How many revolutions will the smaller coin make when rolling around the bigger one?

Imagine a circle with radius 1 cm rolling completely along the circumference of a circle with radius 4 cm. How many rotations did the smaller circle make?

The circumference of a circle with radius $r$ is $2\pi r$, so the circumference of a circle with radius $4r$ would be $8\pi r$. Since $8\pi r \div 2\pi r = 4,$ I figured the answer must be four revolutions. So imagine my surprise when I saw that the answer was given to be five! I read the explanation of why this was indeed the correct answer, and although the reasoning seemed sound, it took some time before I could truly convince myself that my solution was in error. It's an interesting problem, so I presented it to a number of people, most of whom immediately answered "four" as I did, and, like me, were difficult to convince otherwise; only a very few could intuitively see "five" as the correct answer. Here's a better way to think of this problem: rather than rolling along the larger circle, start by imagining the smaller circle as rolling on a line the same length as the circumference of the larger circle. In this case it is straightforward to think of the line as being $8\pi r$ units long, and so the smaller circle clearly must have made $8\pi r \div 2\pi r = 4$ rotations. Next, think of the circle as sliding along the line, without rolling, so that the point on the coin at which it touches the line remains the same. Now consider the difference between sliding along a straight line and doing the same along the circumference of a circle; if you slide units $8\pi r$ along a straight line, you arrive at your destination just as you started, without having ever changed orientation. But if you do the same along the circumference of a circle you will have made a full rotation at the time you return to your starting point. When \emph{rolling} along the same circumference, therefore, you will have made the four rolling rotations plus the one sliding revolution, for a total of five!Put differently: when the small circle rolls along the circumference of the larger circle, two kinds of movement simultaneously occur, revolution and rotation. The four movements one initially considers are the four revolutions, perhaps because these are readily seen. Rotation, on the other hand, is more difficult to grasp.

If you find it difficult to grasp how things work on a circle, you might also want to imagine what would happen on a square. When a circle rolling along the outer periphery of a square encounters the first corner, it will have to rotate an "extra" 90° to continue along the next side. This will happen again at each corner, and since 90° x 4 = 360°, this accounts for an additional full revolution.

Each of the above explanations describes the circle's movement as a decomposition into rotation and revolution, but in reality no such decomposition is taking place. Just as a human's heart and lungs work simultaneously, rotation and revolution take place together. Separating revolution from rotation is helpful for understanding, but doing so does not provide a fundamental solution. Some say that the makeup of our brains does not allow for multitasking, but learning to simultaneously comprehend such phenomena would be of great value.

*A similar problem appeared in Aha! Gotcha: Paradoxes to Puzzle and Delight by Martin Gardner and also in Scientific American in 1868. If you can think of an alternative proof or explanation for this problem, please post a comment or email us!*

### About the author

Yutaka Nishiyama is a professor at Osaka University of Economics, Japan. After studying mathematics at the University of Kyoto he went on to work for IBM Japan for 14 years. He is interested in the mathematics that occurs in daily life, and has written ten books about the subject. The most recent one is *The Mysterious Number 6174: One of 30 mathematical topics in daily life*, published by Gendai Sugakusha in July 2013 (ISBN978-4-7687-6174-8). You can visit his website here.

## Comments

## Another similar way of explaining...

Think of the circle at the end of the 8xpixr line… It rotated four times. Now bend the line to make it a circle. The small circle has to rotate one more time to follow the line.

## Food for thought

I prefer to think of this problem as a coin rotating around a spot; i.e. the 'big' circle has a radius of zero.

It's easy to see that the 'small' circle will rotate once, even though it's travelled a circumference on 0cm.

Then, if the 'big' circle has any additional circumference (in this case 4cm), the 'small' circle will rotate that once, then travel the additional circumference.

I hope this makes sense!

## Arrived at the same reasoning.

I arrived at your reasoning independently before looking at the comments.

## circles rolling on circles

Could one also consider that the edge of the circle are always curved away from the circle rolling on it, and because of this the circle must roll extra to meet this edge verses a straight line?

