Durable Rims for Road Wheels - Recommendations?

[[email protected]]

On Mon, 21 Aug 2006 22:12:09 -0400, RonSonic
<[email protected]> wrote:

>On Mon, 21 Aug 2006 19:18:08 -0600, [email protected] wrote:
>
>>On 21 Aug 2006 16:06:43 -0700, [email protected] wrote:
>>
>>>[email protected] wrote:
>>>> On 21 Aug 2006 15:32:26 -0700, [email protected] wrote:
>>>>
>>>> >[email protected] wrote:
>>>> >> On Mon, 21 Aug 2006 21:24:38 +0000 (UTC), Booker C. Bense
>>>> >> <bbense+rec.bicycles.misc.rec.bicycles.tech.Aug.21.06@telemark.slac.stanford.edu>
>>>> >> wrote:
>>>> >>
>>>> >> >-----BEGIN PGP SIGNED MESSAGE-----
>>>> >> >
>>>> >> >In article <[email protected]>,
>>>> >> > <[email protected]> wrote:
>>>> >> >>[email protected] wrote:
>>>> >> >>> On 17 Aug 2006 16:16:09 -0700, [email protected] wrote:
>>>> >> >>>
>>>> >> >>
>>>> >> >>Back to the first point I raise, you still claim that because it's only
>>>> >> >>a 3 or 5% change, it's a surprise that Jim feels it, an argument which
>>>> >> >>has a huge gap
>>>> >> >>
>>>> >> >
>>>> >> >_ It's also pointless unless Jim is using exactly the same rims
>>>> >> >and tires. My guess is that the difference he notices has way
>>>> >> >more to do with rim depth and shape than the number of spokes.
>>>> >> >
>>>> >> >_ Booker C. Bense
>>>> >
>>>> >Dear Carl,
>>>> >
>>>> >> A slightly wider or thinner ring (different rims) might not make much
>>>> >> difference to this fan in a side wind.
>>>> >
>>>> >What are your numbers behind this statement?
>>>> >
>>>> >Doug
>>>>
>>>> Dear Doug,
>>>>
>>>> A ring/rim has the effectively same side resistance, whether it's
>>>> spinning or not.
>>>>
>>>
>>>Yes, Dear Carl, but... Jim is feeling the resistance that the wheel
>>>would have even were it not spinning. Which is five times the
>>>resistance that comes from the spokes, according to your own
>>>calculations of surface area. Added to that, he is also feeling any
>>>increases or differences due to forward or rotational speed.
>>>
>>>You've come full circle in your arguments - you yourself counted that
>>>79,800 mm^2 in your argument that the change is only 3-5%.
>>>
>>>Since the area of the rim+tire is approx 4 times that of the spokes,
>>>Jim will surely find it (a change in tire or rim size) affecting how
>>>much harder it is to control in a crosswind.
>>>
>>>Remember, it was a change of wheel resulted in whatever that Jim felt,
>>>not a change in bike speed.
>>>
>>>Cheers,
>>>Doug
>>
>>Doug,
>>
>>Jim feels a strong difference at roughly the same speed between two
>>front wheels in gusty side winds.
>>
>>If the effect of the side wind is just a matter of surface area, then
>>the feeling is likely to an illusion--expection coupled with highly
>>variable winds. After all the difference in surface area appears to be
>>only 3% to 5%.
>>
>>But if the effect of the side wind is more a matter of the number of
>>blades whirling in the wind like a fan, then the rim surface becomes
>>trivial and the 25% reduction in the number of blades suggests that
>>Jim is feeling a clear effect.
>>
>>Viewed from the side, whirling fan blades are obviously different than
>>a smooth, spinning rim surface, so the question is whether thin spokes
>>in the peculiar turbulence of the wheel generate enough force to have
>>a significant effect.
>>
>>The blades on the fan in your computer have an effect, but not enough
>>to affect a rider's balance. The "blades" in a wheel are much thinner,
>>are shaped differently, sweep a much larger area, and are very
>>peculiar in that they are rolling instead of stationary.
>>
>>Do they have a big enough aerodynamic effect to explain what Jim
>>feels? I don't know. But those are the two possible explanations that
>>I can see.
>>
>>If you find a wind tunnel test for gusty wind effects, let us know.
>
>Very simply, because that's all I can do, for all practical purposes the area of
>an object includes the mass of roiled air attached to it. If the spoke has less
>drag going forward it will have a much smaller profile to the side winds. The
>crosswind "sees" the wheel components as a single piece with the wad of dirty
>air in and around it.
>
>Need more, ask an aerodynamicist and be prepared to take on a new study.
>
>Ron

Dear Ron,

A solid disk front is universally agreed to be a bad idea in a
crosswind.

