Friday head melter

Keep_Her_Lit

Boost your workplace productivity by starting the day with a little mental stimulation, courtesy of this cycling related teaser ...

Is the following claim true or false? Explain your answer.

Claim : For a bicycle travelling at a road speed of x kph, the forward velocity of the top of each wheel is greater than 2x kph.

Example : If you are hurtling towards the bottom of a hill at 70kph, the tops of your wheels are each headed there at more than 140kph.

True or false?

Hints

• The claim contains no semantic trickery or other obscure sneakiness.
• Just take it at face value.
• Don't google it, think about it!

The prize for the first correct answer with a satisfactory explanation will be a c.20 year old used Sam Browne reflective sash with an irritating buckle that tends to open while riding.

Gotta dash now. Will pop back at lunchtime.

triggermortis

false.

top of the wheel is going in the same direction and at the same speed as you, ie 70kmh

niceonetom

triggermortis wrote: »

false.

top of the wheel is going in the same direction and at the same speed as you, ie 70kmh

Then how does it become the front of the wheel? It must be going faster than you!

Ands how's this one? The part of the tyre that is in contact with the ground is stationary, no matter how fast you go (presuming you're not skidding). So any point on the surface of the centre of the tyre goes from 0 to 2x(your speed) and back to 0 with every revolution.

triggermortis

niceonetom wrote: »

Then how does it become the front of the wheel? It must be going faster than you!

Ands how's this one? The part of the tyre that is in contact with the ground is stationary, no matter how fast you go (presuming you're not skidding). So any point on the surface of the centre of the tyre goes from 0 to 2x(your speed) and back to 0 with every revolution.

But it is rotating around an axis and you are not.
Pick a point on the tyre. When it is at the top it is going the same direction and speed as you. As it rolls forward it starts to descend, so forward speed is slowing. At 90 the degree point it has stopped moving forwards and starts to roll under the level of the axis. It then rolls back under the axis and after touching the floor, it will start to rise and also move forwards again.

Myksyk

Hmm interesting ... I suppose any particular point on the wheel is travelling back towards you at for at least half of its revolution before heading away from you again. Does this mean that it has to go twice as fast to keep up with you?

smacl

False, the distance covered (and hence speed) is covered by the surface of your tyre. You cover 70 kilometres of road in an hour, so does the contact point of the tyre.

Note if you start on the top of a hill, and finish on the flat, the tyre actually covers very slightly more distance than you do, as you're travelling on the inside of a vertical curve (sag). This gets balanced out as you climb up and go over the top of the next hill (hog).

triggermortis

triggermortis wrote: »

But it is rotating around an axis and you are not.
Pick a point on the tyre. When it is at the top it is going the same direction and speed as you. As it rolls forward it starts to descend, so forward speed is slowing. At 90 the degree point it has stopped moving forwards and starts to roll under the level of the axis. It then rolls back under the axis and after touching the floor, it will start to rise and also move forwards again.

Forward speed is slowing, but the speed itself isn't. Even though it's changing direction, it's still spinning at the same speed... no?

serendip

The exact top of the wheel is travelling at 2x. The exact bottom of the wheel is stationary.

First think of the wheel rotating at the requisite speed but stationary. Then add on the x km/h of the forward movement.

Beasty

If both "road speed" and "forward velocity" are determined on the same basis by reference to "horizontal velocity", because you are travelling downhill the "top" of the wheel is not opposite the bit touching the road - it's slightly behind it. Hence although I agree with serendip's analysis that the bit exactly opposite where the wheel touches the ground is travelling at 140kph, the top would appear to be to be going a bit slower (although accelerating towards 140kph)

Diarmuid

If there's a reflective sash in the mix, any answer which doesn't reference special relativity is not accurate enough

Lumen

There is no such thing as "forward velocity".

Velocity is a vector and therefore includes a direction component.