## Yes Your explaination is the only one which I could understand

Thanks

## Ok I get that it's more

Ok I get that it's more complex than it first appears, but what about circles, with circles within them rotating. Do the maths, my head hurts.

http://math.ucr.edu/home/baez/mathematical/tusi_couple_animation.gif

## Mathematics

This is also an interesting problem, like with a circle rotating on the exterior of another circle, the answer is not simply R/r, that is the ratio of the larger radius to the smaller radius, but is actually R/r - 1. By rolling along the interior of a circle, one revolution is lost. An intuitive way of imagining why this happens is by first approximating a circle with a regular polygon, say a hexagon. Imagine rolling a circle around the interior of the hexagon, while the circle rolls along each side of the hexagon, it does not get to roll over the complete length of each side, due to the convexity of the interior of the hexagon, namely, a circle will have completed rotation along one side when it is tangent to two sides of the hexagon, it is clear simply by drawing a picture that the circle does not need to be displaced the entire side length of each side of the hexagon for this to happen, and so some rotation is essentially *lost* because to this.

## coins rolling around larger coins

context not clear but should not "R/r - 1" be "R/r +1" (exterior rotation) ?

whereas "R/r -1" for interior rolling?

## Inside and out

I liked your intuitive explanation, but I'm not sure I agree with it as it stands.

Yes, when rolling a circle round the outside of a regular polygon like a hexagon one can expect the circle to roll along the complete length of each side, and so make more rotations than it does inside, where its travel is limited by not fitting into the corners. But surely this problem doesn't arise when rolling a circle inside a larger circle. Its roll brings it into contact with the entire circumference of the larger. So failure to fit in everywhere inside doesn't account for the lost rotation.

This also applies to rolling a straight sided polygon which fits snugly into the corners of a larger version of itself, such as a small square into a bigger square. I tried it with a cardboard square inside a square hole of 2x the side length, also cut out of a piece of cardboard. Inside, just 1 rotation when rolled right round, but outside 3. So it's the same principle, rolling inside gets R/r - 1 rotations, and outside R/r + 1 rotations. 1 and 3 respectively when R/r = 2.

I think this formula suggests the simplest explanation. When R/r = 1, that is when they are the same size, then rolling outside results in 1+1 = 2 rotations. but inside gets you 1 - 1 = 0 rotations. The latter is hardly surprising since you can't really roll a circle, or indeed any polygon, inside another of exactly the same dimensions. It fits everywhere too snugly. So there's your lost rotation.

of a regular polygon like a hexagon one can expect the circle to roll along the complete length of each side, and so make more rotations than it does inside, where its travel is limited by not fitting into the corners. But surely this problem doesn't arise when rolling a circle inside a larger circle. Its roll brings it into contact with the entire circumference of the larger. So failure to fit in everywhere inside doesn't account for that lost rotation.

This also applies to rolling a straight sided polygon which fits snugly into the corners of a larger version of itself, such as a small square into a bigger square. I tried it with a cardboard square inside a square hole of 2x the side length, also cut out of a piece of cardboard. Inside, just 1 rotation when rolled right round, but outside 3. So it's the same principle, rolling inside gets R/r - 1 rotations, and outside R/r + 1 rotations. 1 and 3 respectively when R/r = 2.

I think this formula suggests the simplest explanation. When R/r = 1, that is when they are the same size, then rolling outside results in 1+1 = 2 rotations. but inside gets you 1 - 1 = 0 rotations. The latter is hardly surprising since you can't really roll a circle, or indeed any polygon, inside another of exactly the same shape and dimensions. It fits everywhere too snugly.

So there's your lost rotation.

## These formulae come from applying basic geometry

In the two cases, consider a small partial roll and the triangle made by the old and new radii of the large circle extended (or cropped, for interior roll) through the centre of the small circle and the new radius of the smaller circle to its initial tangent point extended to meet the extended or cropped original radius of the large circle.

Take the angle of the third vertex of the triangle, not one of the two centres.