So the question still remains: does a 32-spoke wheel look
significantly more turbulent to a gusty crosswind than a 24-spoke
wheel as Jim rides across the Golden Gate?

That is, how much dirtier is the wad of air whipped up inside a 32
spoke wheel, versus a 24-spoke wheel?

Cheers,

Carl Fogel

 

[[email protected]]

> Dear Doug,
>
> There's a nice graph on page 3.
>
> What happens at 30 mph to the drag force in pounds as the angle of the
> wind changes from 0 to 20 degrees?
>
> What happens after that, as the wind angle changes from 20 to 30
> degrees?
>
> Those two contradictory trends surprised me.

Dear Carl,

Well, if the wind was all coming from the front, then it would be
hindering the spokes moving forward, but helping the spokes which are
on the bottom half of the rotatation. So it will tend to slow down
the spokes in the top half of the wheel, but not speed up the spokes in
the bottom half, since they are not free to move backwards, being fixed
to the road. So it hinders more than it helps.

A wind coming from the side, however, cares not about the direction of
the movement of the spokes in the inline plane, and helps or hinders
all 36 the same. Apparently it hinders. Since more spokes are visible
from the side then from the front, there are more spokes affected by
wind from 90 than from 0 degrees.

As the angle of attack of the wind changes from 0 to 20 degrees, the
effect of hindrance on the spokes in the top half of the wheel drops
off, and the graph dips. As the angle of attack increases past 20, the
effect it has on all spokes increases, and the graph rises.

Cheers,
Doug

 

[[email protected]]

On 21 Aug 2006 22:35:28 -0700, [email protected] wrote:

>> Dear Doug,
>>
>> There's a nice graph on page 3.
>>
>> What happens at 30 mph to the drag force in pounds as the angle of the
>> wind changes from 0 to 20 degrees?
>>
>> What happens after that, as the wind angle changes from 20 to 30
>> degrees?
>>
>> Those two contradictory trends surprised me.
>
>Dear Carl,
>
>Well, if the wind was all coming from the front, then it would be
>hindering the spokes moving forward, but helping the spokes which are
>on the bottom half of the rotatation. So it will tend to slow down
>the spokes in the top half of the wheel, but not speed up the spokes in
>the bottom half, since they are not free to move backwards, being fixed
>to the road. So it hinders more than it helps.
>
>A wind coming from the side, however, cares not about the direction of
>the movement of the spokes in the inline plane, and helps or hinders
>all 36 the same. Apparently it hinders. Since more spokes are visible
>from the side then from the front, there are more spokes affected by
>wind from 90 than from 0 degrees.
>
>As the angle of attack of the wind changes from 0 to 20 degrees, the
>effect of hindrance on the spokes in the top half of the wheel drops
>off, and the graph dips. As the angle of attack increases past 20, the
>effect it has on all spokes increases, and the graph rises.
>
>Cheers,
>Doug

Dear Doug,

Could be.

But look at the next graph down for the same 30 mph wind at angles
increasing to 30 degrees.

The 404 deep aero rim wheel shows the dip with 16, 20, 24, 0r 28
spokes.

A non-aero rim with 28 shows what might be a faint dip.

The non-aero rim with 32 spokes shows no dip.

Beastly tricky and surprising, like lots of aerodynamics.

A deeper rim is associated with an odd dip in straight-back drag as
the wind angle increases at 30 mph. The dip disappears with more
spokes on shallower rims. I'm baffled as to what the side forces must
be, but it seems reasonable to deduce that they're present.

Cheers,

Carl Fogel

 

[Donga]

daveornee wrote:
> Velocity Aerohead with Aerohead OC in rear makes a good combination for
> someone your weight.

I have dented two Aerohead OC on the drive side, where the rim is near
vertical and close to the edge of the tire. I'm guessing I hit a
'cateye' on the edge. Perhaps I didn't have my tires pumped up enough?
I don't think so - I pump my tires before most rides to 120lb, tho'
this was with an old Silca that might not be too accurate. Has anyone
else had this experience with the OC rims? I won't use them any more.

