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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.
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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] btinternet com> 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.
>
>
>
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"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.
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> 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
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> > 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/ ]
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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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