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

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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

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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.

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> 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.

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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

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"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.

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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 -

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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

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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 -

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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

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