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.