Meters per second. It matches exactly what the formula of Acceleration x time equals.

So, yes, I was wrong about not using A * 1000 * 1000. That turns out to be right.

After I extended the time from 50 turns to 350 turns, It ended up that the loss in accuracy of the "approximation" of 100mm per turn movement, and real world displacement, ended up being only a loss of .2%

so, MY mistake was that I didn't take into account that it was time^2 * Acceleration.

That makes me feel a little bit better then.

Thanks Aramis.

So now, all that remains is for me to finish my "dream" of a virtual table top for vector movement based Traveller - is to finish with...

Add vector of acceleration to get final location. I already have the code such that I can have a ship's built up velocity determine where it will end up if there is no acceleration applied. All that remains if the final step of adding where the ship will be after the acceleration is computed.

:)
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Old 10-08-2018, 11:59 PM   #7
ak_aramis
 
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Default Re: CT rules, or what am I missing here? Help???

Quote:
Originally Posted by hal View Post

so, MY mistake was that I didn't take into account that it was time^2 * Acceleration.

That makes me feel a little bit better then.

Thanks Aramis.
You're more than welcome.

Don't worry, you're not the first to make that error... I did, too, at a point many years ago.
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