In the exterior case, the supplement of this angle along the original radius of the large circle extended through the centre of the small circle measures the total rotation of the small circle, and its opposite angles are the angles swept out by the original tangent points in each circle. So the rotation is

`π - (π - (d/r + d/R)) = d(1/r + 1/R)`

, where`d`

is the partial circumference swept and`r`

,`R`

are the radii of the smaller and larger circles.In the interior case, this angle actually measures the total rotation of the small circle. Its opposite angles are the angle swept out by the original tangent point in the larger circle and

the supplementof the angle swept out by the original tangent point in the smaller circle. So the rotation is`π - ((π - d/r) + d/R)) = d(1/r - 1/R)`

.Although the construction above applies only to a sufficiently small rolled distance, the circumference of the large circle can be divided into arbitrarily small arcs of equal size and the results summed.

When the whole of circumference of the larger circle has been rolled, the total rotation of the smaller circle is

`2πR(1/r +/- 1/R) = 2π(1 +/- R/r)`

. So, in each case the problem is soluble if`r`

is a divisor of`R`

, and the number of rotations is the integer`1 + R/r`

for exterior rolling, or`1 - R/r`

for interior: in this example, 5 or 3.## These formulae come from applying basic geometry

In the two cases, consider a small partial roll and the triangle made by the old and new radii of the large circle extended (or cropped, for interior roll) through the centre of the small circle and the new radius of the smaller circle to its initial tangent point extended to meet the extended or cropped original radius of the large circle.

Take the angle of the third vertex of the triangle, not one of the two centres.

In the exterior case, the supplement of this angle along the original radius of the large circle extended through the centre of the small circle measures the total rotation of the small circle, and its opposite angles are the angles swept out by the original tangent points in each circle. So the rotation is

`π - (π - (d/r + d/R)) = d(1/r + 1/R)`

, where`d`

is the partial circumference swept and`r`

,`R`

are the radii of the smaller and larger circles.In the interior case, this angle actually measures the total rotation of the small circle. Its opposite angles are the angle swept out by the original tangent point in the larger circle and

the supplementof the angle swept out by the original tangent point in the smaller circle. So the rotation is`π - ((π - d/r) + d/R)) = d(1/r - 1/R)`

.Although the construction above applies only to a sufficiently small rolled distance, the circumference of the large circle can be divided into arbitrarily small arcs of equal size and the results summed.

When the whole of circumference of the larger circle has been rolled, the total rotation of the smaller circle is

`2πR(1/r +/- 1/R) = 2π(R/r +/- 1)`

. So, in each case the problem is soluble if`r`

is a divisor of`R`

, and the number of rotations is the integer`1 + R/r`

for exterior rolling, or`R/r - 1`

for interior: in this example, 5 or 3.## alternative solution

The CENTRE of the small coin needs to travel around a circle of radius 4+1=5cm. So the CENTRE needs to travel 10pi, as though it is traveling along a straight line of length 10pi while revolving. This gives the solution.

## different rotations

The centre of rotation of a point on the smaller circle is the centre of the smaller circle. Therefore for a full rotation the smaller circle will travel a distance of its circumference around the larger circle.

But the centre of rotation of the centre of the smaller circle is the centre of the larger circle. Hence it rotates around the circle by the larger radius as suggested.

## Max

N revolution + 1 rotation

## This is the best explanation mathematically.... 👌

This is the best explanation mathematically.... 👌

## David Van Leeuwen's animation

4 or 5? David Van Leeuwen's animation by GeoGebra is a good explanation.

http://www.geogebratube.org/student/m107691

Yutaka Nishiyama

## Circles rolling on circles

What your demo indicates is actually the location of the CENTER of small circle, which is causing illusion. How about the tip of the red arrow, which is tangential point of the two circles. Suppose you put a small ant at the tangential point (assuming ant not moving by itself). How many time this ant will be in contact with the large circle? Appearantly, the answer is 4, not 5. We are not interested in the movement of the center of small circle at all...

## Circles Rolling on Circles

In the case of coins/circles of equal radius, the moving coin rotates once with respect to the static coin/circle but twice with respect to the observer.

The important point is that were it different, a driven planet gear of radius ‘r’, would rotate more than once when powered by a drive-cog of radius ‘r’. This would involve a creation of energy.

In the case of a coin/circle of radius ‘r’ moving along a line of length 2πr, the moving coin rotates once with respect to the static coin/circle and also once with respect to the observer.

