Author Topic: Newtonian Aurora  (Read 61368 times)

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Offline ndkid

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Re: Newtonian Aurora
« Reply #30 on: August 30, 2011, 12:51:57 PM »
I wouldn't mind knowing how Attack Vector: Tactical handles it. My research online hasn't been very productive on this particular question yet. My intention though is to go with something that reflects the difficulty in changing course at high speed, is easy for players to understand and generally follows the applicable physical laws, without the requirement to follow those laws exactly

Steve

AT:V simplifies, in the first place, by using a hex grid. Ship vectors have three components... one on the z axis, and two on the xy axes, that can be 60-degrees apart.  This allows it to deal with vector math relatively simply: opposing vectors (180) cancel out. Vectors 120 apart "meet in the middle": add the 120-degree off vector to the 60-degree off vector, then subtract it from the 0-degree vector.
 

Offline Charlie Beeler

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Re: Newtonian Aurora
« Reply #31 on: August 30, 2011, 01:18:24 PM »
For reference this is extracted from GZG's Full Thrust/Fleet Book One where vector movement is introduced to that minitures system.

Code: [Select]
COURSE AND FACING

Under the standard Cinematic FT movement, a ship will always be facing in the same
direction that it is moving; under the VECTOR system the ship may be moving one way
and facing another. The direction in which the ship is actually MOVING is termed its
COURSE, while the direction in which the ship model is actually pointing is called its
FACING. The current COURSE is indicated by a small arrow marker placed next to the
ship’s stand, and this marker is also used as a reference point during the process of
moving the model. It should be noted that the FACING of a model should always be one
of the 12 “clockface” points, though the mechanics of the vector movement mean that
the COURSE will usually NOT correspond exactly to a clockface direction.
MAIN DRIVE THRUST

The THRUST RATING of any ship is the amount of thrust that can be produced by its
MAIN DRIVE - the “big engine” at the back. Each point of thrust applied in a turn will
accelerate the ship by 1 inch (or other movement unit) ALONG THE AXIS OF THE SHIP,
so if a ship that is facing in its direction of travel (i.e. its “course” and “facing” are the
same) and currently moving 6" per turn applies 4 points of thrust from its main drive,
it will end up moving at 10" per turn. If the ship’s facing and course are NOT the same
(i.e.: the model is pointing one way and moving another) then the application of thrust
from the main drive will alter the ship’s course AND velocity. To DECELERATE using the
main drive (as opposed to using the forward “retro” thrusters), the ship must be turned
so that it is pointing “backwards” relative to its current course. When writing orders for
your ship, Main Drive thrust is written as MD followed by the number of thrust points
being applied - so MD4 will move the ship 4" in the direction of its present facing.
If using existing ship designs (whether from this book, from FT2 or elsewhere) then the
thrust level shown in the ship’s drive icon is the rating used for the main drive.
MANOEUVRING THRUSTERS

In addition to the main drive, all ships have THRUSTERS - small drives positioned in
clusters around the ship, pointing forward, port, starboard etc. (in reality ships would
also of course have “up” and “down” orientated thrusters, but as we are not concerned
with 3D movement in FT we can ignore these except for their use in rolling the ship).
The thrusters may be used to “push” the ship to alter its course, or to rotate the ship
onto a new facing. The power available to the ship’s thrusters is equal to half the
thrust rating of the main drive - so a ship with a main drive TR of 6 would have 3
manoeuvre points available from its thrusters; unlike the Cinematic movement rules,
thruster use is allowed in addition to applying full available thrust with the main drive
- so that a ship with a Thrust Rating of 4 could apply 2 points of thruster use and still
use all 4 thrust points from its main drive.
We have not depicted the Thruster systems as separate icons on the ship diagrams, in
order that any design may be used with either movement system without alteration.
For the purposes of damage, assume that the thrusters are driven by the same power
systems as the main drives - when the main drive takes damage, thruster power is
halved or lost accordingly.
ROTATION

