I'm trying to get my Chaos Pendulum scene to work, but it ain't happening. I
can't get the math right!

Apparently, both gravity and magnetism work using the "inverse square law of
attraction". I don't actually *know* what that means, but I've given it my
best guess...

Forget gravity, let's talk magnetism. Suppose I have a magnetic ball at
point P, and a magnet of strength S located at point Q (and fixed in place).
Now, I recon the force experienced by the ball is in the direction Q - P,
and it's strength is equal to the reciprocol of the square of the distance
from P to Q; i.e., force = S / vlength(P - Q) * vlength(P - Q).

But hang on... wouldn't that mean that the way the force drops off as we
move away from Q is dependent on our units of measurement? Suppose P is 1
meter from Q. Then we have S / 1 * 1 = S. But if we write this as 100
centimeters instead, we have S / 100 * 100 = S / 10000. Um... a little
confused here!

Wouldn't that also mean that if P = Q, then the force of the magnet is
infinite? (Or more importantly, if P *almost* equals Q, the force would be
astronomically large.) Now I'm *really* confused!

By the way... is the force excerted by a *real* magnet dependent on the mass
of the magnetic ball?

OK, my head *really* hurts now...

Help!
Andrew.

Post a reply to this message

You have the right idea about what the inverse-square law means, but...

The P=Q problem is easily solved since P and Q are points, but the
objects in question are not.  If they are spheres, then P and Q
represent the centres, so unless the spheres are infinitely small, then
P and Q can never be equal.

The problem with units arises because the force is *proportional* to,
not equal to, S / vlength(P-Q)^2.  A constant of proportionality is
needed, which obviously depends on the units.  For example,
gravitational force is F = (G * m(1) * m(2)) / r^2.  Here r is the
distance between the two masses, and m(1) and m(2) are the masses of
each object.  G is the constant of proportionality, equal to 6.67E-11
when the masses are in kilos and the distance is in metres.

PS magnets are nasty to work out, as they must have both poles, and
therefore a complicated field.  Use a charged ball to think about
instead - it only has one polarity to worry about!



"Andrew Coppin" <orp### [at] btinternetcom> wrote in message
news:3d95a3fe@news.povray.org...
> I'm trying to get my Chaos Pendulum scene to work, but it ain't
happening. I
> can't get the math right!
>
> Apparently, both gravity and magnetism work using the "inverse square
law of
> attraction". I don't actually *know* what that means, but I've given
it my
> best guess...
>
> Forget gravity, let's talk magnetism. Suppose I have a magnetic ball
at
> point P, and a magnet of strength S located at point Q (and fixed in
place).
> Now, I recon the force experienced by the ball is in the direction Q -
P,
> and it's strength is equal to the reciprocol of the square of the
distance
> from P to Q; i.e., force = S / vlength(P - Q) * vlength(P - Q).
>
> But hang on... wouldn't that mean that the way the force drops off as
we
> move away from Q is dependent on our units of measurement? Suppose P
is 1
> meter from Q. Then we have S / 1 * 1 = S. But if we write this as 100
> centimeters instead, we have S / 100 * 100 = S / 10000. Um... a little
> confused here!
>
> Wouldn't that also mean that if P = Q, then the force of the magnet is
> infinite? (Or more importantly, if P *almost* equals Q, the force
would be
> astronomically large.) Now I'm *really* confused!
>
> By the way... is the force excerted by a *real* magnet dependent on
the mass
> of the magnetic ball?
>
> OK, my head *really* hurts now...
>
> Help!
> Andrew.
>
>
>

Post a reply to this message

"Andrew" <ast### [at] hotmailcom> wrote in message
news:3d95a832@news.povray.org...
> You have the right idea about what the inverse-square law means,

Well, that's *something*!

> The P=Q problem is easily solved since P and Q are points, but the
> objects in question are not.  If they are spheres, then P and Q
> represent the centres, so unless the spheres are infinitely small, then
> P and Q can never be equal.