Donga

 

[RonSonic]

On Mon, 21 Aug 2006 20:30:48 -0600, [email protected] wrote:

>On Mon, 21 Aug 2006 22:12:09 -0400, RonSonic
><[email protected]> wrote:
>
>>On Mon, 21 Aug 2006 19:18:08 -0600, [email protected] wrote:
>>
>>>On 21 Aug 2006 16:06:43 -0700, [email protected] wrote:
>>>
>>>>[email protected] wrote:
>>>>> On 21 Aug 2006 15:32:26 -0700, [email protected] wrote:
>>>>>
>>>>> >[email protected] wrote:
>>>>> >> On Mon, 21 Aug 2006 21:24:38 +0000 (UTC), Booker C. Bense
>>>>> >> <bbense+rec.bicycles.misc.rec.bicycles.tech.Aug.21.06@telemark.slac.stanford.edu>
>>>>> >> wrote:
>>>>> >>
>>>>> >> >-----BEGIN PGP SIGNED MESSAGE-----
>>>>> >> >
>>>>> >> >In article <[email protected]>,
>>>>> >> > <[email protected]> wrote:
>>>>> >> >>[email protected] wrote:
>>>>> >> >>> On 17 Aug 2006 16:16:09 -0700, [email protected] wrote:
>>>>> >> >>>
>>>>> >> >>
>>>>> >> >>Back to the first point I raise, you still claim that because it's only
>>>>> >> >>a 3 or 5% change, it's a surprise that Jim feels it, an argument which
>>>>> >> >>has a huge gap
>>>>> >> >>
>>>>> >> >
>>>>> >> >_ It's also pointless unless Jim is using exactly the same rims
>>>>> >> >and tires. My guess is that the difference he notices has way
>>>>> >> >more to do with rim depth and shape than the number of spokes.
>>>>> >> >
>>>>> >> >_ Booker C. Bense
>>>>> >
>>>>> >Dear Carl,
>>>>> >
>>>>> >> A slightly wider or thinner ring (different rims) might not make much
>>>>> >> difference to this fan in a side wind.
>>>>> >
>>>>> >What are your numbers behind this statement?
>>>>> >
>>>>> >Doug
>>>>>
>>>>> Dear Doug,
>>>>>
>>>>> A ring/rim has the effectively same side resistance, whether it's
>>>>> spinning or not.
>>>>>
>>>>
>>>>Yes, Dear Carl, but... Jim is feeling the resistance that the wheel
>>>>would have even were it not spinning. Which is five times the
>>>>resistance that comes from the spokes, according to your own
>>>>calculations of surface area. Added to that, he is also feeling any
>>>>increases or differences due to forward or rotational speed.
>>>>
>>>>You've come full circle in your arguments - you yourself counted that
>>>>79,800 mm^2 in your argument that the change is only 3-5%.
>>>>
>>>>Since the area of the rim+tire is approx 4 times that of the spokes,
>>>>Jim will surely find it (a change in tire or rim size) affecting how
>>>>much harder it is to control in a crosswind.
>>>>
>>>>Remember, it was a change of wheel resulted in whatever that Jim felt,
>>>>not a change in bike speed.
>>>>
>>>>Cheers,
>>>>Doug
>>>
>>>Doug,
>>>
>>>Jim feels a strong difference at roughly the same speed between two
>>>front wheels in gusty side winds.
>>>
>>>If the effect of the side wind is just a matter of surface area, then
>>>the feeling is likely to an illusion--expection coupled with highly
>>>variable winds. After all the difference in surface area appears to be
>>>only 3% to 5%.
>>>
>>>But if the effect of the side wind is more a matter of the number of
>>>blades whirling in the wind like a fan, then the rim surface becomes
>>>trivial and the 25% reduction in the number of blades suggests that
>>>Jim is feeling a clear effect.
>>>
>>>Viewed from the side, whirling fan blades are obviously different than
>>>a smooth, spinning rim surface, so the question is whether thin spokes
>>>in the peculiar turbulence of the wheel generate enough force to have
>>>a significant effect.
>>>
>>>The blades on the fan in your computer have an effect, but not enough
>>>to affect a rider's balance. The "blades" in a wheel are much thinner,
>>>are shaped differently, sweep a much larger area, and are very
>>>peculiar in that they are rolling instead of stationary.
>>>
>>>Do they have a big enough aerodynamic effect to explain what Jim
>>>feels? I don't know. But those are the two possible explanations that
>>>I can see.
>>>
>>>If you find a wind tunnel test for gusty wind effects, let us know.
>>
>>Very simply, because that's all I can do, for all practical purposes the area of
>>an object includes the mass of roiled air attached to it. If the spoke has less
>>drag going forward it will have a much smaller profile to the side winds. The
>>crosswind "sees" the wheel components as a single piece with the wad of dirty
>>air in and around it.
>>
>>Need more, ask an aerodynamicist and be prepared to take on a new study.
>>
>>Ron
>
>Dear Ron,
>
>A solid disk front is universally agreed to be a bad idea in a
>crosswind.
>
>So the question still remains: does a 32-spoke wheel look
>significantly more turbulent to a gusty crosswind than a 24-spoke
>wheel as Jim rides across the Golden Gate?
>
>That is, how much dirtier is the wad of air whipped up inside a 32
>spoke wheel, versus a 24-spoke wheel?

