 |
|
|
|
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Mike Williams wrote:
>
> Wasn't it Bill DeWitt who wrote:
> >
> >"Hugo" <hua### [at] post3 teledk> wrote :
> >>
> >> Could you please tell about vector cross & dot products too, then? :o)
> >
> > I finally got what dot products were about a year ago. I don't know how
> >ignorant of them you are, but I had no clue. Here's what I finally figured
> >out.
> >
> > If you have a vector #declare v = <1,2,3>;
> > Then v.x = 1, v.y = 2, v.z = 3.
>
> I guess that's a joke, but just in case your serious, the POV dot
> operator (which is what you have there) is a completely different thing
> from the vector dot product (which you can calculate with the "vdot()"
> vector function).
Well, it's not a completely different thing... ;)
v.x does give exactly the same result as vdot(v, x) does
Tor Olav
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
I seek solice in a quote I seem to remember being attributed to Van Nuemann (sp?
the guy who was in on the very first computers and the a-bomb )
The sense of it was roughly: "You don't so much come to understand mathematics ,
rather, you just sort of get used to it"
PS. When I laid this sage nugget on my son, then twelve, he simply responded, "
I don't what to learn math and I don't want to get used to it either!"
>
>
> Most of the people here cannot imagine how ignorant of math us non-math
> types can be. Someone just posted an explanation of those terms that I was
> completely unable to follow. It makes assumptions of knowledge that I just
> don't have. I am sure it is clear to someone, but not me. I may never
> understand vector math if I have to start at that level.
>
>
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
>
> How old are you?
Much too old to go back and start over. I would try if I didn't have so
much to do in other areas.
> written an excellent and very detailed explanation of vectors. You should
> try to read it!
I saw that and saved it, but don't have much confidence that my ability
to understand higher math will improve even if I can make my way through it.
I have been struggling with remedial geometry for years now and have made
more progress with learning Chinese than with that...
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
"Jamie Davison" <jam### [at] ntlworldcom> wrote in message
news:MPG.16ba7830dff95f1e989ab4@news.povray.org...
> > Most of the people here cannot imagine how ignorant of math us
non-math
> > types can be.
>
> Hell, I was trained as a biologist,
Same here. Beyond a little "plug in the numbers" algebra, I never really
needed math until Povray. I doubt that I ever will beyond my hobbies.
But my real math problem started in grade school. I remember being left
behind during quadratic equations and telling my teacher that I didn't
understand. That was 7th grade and my math scores never got above a 'd'
after that. I had been getting all 'a's in every subject before that.
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Hi,
the one looking for cross and dot products.. Fernando gave a short
description.. I liked how you start by pointing out the VECTOR input and
VECTOR / NUMBER output of each function.. A good start.. You also tell about
the usefullness of the functions, which increase our attention.
Regarding the dot product, I had to skip the sentence "the shadow of a
vector in a plane".. It's still nonsense to me, but alright..You go on to
formulas, with the most simple being at first: A dot B = a1*b1 + a2*b2 +
a3*b3 Thank you! This leaves no confusion in my mind, except that I can't
visualize how the result relates to the original vectors A and B.. I just
see multiplication and addition going on (things I know of).. You proceed
with calculating a dot product by using other formulas, involving radians,
theta, sqrt and exponentation.. This mostly goes beyond me, I still can't
see the "result".. You end the lesson with a good sentence, saying "if A and
B are orthogonal (perpendicular), then cos(theta)=0 and then the dot product
is equal to zero".. Very good.. This brings back a feeling of sense, order
and logic in my mind about the whole thing.. Order, because it seems there's
a formula behind this, that I WILL be able to undestand someday.. All in
all, I wasn't left in the land of strangers, thank you. :o)
for your long explanation! I like that you began with simple things like
translation.. It's good to begin with something most people already knows
and understand, because it kind of makes us relax.. We're travelling with
you in the same train, whatever ... on the same level.. Then you came to
explaining dot and cross products:
I admit the landscape changed from familiar to unknown territory.. So it's
necessary for me to slow down, to follow.. Fortuntately I have your lesson
on text so I can read it as many times as I like.. :o) I like how you
sticked to 2 dimensions on a piece of paper.. You tell that cross products
face either upwards or downwards, depending if the calculation is done
clockwise or counter-clockwise (something all the clever people talk about
when they compute smooth_triangles).. What I don't really understand,
however, is how a cross product can be perpendicular to TWO vectors, and how
this can bring different results - in 2 dimensions.. If both vectors lie
flat, because they are on a piece of paper, it's possible to put a third
vector *everywhere* on the paper, being perpendicular to them.. hhmmm..
Again, what is the visual ... purpose of cross products.
Anyway you used plain english, and I really appriciate your efforts!
