>> Won't speeding up and slowing down make the train swing out more or
>> less? (Like going round a corner in a car, if you speed up it rolls
>> more).
>>
>> In a rotating reference frame you'd be essentially turning even when on
>> a straight piece of track, so I would not say that changes in forward
>> motion won't have any effect.
>>
> Yes, to an extent, this is true. But, since the only angle its changing
> at is side to side, forward motion isn't going to have an affect
> *except* when making turns.

But as I said, in a rotating reference frame the train will be "making 
turns" all the time, even when on a straight track. So any forward/back 
acceleration will, in general, affect the side to side swinging.

You can easily see this by imagining a straight piece of track rotating 
and plotting out the course the train will take as it goes along the 
track - it will be a curve, so some sideways forces must be involved.

Post a reply to this message

>> In that case why not just model it as a point mass at B connected by a
>> suitably stiff spring to point A? Keep track of the position and
>> velocity of B and you can use normal numerical integration to update B
>> (there will be a gravity force and a spring force towards A). In your
>> graphics you can use the angle between the vertical and the line A-->B
>> to draw your train.
>>
>> It may not be 100% physically accurate but much simpler than trying to
>> work through all the maths related to dynamics in a rotating reference
>> frame...
> Uh, yeah, so how do you do that? lol Seriously, the closest I have come
> to differentials is a book promising to take you from basic math up to
> basic calculus, and I got lost like 3/4 of the way through it. :p But,
> yeah, it sounds good...

This is a good guide, there's some other good articles on his site too.

http://gafferongames.com/game-physics/integration-basics/

I don't know how much you've done already on this kind of stuff, but 
even after 4-5 years of learning calculus the rotating reference stuff 
was scary hard. From what I remember, mainly because your unit vectors 
are no longer constant, so no longer disappear when you differentiate 
them...

This is the demo we had in one of our lectures:

http://www.youtube.com/watch?v=kc88SrMG5fA

Fun to watch, but try to prove the physics :-)

Post a reply to this message

Patrick Elliott wrote:
> So.. guess I am trying to work this out myself after all. lol

My apologies.  Real life conspired to take from me all but about five hours of
sleep for a while.

If I understand what you are trying to do, I think you can boil it all down to
figuring the "swing angle" (theta) of the car.  I'm assuming a rigid connection
(a "bar" with length L) between the center of mass (m) of the car and the track.
 The position of the track-end of the bar is given at any point in time, and
from this you can figure the car's velocity and acceleration in the direction
tangent to the track.

You have five "forces" acting on the car.

 * Gravity pulls it downward.
   |F_g| = m*g

 * Centrifugal force pulls it away from the center of the disc.
   |F_disc| = m*r_disc*omega_disc^2

 * Centrifugal force pulls it away from the center of curvature of the track.
   |F_track| = m*|v|^2/r_track

 * Centrifugal force pulls it away from the track-end of the bar as it swings.

 * The bar pulls it toward the track.

This can be further simplified, since the bar will counteract all force
components EXCEPT the component that can cause the car to swing from
side-to-side.  This component is tangent to the swinging arc at the car's
current position.  So, figure out the component of F_g, F_disc, and F_track that
is in this direction.  (You can use trig or direction vector dot products for
this, whichever you prefer.)

Now that you have the total tangential force (F_tan), calculate the tangential
acceleration (a_tan) it would cause from F_tan = m*a_tan.  Then, figure the
angular acceleration (alpha = a_tan/L).  Integrate over time to get the angular
velocity (omega), and again to get the angular position (theta).

Don't forget to calculate and then add the tangential velocity (v_tan) vector to
the total velocity (v) vector when computing F_track.  You will also probably
want to include some kind of frictional force (angular velocity damping) so the
car does not keep swinging forever.

I hope this helps.

http://en.wikipedia.org/wiki/Centripetal_force
http://en.wikipedia.org/wiki/Angular_acceleration

P.S. Your intuition that "forces" don't exist as such is a good one, especially
at the macro scale of everyday life.  They are, however, extremely useful
abstractions.

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"waggy" <hon### [at] handbasketorg> wrote:
>
> Don't forget to calculate and then add the tangential velocity (v_tan) vector to
> the total velocity (v) vector when computing F_track.

I forgot to mention that I usually make *lots* of mistakes as I work through a
problem, and this looks like one of them.  The component of the side-to-side
swinging velocity in the direction of the track should be zero, so forget about
this.

Post a reply to this message

> You have five "forces" acting on the car.
>
>   * Gravity pulls it downward.
>     |F_g| = m*g
>
>   * Centrifugal force pulls it away from the center of the disc.
>     |F_disc| = m*r_disc*omega_disc^2
>
>   * Centrifugal force pulls it away from the center of curvature of the track.
>     |F_track| = m*|v|^2/r_track
>
>   * Centrifugal force pulls it away from the track-end of the bar as it swings.
>
>   * The bar pulls it toward the track.

Do you need the coriolis force in there too?