Empirically, enough.

Is it just the spoke count? Apparently not.

Ron

 

[[email protected]]

On Tue, 22 Aug 2006 08:58:25 -0400, RonSonic
<[email protected]> wrote:

>On Mon, 21 Aug 2006 20:30:48 -0600, [email protected] wrote:
>
>>On Mon, 21 Aug 2006 22:12:09 -0400, RonSonic
>><[email protected]> wrote:
>>
>>>On Mon, 21 Aug 2006 19:18:08 -0600, [email protected] wrote:
>>>
>>>>On 21 Aug 2006 16:06:43 -0700, [email protected] wrote:
>>>>
>>>>>[email protected] wrote:
>>>>>> On 21 Aug 2006 15:32:26 -0700, [email protected] wrote:
>>>>>>
>>>>>> >[email protected] wrote:
>>>>>> >> On Mon, 21 Aug 2006 21:24:38 +0000 (UTC), Booker C. Bense
>>>>>> >> <bbense+rec.bicycles.misc.rec.bicycles.tech.Aug.21.06@telemark.slac.stanford.edu>
>>>>>> >> wrote:
>>>>>> >>
>>>>>> >> >-----BEGIN PGP SIGNED MESSAGE-----
>>>>>> >> >
>>>>>> >> >In article <[email protected]>,
>>>>>> >> > <[email protected]> wrote:
>>>>>> >> >>[email protected] wrote:
>>>>>> >> >>> On 17 Aug 2006 16:16:09 -0700, [email protected] wrote:
>>>>>> >> >>>
>>>>>> >> >>
>>>>>> >> >>Back to the first point I raise, you still claim that because it's only
>>>>>> >> >>a 3 or 5% change, it's a surprise that Jim feels it, an argument which
>>>>>> >> >>has a huge gap
>>>>>> >> >>
>>>>>> >> >
>>>>>> >> >_ It's also pointless unless Jim is using exactly the same rims
>>>>>> >> >and tires. My guess is that the difference he notices has way
>>>>>> >> >more to do with rim depth and shape than the number of spokes.
>>>>>> >> >
>>>>>> >> >_ Booker C. Bense
>>>>>> >
>>>>>> >Dear Carl,
>>>>>> >
>>>>>> >> A slightly wider or thinner ring (different rims) might not make much
>>>>>> >> difference to this fan in a side wind.
>>>>>> >
>>>>>> >What are your numbers behind this statement?
>>>>>> >
>>>>>> >Doug
>>>>>>
>>>>>> Dear Doug,
>>>>>>
>>>>>> A ring/rim has the effectively same side resistance, whether it's
>>>>>> spinning or not.
>>>>>>
>>>>>
>>>>>Yes, Dear Carl, but... Jim is feeling the resistance that the wheel
>>>>>would have even were it not spinning. Which is five times the
>>>>>resistance that comes from the spokes, according to your own
>>>>>calculations of surface area. Added to that, he is also feeling any
>>>>>increases or differences due to forward or rotational speed.
>>>>>
>>>>>You've come full circle in your arguments - you yourself counted that
>>>>>79,800 mm^2 in your argument that the change is only 3-5%.
>>>>>
>>>>>Since the area of the rim+tire is approx 4 times that of the spokes,
>>>>>Jim will surely find it (a change in tire or rim size) affecting how
>>>>>much harder it is to control in a crosswind.
>>>>>
>>>>>Remember, it was a change of wheel resulted in whatever that Jim felt,
>>>>>not a change in bike speed.
>>>>>
>>>>>Cheers,
>>>>>Doug
>>>>
>>>>Doug,
>>>>
>>>>Jim feels a strong difference at roughly the same speed between two
>>>>front wheels in gusty side winds.
>>>>
>>>>If the effect of the side wind is just a matter of surface area, then
>>>>the feeling is likely to an illusion--expection coupled with highly
>>>>variable winds. After all the difference in surface area appears to be
>>>>only 3% to 5%.
>>>>
>>>>But if the effect of the side wind is more a matter of the number of
>>>>blades whirling in the wind like a fan, then the rim surface becomes
>>>>trivial and the 25% reduction in the number of blades suggests that
>>>>Jim is feeling a clear effect.
>>>>
>>>>Viewed from the side, whirling fan blades are obviously different than
>>>>a smooth, spinning rim surface, so the question is whether thin spokes
>>>>in the peculiar turbulence of the wheel generate enough force to have
>>>>a significant effect.
>>>>
>>>>The blades on the fan in your computer have an effect, but not enough
>>>>to affect a rider's balance. The "blades" in a wheel are much thinner,
>>>>are shaped differently, sweep a much larger area, and are very
>>>>peculiar in that they are rolling instead of stationary.
>>>>
>>>>Do they have a big enough aerodynamic effect to explain what Jim
>>>>feels? I don't know. But those are the two possible explanations that
>>>>I can see.
>>>>
>>>>If you find a wind tunnel test for gusty wind effects, let us know.
>>>
>>>Very simply, because that's all I can do, for all practical purposes the area of
>>>an object includes the mass of roiled air attached to it. If the spoke has less
>>>drag going forward it will have a much smaller profile to the side winds. The
>>>crosswind "sees" the wheel components as a single piece with the wad of dirty
>>>air in and around it.
>>>
>>>Need more, ask an aerodynamicist and be prepared to take on a new study.
>>>
>>>Ron
>>
>>Dear Ron,
>>
>>A solid disk front is universally agreed to be a bad idea in a
>>crosswind.
>>
>>So the question still remains: does a 32-spoke wheel look
>>significantly more turbulent to a gusty crosswind than a 24-spoke
>>wheel as Jim rides across the Golden Gate?
>>
>>That is, how much dirtier is the wad of air whipped up inside a 32
>>spoke wheel, versus a 24-spoke wheel?
>
>Empirically, enough.
>
>Is it just the spoke count? Apparently not.
>
>Ron