Wishbone Ash

Beasty wrote: »

If both "road speed" and "forward velocity" are determined on the same basis by reference to "horizontal velocity", because you are travelling downhill the "top" of the wheel is not opposite the bit touching the road - it's slightly behind it. Hence although I agree with serendip's analysis that the bit exactly opposite where the wheel touches the ground is travelling at 140kph, the top would appear to be to be going a bit slower (although accelerating towards 140kph)

serendip

I take the top to be the point on the wheel opposite that which is touching the ground. The question is more interesting with that definition.

How might you account for the OPs proposal that the top may have a speed greater than 2x?

The best I can come up with is ...

It is not a point that is in contact with the road, but a short strip of the tire, perhaps a couple of centimetres long. That strip is flattened into a straight line. It's like a chord on a circle, and it's length is less than that of the corresponding arc. Therefore, the effective diameter of the tire is actually slightly reduced, so the wheel has to rotate slightly faster than you would expect if you only considered the uncompressed diameter. So the uncompressed top of the tire is travelling slightly faster than 2x.

kenmc

has it anything to do with dx/dy? how about the big "S" or the giant "E" things?
I lost interest in maths when the only numbers visible were the page numbers

The tax man

How quick did my finger hit the back button on my keyboard.

seamus

It's all relative innit.

As smacl says, the tyre itself is spinning at 70km/h. But that's the velocity of the point as it orbits the hub. I

If you think of it in terms of X and Y vectors relative to the hub, then the hub is point zero, motionless, while the wheel spins at 70km/h. Note my terminology and notation may be completely wrong here.

At the top of the wheel, the X vector is at a maximum (i.e. all of the motion is positive in the X plane, motion in the Y plane is zero). At the bottom of the wheel, the X vector is at a minimum (it's negative) and Y is again zero. So at the top of the wheel, relative to the hub, the point's velocity is (70,0). At the bottom of the wheel it's (-70, 0).

Equally, at the "back" of the wheel, the X component is at zero (0,-70) because the tyre is not moving in the X plane.

So we can agree that when the hub is motionless and the wheel is spinning at 70km/h, then the point at the top of the tyre is moving "forward" at 70km/h.

Now, set the hub moving at 70km/h, relative to the ground. The wheel remains moving at 70km/h relative to the hub, but now the point at the top is moving at 140km/h relative to the ground.

Lumen

serendip wrote: »

It is not a point that is in contact with the road, but a short strip of the tire, perhaps a couple of centimetres long. That strip is flattened into a straight line. It's like a chord on a circle, and it's length is less than that of the corresponding arc. Therefore, the effective diameter of the tire is actually slightly reduced, so the wheel has to rotate slightly faster than you would expect if you only considered the uncompressed diameter

The diameter may be reduced but the circumference isn't. Rotation speed depends on circumference.

serendip

Lumen wrote: »

The diameter may be reduced but the circumference isn't. Rotation speed depends on circumference.

Claim: The effective radius is the distance from the centre of the hub to the centre of the chord formed by the contact with the road. This is less than the distance from the centre of the hub to the top of the wheel (which you might call something like the "apparent radius").

Imagine an extreme and silly example: a regular wheel of some size surrounded by some soft foam forming, if you like, an outer wheel edge (with a greater radius). For a fixed speed, the rate of rotation is determined by the inner, real wheel. The foam just collapses under the weight of the bike/rider into a flat chord in contact with the ground.

So, what may seem like a big wheel, actually behaves like a smaller wheel.

Returning to the real world, we see the same effect (just a lot smaller) due to the chord in contact with the ground. So the effective radius is smaller. So the wheel rotates faster than would appear necessary given its apparent radius. So the top of the wheel (which is the apparent radius from the centre) moves faster.

Edit: If the diameter is reduced, then the circumference is reduced.

ednwireland

are helmets compulsory on this thread ?

marketty

ednwireland wrote: »

are helmets compulsory on this thread ?

Too late my head already hurts

Lumen

serendip wrote: »

If the diameter is reduced, then the circumference is reduced.

Only for a circular "wheel".

What you're suggesting is the equivalent of "rolling circumference depends on inflation pressure". It doesn't, because the tread of a tyre does not stretch to any significant degree when you inflate it.

