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February 25th, 2019, 01:01 PM   #1
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Parametric equations!

Let's be honest, equations of curves don't become interesting until you start to throw parameters in there. Nothing satisfies me more than a point travelling along a curve in some defined manner.

But... I run into problems of my own with what the equations of motion should be.

For example, let's say I have a sine curve defined from zero to pi. I have a particle (object), that travels from zero to pi, but I want it to come back.

How do you accomplish this?
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February 25th, 2019, 02:21 PM   #2
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With vectorial calculus .
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February 25th, 2019, 02:55 PM   #3
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Quote:
Originally Posted by NineDivines View Post
Let's be honest, equations of curves don't become interesting until you start to throw parameters in there. Nothing satisfies me more than a point travelling along a curve in some defined manner.

But... I run into problems of my own with what the equations of motion should be.

For example, let's say I have a sine curve defined from zero to pi. I have a particle (object), that travels from zero to pi, but I want it to come back.

How do you accomplish this?
Do you mean that you have a particle at x such that $\displaystyle x = \sin( \theta )$ and you need a $\displaystyle \theta (t)$ such that $\displaystyle \theta$ goes from 0 to $\displaystyle \pi$ to 0? Any function $\displaystyle \theta (t)$ that goes from 0 to $\displaystyle \pi $ to 0 (such as a parabola, or even a sinusoidal curve if you have that kind of sense of humor) will do.

If you need it to do more than that, you'll need to give us more information.

-Dan
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Last edited by skipjack; February 26th, 2019 at 09:02 AM.
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February 26th, 2019, 03:27 AM   #4
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I'm just using a basic example to say at time t where does the particle lie on the curve?

Here's another example. Let's say I have a helix curve in the form of a spring. The particle travels forward 3 coils and travels backwards one coil at a constant velocity then repeats so at time t it's making its way along the spring but not in the usual way.

Presumably it will have a form f(t) = h(t) - g(t)

Last edited by NineDivines; February 26th, 2019 at 03:32 AM.
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February 26th, 2019, 07:13 AM   #5
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A helix, with vertical axis and radius r, can be written as x= r cos(t), y= r sin(t), z= at. That will up three "coils" with t from 0 to $\displaystyle 6\pi$. If you want the particle to stop and go back to the initial point in time T, multiply each of those by $\displaystyle \sin(2\pi t/T)$.
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Last edited by skipjack; February 26th, 2019 at 09:02 AM.
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February 26th, 2019, 11:43 AM   #6
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Quote:
Originally Posted by Country Boy View Post
A helix, with vertical axis and radius r, can be written as x= r cos(t), y= r sin(t), z= at. That will up three "coils" with t from 0 to $\displaystyle 6\pi$. If you want the particle to stop and go back to the initial point in time T, multiply each of those by $\displaystyle \sin(2\pi t/T)$.
Yeah I see, that makes sense.

But how do you account for the points on the curve where the velocity is zero? I could list them for obvious reasons by what you've shown say $\displaystyle 6 \pi n $ and $\displaystyle \sin(2n\pi t/T)$ but it would be much more concise if the derivative gave a magnitude of zero for the velocity vector.

Last edited by skipjack; February 26th, 2019 at 01:45 PM.
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February 28th, 2019, 12:04 PM   #7
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I’ve given this some thought and it requires a transformation to make the derivative for velocity equate to zero at these points. Any point I choose on whatever curve I choose, if the particle stops wherever it chooses only a transformation can handle it. I will figure this out.
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