Need Physics Help!

9 posts ยท Jul 14 1998 to Jul 17 1998

From: Noah Doyle <nvdoyle@m...>

Date: Tue, 14 Jul 1998 17:50:22 -0500

Subject: Need Physics Help!

OK, in the course of trying to get my (pseudo) realistic FT rules put to

gether, I have run into an obstacle: I need equations! Many SF games have
some, but not all, and I'm not sure that they are right. I've got an OK

physics text, but it's more concerned with cars and inclined planes -
I'm still hashing thru to find what I need, but I'll appeal for help here.

I have, variously, the numbers for acceleration, distance and time. I may need
to solve for any one of these. I need equations for:
Rest-to-rest with constant accel/decel.
Rest-to-point with constant accel (& vice versa: point-to-rest with
constant decel).
And both of those again, with only a certain amount of accel/decel
(limited thrust manuvers)

The first two are most important. If these could be in MKS, that would be
appreciated. CGS I just use to figure gravitational attraction of planets. So
c'mon, you physicists & engineers, please help me make interplanetary travel
difficult! Thanks in advance,

From: Nyrath the nearly wise <nyrath@c...>

Date: Tue, 14 Jul 1998 21:59:08 -0400

Subject: Re: Need Physics Help!

> Noah Doyle wrote:

Here are some equations I've managed to accumulate:

* WARNING * The below equations assume a constant acceleration, which is not
true for a ship expending mass (for instance,
propellant). Ai = F/Mc so as the ship's mass goes down, the acceleration
goes up
( where Ai = "instantaneous" acceleration (m/sec^2),
F = Thrust (Newtons or kg m/sec),
and Mc = Ship's "current" mass (at this moment in time) (kg))

============================================
When you have two out of three of
average velocity (Va) in m/sec,
change in distance (S) in meters or time (T) in seconds
Va = S / T
S = Va * T
T = S / Va
============================================
When you have two out of three of
acceleration (A) in m/sec^2,
change in velocity (V) in m/sec
or time (T) in seconds
A = V / T
V = A * T
T = V / A
============================================
When you have two out of three of change in distance (S) in meters,
acceleration (A) in m/sec^2,
or time (T) in seconds
plus Initial Velocity (Vi) in m/sec
(Note: if deaccelerating, acceleration A is negative)
S = (Vi * T) + ((A * (T^2)) / 2)
A = (S - (Vi * T)) / ((T^2) / 2)
T =  (sqrt[(Vi^2) + (2 * A * S)] - Vi) / A
If Vi = 0 then
S = (A * (T^2)) / 2
A = (2 * S) / (T^2)
T = sqrt[(2 * S) / A]
============================================
When you have two out of three of change in distance (S) in meters,
acceleration (A) in m/sec^2,
or final velocity (Vf) in m/sec
plus Initial Velocity (Vi) in m/sec
(Note: if Vf < Vi, then A will be negative (deacceleration))
S = (Vf^2 - Vi^2) / (2 * A)
A = (Vf^2 - Vi^2) / (2 * S)
Vf = sqrt[Vi^2 + (2 * A * S)]
If Vi = 0 then
S = (Vf^2) / (2 * A)
A = (Vf^2) / (2 * A)
Vf = sqrt[2 * A * S]

From: Mikko Kurki-Suonio <maxxon@s...>

Date: Wed, 15 Jul 1998 22:24:14 +0300 (EEST)

Subject: Re: Need Physics Help!

> On Tue, 14 Jul 1998, Noah Doyle wrote:

> OK, in the course of trying to get my (pseudo) realistic FT rules put

Get Alonso-Finn University Physics Volume I.

And take some time to understand the equations. What you need is basic
dynamics and kinetics, but all things basic come with huge caveats --
understand them.

From: Mikko Kurki-Suonio <maxxon@s...>

Date: Wed, 15 Jul 1998 22:27:44 +0300 (EEST)

Subject: Re: Need Physics Help!

> On Tue, 14 Jul 1998, Noah Doyle wrote:

> appreciated. CGS I just use to figure gravitational attraction of

Oh, I forgot to mention: If you have thrust anywhere near the levels needed
for FT style combat manuevers, planetary gravitation becomes a
non-issue -- it's effect is about the same as a small rounding error.

