BBS: Inland Empire Archive
Date: 01-03-93 (16:59) Number: 244
From: WES GARLAND Refer#: NONE
To: JOHN GALLAS Recvd: NO
Subj: calculations.. Conf: (2) Quik_Bas
Hi John!
JG> I need some help with some calculations I have to do.
JG> Lets say my spaceship has coordinates of 1550 and
JG> 1438, and is facing 348 degrees (0 being straight
JG> up), and is traveling 3 units per "update".
Was that you asking about this stuff about a month? I wrote
you a two-hour reply, explaining the practical applications
of vector math.. I hope you got it....
JG> How do I calculate how much to add (or subtract) to the X and
JG> Y? And would it still work if I wasn't using single
JG> precision (no decimals)?
First off all, in any calculation, keep the decimals until
you absolutely can't...
ANYHOW. The easiest way to do your calculation is this...
ActualX = <your X coordinate>
ActualY = <your Y coordinate>
Set up a grid like this though:
^
|___
| /|
| / |
|/@ |
<--+----------->
|
V
Your ship is at the plus. The diagonal line is R units
long, which is the amount of space covered per update. @ is
theta, the angle between the X-axis and your ship.
So, using good ol' SOHCAHTOA,
Sin @ = NewY/R
Cos @ = NewX/R
but remember, the computer uses radians for its trig
calculations (I think... I KNOW earlier MSBasics did, and I
don't imagine they've changed it...)
So, the angle theta you plug in has to be converted from
your coordinate system to this one, so you add ninety
degrees (PI/2 rads) to it. Then you convert it to radians.
Remember, theta is the angle between the line and the x
axis.
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