Dear RonSonic

Hard to keep the Ron's straight here.

Elsewhere in this thread, Ron Ruff found a very nice link to a side
force article with a graph well down that suggests that spoke count is
much less important than rim differences:

http://groups.google.com/group/rec.bicycles.tech/msg/c01ecef22aeb16cb

The link above to my post includes Ron Ruff's post, link, and the
related link.

Cheers,

Carl Fogel

 

[Ron Ruff]

jim beam wrote:
> that may well be the case, but if i have two wheels, same rim profile,
> one high spoke count, the other low, the comparison for /spokes only/ is
> direct and easy.

I was just thinking... side force is one parameter, but wouldn't it be
more important to know where it is relative to the steering axis? Seems
like that is the thing that would upset your handling. If the side
force is equal fore and aft of the steering axis it wouldn't turn the
wheel, but if the center of pressure was towards the front it sure
would!

So the question is... what configuration is most likely to give you the
biggest *unbalanced* side force... or net moment on the steering axis?
I'm still thinking the spokes won't matter much... but it is much too
late to think...

 

[jim beam]

Ron Ruff wrote:
> jim beam wrote:
>> that may well be the case, but if i have two wheels, same rim profile,
>> one high spoke count, the other low, the comparison for /spokes only/ is
>> direct and easy.
>
> I was just thinking... side force is one parameter, but wouldn't it be
> more important to know where it is relative to the steering axis? Seems
> like that is the thing that would upset your handling. If the side
> force is equal fore and aft of the steering axis it wouldn't turn the
> wheel, but if the center of pressure was towards the front it sure
> would!
>
> So the question is... what configuration is most likely to give you the
> biggest *unbalanced* side force... or net moment on the steering axis?
> I'm still thinking the spokes won't matter much... but it is much too
> late to think...
>
i haven't considered that because it's not a variable for me - this all
on the same bike/fork. on the spoke count is different between wheels.

 

[dvt]

Ron Ruff wrote:
> jim beam wrote:
>> that may well be the case, but if i have two wheels, same rim profile,
>> one high spoke count, the other low, the comparison for /spokes only/ is
>> direct and easy.
>
> I was just thinking... side force is one parameter, but wouldn't it be
> more important to know where it is relative to the steering axis?

Yes... in other words, what is the steering *torque* induced by the
wind? I don't know how to estimate this.

--
Dave
dvt at psu dot edu

Everyone confesses that exertion which brings out all the powers of body
and mind is the best thing for us; but most people do all they can to
get rid of it, and as a general rule nobody does much more than
circumstances drive them to do. -Harriet Beecher Stowe, abolitionist and
novelist (1811-1896)

 

[[email protected]]

On Thu, 24 Aug 2006 13:10:29 -0400, dvt <[email protected]> wrote:

>Ron Ruff wrote:
>> jim beam wrote:
>>> that may well be the case, but if i have two wheels, same rim profile,
>>> one high spoke count, the other low, the comparison for /spokes only/ is
>>> direct and easy.
>>
>> I was just thinking... side force is one parameter, but wouldn't it be
>> more important to know where it is relative to the steering axis?
>
>Yes... in other words, what is the steering *torque* induced by the
>wind? I don't know how to estimate this.