If the tread of a tyre is 2m long, it doesn't matter what shape (diameter) you contort it into, it will still take 2m to complete one revolution.

smacl

seamus wrote: »

So we can agree that when the hub is motionless and the wheel is spinning at 70km/h, then the point at the top of the tyre is moving "forward" at 70km/h

In a manner of speaking. It's moving in an orbit around the hub. The horizontal component of that rotational movement is the sine of the bearing of the point on the wheel from the hub. i.e. when the point on the wheel is directly above the hub the rotational velocity and horizontal velocity are the same. So for a stationary wheel, a point on the tyre moves horizontally backwards and forwards by the radius of the tyre.

(This was so much easier when we just argued about helmets and hi-viz on a Friday. Come back monument, all is forgiven).

hardCopy

This is getting more complicated than the pedalling backwards thread.

Crippens1

hardCopy wrote: »

This is getting more complicated than the pedalling backwards thread.

The pedalling backwards thread was funny.

This one is painful.

Is it too early for cake?

serendip

Lumen wrote: »

What you're suggesting is the equivalent of "rolling circumference depends on inflation pressure". It doesn't, because the tread of a tyre does not stretch to any significant degree when you inflate it.

I think it does become a little scrunched up.

Wishbone Ash

ednwireland wrote: »

are helmets compulsory on this thread ?

Bunnyhopper

Lusk Doyle

The only way for me to figure this out is to go for a cycle. I'll report back with my findings.

sy

As already described by niceonetom

(I am a visual learner but no PowerPoint in my day )

ford2600

Keep_Her_Lit wrote: »

Boost your workplace productivity by starting the day with a little mental stimulation, courtesy of this cycling related teaser ...

Is the following claim true or false? Explain your answer.

Claim : For a bicycle travelling at a road speed of x kph, the forward velocity of the top of each wheel is greater than 2x kph.

Example : If you are hurtling towards the bottom of a hill at 70kph, the tops of your wheels are each headed there at more than 140kph.

True or false?

Hints

• The claim contains no semantic trickery or other obscure sneakiness.
• Just take it at face value.
• Don't google it, think about it!

The prize for the first correct answer with a satisfactory explanation will be a c.20 year old used Sam Browne reflective sash with an irritating buckle that tends to open while riding.

Gotta dash now. Will pop back at lunchtime.

Slightly confused question.

The linear velocity (a vector with a magnitude and direction)of a point on tire is constantly changing as wheel rotates.
This has a horizontal and vertical component.

At top of wheel the horizontal component is xkm/h and the vertical component is zero.

When it reaches bottom the hirizontal component is -xkm/h.

The rotational velocity never changes.

Hope this helps

Lumen

ford2600 wrote: »

The linear velocity (a vector with a magnitude and direction)of a point on tire is constantly changing as wheel rotates...The rotational velocity never changes.

There are no such things as "linear velocity" and "rotational velocity". There is only velocity.

Also, your analysis uses within the reference frame of the bicycle, whereas the original puzzle uses the reference frame of the ground.

In the bicycle reference frame, the velocity of the rim constantly changes as it rotates. Changes to velocity are known as "acceleration". The rim is being accelerated around the hub due to the combination of forward force from the dropouts, tension in the spokes and resistance/reaction forces at the contact patch.

Beasty

sy wrote: »

As already described by niceonetom

(I am a visual learner but no PowerPoint in my day )

But serendip's point is it's not quite a circle, with the bit touching the road having a marginally shorter "radius" than the bit exactly opposite, which is where I presume the OP is coming from.

However I would still argue that the bit he is referring to is not actually the top of the wheel when cycling downhill. Obviously the calculation then depends on the angle of slope and degree of tyre compression, meaning it's possible to envisage scenarios where the top is going at less than 140kph, or more than 140kph or indeed exactly 140kph.

So the answer is yes, it can be true or false

dave_o_brien

Lumen wrote: »

There are no such things as "linear velocity" and "rotational velocity". There is only velocity.