Thinking about it, I realised that in my particular scene the ball is
prevented from reaching the magnet anyway - I just forgot to put this into
the math! (Ooops...)

> The problem with units arises because the force is *proportional* to,
> not equal to, S / vlength(P-Q)^2.  A constant of proportionality is
> needed, which obviously depends on the units.

Now we're making sense...

>  For example,
> gravitational force is F = (G * m(1) * m(2)) / r^2.  Here r is the
> distance between the two masses, and m(1) and m(2) are the masses of
> each object.  G is the constant of proportionality, equal to 6.67E-11
> when the masses are in kilos and the distance is in metres.

Right... So since the Earth's mass is so large, the force of gravity is more
or less exactly equal to 9.8006 * m(2)? (Well, at sea level anyway!)

Interestingly, that seems to answer another of my questions - if this is
right, and if magnets are similar, then the force of the magnet would indeed
depend on the size of the magnetic ball it's pulling... (Of course, since F
= M * A, that means that at a given distance the acceleration due to the
magnet is more or less constant... I think! Maybe I'm getting out of my
depth here 8-|... )

> PS magnets are nasty to work out, as they must have both poles, and
> therefore a complicated field.  Use a charged ball to think about
> instead - it only has one polarity to worry about!

Ooo blimey, there's no way I'm gonna bother with putting poles on them!
LOL... I just wanted a fairly abstract concept of a point in space that
pulls matter towards it acording to the inverse square law.

Let's try again...
Andrew.

Post a reply to this message

> Right... So since the Earth's mass is so large, the force of gravity is
more
> or less exactly equal to 9.8006 * m(2)? (Well, at sea level anyway!)
>

I think (?) you've got the wrong 'G' here  - the 9.8 one is the acceleration
due to gravity - the one you need is the universal gravitational consant.

Also for the point charge thing there is a very simple equation (which
totally escapes me at the mo) but I try a web search for "coulombs law"

jim

Post a reply to this message

> >  For example,
> > gravitational force is F = (G * m(1) * m(2)) / r^2.  Here r is the
> > distance between the two masses, and m(1) and m(2) are the masses of
> > each object.  G is the constant of proportionality, equal to 6.67E-11
> > when the masses are in kilos and the distance is in metres.
>
> Right... So since the Earth's mass is so large, the force of gravity is
more
> or less exactly equal to 9.8006 * m(2)? (Well, at sea level anyway!)

Don't confuse 'g' with 'G'. G is the constant of proportionality for the
above equation, g is what the equation happens to give for acceleration when
r = the radius of the earth.

> Interestingly, that seems to answer another of my questions - if this is
> right, and if magnets are similar, then the force of the magnet would
indeed
> depend on the size of the magnetic ball it's pulling... (Of course, since
F
> = M * A, that means that at a given distance the acceleration due to the
> magnet is more or less constant... I think! Maybe I'm getting out of my
> depth here 8-|... )

Magnets work differently; it's very similar to the above equation, but the
masses (m1, m2) are replaced with the charges of the magnets (q1, q2) and
the constant is K = 9E9.

 - Slime
[ http://www.slimeland.com/ ]

Post a reply to this message

In article <3d95a3fe@news.povray.org>, orp### [at] btinternetcom says...

> But hang on... wouldn't that mean that the way the force drops off as we
> move away from Q is dependent on our units of measurement?

No.

> Suppose P is 1
> meter from Q. Then we have S / 1 * 1 = S.

When solving physical equations ALWAYS put in the units, instead of just 
the numbers.

We now have S / (1m * 1m) = S*1 m^-2 = S*100cm-^2 etcpp.

If you want the force, you probably want it in Newton, that is 
1N = 1kg*1m*1sek^-2.

If you use cemtimeters instead of meters when calculating the force your 
result will be the same force, but it's unit won't be Newton. If you 
convert it to Newton, it will be the same as if you put in meters in the 
first place.

> By the way... is the force excerted by a *real* magnet dependent 
> on the mass of the magnetic ball?