Dear Dave,

It's going to be tricky, even when idealized.

Let's say the wheel is going 20 mph forward.

A sudden gust of wind at 20 mph hits it at 90 degrees.

A flare of turbulence should erupt from the far side of the wheel,
trailing backward sharply.

Some of the effect will be from torque trying to turn the wheel.

Some of the effect will be from the whole wheel being tilted over
sideways without turning--keep the wheel perfectly straight, tilt the
bike slightly, and the rider must either turn or topple.

At the moment the wind hits the wheel, the bottom of the bottom spoke
is motionless relative to the road, while the top of the top spoke is
doing 40 mph to one side.

The ends of the horizontal spokes are going forward at 20 mph and also
going either up or down at 20 mph.

Everything behind the leading point of the front tire is already is a
turbulent mess.

The mess in roughly the upper rear quarter of the wheel is complicated
by the turbulence off both fork legs.

The slower-moving hub ends of the dished spokes stick out past the
plane of the tire, while the faster-moving rim ends are tucked in.

At a certain angle, two of the four roughly horizontal spokes will be
pointing straight into the apparent wind, one leading, one trailing.

And so on . . . even a wind tunnel test will give only the side force
under fairly steady conditions, not what happens when the wind gusts
back and forth erratically with real-life abruptness.

But it's a fascinating question.

Cheers,

Carl Fogel

 

[[email protected]]

[email protected] wrote:
> On Thu, 24 Aug 2006 13:10:29 -0400, dvt <[email protected]> wrote:
>
> >Ron Ruff wrote:
> >> jim beam wrote:
> >>> that may well be the case, but if i have two wheels, same rim profile,
> >>> one high spoke count, the other low, the comparison for /spokes only/ is
> >>> direct and easy.
> >>
> >> I was just thinking... side force is one parameter, but wouldn't it be
> >> more important to know where it is relative to the steering axis?
> >
> >Yes... in other words, what is the steering *torque* induced by the
> >wind? I don't know how to estimate this.
>
> Dear Dave,

Dear Carl,

> Let's say the wheel is going 20 mph forward.
OK
> A sudden gust of wind at 20 mph hits it at 90 degrees.
OK
> A flare of turbulence should erupt from the far side of the wheel,
> trailing backward sharply.
OK
> Some of the effect will be from torque trying to turn the wheel.
OK
> Some of the effect will be from the whole wheel being tilted over
> sideways without turning--keep the wheel perfectly straight, tilt the
> bike slightly, and the rider must either turn or topple.
OK
> At the moment the wind hits the wheel, the bottom of the bottom spoke
> is motionless relative to the road, while the top of the top spoke is
> doing 40 mph to one side.

To the side? Not forward?

> The ends of the horizontal spokes are going forward at 20 mph and also
> going either up or down at 20 mph.

Are you sure? Forward at 20, yes, but up and down too?

Cheers,
Doug

 

[[email protected]]

On 24 Aug 2006 12:19:02 -0700, [email protected] wrote:

>
>[email protected] wrote:
>> On Thu, 24 Aug 2006 13:10:29 -0400, dvt <[email protected]> wrote:
>>
>> >Ron Ruff wrote:
>> >> jim beam wrote:
>> >>> that may well be the case, but if i have two wheels, same rim profile,
>> >>> one high spoke count, the other low, the comparison for /spokes only/ is
>> >>> direct and easy.
>> >>
>> >> I was just thinking... side force is one parameter, but wouldn't it be
>> >> more important to know where it is relative to the steering axis?
>> >
>> >Yes... in other words, what is the steering *torque* induced by the
>> >wind? I don't know how to estimate this.
>>
>> Dear Dave,
>
>Dear Carl,
>
>> Let's say the wheel is going 20 mph forward.
>OK
>> A sudden gust of wind at 20 mph hits it at 90 degrees.
>OK
>> A flare of turbulence should erupt from the far side of the wheel,
>> trailing backward sharply.
>OK
>> Some of the effect will be from torque trying to turn the wheel.
>OK
>> Some of the effect will be from the whole wheel being tilted over
>> sideways without turning--keep the wheel perfectly straight, tilt the
>> bike slightly, and the rider must either turn or topple.
>OK
>> At the moment the wind hits the wheel, the bottom of the bottom spoke
>> is motionless relative to the road, while the top of the top spoke is
>> doing 40 mph to one side.
>
>To the side? Not forward?
>
>> The ends of the horizontal spokes are going forward at 20 mph and also
>> going either up or down at 20 mph.
>
>Are you sure? Forward at 20, yes, but up and down too?
>
>Cheers,
>Doug

Dear Doug,

Let's leave the side wind for further down.