Also, your analysis uses within the reference frame of the bicycle, whereas the original puzzle uses the reference frame of the ground.

In the bicycle reference frame, the velocity of the rim constantly changes as it rotates. Changes to velocity are known as "acceleration". The rim is being accelerated around the hub due to the combination of forward force from the dropouts, tension in the spokes and resistance/reaction forces at the contact patch.

There is such a thing as linear velocity. There is also a thing called angular velocity, which is what I'm sure he meant by rotational velocity.

And the frame of reference thing is misleading. The velocity of the wheel-bicycle-cyclist system is easily calculated. The speed of the velocity of the wheel at any point in the rotation is also easily calculated. They are different things though.

flatface

do I win the sash?

el tel

Statement is False.

The the forward velocities (i.e. foward vectors) of rotation and motion (of the wheel/bike) are summed i.e. 2x kph

However it won't be going greater than 2x kmh.

It will be close, but not greater.

dave_o_brien

flatface wrote: »

do I win the sash?

Clearly not if she's around...

velo.2010

Reminds me of the airplane on a threadmill scenario. Maybe give those guys at Mythbusters a call to explain this riddle to those who don't grasp it!

Keep_Her_Lit

It seems there's life in this thread yet ... carry on!

BTW, the contributions are very good, for a variety of reasons

Beasty

Keep_Her_Lit wrote: »

It seems there's life in this thread yet ... carry on!

BTW, the contributions are very good, for a variety of reasons

Are you getting us to do your homework for you??

tfrancer

Beasty wrote: »

Are you getting us to do your homework for you??

I think that this problem depends on the semantics. The "top" of the wheel is only the "top" instantaneously so the forward velocity at that point is the same as the other parts of the bike and the rider. Alternatively it could be considered as zero since since in theory at any point in time the velocity measured instantaneously is zero!

serendip

Beasty wrote: »

However I would still argue that the bit he is referring to is not actually the top of the wheel when cycling downhill.

OP: Could you clarify this?

el tel

Keep_Her_Lit wrote: »

The prize for the first correct answer with a satisfactory explanation will be a c.20 year old used Sam Browne reflective sash with an irritating buckle that tends to open while riding.

Gotta dash now. Will pop back at lunchtime.

Right.

You said there'd be a "prize for the first correct answer with a satisfactory explanation".

There have been a number of correct answers, with seeming satisfactory explanations.

Can you explain why the explanations given are not satisfactory and why you have not identified the first correct answer.

Yeah, answer that saying as you are so clever! :D

tfrancer

tfrancer wrote: »

I think that this problem depends on the semantics. The "top" of the wheel is only the "top" instantaneously so the forward velocity at that point is the same as the other parts of the bike and the rider. Alternatively it could be considered as zero since since in theory at any point in time the velocity measured instantaneously is zero!

I now think that the correct answer is that it depends on the size of the wheel. And you cannot really compare the forward velocity of the rider, measured in kph, with the rotational velocity of a point on the wheel, measured in radians or degrees. Another relevant question to which I do not know the answer: are the rotational velocities the same on the front and back wheels of a penny-farthing bike?

ford2600

V= wr
V in m/s
W radians/sec
R in m
V is same for front and back wheel and angular velocity proportional to rsdii

tfrancer wrote: »

I now think that the correct answer is that it depends on the size of the wheel. And you cannot really compare the forward velocity of the rider, measured in kph, with the rotational velocity of a point on the wheel, measured in radians or degrees. Another relevant question to which I do not know the answer: are the rotational velocities the same on the front and back wheels of a penny-farthing bike?

Keep_Her_Lit

el tel wrote: »

Right.

You said there'd be a "prize for the first correct answer with a satisfactory explanation".

There have been a number of correct answers, with seeming satisfactory explanations.

Can you explain why the explanations given are not satisfactory and why you have not identified the first correct answer.

Yeah, answer that saying as you are so clever! :D

Sorry el tel, I think you got the wrong end of the stick there. I didn't mean to imply that the correct answer and explanation hadn't been provided. I just observed that the discussion was still lively and entertaining and that there was no need to conclude it just yet. But you're right, I should honour my original committment and respond when I said I would. I can't get into a long discussion right now, as I have to get back to work. I'll return this evening if needed.