No, the force will be the same, but not the acceleration of the ball 
(since a=F/m).

Lutz-Peter

Post a reply to this message

In article <3d95c51c@news.povray.org>, jim### [at] blueyondercouk says...

> Also for the point charge thing there is a very simple equation (which
> totally escapes me at the mo) but I try a web search for "coulombs law"

F = -1/(4*Pi*Epsilon_0) * q*Q/r^2

q,Q are the charges of the balls
Epsilon_0 is a physical constant (8.854 * 10^-12 kg*m^3/(s^2*C^2))
r is the distance between the centers of the two balls.

Lutz-Peter

Post a reply to this message

Thankyou to all the people who replied ;-)

This is turning out to be *much* harder than I thought... Currently, the
ball approaches on of the magnets, accelerates to implausibly heigh speed,
and then ends up so far away from the magnets that is just demonstrates
Newton's 1st - it travels in a straight line forever. Bum!

Well anyway, at this point it was about 9pm, so I gave up and sulked off to
read a book instead. I haven't tried it again in light of the latest posts -
hopefully I might actually get somewhere! Well let folks know once I work
out exactly what's happening and can ask more *specific* questions.

Weary Andrew.

Post a reply to this message

> F = -1/(4*Pi*Epsilon_0) * q*Q/r^2

Isn't that the same as F = -r^2 / (4 * Pi * Epsilon_0 * q * Q)?

> q,Q are the charges of the balls
> r is the distance between the centers of the two balls.

I'm with you. (I think...)

> Epsilon_0 is a physical constant (8.854 * 10^-12 kg*m^3/(s^2*C^2))

So... 8.854e-12 is the number... kg*m^3 is the unit... what's the s^2 * C^2
bit about?

Andrew.

Post a reply to this message

In article <3d96d83d@news.povray.org>, orp### [at] btinternetcom says...

> Isn't that the same as F = -r^2 / (4 * Pi * Epsilon_0 * q * Q)?

Hu? No (beware: y/x*z = y*z/x and not =y/(x*z)).

> > Epsilon_0 is a physical constant (8.854 * 10^-12 kg*m^3/(s^2*C^2))
> 
> So... 8.854e-12 is the number... kg*m^3 is the unit... what's the s^2 * C^2
> bit about?

s^2 * C^2 also belongs to the unit (s: seconds, C: Coloumb).

Lutz-Peter

Post a reply to this message

"Lutz-Peter Hooge" <lpv### [at] gmxde> wrote in message
news:3d96e745$1@news.povray.org...
> In article <3d96d83d@news.povray.org>, orp### [at] btinternetcom says...
>
> > Isn't that the same as F = -r^2 / (4 * Pi * Epsilon_0 * q * Q)?
>
> Hu? No (beware: y/x*z = y*z/x and not =y/(x*z)).

Are you beginning to see why my math doesn't work? <sob!>

> > > Epsilon_0 is a physical constant (8.854 * 10^-12 kg*m^3/(s^2*C^2))
> >
> > So... 8.854e-12 is the number... kg*m^3 is the unit... what's the s^2 *
C^2
> > bit about?
>
> s^2 * C^2 also belongs to the unit (s: seconds, C: Coloumb).

Right... So if I let q and Q be the charge of my ball and magnet (what's the
correct unit for charge?), and I measure r in meters, then F will come out
in Netwons? While we're on the subject, what would be a suitable range of
magnitude for q and Q? (The ball is 80g in mass.)

Thanks!
Andrew.

Post a reply to this message

Andrew Coppin <orp### [at] btinternetcom> wrote:
> This is turning out to be *much* harder than I thought... Currently, the
> ball approaches on of the magnets, accelerates to implausibly heigh speed,
> and then ends up so far away from the magnets that is just demonstrates
> Newton's 1st - it travels in a straight line forever. Bum!

