Brachistochrome Problem

G

Goistein

Junior Member
Joined
Oct 8, 2006
Messages
109
OK, I've just learned a few things in Diff EQ, and I came across this neat problem. It asked about the curve at which a particle can frictionlessly slide and go from one point to another in the least possible time. Now after reading the solution and seeing it solved with the differential equations I saw that is was a cycloid.

However, wouldn't the fastest curve be this one:
|
|
|
L________

(The corner should actually be rounded so it slides to the right instead of bouncing, but using an L was quicker.)

Because at the first part it drops the full height and accelerates at the fastest possible velocity, 9.8m/s^2. Then it reaches the bottom moving at it's fastest speed so it will clear the horizontal part quickly too. Isn't this technically the fastest curve instead of the cycloid?

Should this have been posted in the Differential Equations section?
 
It seems counter intuitive that the time saving curve does not have the particle only moving down, but I'm also surprised that the proof uses almost no calculus.

Thanks for the link!
 
Goistein said:
It seems counter intuitive that the time saving curve does not have the particle only moving down, but I'm also surprised that the proof uses almost no calculus.

Thanks for the link!
Click to expand...

Well, you could always tell your intuition that the right angle is a longer path than the cycloid... Since it has a longer distance to travel, it loses some of the advantage of its speed....
 
The Brachistochrome is also known as an Isochronic ("equal time") curve.

Place a ball bearing anywhere on the cycloidal track.
It will take the same time to reach the bottom.
Code: 
      *
      |
      |*o
      |:*
      |:::*.
      |:::::::*..
      |:::::::::::::*...
      |:::::::::::::::::::::*....       |
      * - - - - - - - - - - - - - - - - *

Start the ball near the top.
It will take \(\displaystyle t\) seconds to reach the end.

Code: 
      *
      |
      |*
      |:*
      |:::*.
      |:::::::*..   o
      |:::::::::::::*...
      |::::::::::::::::::::::*....      |
      * - - - - - - - - - - - - - - - - *

Start the ball anywhere else on the track.
It will still take exactly \(\displaystyle t\) seconds to each the end.


This is the basis of a "sucker bet".

Set up two cycloidal tracks and tell your victim,
"You place your marble anywhere on your track.
I'll place mine on my track and beat you to the bottom."

The apparatus should have starting gates that guarentee simulataneous starts,
and a finish-line device to determine the winner.

Just before you begin the first race, you say,
"Remember, you have to beat me to win,"
subtly implying that ties don't count.


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Here's a bit of obscure trivia about the cycloid.

A pendulum does not actually keep accurate time.
To do so, the weight must swing in a cycloidal arc.

The usual pendulum swings in an arc of a circle,
. . which, for small arcs, is accurate enough.

So we need a pendulum arm that changes lengths?
There is a way around this problem.
Suspend the weight on a flexible arm (a string).

At the pivot, place "bumpers" on both sides.
This will bend the path of the weight into a cycloidal arc.
Code: 
  *:::::::::::::*:::::::::::::*
   *:::::::::::* *:::::::::::*
     *:::::::*     *:::::::*
         * /         \ *
          /           \
         /             \
        *               *
           *         *
                *

Can you guess the shape of these "bumpers"?

That's right . . . cycloids!

 
DrMike said:
Goistein said:
It seems counter intuitive that the time saving curve does not have the particle only moving down, but I'm also surprised that the proof uses almost no calculus.

Thanks for the link!
Click to expand...

Well, you could always tell your intuition that the right angle is a longer path than the cycloid... Since it has a longer distance to travel, it loses some of the advantage of its speed....
Click to expand...

OK, well, I was thinking of the greedy method of solvation.

That would be a nice way to make some money... betting on roll time. I should try that.
 
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