Rotation of a ship around its axis requires much less power than actually changing its
vector. When the thrusters are used to rotate a ship onto a new heading, ONE
manoeuvre point from the thrusters allows the ship to be rotated by any desired
number of facing points. Thus, for the expenditure of one point of thruster power a
ship can be rotated to face in any of the 12 possible facing directions, regardless of
the thrust rating of its drives (the only difference between rotating 30 degrees and
rotating 180 degrees is simply that, once the thrusters have started the ship spinning,
the ship is allowed to rotate for longer before the thrusters burn again to cancel the
spin). Note that a ROTATION changes the ship’s FACING only, and never its COURSE.
Note: when thrusters are used to rotate the ship onto a new facing, it is assumed that
several of the ship’s thrusters are fired in unison to achieve the desired effect - for
example, to rotate the ship to starboard it would fire the PORT FORWARD thrusters and
the STARBOARD REAR ones simultaneously to spin the ship around its centre of mass. It
is assumed that, in the same turn, a compensating burst is applied as the desired new
facing is reached in order to stop the ship’s rotation - the combined effect of these
operations constitutes one “rotation” action.
ROTATION orders should be written down as TP (Turn Port) or TS (Turn Starboard),
followed by the number of points of heading change - thus TP2 indicates a rotation to
port of 2 clockface points (ie: 60 degrees).
THRUSTER PUSHES

A thruster “push” is firing a combination of manoeuvre thrusters to alter the course
and/or velocity of the ship, WITHOUT affecting its actual facing (i.e.: the ship ends the
turn with its model pointing the same way it started, although its course may have
changed). Pushes may be made to PORT, STARBOARD or REVERSE (using the forward
“retro” thrusters to slow the ship down without having to spin it round and use the
main drive). It requires ONE manoeuvre point of thrust applied to displace the ship by
one movement unit; a push of 3 with the port-side thrusters will shift the ship 3" to
starboard (for simplicity of play, this is referred to as a STARBOARD PUSH - to avoid
confusing of orders we always use the direction of the EFFECT rather than the location
of the thrusters being used). Note that a PUSH changes the ship’s COURSE (and/or
VELOCITY) only, and never its FACING.
PUSH orders should be written as PP (Push to Port), PS (Push to Starboard) or PR (Push
in Reverse), again followed by the number of thrust points applied - so PR3 would be
using 3 manoeuvre points from the retros to push the ship 3 units “backwards” relative
to its current heading. Pushes may only be applied directly to port, starboard or
rearward relative to the ship’s facing at that moment.
COMBINING MANOEUVRES

If desired, a ship may combine both ROTATION and PUSH uses of its manoeuvring
thrusters in a single game turn, but no more than ONE of each, provided the TOTAL of
manoeuvre points expended does not exceed the total available. It is quite acceptable
for a ship with (say) 3 manoeuvre points of thruster power available to make a rotation
(using up 1 thruster point), then apply a main drive burn, then use the remaining 2
manoeuvre points for a 2" thruster push to port, starboard or aft as desired. The final
position, course and velocity would be measured after ALL manoeuvres are completed.
ORDER SEQUENCE

The actual sequence in which thruster and main drive burns are applied in a single turn
will make a difference to the final course and velocity of the ship, so it is necessary to
rule on what order things are done in. Each effect is applied to the ship strictly IN THE
ORDER THEY ARE WRITTEN DOWN BY THE PLAYER. If the player writes TP2, MD6 then
the ship will first be moved according to its starting vector (as always), then turned 2
points to port (TP2) and then moved 6" along its new facing (MD6). If, on the other
hand, the order is written MD6, TP2 (thus applying the main drive burn BEFORE
rotating the ship to its new facing) then the result will be VERY different in terms of
the ship’s final vector and position - plot each one out and you’ll see what we mean!
Amateurs study tactics, Professionals study logistics - paraphrase attributed to Gen Omar Bradley
 

Offline Steve Walmsley

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Re: Newtonian Aurora
« Reply #32 on: August 30, 2011, 03:39:09 PM »
Congratulations for getting me well and truly distracted from work today by the way!

