Revised Agility Calculation Formula

A thread of 3 messages, 6 August 1991 to 8 August 1991, in the order they were sent.

#2729 Revised Agility Calculation Formula

[unnamed sender] · 6 August 1991

Hello Folks,

Rob Dean and I have been discussing the rules concerning the
calculation of agility in MegaTraveller.  We thought after
coming to some conclusions we should open this up to the
TML list.  Thanks go to George William Herbert for his thoughts
on the subject of aircraft agility.  (Maybe this should have
3 authors on it)  Please note that the following will be the
basis of all My agility calculations in future designs posted
to the TML.

Opinions will be warmly welcome.
Thank you,

Scott Kellogg
- - -------------------------------------------------------------------
The following is a summary of a discussion between Scott S. Kellogg
and Robert S. Dean.

We have noticed that the agility calculations as presented in the
MegaTraveller Referee's Manual make little sense when compared
with the Starship Operator's Manual, High Guard, and for that matter,
the Referee's Manual.

Problem:  A ship has 2 G acceleration.  It has no additional power
for weapons, just power for engines, sensors, environment, computer
etc.  Excess power is zero.  Result:  Agility=0.  This is assuming,
of course, that the undefined term 'excess power' refers to power
which is not allocated to any other operation aboard the ship apart
from increasing agility.  Ok, now when the ship is in emergency status,
the ship suddenly goes to agility=2.
Question:  Where did this "extra" power come from?

You see the problem? Is this what excess power really means? Is this the
result intended by the rules?

Ok, High Guard functioned like this:  A ship with power plant-2 and
maneuver drive-2 automatically becomes agility=2.  In order to have
a functioning maneuver drive, you had to have a power plant factor
equal to the maneuver drive thus a ship with the above plants will
always be agility=2 when not powering lasers (energy wpns etc). A
hit to the power plant reduces the agility of the ship by that amount.

Now, when a High Guard ship attempts to escape, the ship relies on
agility to do so, not maneuver drive.  So a 3-G ship with agility=3
can escape from a 6-G ship with agility=1.

Rationalization:  The 6G ship's power plant is being drained away
from the maneuver drive and it is not putting out 6-G's, it is putting
out only 1G.

In view of the MegaTrav design rules this does not translate unless
we do the following:

	(Power required for maneuver drive + "excess" power)
Agility=--------------------------------------------------- * 5.4
			Unloaded Weight

Agility may not exceed maneuver drive rating.
(No more agility=6 1-G drive ships...)
If you read the section of the Ref's Handbook, the rule there is a
little unclear.  I therefore think that the above clarification
(adding the power required by the drive to the equation as stated)
does make sense.

Note that if a ship is powering weapons the "excess" power can be negative
thus subtracting from the ship's ability to maneuver.  This is what Rob
refers to as "Double Dipping".

Please note that the above equation closely resembles Rob Dean's Thrust
based maneuver drive equations.

His opinion is that we should use the loaded weight for the above
calculations and I concur.  How often are ships flying around with
no fuel and no cargo?  He also suggests that the 5.4 in the above
be 10 as would occur in his thrust based maneuver drive calculations
(for grav based not thruster based vehicles).  That is, in a small
vehicle using standard grav propulsion, such as a 10 ton fighter, the
acceleration is the thrust divided by the loaded weight, and the
thrust is conveniently equivalent to the power input to the grav
units in MW multiplied by 10.

Thus Rob's equation:

	(Power req for maneuver drive + "excess" power)
Agility=----------------------------------------------- * 10
			Loaded Weight

The above should be multiplied by 13/14 if using thruster plates as
opposed to grav plates. This will not only give you the agility, but
the true acceleration of the ship as given by the thrust based
maneuver arguments.

Note that when a ship is dipping into it's maneuver power it's
maneuver drive performance will be affected in the following manner:

		(Power req for maneuver drive + "excess"pwr)
Combat Maneuver=--------------------------------------------* rated maneuver
			(Pwr req for manuever drive)

Note:  If "excess" power is positive, combat maneuver is not to exceed
rated manuever by more than 400% for more than 5 minutes without risk
of damage.  40% is acceptable for longer periods (days) if there is
suffient engineering skill among the engine crew.  These acceptable
overload factors are taken from the DGP Starship Operator's Manual.

Translation:  Algebra into English:
When a ship does not have power for all its weapons, it may divert
power from the maneuver drive.  Diverting power will reduce the
output of the drive.  It is possible to overload the engines, but
that may result in damaging them.  The time allowed for overloading
them for large increases in performance is considerably less one
starship combat round.

Robert S. Dean and Scott S. Kellogg
- - -------------------------------------------------------------------------
On a different note:

Aircraft Agility
In my opinion, aircraft agilities are far too low.