  The closer you get to a physically correct model, the more
real-life problems you will encounter.
  Orbits caused by gravitational or magnetic forces are very very
unstable. Even the slightest difference in a stable orbit can make it
unstable and the orbiting object will most probably be ejected away 
(unless you have modelled object collision as well, and the orbiting object
happens to fall into the other object :) ).

  (So how come the planets and moons in our solar system are in so nice
stable orbits? Because from the millions and millions of objects very long
time ago floating around the forming Sun, these particular objects happened
to be, by chance, in the right places at the right times and survived. All
the other objects either collided with these or were ejected from the solar
system.)

-- 
#macro N(D)#if(D>99)cylinder{M()#local D=div(D,104);M().5,2pigment{rgb M()}}
N(D)#end#end#macro M()<mod(D,13)-6mod(div(D,13)8)-3,10>#end blob{
N(11117333955)N(4254934330)N(3900569407)N(7382340)N(3358)N(970)}//  - Warp -

Post a reply to this message

In article <3d9710d0@news.povray.org>, war### [at] tagpovrayorg says...

>   Orbits caused by gravitational or magnetic forces are very very
> unstable.

This is only true for systems with three or more bodies.
A system of two bodies will always be stable I think (in real life, a 
simulation of it may be unstable, especially if computed using the Euler 
algorithm).

Lutz-Peter

Post a reply to this message

Lutz-Peter Hooge <lpv### [at] gmxde> wrote:
> This is only true for systems with three or more bodies.
> A system of two bodies will always be stable I think (in real life, a 
> simulation of it may be unstable, especially if computed using the Euler 
> algorithm).

  It may depend on how "stable" is defined.
  I think that an orbit is defined to be stable when the orbiting body
has a permanent and well-defined orbit around the other object
(that is, it will never collide with the other object nor it will be
ejected to infinity).
  In this sense it's perfectly possible to have an unstable orbit in
a two-body system (eg. simply by having them in collision course; it's
also possible that they will escape to infinity with respect to each other
if the minimum escaping speed is reached).

  Perhaps you confused this with the fact that a two-body
system can be modelled analytically while a three-(and higher) body
system cannot (but must be approximated numerically)?

-- 
#macro N(D)#if(D>99)cylinder{M()#local D=div(D,104);M().5,2pigment{rgb M()}}
N(D)#end#end#macro M()<mod(D,13)-6mod(div(D,13)8)-3,10>#end blob{
N(11117333955)N(4254934330)N(3900569407)N(7382340)N(3358)N(970)}//  - Warp -

Post a reply to this message

In article <3d97179b@news.povray.org>, war### [at] tagpovrayorg says...

>   In this sense it's perfectly possible to have an unstable orbit in
> a two-body system (eg. simply by having them in collision course; it's
> also possible that they will escape to infinity with respect to each other
> if the minimum escaping speed is reached).

But then it is no orbit at all. Of course it can collide, or escape to 
infty, but if it orbits at all, it will orbit forever (that is what I 
mean with stable).

Lutz-Peter

Post a reply to this message

>
> Right... So if I let q and Q be the charge of my ball and magnet (what's
the
> correct unit for charge?),

Coulombs (C)

> and I measure r in meters, then F will come out
> in Netwons?

Newtons. Yup.

>While we're on the subject, what would be a suitable range of
> magnitude for q and Q? (The ball is 80g in mass.)

Quite small. The electrostatic forces are very strong

eg. If you give one ball a charge of 5*10^-5 C and the other a charge
or -5*10^-5C and put them a metre apart the force between them is 22.5 N
That's equivalent to the gravitational force exerted by the earth on an
object of
mass 2.2kg.

