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Newcomers DBPro Corner / circular motion

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Zokomoko
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Posted: 21st Feb 2005 00:23
Hi.
I'm a bit new to the circular motion physics and how to implement it into DBP, and I hope you could help me with a little problem I have.

I'm trying to create a yellow object that is going in a cirular like path towards a green object that doesn't move.

Yellow object has acceleration on the T axis, and of course the radial acceleration.
Green object has nothing, it's just a point in 3d space.

so I've ran into a couple of problem:
(1) How can the radius be calculated, given the positions of the 2 objects. I could use intersect object, but that's too slow.
(2) Almost everything is a given (acceleration on both axis, start velocity, end velocity, radius which was calculated previously, and the angle between the objects).
How could you possibly find the time it gets the yellow object to reach the green object ?




Thx in advance, I'm totaly lost about this.
Baggers
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Posted: 21st Feb 2005 01:16
If you have the position of the centre of the circle and of a point on the circle (eg: the green dot) then you can use these.


Any help?
Zokomoko
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Posted: 21st Feb 2005 01:28
no, because I don't have the position of the centre of the circle.
Monsieur Mort
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Posted: 21st Feb 2005 01:50
I think you can calculate the radius like this:

from a level physics, the velocity of the object = radius * angular velocity (in radians per second)

also from a level physics, the angular velocity = pi * frequency (number of times a complete circle is made per second).

therefore v=pi*r*f

now I think the frequency has to be the circumference of the circle divided by the velocity, in which case f = 2*pi*r/v

so v=2*pi*pi*r*r/v

so v^2=2*pi^2*r^2

and taking roots, v=sqrt(2)*pi*r

so r=v/(sqrt(2)*pi)

You might want to check through that though, I'm not entirely confident it's right.




For the second point, you want the time taken for the yellow object to turn through a certain angle.

Time taken = distance (angle in this case)/speed (or in this case angular velocity)

As before, angular velocity = pi*f = 2*pi^2*r/v

so t=x/(2*pi^2*r/v)

so t=xv/2pi^2r

where x is the angle you want to turn through in radians. Again you might want to check through that

To convert from degrees to radians, divide 2pi by 360 and multiply by the angle you want. Similarly to convert from radians to degrees divide 360 by 2pi and multiply by the angle you want. hope that helps (is right!)

Why do you sink I ave zees outRAGEOUS accente?!
Zokomoko
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Posted: 21st Feb 2005 03:58
Thx for the long post, I'll look into that tommorow.
but I think f =v/(2*pi*r).
Monsieur Mort
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Posted: 21st Feb 2005 04:00
Actually I think you're right - angular velocity = 2*pi*f not pi*f.

Still, I think you should be able to work it out.

Why do you sink I ave zees outRAGEOUS accente?!
Ric
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Location: object position x
Posted: 21st Feb 2005 10:50 Edited at: 21st Feb 2005 11:19
@MM:
Quote: "now I think the frequency has to be the circumference of the circle divided by the velocity, in which case f = 2*pi*r/v
"


In fact, it's the other way (upside down!). The frequency is the velocity divided by the circumference (f=v/2piR). This means that the v's, and everything else cancel when you put f=v/2piR into v=2piRf, and you'd actually end up with 1=1 - not surprising, as the two equations are actually just different arrangements of the same thing.

@Zokomoko - looks like an old problem we never solved! Anyway, so long as the direction of the object forms a tangent to the circular path it is following, the radius would be given, I think, by

radius=D/(2*sin(theta))

where D is the distance between the two objects, and theta is the angle between the object's direction and the target. Anyone who wants to double check that for me, here's how I came up with it:



As for the time, I get:

t=(2*theta*r)/v

where theta is the same as before. Again, anyone who wants to double check, my derivation was:

angularvelocity=angle/time
>
time=angle/angularvelocity
>
time=angle/(velocity/radius)
>
time=(angle*radius)/velocity
and the angle (alpha in your diagram) is twice theta (in my diagram) so:

t=(2*theta*r)/v




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Coding Fodder
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Posted: 21st Feb 2005 11:00 Edited at: 21st Feb 2005 11:02
Ric I think you meant to write R=D/(2*sin(theta))

[edit] you edited while I was posting!! good work though

Something really catchy that makes people stop and think about the meaning of life and say to themselves "My but thats clever"
Ric
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Posted: 21st Feb 2005 11:05
lol - I always get caught out mid-edit! I should learn to write posts in one go!


Coding Fodder
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Posted: 21st Feb 2005 11:15
angular velocity= velocity/radius.

But somehow you get to the correct equation the next step I am forced to assume a typo

Something really catchy that makes people stop and think about the meaning of life and say to themselves "My but thats clever"
Ric
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Posted: 21st Feb 2005 11:20
Well spotted - indeed a typo. Now edited - again!


Zokomoko
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Posted: 22nd Feb 2005 19:27 Edited at: 22nd Feb 2005 19:30
Thx for the replies.

But I still can't seem to find out what the radius is, since I don't know what's the position of the center of the circle, or the Theta angle.


Here's a little code I've made.
The radius there is a given.



wmf
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Posted: 23rd Feb 2005 01:45
this has random values for both objects. it's kinda simple so I don't know if it helps.
Zokomoko
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Posted: 23rd Feb 2005 02:02
Not quite.
I will try to explain my problem a little better.
Object A's positions are: 50,0
Object B's positions are: 0,50
How would I know the coordinates of the center of the circle ?

In the example I gave above, the coordinates of the center of the circle is simply 0,0 and the radius is obviously 50.
But I'ad like to know the formula.

These are givens:
X1,Z1 X2,Z2
those are the positions of object 1 and 2 accordingly.
What I need to know:
Radius=? Coordinates of the center of the circle = ?

I hope this is possible mathemtically.
wmf
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Posted: 23rd Feb 2005 03:34 Edited at: 23rd Feb 2005 03:39
the random numbers are for testing. if you have two objects on the screen then you know their positions. I expanded this line:
radius# = sqrt((object position x(2)-x)^2 + (object position z(2)-z)^2)

replacing x and y with the obj positions. it prints the radius. sure looks mathematical to me.
Zokomoko
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Posted: 23rd Feb 2005 03:43
Thx.
At first I was a little confused as for the random values.
But now I understand, thx alot for the help!
RiiDii
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Posted: 23rd Feb 2005 04:51 Edited at: 23rd Feb 2005 04:54
Edited: Posted too late.

"Droids don't rip your arms off when they lose." -H. Solo

REALITY II

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