In the animation at http://www.geogebratube.org/student/m107691 the black arrow is “with respect to the observer” and the red arrow is “with respect to the static circle.”

The mathematical formula Xπr / Yπr = X/Y remains true in all cases: anything else is an illusion.

## Orbital motion

The article reminds me of a surprising fact I learnt recently about the Moon's orbit round the Earth. It keeps the same face towards us all of the time (except for a relatively small effect known as libration) because it does one complete rotation for every revolution. I believe this synchronous orbit is explained as a gravitational effect called tidal locking.

This motion seems to be similar to what's happening when, as described in the article, you slide but don't roll a coin A round the edge of another B so that the same point on A's edge is always in contact with B. Coin A then does one full rotation, but not two.

It's almost as if gravity answers the reverse of the question posed in the article by explaining why we get one less rotation than we should.

Chris G

## Something physicsy?

Looking at the picture of the £2 coins, imagine if the coin on top rotated one full time, but the bottom coin also rotated in the opposite direction - as if they were interlocking cogs, or a bit like the bottom coin is a treadmill on which the top coin turns.

In order for the top coin to rotate clockwise once, (i.e. 2 * pi * r), the bottom coin must also rotate anti-clockwise once. That is two full rotations. Now, if the bottom coin stays fixed (like a pavement and not a treadmill) as in the original statement of the problem, then is there some principle of physics that dictates the missing revolution of the bottom coin - because this bottom coin is now static - that says that revolution must come out somewhere? And that is why the top coin actually rotates twice? I'm thinking about a conservation law, but I do not know physics.

Michael B.

## clarity of question

The problem here lies not in the maths used to get the answer or in incorrect thinking on the part of the problem solver - it lies in asking a vague question. There are several versions of this puzzle, all of them relying on vagueness to trick the reader. There may be several correct answers to the general question "How many rotations did the smaller circle make?":

1: the smaller circle only rotates around the centre of the larger circle once.

4: the smaller circle rotates around its own centre four times.

5: the smaller circle makes two types of rotation as above, totalling five rotations.

undetermined/zero/infinity: no start or end point for either circle has been defined, so the smaller circle rotates indefinitely, or does not rotate at all, because no rotation point has been defined either.

It can be argued that the standard definition for a rotation of a circle is a rotation around its own centre. A

revolution, however, is a circular movement around an external point (external to the circle). So, in the above problem, the smaller circle makes 4 rotations and one revolution. (Of course, definitions of rotation and revolution can change depending on the context).The question cannot be answered - fairly and mathematically - with a single answer, without first defining the rotation point and clarifying the definition of 'rotation' in this instance.

It does not help matters that, in the above example, 'revolution' is used in the diagram, while 'rotation' is used in the general text.

I detest these kinds of 'trick' questions, with the 'gotcha!' at the end. They make people feel foolish when they have only tried to think about the question and answer it with the information provided. If you fail to get the correct answer without being given all the information or when the question is badly or incorrectly phrased, then it is clearly not your fault.

Questions such as these do not support the use of mathematical skills (especially logic) in problem solving - they are essentially interpretative grammar tests, and the correct answer should be spotting the grammar or logic mistakes in the question.

Evidence for the above assertions lie in the fact that even people as intelligent as Yutaka Nishiyama can be duped by them. It is not a difficult problem, but a poorly written one.

It is exactly like the 'Deep Thought' computer in Hitchiker's Guide to the Galaxy providing the answer to the 'Ultimate Question' as 42. When confronted about providing such a flippant, simple answer, the computer states: "I think the problem, such as it was, was too broadly based. You never actually stated what the question was." All poorly stated questions in mathematics should be treated with similar contempt!

## more clarity

N.B.: Just to be clear, I am in no way criticising the esteemed Mr. Yutaka Nishiyama - I am only criticising the original question and all similarly badly written questions!

## the question is very clear... read it carefully

I am sorry but your argument is flawed. The question is very clear. It says the small circle is 'rolling' completely, which means it has to rotate around it's centre forever while moving, there's absolutely no sliding motion. Then the question asks, how many 'rotations' did the small circle make, not how many 'times' the small circle had to roll completely. If the question was the latter the answer wound have been 4. But given the change in orientation of the initial position that occurs in the small circle owing to moving on the big circle, we must consider a 5th rotation!