Obviously you have done efforts, and english is not your native language (or
mine).. I will read your description again and try to get a picture in my
head.. Maybe it's possible to make a simple illustration of a cross product,
not in 3 dimensions with 3 cylinders with each their color and direction,
it's nonsense.. :o) I still haven't found a place on the net, where a good
illustration is found.. For example, a problem on paper that leads everyone
to think the solution is obvious, "just do this!" but then, they can't find
a mathematical way of doing it, until cross products suddenly are the magic
stick.
Annnnd, thanks for reading my long explanation of how I felt, being in your
class room.. :o) Don't take me wrong please.. I enjoyed it, and I'll come
back.
Regards,
Hugo
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Hi Hugo,
was
You are welcome!
> however, is how a cross product can be perpendicular to TWO vectors, and
how
> this can bring different results - in 2 dimensions.. If both vectors lie
> flat, because they are on a piece of paper, it's possible to put a third
I see your problem. The example I used for the cross product uses two
vectors within a flat plane as input. That's right. But the resulting vector
( which is the cross product) extents into the -third- dimension. Although
the input vectors lie flat on the floor, the result vector goes either up or
down. This is the only way to be perpendicular on both input vectors. No
other chance (except the input vectors are parallel or antiparallel [simply
spoken both lie along a straight line]).
> vector *everywhere* on the paper, being perpendicular to them.. hhmmm..
No you can't. Remember my very first line: vectors start always in the
origin. So all THREE vectors (input and result) share one common point: the
origin. Now you know where your vector has to start from. Try to find a
place for it's tip so that the whole thing is perpendicular to both vectors
(which form a "V" for example). The only way is up or down. Draw a big V
onto your sheet of paper (which symbolise two vectors starting at the
origin) and put your pencil onto the basepoint of the V. The pencil stays
with its lower end at this point. Move the tip of it to find a situation so
that the pencil is perpendicular to both vectors.
I think this makes it clear:
Since both vectors lie in one plane (your sheet of paper), the result vector
can only be perpendicular to both vectors at the same time if it is
perpendicular to exactly this PLANE. So the cross product will always be
perpendicular to your sheet. Regardless in which directions you rotate or
move it. The pencil rotates and moves as well and stays perpendicular to the
paper.
> Again, what is the visual ... purpose of cross products.
It's advantage is that you can define a plane by using two vectors and get
easily a normal vector onto it. This is very important to compute the angle
of an incoming light ray hitting our plane (as used for reflection) for
example. It depends on the problem you want so solve mathematically but
believe me there are a lot of situations where a vector which is standing
normal to a plane is very helpful. It's usage in computer graphics might be
to determine if you "see" the front side or the flip side of a flat piece of
the surface (e.g. on triangle of a mesh). This could be solved by using the
cross product since it direction relative to the plane is well defined and
you don't have to care about the individual input vectors.
Additionally you know the content of the area defined by the input vectors
by just measuring the cross product vector's length.
> head.. Maybe it's possible to make a simple illustration of a cross
product,
> not in 3 dimensions with 3 cylinders with each their color and direction,
I think a three dimensional model is the only way to get the cross product
right. I don't know if a cross product is even defined for two-dimensional
vectors. I don't think so. Some operations just need explicit prerequisites.
You can't explain addition or subtraction by just using one number....
An animation would be the best but I'm afraid I haven't time enough to make
a good movie. I have an idea but it takes it's time.
Anyway, if all failes I still can do that.
I'll try to describe it:
Imagine a big L. Take this L and rotate it's long axis. Watch the short
axis: it moves in a circle, creating a disk if rotated fast enough. The long
axis of this L would be a vector. The rotating short axis is the SUMMARY of
all possible vectors perpendicular to the it. (Remember all vectors start at
the origin therefore you can't move the connection point of the two axes).
Now imagine you replace the rotating axis by a disk. You get a thing similar
to a flat umbrella. This thing is our vector and a plane perpendicular to
it. ALL vectors perpendicular to our vector MUST lie within this plane,
within this two dimensions.
Create a second thing like this. This would be the second vector. Now you
got two umbrellas. Since all vectors start in the same point we have to
bring our vectors (together with their normal planes) together. Where should
they join? Exactly in the point where the vector hits the plane (since this
is the common point of the vector and all its normal vectors. Therefore it
must be the same point for the other vector!). Bring the umbrellas together
so that they share this point. What happens to the two planes? Correct, they
intersect. The intersection of two planes is a line. This line is the only
intersection of the normal vectors of BOTH input vectors. This intersection
is perpendicular to BOTH vectors since both normal planes share it. It is
our cross product!
Is this scenario more or less confusing? :o)
> to think the solution is obvious, "just do this!" but then, they can't
find
> a mathematical way of doing it, until cross products suddenly are the
magic
> stick.
