The Rolling Shutter Effect

Geometry Level 5

You might want to know what the Rolling Shutter Effect is about before attempting this question.

A Ball of radius 1/2 is spinning around a point (and always touching that point), which is at the center of the (blue) square, of side 2, as shown in the GIF above. The Black Line, or the Shutter, that moves from the bottom to the top of the screen, takes 2 seconds to travel across the square. The Blue Lines, which form a shape when the Shutter passes over it, is the image. At the start, the Black Line touches the bottom of the square while the Ball touches the top of the square.

Given that the fastest possible speed of the ball (In revolutions per second) such that the image shows a single, egged-shaped shape is A A , find

⌊ 100000 A ⌋ \left\lfloor 100000A \right\rfloor

(The first GIF from the top shows the image generated at speed A A )

This is the image generated at speed > A >A

This is the image generated at speed < A <A

A V V shaped shape, which occurs in the case when 2 egged shaped shapes combine is not counted.


The answer is 47197.

Julian Poon by Julian Poon , 20, Singapore

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1 solution

Michael Mendrin
Jun 16, 2015

Let t t be time, R R be the rate of rotation of the ball, and T T be the time to the when the circle makes its last contact with the black line. Then it's about solving the equation

1 2 C o s ( 2 π t R ) + 1 2 = t − 1 \dfrac { 1 }{ 2 } Cos\left( 2\pi tR \right) +\dfrac { 1 }{ 2 } =t-1

where the right side is a line that is tangent to the Cosine curve. The slope of the Cosine curve at t = T t=T is

− π R S i n ( 2 π R T ) -\pi RSin\left( 2\pi RT \right)

from which we can immediately get T T in terms of R R , because the slope is 1 1 , which is (with a little bit of hand waving with regards to multiples of π \pi )

T = 1 R − 1 2 π R A r c S i n ( 1 π R ) T=\dfrac { 1 }{ R } -\dfrac { 1 }{ 2\pi R } ArcSin\left( \dfrac { 1 }{ \pi R } \right)

The equation of the tangent to the Cosine curve at t = T t=T is

− π R ( t − T ) S i n ( 2 π R T ) + 1 2 C o s ( 2 π R T ) + 1 2 -\pi R(t-T)Sin\left( 2\pi RT \right) +\dfrac { 1 }{ 2 } Cos\left( 2\pi RT \right) +\dfrac { 1 }{ 2 }

which, at t = 0 t=0 has to equal to 1 1 . Using the expression for T T in terms of R R and simplifying, we finally end up this equation to solve for R R

A r c C s c ( π R ) + R 2 π 2 + 1 + 3 π R − 2 π = 0 ArcCsc\left( \pi R \right) +\sqrt { { R }^{ 2 }{ \pi }^{ 2 }+1 } +3\pi R-2\pi =0

which can be done by numerical means to get

R = 0.471971517... R=0.471971517...

and so the answer is 47197. Not too terribly elegant, but it gets the job done.

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Thanks for improving the question!

Thanks!

Julian Poon - 5 years, 12 months ago

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