## I think you're missing the…

I think you're missing the point of the question. This isn't some gotcha riddle where you have to add two unrelated numbers to get the trick answer. The smaller circle truly rotates around its own center five times, not four.

## 1) Radius:1 - 2(Pi)r =

1) Radius:1 - 2(Pi)r = 2.68

2) Radius:4 - 2(Pi)r = 25.12

25.12/2.68 = 9.3 revolutions!?

Where I am I going wrong?

## Number theory

Your error is in solving your first equation: when r = 1 then 2(pi)r = 6.28 (to 2 d.p.), not 2.68.

You'll then find that 25.12 / 6.28 = 4, as expected.

## Mathematics

I think it is possible to get a better picture by first thinking of a regular polygon, take a hexagon, for example. The hexagon has six corners, now if we initially started rotating a coin along the sides of the hexagon, once it reaches a corner, the coin rotates an additional 60 degrees ''around'' this corner, try and visualise this and hopefully it makes sense. Therefore, along with the rotation induced due to the coins motion around the sides of the hexagon, there is additional rotation along each of the six corners, and so one additional revolution is made once the coin reaches its starting position. Now if you think of a circle as the limit of regular polygons as number of sides tends to infinity, you can extend this logic to this problem to perhaps get a better intuitive sense as to why an additional revolution takes place.

## Mathematics

I completely agree, this is a nice pure mathematical proof.

Better still, and to formalise the argument into neat algebra: Make the both coins hexagons. If the large coin has a radius say 3 times the size of the small one then you can easily see that you for the small hexagon to traverse the large one it has to be moved from one side touching the larger hexagon to the next touching the larger hexagon 6*3 times. In 6*(3-1) of these moves the small hexagon rotates through 360/6 degrees. In the other 6 (when it his a corner of the large hexagon) it rotates through 360/6+360/6. Adding all these up gives 6*(3-1)*(360/6)+6*(360/6+360/6) = (6+1)*360 i.e. 6+1 rotations.

Think of N sides and the large coin having x times the radius of the small. The case above is N=6, x=3. The argument works though for any N and any x. And it works for circles too: Just let N tend to infinity while holding the radius of the polygon fixed. It also, with small modification, works for non-integer x and so is completely general.

## Totally Confused

Hi, okay, I'm really confused. What qualifies a revolution? How do we know that it has done one by the time it is at the bottom of the coin? Just because the coin is right side up at the bottom doesn't mean it has revolved completely (does it?). The way I see it is from the center coin's perspective; first it is being touched by the Queen's neck, then her face, then her hair, then the back of her head, and finally her neck again. But that leaves us at the start, causing me to think that it has really only performed one revolution. But obviously I'm wrong. Please can someone explain where I've gone wrong. Thanks.

## Translation

No need to be confused or see any ambiguity. The coin translates and rotates as it moves around. Translation is measured by the movement in the (x,y) co-ordinates if its centre. Rotation is measured by the angle theta a line that starts out going from the centre to the right-most point of the coin (imagine it drawn on) makes me with the radius that currently has the orientation.

X,y and theta completely define the position of the circle and are sufficient framework for thinking about the problem. Movements in theta form the revolutions of interest.

## Circles rolling on circles

Thank you very much for your explanation. I was getting a little frustrated that no source I found went into detail explaining exactly why an extra "rotation" happened to be but your "sliding revolution" reasoning cleared the air beautifully.

'Sometimes, logic needs to take a back seat.'

-Me, probably.

## Space & Co-ordinates

My 16 year old son caught me with this last night & I marched straight into the incomplete response, but fairly quickly rebutted that the smaller circle does in fact revolve only [ratio] times in the radial space, but I conceded that a complete response would be to add that it revolves [ratio] + 1 times in the Cartesian space (after he clarified my assumption for me). I'll also concede that my retort was a complete reaction, but it now has me wondering if there are legitimate space-transformations to clarify the distinction?