Was just thinking a little more about tactical implications. Looking at your example cruiser, using rounded numbers and some noddy maths - if the ship is doing 4000 km/s its going to take it about 576m k to slow down to zero. That's about the range of my best size sensor so if something is coming the other way at the same speed on an intercept I can expect to be getting up close and personal with them whilst still at significant speed. It also means it's going to take a couple of days for the fleet to come round and make a second approach. That feels like maybe too little reaction time and too long on any re-engagement time.

It occurs to me that all the current games with some element of Newtonian mechanics are all tactical. So the timespan for acceleration is very limited and thus course changes are also possible in fairly short timespans. In Aurora FTL, ships may have been accelerating for days or weeks so it is only reasonable that they would require a long time to change course. I think intelligence gathering and scouting will be even more important. Probes will become much more common as you could send one coasting into an inner system after it bulds velocity a long way out.

Given accelerations of 1-2G, I think the 'commit yourself to a course' situation is how combat would turn out in reality.

Quote
Weapons wise it also looks to be leaning very heavily towards the massive volley strikes with missiles once in range and also maybe lots of energy weapons with reduced size and recharge rate.

Maybe some option for limited higher g manoeuvres would be a good way to help with this.

As with Aurora, I want to stay within the physics model for the game. The only way I can think of adding higher G maneuvers is to have some strap-on boosters with high-G and incredibly low fuel efficiency. Missiles will follow the fuel efficiency model as ships but I may extend it up to 500% boost and 10,000x fuel (which is the same as Aurora), so the strap-on booster could be modelled as a huge missile. A ship would have hardpoints for boosters or disposable fuel tanks. That type of things is some way in the future though after I have the basic movement model working.

Steve
 

Offline Steve Walmsley

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Re: Newtonian Aurora
« Reply #33 on: August 30, 2011, 03:44:53 PM »
AT:V simplifies, in the first place, by using a hex grid. Ship vectors have three components... one on the z axis, and two on the xy axes, that can be 60-degrees apart.  This allows it to deal with vector math relatively simply: opposing vectors (180) cancel out. Vectors 120 apart "meet in the middle": add the 120-degree off vector to the 60-degree off vector, then subtract it from the 0-degree vector.

The model I am planning to use is just a more detailed version of that, except in 2D not 3D. Aurora FTL will track headings to 1/10,000th of a degree so I should be able to get the right feel for maneuvers. I think I am going to go with 1% of velocity as the required delta-V for 1 degree of course change. That is about 10% better than reality but it is very easy for players to visualise and make calculations in their head. The fleet window will show the delta-V budget for each fleet/ship, along with estimated time and distance to decelerate to rest and other useful stats as I think of them.

Steve
 

Offline Beersatron

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Re: Newtonian Aurora
« Reply #34 on: August 30, 2011, 03:46:28 PM »
Side question: are you finished with the latest campaign since you are devoting more time to FTL? I was kinda rooting for the Chinese to come back with gusto!
 

Offline Steve Walmsley

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Re: Newtonian Aurora
« Reply #35 on: August 30, 2011, 03:49:49 PM »
Side question: are you finished with the latest campaign since you are devoting more time to FTL? I was kinda rooting for the Chinese to come back with gusto!

No, I have already done a little more with that campaign since the last AAR and I do intend to come back to it at some point. I just haven't had time due to my latest distraction :)

Steve
 

Offline MattyD

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Re: Newtonian Aurora
« Reply #36 on: August 30, 2011, 05:20:41 PM »
All this talk of Newtonian movement and delta v is really giving me a Larry Niven Protector vibe, I might read that again, its well worth another bash.

If your current plans make combat less black and white then I'm all for it, currently we are facing the same problem as the admirals who squared off at Jutland - its safer not risking everything on a single combat which invariably is devastating for one side.