If you charge a plastic rod by rubbing it with fur you can typically get a
charge
of 10^-9 C

Gail
--
#macro G(H,S)disc{0z.4pigment{onion color_map{[0rgb<sin(H/pi)cos(S/pi)*(H<6)
cos(S/pi)*(H>6)>*18][.4rgb 0]}}translate<H-5S-3,9>}#end G(3,5)G(2,5.5)G(1,5)
G(.6,4)G(.5,3)G(.6,2)G(1,1)G(2,.5)G(3,.7)G(3.2,1.6)G(3.1,2.5)G(2.2,2.5)G(9,5
)G(8,5.5)G(7,5)G(7,4)G(7.7,3.3)G(8.3,2.7)G(9,2)G(9,1)G(8,.5)G(7,1)///GS

Post a reply to this message

> Right... So if I let q and Q be the charge of my ball and magnet (what's
the
> correct unit for charge?), and I measure r in meters, then F will come out
> in Netwons? While we're on the subject, what would be a suitable range of
> magnitude for q and Q? (The ball is 80g in mass.)
>

btw, I wouldn't mix magnets and electrostatic charges. The math can get very
complex.
Stick to two charges balls and it's not that hard.

Gail
--
#macro G(H,S)disc{0z.4pigment{onion color_map{[0rgb<sin(H/pi)cos(S/pi)*(H<6)
cos(S/pi)*(H>6)>*18][.4rgb 0]}}translate<H-5S-3,9>}#end G(3,5)G(2,5.5)G(1,5)
G(.6,4)G(.5,3)G(.6,2)G(1,1)G(2,.5)G(3,.7)G(3.2,1.6)G(3.1,2.5)G(2.2,2.5)G(9,5
)G(8,5.5)G(7,5)G(7,4)G(7.7,3.3)G(8.3,2.7)G(9,2)G(9,1)G(8,.5)G(7,1)///GS

Post a reply to this message

> Coulombs (C)

OK, so tell me people - *is* there a "u" in that or not? ;-) Some seem to
think there is, others not...

> >While we're on the subject, what would be a suitable range of
> > magnitude for q and Q? (The ball is 80g in mass.)
>
> Quite small. The electrostatic forces are very strong

So I recall... "A person jumps off the top of a building. It takes him 30
seconds to accelerate down to the bottom under gravity, but only a fraction
of a second for electrostatic forces to bring his bode to a half again.
[Presumably rearranging it beyond recognition in the process!]"

> eg. If you give one ball a charge of 5*10^-5 C and the other a charge
> or -5*10^-5C and put them a metre apart the force between them is 22.5 N
> That's equivalent to the gravitational force exerted by the earth on an
> object of mass 2.2kg.

Ah... yes, *charge balls*... I had _better_ remember to give them OPPOSITE
charges... presumably they'll repell instead of attract otherwise? (Hmm...
that might actually be useful later on...)

> If you charge a plastic rod by rubbing it with fur you can typically get a
> charge
> of 10^-9 C

So I take it a 1C is a fairly large charge then? (I remember hearing that 1
Farrid is larger than any capacitor ever built - "built" being the word!)

> Gail

Thankyou very much!

So, in summary, I have three "magnets" (at least, fixed points which I want
to "attract" a moving particle). Right... so I need r in meters, q and Q
with opposite sign and at around about 10^-7 C or so, and the formula will
give me an answer in Newtons which should be halfway sane (assuming I make
sure that r stays away from zero!) Right, will try...

Thanks again for all the people who bothered to help a hapless half-brain!
Andrew.

Post a reply to this message

> btw, I wouldn't mix magnets and electrostatic charges. The math can get
very
> complex.

Mmm... complex is bad... (Unless it involved the square root of -1 ;-)

> Stick to two charges balls and it's not that hard.

Sounds good to me...

Post a reply to this message

Andrew Coppin wrote:

> So I take it a 1C is a fairly large charge then? (I remember hearing that 1
> Farrid is larger than any capacitor ever built - "built" being the word!)

http://www.partsexpress.com/pe/showdetl.cfm?&User_ID=8556534&St=3685&St2=71679322&St3=65133449&DS_ID=3&Product_ID=118270&DID=7

-- 
Ken Tyler

Post a reply to this message

> Andrew Coppin wrote:
>
> > So I take it a 1C is a fairly large charge then? (I remember hearing
that 1
> > Farrid is larger than any capacitor ever built - "built" being the
word!)