Consider:	A disp=40 6-G grav fighter: Agility=6, vs.
		F-14 Tomcat: Agility=6

In a turn, A Tomcat uses its wings and lifting surfaces to pull
a high G turn at approximately 9 G's

The Grav fighter (assuming it has wings) has 1/3 the wing area of a
jet fighter and is many many many times it's weight.  (In 101 vehicles
there is a Zho grav sled that weighs close to 180 tons, a displacement
of 2, and wings? It weighs close to what a B-52 does and is smaller
than a Cessna and they expect to get lift?)  It can get little if any
lift and must rely on it's grav units to turn it.  These as we said
before are only 6G's.

Result:  In a fast turning Dogfight the Tomcat can run rings around
the Grav fighter.  Of course the Tomcat has no inertial comps and so
the pilot takes a beating on the inside, but he will always be able
to get the advantage of the grav fighter.

Personal experience:
A Cessna 152 light utility trainer has a 110 horsepower Lycoming
engine, but it is rated for 4.4G's  No air/raft will be able to keep
up with it in a turn.  If you look at many of the designs in 101
vehicles, the Cessna might not be able to out run them, but it can
Really out turn them.  There are 5 designs in there that can out turn
the Cessna.  Two are civilian with +5 G MaxAcceleration, and there
are 3 military.  All are TL 13 or higher.

Conclusion:  I believe that aircraft agility should not be limited to 6,
but 9 (the physical limit of a pilot).  At TL 10, when inertial
compensators are used the agility could go even higher.

(This is assuming there is a pilot on board, George Herbert informs me
there are pilotless drones now flying around capable of making 15 G
turns)

Scott Kellogg

#2731 Re: Revised Agility Calculation Formula

Adrian Hurt · 7 August 1991 · reply to #2729

[email removed] writes:
> We have noticed that the agility calculations as presented in the
> MegaTraveller Referee's Manual make little sense when compared
> with the Starship Operator's Manual, High Guard, and for that matter,
> the Referee's Manual.

What a surprise.  Something else in the MT Manuals which doesn't make sense.
:-)

> Problem:  A ship has 2 G acceleration.  It has no additional power
> for weapons, just power for engines, sensors, environment, computer
> etc.  Excess power is zero.  Result:  Agility=0.  This is assuming,
> of course, that the undefined term 'excess power' refers to power
> which is not allocated to any other operation aboard the ship apart
> from increasing agility.  Ok, now when the ship is in emergency status,
> the ship suddenly goes to agility=2.
> Question:  Where did this "extra" power come from?

>From what I remember of High Guard, the calculation was made in a different
order.  Let's assume this is a 1000 ton ship, with USP power rating 2.  Its
power plant produces 0.01 * displacement * power plant rating = 20 Energy
Points.  From this, deduct the power used by ship's systems (weapons, computer
and screens).  The Energy Points left over go to the manoeuvre drives to supply
the ship's Agility as follows:
Agility = Remaining Energy Points / (Displacement * 0.01).

So, if the ship uses 20 EP on its systems, its agility is 0.  If it uses 10 EP
or less for ship's systems, its agility is 1.

For emergency agility, assume that ship's systems are turned off; the only
thing drawing power is the manoeuvre drive.  The "extra" power is the power
previously used by everything else.

Note that neither type of agility was allowed to exceed the ship's manoeuvre
drive rating; if it only had manoeuvre drive 1, its emergency agility would
be 1.

> Ok, High Guard functioned like this:  A ship with power plant-2 and
> maneuver drive-2 automatically becomes agility=2.

Only if it has no energy-consuming weapons, screens or computer (i.e.
only missile and sandcaster armament, computer Model 2 or less, no
screens except a black globe).

>						    In order to have
> a functioning maneuver drive, you had to have a power plant factor
> equal to the maneuver drive ...

At least equal to the manoeuvre drive.  The power plant rating could be
greater than the manouvre drive rating, so you could use weapons, screens,
computer and manoeuvre drive.

> Now, when a High Guard ship attempts to escape, the ship relies on
> agility to do so, not maneuver drive.  So a 3-G ship with agility=3
> can escape from a 6-G ship with agility=1.
>
> Rationalization:  The 6G ship's power plant is being drained away
> from the maneuver drive and it is not putting out 6-G's, it is putting
> out only 1G.

Correct, unless the 6G ship uses its emergency agility of 6G (and has a large
enough power plant to supply its 6G manoeuvre drive).  (In which case, the
3G ship will probably turn round and fight, but that's not important to this
argument.)

> In view of the MegaTrav design rules this does not translate unless
> we do the following:
>
> 	(Power required for maneuver drive + "excess" power)
> Agility=--------------------------------------------------- * 5.4
> 			Unloaded Weight

I'd substitute "Power available for manoeuvre drive" for the entire top line.
Power available for manoeuvre drive = output of power plant - power used by
other things.  This is presumably what the MT manual means by "excess power".
In fact, this is precisely what it means.  In my manual, it says "Excess power
output is the power left over from the power plant after all of the craft's
other components have been powered".  This is a big surprise - something in the
MT Referee's manual does make sense! :-)

> Agility may not exceed maneuver drive rating.
> (No more agility=6 1-G drive ships...)