Ohh and +1 on a Chinese comeback.
My Newbie AAR
 

Offline chrislocke2000

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Re: Newtonian Aurora
« Reply #37 on: August 30, 2011, 05:25:44 PM »
As with Aurora, I want to stay within the physics model for the game. The only way I can think of adding higher G maneuvers is to have some strap-on boosters with high-G and incredibly low fuel efficiency. Missiles will follow the fuel efficiency model as ships but I may extend it up to 500% boost and 10,000x fuel (which is the same as Aurora), so the strap-on booster could be modelled as a huge missile. A ship would have hardpoints for boosters or disposable fuel tanks. That type of things is some way in the future though after I have the basic movement model working.

Steve

Would you be able to do this through an extension of the fuel efficiency of engines when in use on top of fuel efficiency for size? I.e. make the current fuel consumption and thrust levels set for 50% max power which becomes the most efficient means of acceleration and deceleration. Then make decreasing returns on increasing thrust and increasing fuel cost up to 100% engine power. This allows ships to make relatively brief heavier accelerating or breaking manoeuvres at the cost of more rapid use of available Delta V?
 

Offline Erik Luken

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Re: Newtonian Aurora
« Reply #38 on: August 30, 2011, 07:11:50 PM »
AT:V simplifies, in the first place, by using a hex grid. Ship vectors have three components... one on the z axis, and two on the xy axes, that can be 60-degrees apart.  This allows it to deal with vector math relatively simply: opposing vectors (180) cancel out. Vectors 120 apart "meet in the middle": add the 120-degree off vector to the 60-degree off vector, then subtract it from the 0-degree vector.


Very similar to how I do it in Astra Imperia, except I stick with 2 dimensions. A third would be hard without some form of aid.
 

Offline sloanjh

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Re: Newtonian Aurora
« Reply #39 on: August 30, 2011, 11:23:16 PM »
For the moment I am working with a one degree course change requiring delta-V equal to 1/90th of your speed. This is very simplistically based on the fact that reversing course and achieveing the same speed requires a delta-V equal to double your speed, so if 180 degrees = 2x speed then 1 degree = 1/90th your speed. So if you are moving at 2000 km/s, a five degree course change will require 5/90 x 2000 km/s = 111.11 km/s of delta-V.

Umm actually the trigonometric answer would be (2*sin(theta/2)*V), where theta is the angle turned.  This is gotten by drawing a circle of radius V, drawing radii in the direction of the old course and the new course, then drawing a line between the two endpoints (giving you an isoscelese triangle).  If you split the triangle down the middle, you get two right triangles, with angle at the center of theta/2. The length of the split side is 2*(V*sin(theta/2)).  Note that for theta = 180 degrees, this gives the 2*V answer you have.  For small theta, you can use sin(angleInRadians) ~ angleInRadians, to get dV = V*thetaInRadians = V*(thetaInDegrees)*(pi Radians/180 degrees) ~ (V/60)*theta.  Note that the speed drops during the burn.  If you want to keep constant speed during the burn, then you basically want to keep the small angle formula for the full angle change, so the constant-speed equation is just V*(pi/180)*thetaInDegrees.  This makes sense, since to do the constant speed course reversal the velocity arrow would need to travel 1/2 circumference of the circle, while to do the "turn over and start blasting" reversal the velocity arrow travels the length of the diagonal.  By definition the ratio of these two distances is pi/2, which is the ratio of the two answers.

To put it a different way, your formula is about 50% (pi/2) too generous for small-angle course changes.

John
« Last Edit: August 31, 2011, 08:39:56 AM by sloanjh »
 

Offline James Patten

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Re: Newtonian Aurora
« Reply #40 on: August 31, 2011, 06:18:29 AM »
All this talk of Newtonian movement and delta v is really giving me a Larry Niven Protector vibe, I might read that again, its well worth another bash.