The claim is ... TRUE
And the Lucky Winner is ... serendip

And yes, that delapidated Sam Browne belt really does exist and I will happily hand it over. serendip, please PM me to agree details for the prize-giving ceremony.:D

niceonetom also deserves an honourable mention for being so quick off the mark with his lucid and concsie explanation. Pity he didn't follow up to deal with the "more than" element of the puzzle.

But serendip demonstrated a clear understanding of the issue from the outset and quickly recognised that the loaded and unloaded radii of the wheel differ and that this necessarily produces higher surface speeds for the unloaded portion of the tyre.


So for the example given, just how much faster than 140 kph might the top of the wheel be travelling?

• Let's assume a 700 x 25 tyre, which might have an unloaded radius of approximately 340mm.
• If it's properly inflated, the loaded radius will be less than the unloaded radius by roughly 15% of the tyre width, which is 3.75mm.
• So the radius which determines the wheel's angular velocity (units: rad/s) is 336.25mm.
• Consequently, the surface speed of the unloaded portion of the tyre will be 1.011 (340.0/336.25) times that of the loaded portion, i.e. 70.78kph
• So, when a point on the tyre is directly opposite the centre of the contact patch below, it will be moving in the same direction as the bike at a speed of 140.78kph.

Regarding the issue of gradient vs flat, I didn't mean to muddy the waters by introducing a downhill example. The bigger the numbers, the more impressive the claim sounds ... 140 kph sounds better than 80kph ... hence I chose downhill speeds. If anyone here can do 70 kph on the flat, hats off to you!

So, by "forward", all that is meant is "parallel to the surface on which the bike is travelling".

Beasty

As an additional "prize", serendip gets to start next Friday's thread - it must be cycling related, and posted in After Hours

Lusk Doyle

I think I want to kill myself after this thread. Can I get an upgrade?

el tel

Keep_Her_Lit wrote: »

Sorry el tel, I think you got the wrong end of the stick there. I didn't mean to imply that the correct answer and explanation hadn't been provided.

Ahh was just yanking your chain, thanks for your answer!

HOWEVER.

Nowhere in the original riddle did you mention a bike tyre.

The question posed only asked re. the speed of the top of the bike WHEEL moving at 2x speed of the bike.

The answer given is the answer to a different question: i.e. what is the speed of the top of the tyre.

In particular, the answer is includes the following statement:

"So, when a point on the tyre is directly opposite the centre of the contact patch below, it will be moving in the same direction as the bike at a speed of 140.78kph".

A bike wheel is quite distinct from a bike tyre. The top of a tyre fitted to the wheel is not the same as "the top of the wheel".

I rest my case.:p

Beasty

el tel wrote: »

Ahh was just yanking your chain, thanks for your answer!

HOWEVER.

Nowhere in the original riddle did you mention a bike tyre.

The question posed only asked re. the speed of the top of the bike WHEEL moving at 2x speed of the bike.

The answer given is the answer to a different question: i.e. what is the speed of the top of the tyre.

In particular, the answer is includes the following statement:

"So, when a point on the tyre is directly opposite the centre of the contact patch below, it will be moving in the same direction as the bike at a speed of 140.78kph".

A bike wheel is quite distinct from a bike tyre. The top of a tyre fitted to the wheel is not the same as "the top of the wheel".

I rest my case.:p

Under the OP's analysis you get to the same answer - the "top" of the tyre is going at the same speed as the "top" of the wheel

el tel

Beasty wrote: »

Under the OP's analysis you get to the same answer - the "top" of the tyre is going at the same speed as the "top" of the wheel

But what if you take away the tyre. There was no tyre in the question, only a wheel.

Or something.

Beasty

el tel wrote: »

But what if you take away the tyre. There was no tyre in the question, only a wheel.

Or something.

Same analysis if there is any "give" whatsoever in the wheel ...