## Optical Illusion

Using the standard definition the one revolution is 360 degrees. Look carefully at the rotation of the coin about the other coin. Though it appears that the outer coin rotates twice as it goes around the center coin, it does not. It does one half a revolution from the top to the bottom, then completes another half from the bottom to the top. From the top coin as a reference, the right side of the coin is engaged in the first half of the rotation and the left side in the second half. A third way to see this is the contact point of the top coin is at the bottom of the coin, this point does not come back into contact with the center coin until it reaches the top position. The optical illusion comes from the orientation of the coin when it is at the bottom. Though the orientation is the same as when it is at the top, it is the top of the coin in contact the the center coin not the bottom point. The outer coin only does one 360 rotation as it rolls around the center coin.

## circle round a circle

Imagine a coin rolling around a pentagon, the coin will rotate in regular angular increments along each edge with an additional fifth of a rotation at each vertex. Now imagine increasing the number of edges. When the edges shorten and approach the vertices the Archimedean principle of exhaustion holds even when infinitely short edge and vertex are one. There is therefore an extra - you could say hidden - rotation. Of sorts.

## Selecting Reference Points

A CIRCLE ( RADIUS = 1 ) ROLLING ( 1 REV ) AROUND THE CONVEX SIDE OF ANOTHER CIRCLE ( RADIUS = 4 ).

THE DISTANCE THE CENTER REFERENCE POINT OF THE SMALLER CIRCLE TRAVELS IS 2Pi(4+1)=10Pi

ANY ONE CHOSEN REFERENCE POINT, ON THE CIRCUMFERENCE OF THE SMALLER CIRCLE, WILL TRAVEL A DISTANCE OF

2Pi(1) x 5REV = 10Pi.

THE CIRCUMFERENCE OF THE LARGER CIRCLE IS 2Pi(4)=8Pi.

THE CIRCUMFERENCE OF THE SMALLER CIRCLE IS 2PI(1)=2Pi.

The emphasis on reference points is to keep your attention on what the reference points are doing and not on the length of the two circumferences, which are not reference points.

The point of contact between the two circles is not a reference point,

that point of contact is a start-stop reference.

Any point on the circumference of the smaller circle is a reference point.

The two circle centers are reference points.

Will any one point anywhere on the

unit circle travel 10Pi after 5 revolutions?

## How about 3-space or higher dimensional spaces?

Do you know whether there are any surprises with regard to this sort of thing when you transfer the problem into space? Maybe for starters, if you consider a ball rotating around the surface of another ball?

## Alernative solution to puzzle

Here is an alternative solution to the problem of a smaller circle (of radius 1 cm) rotating along the circumference of a larger one (of radius 4 cm). Let us stretch out both the circles into straight lines, placed side by side, with their left ends coinciding. It is easy to see that the shorter line is contained four times within the longer line. This is exactly equal to four rotations of the smaller circle around the larger circle. But, the shorter line is still within the length of the longer line (circumference of the larger circle) since its left end has not coincided with the right end of the longer line. The fifth rotation of the smaller circle is equivalent to the left end of the shorter line coinciding with the right end of the longer line.

## This is an odd way to look at it

Calling the example of only one coin rotating as the other revolves around it as "rotating" feels like a bit of a stretch. The coin isn't *just* rotating, it is also translating in space. If you eliminate the translation and fix the center point of both coins, you will get 4 and 1 rotation respectively. That is because in this example, both coins are a part of the same body, so when one rotates, the other must counter rotate else one of the coins is forced to move through space.

The coin is *not* rotating 5 and 2 times, respectively, it is *seeming* to rotate from a fixed point perspective while also counter rotating through space an entirety of its circumference by the time it arrives back at its origin.

## The 1/4 rotation problem (posted some years ago)

Just use an O-scope in any scale to solve.

One "rotation" = 2 segments, the positive and negative peaks, of a sine wave.

The wave form is standard, of course. Any length of unit 1 is displayed as 2 unit lengths of 1 merely as the form of the wave. (+ : -) = 0 = 1 cycle. This works for any ratio where the unit length = the "set" which is to generate the "number of sets" corresponding to the larger value of the respective radius length to be used.