From: Richard Slattery <richard@m...>

Date: Wed, 15 Jul 1998 23:33:04 +0000

Subject: Re: Need Physics Help!

> On 15 Jul 98 at 22:27, Mikko Kurki-Suonio wrote:

> Oh, I forgot to mention: If you have thrust anywhere near the levels

.. you may have missed an earlier thread...

Taking the 15 minute turn/1,000km per move unit scale, full thrust
ships have a puny thrust of a 1/4g to 1/8g per thrust point, so
gravity really sucks. <yes! I worked that pun in at last> Of course, if you
change the scales it changes everything. I'm heading for 7,500km per move unit
to give about 1g per thrust point.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

From: Noah Doyle <nvdoyle@m...>

Date: Thu, 16 Jul 1998 01:24:46 -0500

Subject: RE: Need Physics Help!

First - Mikko, thanks for the suggestions - I lost your email.
Second, at 7500 km/MU, Terra is about 2 inches wide (12,9xx km dia.) and

Jupiter is 10 inches (71,000 kmdia.)? Does that sound about right? That
is actually starting to sound better & better- I would like to have gas
giants on-table, and I have been balking at the idea of apinting a 13"
sphere into a good-looking Terra-type planet.  But with 7500km/MU, your
railguns & Pulse Torpedoes had better have some real oomph to them - you

need a lot of velocity to make up for those distances. Maybe gravitic accel?
You could far outdistance EM accel that way. And with the inertial
compensation that would entail, you could orbit however you wanted.

Noah

[quoted original message omitted]

From: Mikko Kurki-Suonio <maxxon@s...>

Date: Thu, 16 Jul 1998 12:15:30 +0300 (EEST)

Subject: Re: Need Physics Help!

> On Wed, 15 Jul 1998, Richard Slattery wrote:

> Taking the 15 minute turn/1,000km per move unit scale, full thrust

I can't agree. I did my campaign with 1 thrust = 1/4g (knowing that MU
and turn length don't matter) and gravity is a minor irritation at best.

Consider that gravity is inversely related to the square of distance. Actually
ON earth, it's 1g. But few spaceships fight on the ground.

About 6500km out, it's only 1/4g
About 20000km out, it's 1/16g (0.25 thrust)

Geostationary orbit is, IIRC, about 30000km out.

Moon, as we all know, is about 300 000km out, which on the solar system scale
is still an insignificant distance.

So, gravity is only a factor if you *choose* to fight *very* close to a
planetary body, amidst all the orbiting satellite junk (including weapon
platforms).

If you were moving at constant acceleration of say, 0.5g from Mars to Venus
and happened to pass within 50 000km of Earth, the impact on your final course
is close to a rounding error.

From: Richard Slattery <richard@m...>

Date: Fri, 17 Jul 1998 00:40:01 +0000

Subject: Re: Need Physics Help!

> On 16 Jul 98 at 12:15, Mikko Kurki-Suonio wrote:

> On Wed, 15 Jul 1998, Richard Slattery wrote:

When I said, gravity sucks, I meant, if you do manage to get /too/
close to something with more gravity than the earth... a gas giant for
example, you could be in serious trouble. Also, those ships that are intended
to land on planets had better have good enough drives to manage it. If they
have atmospheric streamlining, sure, they can glide with the bricklike quality
that the space shuttle does, but taking off again with half a g of thrust is
going to be problematic. Even worse, if the planet has no atmosphere for you
to get lift in...

With regard to fighting near a planet... yup, I'd prefer to if I'm defending
and it has orbiting defenses to help my fleet.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

From: Mikko Kurki-Suonio <maxxon@s...>

Date: Fri, 17 Jul 1998 13:12:00 +0300 (EEST)

Subject: Re: Need Physics Help!

> On Fri, 17 Jul 1998, Richard Slattery wrote:

> When I said, gravity sucks, I meant, if you do manage to get /too/

Yes, but too close is surprisingly close -- especially if have a
substantial initial velocity.

> With regard to fighting near a planet... yup, I'd prefer to if I'm

Well, if 1 MU = 1000km, your defenders around an earth-like planet maybe
2000-10000km away from the planet feel the tug, but their opponents can
take 20"-30" shots with negligible gravity effects -- a bit like taking
your wet navy fleet so close to shore that you risk running aground to
gain the help of shore batteries. In other words, a trade-off now always
worth the risk.