I used to work at the fusion research center here in the UK, and my office
was litterally meters from a room where they had 1000's huge caps - each
about 25cm in diameter. anyway they would charge them up and discharge them
through the plasma to induce currents of 1+ mega-Amps (hence the name MAST -
mega Amp spherical tokomak)

jim

Post a reply to this message

Waheeeeeeeeey! [Dancing round the room like an insane thing possed] It
works! It works! Yipeeeey!

[Calms down for a moment]

The charge ball formula [eventually] worked a treat! I now have everything
[almost] as I want it - except a post in povray.animations as soon as I
finish uploading the results to my website...

Andrew.

["From the ashes of disaster grow the roses of sucess!..." [etc]]

Post a reply to this message

Lutz-Peter Hooge <lpv### [at] gmxde> wrote:
> But then it is no orbit at all. Of course it can collide, or escape to 
> infty, but if it orbits at all, it will orbit forever (that is what I 
> mean with stable).

  Of course it's an orbit. It's an unstable orbit.

  Besides, making a difference between them does not help finding a stable
orbit.

-- 
#macro N(D)#if(D>99)cylinder{M()#local D=div(D,104);M().5,2pigment{rgb M()}}
N(D)#end#end#macro M()<mod(D,13)-6mod(div(D,13)8)-3,10>#end blob{
N(11117333955)N(4254934330)N(3900569407)N(7382340)N(3358)N(970)}//  - Warp -

Post a reply to this message

"Andrew Coppin" <orp### [at] btinternetcom> wrote in message
news:3d9755af@news.povray.org...
> > Coulombs (C)
>
> OK, so tell me people - *is* there a "u" in that or not? ;-) Some seem to
> think there is, others not...

My physics textbook says yes.

> Ah... yes, *charge balls*... I had _better_ remember to give them OPPOSITE
> charges... presumably they'll repell instead of attract otherwise?

That's correct.

> (Hmm...
> that might actually be useful later on...)
>
> > If you charge a plastic rod by rubbing it with fur you can typically get
a
> > charge
> > of 10^-9 C
>
> So I take it a 1C is a fairly large charge then?

Yes, very. Two charges, one 1C and the other -1C seperated by a metre
will attract with a force of  about 9*10^9 N (equivalent to the weight of a
900000000 kg object on earth........)

>(I remember hearing that 1
> Farrid is larger than any capacitor ever built - "built" being the word!)

I've heard of a 1 Farad capacitor, never seen one though.

> Thankyou very much!
>
> So, in summary, I have three "magnets" (at least, fixed points which I
want
> to "attract" a moving particle). Right... so I need r in meters, q and Q
> with opposite sign and at around about 10^-7 C or so, and the formula will
> give me an answer in Newtons which should be halfway sane (assuming I make
> sure that r stays away from zero!) Right, will try...

If you need more help with equations please ask.

Gail
--
#macro G(H,S)disc{0z.4pigment{onion color_map{[0rgb<sin(H/pi)cos(S/pi)*(H<6)
cos(S/pi)*(H>6)>*18][.4rgb 0]}}translate<H-5S-3,9>}#end G(3,5)G(2,5.5)G(1,5)
G(.6,4)G(.5,3)G(.6,2)G(1,1)G(2,.5)G(3,.7)G(3.2,1.6)G(3.1,2.5)G(2.2,2.5)G(9,5
)G(8,5.5)G(7,5)G(7,4)G(7.7,3.3)G(8.3,2.7)G(9,2)G(9,1)G(8,.5)G(7,1)///GS

Post a reply to this message

"Gail Shaw" <gai### [at] mwebcoza> wrote in message
news:3d98b505@news.povray.org...

> I've heard of a 1 Farad capacitor, never seen one though.

http://www.maplin.co.uk/
Enter code JR01B (thats a zero) in the search box.
Size: 8.0mm x 21.5mm dia.

Alf

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