A = 4 units B = 1 unit

Say: 4/1 = 8 segments or 4 cycles / 2 segments or 1 cycle = A/B = 8 + : 8 - = peak to peak / 2+ : 2 - peak to peak = 4.

The number of rotations is given by unit length alone.

Graphic: unit ---- set ---- ---- ---- ---- : 4 : cycles or rotation of B about A. e.g 8 cycles / 2 cycles.

Works for me, anyway. At least it's intuitive and based on "workshop" manipulation rather than other numerical speculation.

SWB 15 APR 2021

## Crisp and clear explanation

I had come across this problem in the past, and had learnt via experimentation using two coins that the answer is (circumference of stationery circle ÷ circumference of moving circle) +1. However your theoretical explanation really helped in understanding the underlying principles.

## Circles rolling on circles

Various folk have made comments which relate to the answer, some have even spelled out the answer but I haven’t yet come across an explanation of the logic and the maths.

Logic tells everyone who first encounters this problem that the distance covered by the small circle (r) is 2pi times the radius of the large circle (R). That seems right but it isn’t, the distance covered by the small circle is 2pi times (R + r) as that is the point which remains “still” throughout the whole rotation.

If we consider a small circle 1/3 the diameter of the large one then

r=R/3 or 3r = R thus radius of rotation is 3r+r, lets call this Rc

Distance covered is circumference of the circle Rc divided by circumference of small circle r.

It is the smaller circle which is moving therefore distance covered is

2pi (3r + r) / 2pi (r) or (2pi*Rc / 2pi*r)

The 2pi elements cross out so we are left with

(3r + r)/ r = 4. QED

I don’t have the ability to draw here, but if you want to sketch it out it makes much more sense.

## try doing the math

No, you haven't proven anything that's relevant.

I have yet to see anyone prove mathematically that the number of rotations of the outer small circle is R/r + 1, where R and r are the radii of the large and small circles respectively and the ratio R/r is an integer. I haven't yet done it but I can see how the math problem should be formulated.

Imagine the large circle with its center at the origin of the x-y axis. Now place the small circle with its center at (0,R+r). Draw an ARROW as the diameter line of the small circle such that the tail of the arrow is at (0,R) and the head of the arrow is at (0, R+2r). This is crucial for counting complete rotations of the small circle. Now imagine rolling the small circle clockwise along the circumference of the large one and STOP when the arrow points STRAIGHT UP again. This corresponds to exactly one rotation of the small circle (orientation of the arrow is the same as at the start). This happens BEFORE the tail point of the arrow again comes into contact with the large circle. But the arc length between consecutive tail points touching the circumference of the large circle is clearly 2*pi*r, so the distance swept along the large circle's circumference when the small circle performs one complete rotation is LESS THAN 2*pi*r. It is 2*pi*r - D, where D is the arc length of the "shortfall".

Because D is not zero, it's already clear that by the time the small circle reaches its starting point back at the top of the large circle it will have completed MORE than R/r rotations. But how many more? To figure that out we need to calculate D.

We KNOW that if R/r = n (an integer) then the small circle will arrive PRECISELY back at its starting point after exactly n rolls of its circumference along the circumference of the large circle. But this also implies that the total number of rotations of the small circle is an integer that is GREATER than n. From cardboard circle experiments we know that the total number of rotations is n+1. This can happen only if n*D = 2*pi*r - D, or

D=2*pi*r^2/(R+r).

If you can derive this formula mathematically, from the geometry, then you've just proven that if R/r = n then the small circle rotates n+1 times as it goes around the big circle.

Good luck!!

## Because after the small…

Because after the small circle has gone round along the big circle, the small circle itself has gone round. If you draw a point at the bottom of the small circle at the starting point, after turning, this point will appear five times on the big circle, that is, the position of the small circle relative to the big circle has changed, and it has also turned around. So it's five laps.

## Circles rolling on circles.

It’s a small though important point. The circles on circles explanation provided by Marianne on 16th May, 2014 has an error repeated several times.

Marianne states “The circumference of a circle with radius r is 2 pi r, ∴ if the radius is 4, then the circumference is 8 pi r.”

In fact, the circumference of a circle with a radius of 4 is 8 pi. not 8 pi r. There is no ‘r’ in the answer.