All figures are slightly high because the spokes don't reach the edge
of the tire--20 mph for the spokes is 20 mph minus a little bit.

Everything on the bicycle is moving forward relative to the ground at
20 mph.

The wheel is also rotating at enough rpm that its outer edge is doing
20 mph.

So the top of the top spoke is going 20 + 20 = 40 mph forward.

The bottom of the bottom spoke is going 20-20 = 0 mph forward.

The entire front horizontal spoke is going 20 mph forward, while its
rim end is also moving 20 mph straight down.

The entire rear horizontal spoke is also going 20 mph forward, but its
rim end is moving 20 mph straight up.

Okay, now look at the wheel from the side, the wind's point of view.

The bottom spoke's bottom end is motionless when viewed from the side,
since it's going 0 mph forward relative to the ground.

The top spoke's top end is doing 40 mph forward (to one side as viewed
by the side wind.)

The two horizontal spokes are moving 20 mph forward (to one side as
viewed by the side wind) and their rim ends are also moving either up
or down at 20 mph.

Of course, the closer you get to the hub, the more all spokes are
moving mostly forward at around 20 mph. The center of the axle is the
only thing that actually moves straight forward at 20 mph.

This is why I keep adding that if the spokes act as propellors, then
it's a very weird propellor. The face of a normal fan is stationary
relative to the wind and we can forget everything above--all the
blades are moving at the same speed relative to the wind that they
generate.

Bad as this explanation is, it could be worse, but I'll spare everyone
attempts at drawing pictures. Maybe someone knows of a nice set of
illustrations of spoke speeds relative to the ground.

Cheers,

Carl Fogel

 

[[email protected]]

> >> The ends of the horizontal spokes are going forward at 20 mph and also
> >> going either up or down at 20 mph.
> >
> >Are you sure? Forward at 20, yes, but up and down too?
> >
> >Cheers,
> >Doug
>
> Dear Doug,

Dear Carl,

> Let's leave the side wind for further down.
OK
> All figures are slightly high because the spokes don't reach the edge
> of the tire--20 mph for the spokes is 20 mph minus a little bit.
Okey Doke
> Everything on the bicycle is moving forward relative to the ground at
> 20 mph.

Disagree. The wheel, where it touches the ground, is not moving
forward at all.

> The wheel is also rotating at enough rpm that its outer edge is doing
> 20 mph.

At two points [on the outer edge] but not at the rest of it.

> So the top of the top spoke is going 20 + 20 = 40 mph forward.
>
> The bottom of the bottom spoke is going 20-20 = 0 mph forward.

The bottom of the bottom spoke is not 0-0, not 20-20.

> The entire front horizontal spoke is going 20 mph forward, while its
> rim end is also moving 20 mph straight down.

Really? I would have said that it is moving approximately 25mph down.
Remember that during one rotation of the wheel, any point on the
circumference moves forward by one circumference, so, 2piR, but up and
down by two diameters, so, 4R. The vertical distance travelled is less
than the horizontal distance, it's 2/pi (0.6366), times it, so it looks
doubtful that the number we are looking for is going to come out to 20
or 40.

The ratio of horizontal movement to vertical movement is 2piR to 4R, or
pi to 2. If you multiply 20 miles times pi/2, you get approximately
12.75. So a point on the circumference has to average 12.75 mph
vertically while averaging 20mph horizontally.

Furthermore, the point on the circumference speeds up and slows down
twice per rotation, as opposed to once in the forward direction. So it
has to accelerate to 25.5mph and then decelerate to zero once on the
way up and then do it all again on the way down.

Correct?