Post a reply to this message

scott wrote:
> Do you need the coriolis force in there too?

Good catch.  It looks like it could be significant for this problem.  As usual,
the Wikipedia page has a useful formula.

http://en.wikipedia.org/wiki/Coriolis_effect#Formula

(If I ever manage to make a real income from math, I plan to send them a
donation for having been an invaluable mathematics reference for years.)

Post a reply to this message

On 9/6/2012 3:10 AM, scott wrote:
>>> In that case why not just model it as a point mass at B connected by a
>>> suitably stiff spring to point A? Keep track of the position and
>>> velocity of B and you can use normal numerical integration to update B
>>> (there will be a gravity force and a spring force towards A). In your
>>> graphics you can use the angle between the vertical and the line A-->B
>>> to draw your train.
>>>
>>> It may not be 100% physically accurate but much simpler than trying to
>>> work through all the maths related to dynamics in a rotating reference
>>> frame...
>> Uh, yeah, so how do you do that? lol Seriously, the closest I have come
>> to differentials is a book promising to take you from basic math up to
>> basic calculus, and I got lost like 3/4 of the way through it. :p But,
>> yeah, it sounds good...
>
> This is a good guide, there's some other good articles on his site too.
>
> http://gafferongames.com/game-physics/integration-basics/
>
> I don't know how much you've done already on this kind of stuff, but
> even after 4-5 years of learning calculus the rotating reference stuff
> was scary hard. From what I remember, mainly because your unit vectors
> are no longer constant, so no longer disappear when you differentiate
> them...
>
I have done.. almost nothing. Like I said, never took calculus at all, 
and the book I read through derailed about where the equations got into 
integration and differentials. I did pick up a few interesting trick 
about raising to powers and negative powers that I had no idea was 
possible, but otherwise.. confusing as hell. I need to sit down one of 
these days and actually try to work out what they are doing.

> This is the demo we had in one of our lectures:
>
> http://www.youtube.com/watch?v=kc88SrMG5fA
>
> Fun to watch, but try to prove the physics :-)

O.o

Ok, yeah, that is.. not something I would have predicted. lol

Post a reply to this message

On 9/6/2012 9:27 AM, waggy wrote:
> scott wrote:
>> Do you need the coriolis force in there too?
>
> Good catch.  It looks like it could be significant for this problem.  As usual,
> the Wikipedia page has a useful formula.
>
> http://en.wikipedia.org/wiki/Coriolis_effect#Formula
>
> (If I ever manage to make a real income from math, I plan to send them a
> donation for having been an invaluable mathematics reference for years.)
>
Hmm. Ok. All that should get me a start. Wouldn't have had a clue what 
all to look for without help. Just need the time to actually write a 
script, test and debug. Unfortunately, no idea when that is going to be. :p

Post a reply to this message

On 9/6/2012 3:00 AM, scott wrote:
>>> Won't speeding up and slowing down make the train swing out more or
>>> less? (Like going round a corner in a car, if you speed up it rolls
>>> more).
>>>
>>> In a rotating reference frame you'd be essentially turning even when on
>>> a straight piece of track, so I would not say that changes in forward
>>> motion won't have any effect.
>>>
>> Yes, to an extent, this is true. But, since the only angle its changing
>> at is side to side, forward motion isn't going to have an affect
>> *except* when making turns.
>
> But as I said, in a rotating reference frame the train will be "making
> turns" all the time, even when on a straight track. So any forward/back
> acceleration will, in general, affect the side to side swinging.
>
> You can easily see this by imagining a straight piece of track rotating
> and plotting out the course the train will take as it goes along the
> track - it will be a curve, so some sideways forces must be involved.
>
Ok, true. Maybe I am phrasing wrong then. What I meant is that, since 
the reference is going to be, in principle, constant, what ever forces 
it is applying are not going to have the same effect as say, 
acceleration or deceleration of the train **on** the track, if the 
direction is directly parallel to the existing force being applied by 
the reference. The result will, thus, only matter when the resulting 
vectors actually change, with respect to the angle it can swing in. I.e.

Possible angle = <0,1,0>
Forces = <4,2,1>
Result = <0,2,0>

This is not the case, obviously, when the track is not directly parallel 
to the force, or if the possible rotations where <1,1,0> (i.e., both X 
and Y, not just Y).

Post a reply to this message

On 9/6/2012 8:59 AM, waggy wrote:
> "waggy" <hon### [at] handbasketorg> wrote:
>>
>> Don't forget to calculate and then add the tangential velocity (v_tan) vector to
>> the total velocity (v) vector when computing F_track.
>
> I forgot to mention that I usually make *lots* of mistakes as I work through a
> problem, and this looks like one of them.  The component of the side-to-side
> swinging velocity in the direction of the track should be zero, so forget about
> this.
>
That's OK. I might actually do a gondola style thing too, maybe, so, in 
that case, the force wouldn't be 0 in the track direction as well. But, 
for the design I intended, yeah, I already figured that. lol

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