Re: Vectors question 2
From: Miro Jurisic (macdev_at_meeroh.org)
Date: 12/05/04
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Date: Sun, 05 Dec 2004 03:07:09 -0500
In article <e95sd.57541$K7.31159@news-server.bigpond.net.au>,
Cassandra Thompson <newsgroups@calebsoftware.com> wrote:
> Miro Jurisic wrote:
> > In article <erZrd.57293$K7.19822@news-server.bigpond.net.au>,
> > Cassandra Thompson <newsgroups@calebsoftware.com> wrote:
> >
> >
> >>So after reading your response I began to think. Ignoring vectors and i & j
> >>etc. If I wanted to figure this out a few weeks ago (prior to learning this
> >>stuff) my method would have been as follows.
> >>
> >>Put the direction of the train on the x-axis, the direction of the stuntman
> >>on the y-axis. (this is I think what you are trying to say).
> >
> >
> > That method is correct, and produces the correct answers... so now my
> > question is: what is it that you learned over the last few weeks that
> > confused you into trying to plot time as well?
> >
> To answer your question, I overthought the question. We are studying
> vectors, including 3-dimensional vectors and projections. Perhaps it was
> because I had only just finished studying projections. Or because I thought
> that question must be more complex then it looked etc.
>
> I really don't know why, however I am sort of glad because it has allowed me
> to gain alot better understanding into using vectors, but then making
> mistakes does usually have that effect.
Well, if you want to consider this and include time as one of your dimensions,
then you have to realize that you have to project the space-time coordinate
system into a space-only coordinate system before trying to describe it in terms
of visible effects.
For example, consider a single point traveling along a straight line. Let the
line of travel be the X axis and let time T be perpendicular to it. Then, in the
XT plane, the point will be describing some curve (continuous, unless your
universe includes teleportation). For example, if the point is oscillating about
the origin, then the space-time curve describing its motion would be a sine wave
on the T axis. But a description of that motion would probably not refer to a
sine wave about the T axis, it would refer to an oscillatory motion about the
origin.
This seems like a trivial point until you start considering more dimensions:
consider an XY plane and a time T axis perpendicular to it, and two points
traveling in the plane -- one at a fixed velocity along the X axis, starting at
the origin, and the other at the same velocity along the Y axis, also starting
at the origin. If you look at them in the XY plane, then their trajectories
(namely, the X and Y axes) are perpendicular. However, if you consider them to
be points in XYT space, then the first one's trajectory is a straight line in
the XT plane, and the second one's trajectory is a straight line in the YT
plane, and those trajectories are not perpendicular! It is only their
projections into the XY place (the space-only coordinate system) that are
perpendicular.
Which brings me to your question; your mistake in considering this with distance
on the X axis and the time on the T axis was that when the description of
physical motion of the stunt man was described as being perpendicular to the
motion of the train, you drew the perpendicular in the XT plane, which doesn't
work.
What you need to do here is count the number of dimensions of space (clearly,
you need at least two, as the man the the train are moving at right angles to
each other. Then you add one for time, and you see that you need a
three-dimensional system. In that system, the trajectory of the train will be a
straight line; if the train travels along the X axis, as it does in your
original graph, then its XYT trajectory will be a straight line in the XT plane,
as it is in your original graph. However, the motion of the man is, according to
the problem statement, perpendicular to that of the train -- but the problem
statement refers to them in the space-only coordinate system, which means that
you have to have the man moving perpendicular not to the XYT time-space
trajectory of the train, but to the XY spatial trajectory of the train. If you
do that, then everything comes out right, but it's somewhat complicated to draw
that.
Now, there is no really good reason to involve time in this, as the movement of
the train (relative to the ground) and the man (relative to the train) are
constant with time, so what you are asked for (movement of the man relative to
the ground) is also constant and you can draw it without involving the T axis at
all. However, I think you should know why your initial solution didn't work and
how you could have made it work.
meeroh
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