Perhaps they should have stated that explicitly.  This was certainly the case
in High Guard.

> Note that if a ship is powering weapons the "excess" power can be negative
> thus subtracting from the ship's ability to maneuver.  This is what Rob
> refers to as "Double Dipping".

Not by the above definition.  Excess power = power output - power used by
systems other than manoeuvre drive.  If this is negative, you need a bigger
power plant or smaller systems.  If it is zero, the ship's agility is zero.

> His opinion is that we should use the loaded weight for the above
> calculations and I concur.  How often are ships flying around with
> no fuel and no cargo?

I agree with this.  In High Guard, the ship's displacement included fuel, cargo
and waste space.  Waste space was any of the ship's tonnage which wasn't used
for something else, and only showed up if the designer didn't use it as cargo
space for some reason.

- - --
 "Keyboard?  How quaint!" - M. Scott

 Adrian Hurt			     |	JANET:  [email removed]
 UUCP: ..!ukc!cs.hw.ac.uk!adrian     |  ARPA:   [email removed]

#2741 Re: (2731) Re: Revised Agility Calculation Formula

Robert S. Dean · 8 August 1991 · reply to #2731

Adrian Hurt writes:

> > Ok, High Guard functioned like this:  A ship with power plant-2 and
> > maneuver drive-2 automatically becomes agility=2.
>
> Only if it has no energy-consuming weapons, screens or computer (i.e.
> only missile and sandcaster armament, computer Model 2 or less, no
> screens except a black globe).

Your are, of course, correct, but I didn't want to make the explanation
any more longwinded than necessary.

> >						    In order to have
> > a functioning maneuver drive, you had to have a power plant factor
> > equal to the maneuver drive ...
>
> At least equal to the manoeuvre drive.  The power plant rating could be
> greater than the manouvre drive rating, so you could use weapons, screens,
> computer and manoeuvre drive.

See above. (-8

> > Rationalization:  The 6G ship's power plant is being drained away
> > from the maneuver drive and it is not putting out 6-G's, it is putting
> > out only 1G.
>
> Correct, unless the 6G ship uses its emergency agility of 6G (and has a large
> enough power plant to supply its 6G manoeuvre drive).
>
> > In view of the MegaTrav design rules this does not translate unless
> > we do the following:
> >
> > 	(Power required for maneuver drive + "excess" power)
> > Agility=--------------------------------------------------- * 5.4
> > 			Unloaded Weight
>
> I'd substitute "Power available for manoeuvre drive" for the entire top line.
> Power available for manoeuvre drive = output of power plant - power used by
> other things.  This is presumably what the MT manual means by "excess power".
> In fact, this is precisely what it means. In my manual, it says "Excess power
> output is the power left over from the power plant after all of the craft's
> other components have been powered". This is a big surprise--something in the
> MT Referee's manual does make sense! :-)

While I agree with you, of course, it is not so obvious as you think it is, at
least to several other eminent members of this group with whom I have had this
discussion.  As a result, I thought it best to spell it out.

> > Agility may not exceed maneuver drive rating.
> > (No more agility=6 1-G drive ships...)
>
> Perhaps they should have stated that explicitly.  This was certainly the case
> in High Guard.

Ah! But they didn't, because they didn't intend for it to be the same.  That's
why we've got Joe Fugate explaining that the agility rating is your ability
to 'alter your heading' for pity's sake, rather than the ability to alter
your maneuver vector.  Since you can rotate all you want (with three degrees
of freedom, no less) and still end up presenting a 'ballistic' rather than
a maneuvering target, I reject this view utterly.

> > Note that if a ship is powering weapons the "excess" power can be negative
> > thus subtracting from the ship's ability to maneuver.  This is what Rob
> > refers to as "Double Dipping".
>
> Not by the above definition.  Excess power = power output - power used by
> systems other than manoeuvre drive.  If this is negative, you need a bigger
> power plant or smaller systems.  If it is zero, the ship's agility is zero.

I use this terminology to make it explicit that I mean exactly what High
Guard meant, as opposed to certain folks around here (-8, who feel that
the power used _by the drive_ should be included, leaving excess power as
anything over and above that normally used by the drive (overloading!).

The whole question of "What is agility?" is tied in pretty closely with the
question of whether or not the combat system is rational as it now stands
(give or take a little).  If we view ships as 'hard to hit', then it would
make sense to spray your weapons all over their potential positions, which
is why a 30 laser battery does the same damage as a 1 laser battery, if
both manage to hit.  Now, if ships are spraying automatic weapons fire (-8
all over, then to be hard to hit, you obviously must need to make a substantive
change in your projected position, which means using the maneuver drive
to accelerate rather than just 'dodging' a little bit in place.  Does that make
any sense to anyone?

Rob Dean