You can solve all these problems without using vectors as well. But doing it
without vectors results usually in a vast amount of sine, cosine, tanges and
their reverse relatives and hyperbolic cousins. Of course you would get the
same thing but the effort is very large, time consuming, easily misleading
and last but not least it takes tons of mathematical operations. Vectors are
just simpler and much nicer for that. All you need for (static) vectors is
basically what I described in my last post. As soon as you got used to the
dot and cross product nothing can stop you anymore ;-)
> class room.. :o) Don't take me wrong please.. I enjoyed it, and I'll come
> back.
Great. As long as you have fun with it keep on asking!
Stupid questions don't exist. There are only stupid answers.
best regards
Post a reply to this message
|
|
| |
| |
|
|
From: Fernando González del Cueto
Subject: Re: ExplainationPlease...??
Date: 25 Jan 2002 12:26:51
Message: <3c51955b@news.povray.org>
|
|
|
| |
| |
|
|
Hi Hugo,
It is difficult to explain these kind of things without using drawings.
About the "shadow of the vector", it is not very important for you to
understand it right now, but let me see if I can give a clearer explanation:
If you have a 3d vector, (imagine an arrow in 3d space) and you have a light
from the top of room, it will cast a shadow of the arrow in the floor. That
shadow, is the projection of the vector in the floor plane. The dot product
allows you to find this projection.
About your doubt concerning the "visualization" of the dot product, I
remember I had the same feeling in my first course in linear algebra. We are
used to the natural interpretation of the usual product between to numbers,
but the dot product fails to have the same natural interpretation. If you
see the second formula for the dot product, |A|*|B|*cos(angle), you can see
that the dot product can be very useful because it contains the angle
between the two vectors. The first formula lets you calculate the dot
product between A and B very easily. Then, |A| and |B|, the norms of the
vectors (the lengths) can be calculated very easily too. Finally, you can
isolate the cosine of the angle, which can be very useful.
Maybe you can visualize the dot product as a "weighted" measure of the angle
between two vectors.
About your other question, in which you cannot see how can the cross product
can be perpendicular to the other two vectors, YOU'RE RIGHT! That is because
the cross product is defined only for 3d vectors.
You cannot have 3 vectors perpendicular to each other in 2D space, but you
can have them in 3D space.
I hope this helps you,
Fernando.
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
> The computation is more cumbersome (although there is a trick to memorize
it
> using a determinant, but I won't state it):
Maybe there's a more easy (?) way to memorize the cross product. It's called
the "cyclic minus anticyclic" rule. The idea is, that you count forward
(cyclic) for the first term and backwards (anticyclic) for the "minus term".
So for the first component you count 1- "2" - "3" using "2" and "3" for the
indices of the vectors minus 1 - "3" - "2" (backwards). Same way for the
second 2 - "3" - "1" minus 2 - "1" - "3" and the third 3 - "1" - "2" minus
3 - "2" - "1".
So this is cyclic: 1 - 2 - 3 - 1 - 2 - 3 and this is anticyclic 3 - 2 - 1 -
3 - 2 - 1.
Hope that nobody's more confused (by my english?), then forget it ;)
Best regards,
Florian
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Hi folks,
I think I finally understand cross products now!! :o) I'll need to
experiment a little more to see how it behaves, but I definitely got the
> Is this scenario more or less confusing? :o)
Oh, a little, but the first part of your letter was very clear.. I
understand now! I hope, this was also helpful to Ryan, Bill and some other
people here.
> An animation would be the best but I'm afraid I haven't time
> enough to make a good movie. I have an idea but it takes
> it's time. Anyway, if all failes I still can do that.
No no, I've taken enough of your time now. :o)
Fernando wrote:
> Then, |A| and |B|, the norms of the
> vectors (the lengths) can be calculated
Does this mean, the word "norms" are not a shortage for "normal" but it
means "length"? That's good to know.. I come across new words in math all
the time, and sometimes I just guess their meaning.. Thanks for explaining
dot products, I understand them better now, although not fully.
I couldn't find these kind of lessons elsewhere on the net.. I browsed many
math sites, but either they assume (1) you're already educated in these
things, (2) you are undergoing an education, or (3) they are for children..
It's probably the same with books, and I don't know what to look for.. I had
a "plain" education in math years ago, but not in english, and I have
forgotten much of it..
Thanks for your help.
Regards,
Hugo
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
that's what i was looking for... =]
thanks... i also got little tid bits on math and such... Thanks all... =]
Josh English wrote:
> I've got a few tutorials at http://www.spiritone.com/~english/cyclopedia
>
> there is one on vaxis_rotate, vrotate, vnormalize, and vcross (These
> should help and will be updated soon with more material)
>
> Josh English
> eng### [at] spiritonecom
> http://www.spiritone.com/~english
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
|
|
