Rotating Wheels and Point Line-up

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Discussion Overview

The discussion centers around the problem of determining the conditions under which points marked on the circumferences of rotating wheels will line up again after an initial alignment. The focus is on the mathematical relationships between the frequencies of rotation of the wheels and the implications of using different numerical values, including rational numbers and irrational numbers like π.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes a method to calculate the alignment of two wheels based on their frequencies and suggests using graphical methods to find intersection points as solutions.
  • Another participant identifies the problem as a number theory issue, stating that the first and second wheels line up when the first has completed an integral number of turns, and questions whether the third wheel can also align based on the ratio of its velocity to the first wheel's velocity.
  • A further inquiry is made about the implications of replacing π with a rational number, asking if the initial line-up would still occur and how such scenarios can be proven.
  • One participant asserts that alignment would occur if the angular velocities of the wheels are expressed as rational numbers, providing a condition involving the least common multiple of certain denominators.

Areas of Agreement / Disagreement

Participants express differing views on the implications of using π versus rational numbers, and whether the conditions for alignment can be satisfied under various scenarios. The discussion remains unresolved regarding the general case of alignment with different wheel speeds.

Contextual Notes

There are assumptions regarding the nature of the frequencies and the mathematical properties of π and rational numbers that are not fully explored. The discussion does not resolve the complexities of the relationships between the wheels' speeds and their alignment.

r731
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Hello all,

I'm trying to obtain solution(s) for problems of the following nature:
Suppose there were three wheels of equal size, rotating on the same axis, one point marked on the circumference of each wheel, and these three points lined up in one straight line. If the second wheel rotated twice as fast as the first, and if the speed of the third wheel was 1/π of the speed of the first, the initial line-up would never recur.

To keep things simple, I begin with two wheels instead, A and B.
If A has the frequency f_A, and B f_B, then the location (in terms of degrees) of each marked point is computed as follows:
f_A * 360 * t % 360;
f_B * 360 * t % 360;

where t is the elapsed time.

And,
360*f_A - 360*floor(f_A*t) = 360*f_B - 360*floor(f_B*t),
which simplifies to,
f_A - floor(f_A*t) = f_B - floor(f_B*t)
can be solved for t; the solutions of which represent the time instants during which the two points (each of which is marked on one wheel) line up.

Is there any simpler way other than using %-operator and floor()?
I was especially thinking of graphs plotted on a coordinate system, where the intersection points would be solutions.

Thanks in advance.
 
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It's a number theory problem.

It is easy to derive that the first and second wheels line up when and only when the first has performed an integral number of turns - call that number n. For the third wheel to also line up it must have also performed an integral number of turns.

Using the ratios of velocities of the 1st and third wheels, work out how many turns the third wheel must have made when the 1st has made n.

Can that number be an integer, given what is known about pi in number theory?
 
Now if pi were replaced by a rational number, say 3.14159, would the initial line-up occur?

How to prove these kinds of stuff?
 
Yes it would occur. Say the angular velocities of the three wheels are ##1, \frac{a}{b}, \frac{c}{d}## where ##a,b,c,d## are integers and ##b,d>0##. Then the dots will line up after every ##kn## revolutions of the first wheel where ##k## is any positive integer and ##n## is the least common multiple of ##b## and ##d##.

Can you see why?
 

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