I'm getting a CJ Cherryh (Earth Company/Union/Merchanter's Alliance) vibe myself.  Although the Merchanter universe has jump not warp, they have to worry about velocity and so forth.
 

Offline ndkid

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Re: Newtonian Aurora
« Reply #41 on: August 31, 2011, 12:11:30 PM »
The model I am planning to use is just a more detailed version of that, except in 2D not 3D. Aurora FTL will track headings to 1/10,000th of a degree so I should be able to get the right feel for maneuvers. I think I am going to go with 1% of velocity as the required delta-V for 1 degree of course change. That is about 10% better than reality but it is very easy for players to visualise and make calculations in their head. The fleet window will show the delta-V budget for each fleet/ship, along with estimated time and distance to decelerate to rest and other useful stats as I think of them.

Steve

Steve, I don't think your math is as good as within 10%. Here's a quick example: the delta-v necessary for a 90-degree course change. In your system, that'd be a delta-v equal to 90% of the velocity, right? What would be necessary in reality is a delta-v of 141% (square root of two, basically).

At 180 degrees, of course, the math is simple... you're costing 180% v rather than 200% v, so you're only off by about 11% there. The smaller the angle, the worse the margin of error gets; at 1%, you charge 1% rather than the real 1.74%.

I take it your current coordinate system is based on speed and angle, rather than x and y vectors?

EDIT: sloanjh got here before me, I just figured I'd crunch the numbers and toss them into a spreadsheet to make sure I wasn't miscalculating.
« Last Edit: August 31, 2011, 10:45:39 PM by ndkid »
 

Offline waresky

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Re: Newtonian Aurora
« Reply #42 on: August 31, 2011, 04:08:17 PM »
Err..am only need one simple thimg: "Where r the Red Engine Button,Sir?"..here..ok..pusch to engine on.

Am a "simple" man,dude..

 

Offline procyon

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Re: Newtonian Aurora
« Reply #43 on: September 01, 2011, 04:19:41 AM »
If this direction actually works out, I may just have to learn how to play this game.

The thought of Newtonian elements makes this game much more attractive for me at least.
There are a number of games out there with 'reactionless' and 'jump points'.

Doing it the hard way, and not just a tactical game, would be incredible.... :)
... and I will show you fear in a handful of dust ...
 

Offline Steve Walmsley

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Re: Newtonian Aurora
« Reply #44 on: September 01, 2011, 05:10:54 AM »
Umm actually the trigonometric answer would be (2*sin(theta/2)*V), where theta is the angle turned.  This is gotten by drawing a circle of radius V, drawing radii in the direction of the old course and the new course, then drawing a line between the two endpoints (giving you an isoscelese triangle).  If you split the triangle down the middle, you get two right triangles, with angle at the center of theta/2. The length of the split side is 2*(V*sin(theta/2)).  Note that for theta = 180 degrees, this gives the 2*V answer you have.  For small theta, you can use sin(angleInRadians) ~ angleInRadians, to get dV = V*thetaInRadians = V*(thetaInDegrees)*(pi Radians/180 degrees) ~ (V/60)*theta.  Note that the speed drops during the burn.  If you want to keep constant speed during the burn, then you basically want to keep the small angle formula for the full angle change, so the constant-speed equation is just V*(pi/180)*thetaInDegrees.  This makes sense, since to do the constant speed course reversal the velocity arrow would need to travel 1/2 circumference of the circle, while to do the "turn over and start blasting" reversal the velocity arrow travels the length of the diagonal.  By definition the ratio of these two distances is pi/2, which is the ratio of the two answers.

To put it a different way, your formula is about 50% (pi/2) too generous for small-angle course changes.

John

I couldn't find the formulas I needed online so I went for a total guess that happened to work well for game mechanics and ease of player understanding :). Even though I now know the correct formulas I am still tempted to go for 1% per degree, although I suppose 1.5% per degree isn't really any harder to visualise.

Steve
 

 

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