From: | Amonchare <mak9@******.EDU> |
---|---|

Subject: | acceleration? |

Date: | Thu, 4 Mar 93 05:43:22 EST |

A summary of these Eqs. has already been encluded in a previous post, so if

you do not want to read a lengthy derivation stop here.

Based on the power Eq.

Power = Mass x Acc. x Vel.

which I have been using, one can develop an Eq. for acceleration by dividing

both sides of the Eq. by mass and Vel giving.

Power

----------- = Acc.

mass x Vel.

showing that acc. at a given time depends upon the Power-to-mass ratio (which

from now on I will abbreviate as P/M) and the velocity the car is traveling at

that time. This acceleration is the actual acceleration of the car plus the

acceleration required to maintain a speed. I have previously define the

acceleration required to maintain a speed as 1 m/s^2. The Eq. for the cars

acceleration is now.

P/M

----- - 1 = Acc.

Vel.

This is really a useless eq in game terms what we really need (IMHO) is an eq.

for vel and/or distance after a given acc. time. The above Eq is a realistic

starting point for these eqs.

First of all Acc. by definition is the change in vel over time. So in order

to get vel from the above eq. one must integrate it. The eq. the way it is is

to complicated for me to integrate, so I'm go to ignore the -1. By remove the

-1 the resulting eq. will give results higher then they would normally be.

And the max. vel cap inherent in the "real" Acc. formula is removes. By the

velocity formula the will result, there will be no max. vel. Other wise the

curve generated should be some what realistic.

Anyway integration of the resulting acceleration formula gives:

vel. (m/s) = Sqr. root( 2 x time(s) x P/M)

And if we measure time in combat turns:

vel. (m/ct) = Sqr. root( 54 x time(ct) x P/M)

Again by definition velocity is the change is distance over time. So to get

the distance eq. one must integrate the the velocity eq. Doing so gives(based

on the m/s velocity eq.:

2M

dis.(m) =---- x [ 2 x time(s) x P/M]^(3/2)

3P

which can be simplified to:

time(s) time(s) x vel(m/s)

dis.(m) = ------- x Sqr. root(2 x time(s) x P/M) or = ----------------

3 3

if time is measured in combat turns:

dis.(m) = time(ct) x vel(m/ct)

So, all we need to do is figure out a suitable velocity formula and the

distance one will fall into place.

For the above eqs. i was assuming an initial vel of 0. If this assumption is

not valid the eqs. get more complicated. I would really like to get a simple

eq. that appromiates the vel eq. but I cannot come up with one. Anyway these

are the eqs. I came up with starting from the ground up. Comments?