2 particles in a 3d corkscrew over surface of a sphere


by T.A. Zenaide
Tags: 3d sine-wave, particles physics, sphere
T.A. Zenaide
T.A. Zenaide is offline
#1
Feb20-12, 02:33 PM
P: 4
Aloha!

This is my first post, and I hope that I am posting in the correct area...

I need help with a math problem that is beyond my math skills. If anyone can answer it, I'd be very grateful!

Here's my problem:

Imagine two particles, say "a" and "b" that are traveling (in positive time -- time = 0 to 1) in a straight line directly towards each other. They encounter a sphere and exchange places via a 3-dimensional sine-wave, like a corkscrew, each maintaining an equal distance on either side of the surface of the sphere. (The line between them would be the diameter of the sphere.) They exit the sphere maintaining their original straight course.

Mahalo nui loa for any help ,

Trevor Avichennya Zenaide
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meldraft
meldraft is offline
#2
Feb20-12, 03:05 PM
P: 280
...so what is your question?? You haven't posed it yet! Also, I don't really understand what you mean by:

"they exchange places via a 3-dimensional sine-wave".
T.A. Zenaide
T.A. Zenaide is offline
#3
Feb20-12, 03:43 PM
P: 4
Oh! I'm sorry I was unclear... What I mean by "exchange places" is like those math problems we did long, long ago when it was: Train A leaves the station at 10:00. Train B leaves another station at 11:00... etc... what time do they meet?

So, imagine that particle "a" and particle "b" are moving in a straight line towards each other, but there is a sphere in the way. The way they "exchange places" is that "a" and "b" hit the sphere on directly opposite sides of the sphere and the way they "exchange places" is by "a" moving one way on a 3d corkscrew path and "b" moving the equal and opposite way on that corkscrew path, and they both maintain "a connection" as an imaginary line that is the diameter of the sphere.

Is that clearer?

GDLY
GDLY is offline
#4
Feb21-12, 09:05 AM
P: 2

2 particles in a 3d corkscrew over surface of a sphere


I think i understand what you mean they are no longer moving in a strait line upon encountering the sphere and they are moving along its surface in a helix. However, you still haven't provided a question and i'm not sure if you mean to say that they ever leave the sphere and if so did they leave out the opposite end traveling strait the way the other one came?

Regardless there is still no question.
meldraft
meldraft is offline
#5
Feb21-12, 09:19 AM
P: 280
I think that you mean that body A will leave the surface of the sphere at the entry point of body B, and vice versa. In any case, by saying that they are headed straight towards each other (I assume in a straight line), your problem is planar, so the bodies are going to travel along 2D space along a circle. In that case you can use simple circular motion equations to derive the kinematics (if that is what you are looking for?).
T.A. Zenaide
T.A. Zenaide is offline
#6
Feb21-12, 12:44 PM
P: 4
Yes, that is what I mean, in that "a" leaves where particle "b" entered and vice-versa... However, they don't travel over a circle, they travel in a helix (thank you for the correct term) over the surface of a sphere.

Am I leaving anything out, now, that makes this an actual question? Whew! I didn't realize I was so imprecise when I started. Thank you both for helping me clarify my needs
T.A. Zenaide
T.A. Zenaide is offline
#7
Feb21-12, 01:05 PM
P: 4
I guess my question is if at time = 0, particle "a" and particle "b" arrive at diametrically opposed places on the sphere, and when time = 1, particle "a" and particle "b" arrive at each other's starting point, what is the equation for both of them traveling in a helix over the surface of the sphere, say "a" going clockwise and "b" going counter-clockwise in the helix, both maintaining an equal and opposite path (separated by the diameter of the sphere)?

Is that a complete question???


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