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Difference between revisions of "Brachistochrone"

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A brachistochrone is the curve of fastest descent from <math>A</math> to <math>B</math>.  
 
A brachistochrone is the curve of fastest descent from <math>A</math> to <math>B</math>.  
 
It is described by parametric equations which are simple to derive.
 
It is described by parametric equations which are simple to derive.
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==From Circle to Cycloid==
 
==From Circle to Cycloid==
 
A cycloid is the path traced by a point on a rolling circle:
 
A cycloid is the path traced by a point on a rolling circle:
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so the y-coordinate of the point is: <math>a(1-\cos t)</math>
 
so the y-coordinate of the point is: <math>a(1-\cos t)</math>
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Latest revision as of 01:41, 27 July 2019

A brachistochrone is the curve of fastest descent from $A$ to $B$. It is described by parametric equations which are simple to derive.

From Circle to Cycloid

A cycloid is the path traced by a point on a rolling circle: How to Create a Cycloid



If the radius of the circle is $a$ and the center of the circle is moving at a speed of $a$ units per second, then it moves $2\pi a$ units, or one revolution, every $2\pi$ seconds (in other words, it revolves 1 radian per 1 second).



Then

the x-coordinate of the center relative to the ground is: $at$

the x-coordinate of the point relative to the center is: $a\cos(-\frac{\pi}{2}-t)=-a\sin t$

so the x-coordinate of the point is: $a(t-\sin t)$



Similarly,

the y-coordinate of the center relative to the ground is: $a$

the y-coordinate of the point relative to the center is: $a\sin(-\frac{\pi}{2}-t)=-a\cos t$

so the y-coordinate of the point is: $a(1-\cos t)$

From Cycloid to Brachistochrone

Since a brachistochrone is an upside-down cycloid, we reverse the sign of y: $x=a(t-\sin t)$ $y=a(\cos t-1)$ This is a brachistochrone starting at $(0,0)$.



If you want it to start at $(x_0,y_0)$ you just shift it: $x-x_0=a(t-\sin t)$ $y-y_0=a(\cos t-1)$



If you want it to go through $(c,d)$ you need to solve for $a$: $c-x_0=a(t-\sin t)$ $d-y_0=a(\cos t-1)$



I recommend first solving for $t$ in terms of $a$ using $d-y_0=a(\cos t-1)$, then substituting into $c-x_0=a(t-\sin t)$.

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