 

[[email protected]]

On 24 Aug 2006 13:56:08 -0700, [email protected] wrote:

>> >> The ends of the horizontal spokes are going forward at 20 mph and also
>> >> going either up or down at 20 mph.
>> >
>> >Are you sure? Forward at 20, yes, but up and down too?
>> >
>> >Cheers,
>> >Doug
>>
>> Dear Doug,
>
>Dear Carl,
>
>> Let's leave the side wind for further down.
>OK
>> All figures are slightly high because the spokes don't reach the edge
>> of the tire--20 mph for the spokes is 20 mph minus a little bit.
>Okey Doke
>> Everything on the bicycle is moving forward relative to the ground at
>> 20 mph.
>
>Disagree. The wheel, where it touches the ground, is not moving
>forward at all.
>
>> The wheel is also rotating at enough rpm that its outer edge is doing
>> 20 mph.
>
>At two points [on the outer edge] but not at the rest of it.
>
>> So the top of the top spoke is going 20 + 20 = 40 mph forward.
>>
>> The bottom of the bottom spoke is going 20-20 = 0 mph forward.
>
>The bottom of the bottom spoke is not 0-0, not 20-20.
>
>> The entire front horizontal spoke is going 20 mph forward, while its
>> rim end is also moving 20 mph straight down.
>
>Really? I would have said that it is moving approximately 25mph down.
>Remember that during one rotation of the wheel, any point on the
>circumference moves forward by one circumference, so, 2piR, but up and
>down by two diameters, so, 4R. The vertical distance travelled is less
>than the horizontal distance, it's 2/pi (0.6366), times it, so it looks
>doubtful that the number we are looking for is going to come out to 20
>or 40.
>
>The ratio of horizontal movement to vertical movement is 2piR to 4R, or
>pi to 2. If you multiply 20 miles times pi/2, you get approximately
>12.75. So a point on the circumference has to average 12.75 mph
>vertically while averaging 20mph horizontally.
>
>Furthermore, the point on the circumference speeds up and slows down
>twice per rotation, as opposed to once in the forward direction. So it
>has to accelerate to 25.5mph and then decelerate to zero once on the
>way up and then do it all again on the way down.
>
>Correct?

Dear Doug,

I think that the difference is that you may be thinking of averaging,
curves, and angles, while I'm talking about simpler straight forward
and straight up rates of motion.

That is, you may be thinking about speed along a curved descent or
ascent, while I'm simply treating it as rate of climb and pure forward
progress

To simplify the terms, treat the ends of the spokes as a metal rim
with no tire.

If the axle moves forward at 20 mph, then the rim must spin at 20 mph.

The exact top of the rim is always doing 20 + 20 = 40 mph straight
forward.

The exact bottom of the rim is always doing 20 - 20 = 0 mph in every
direction relative to the ground.

(If the exact bottom of the rim did anything except 0 mph relative to
the ground, it would leave a skid mark.)

The exact horizontal forward part of the rim is always moving 20 mph
straight down at the same instant that it's doing 20 mph forward.

Same for the exact horizontal rearward part of the rim, except that it
moves straight up at 20 mph.

Viewed from the side, a pinpoint light on the rim proceeds in an
endless series of half-circle leaps, slowing to 0 mph forward speed
each time it "bounces" off the ground, accelerating to 40 mph forward
speed at the top of the half-circle, and slowing back down to 0 mph.

With a rim moving forward and rotating at 20 mph, the maxium straight
upward or straight downward speed is obviously limited to 20 mph,
since the axle does not move up or down.

Cheers,

Carl Fogel

 

[[email protected]]

[email protected] wrote:
> On 24 Aug 2006 13:56:08 -0700, [email protected] wrote:
>
> >> >> The ends of the horizontal spokes are going forward at 20 mph and also
> >> >> going either up or down at 20 mph.
> >> >
> >> >Are you sure? Forward at 20, yes, but up and down too?
> >> >
> >> >Cheers,
> >> >Doug
> >>
> >> Dear Doug,
> >
> >Dear Carl,
> >
> >> Let's leave the side wind for further down.
> >OK
> >> All figures are slightly high because the spokes don't reach the edge
> >> of the tire--20 mph for the spokes is 20 mph minus a little bit.
> >Okey Doke
> >> Everything on the bicycle is moving forward relative to the ground at
> >> 20 mph.
> >
> >Disagree. The wheel, where it touches the ground, is not moving
> >forward at all.
> >
> >> The wheel is also rotating at enough rpm that its outer edge is doing
> >> 20 mph.
> >
> >At two points [on the outer edge] but not at the rest of it.
> >
> >> So the top of the top spoke is going 20 + 20 = 40 mph forward.
> >>
> >> The bottom of the bottom spoke is going 20-20 = 0 mph forward.
> >
> >The bottom of the bottom spoke is not 0-0, not 20-20.
> >
> >> The entire front horizontal spoke is going 20 mph forward, while its
> >> rim end is also moving 20 mph straight down.
> >
> >Really? I would have said that it is moving approximately 25mph down.
> >Remember that during one rotation of the wheel, any point on the
> >circumference moves forward by one circumference, so, 2piR, but up and
> >down by two diameters, so, 4R. The vertical distance travelled is less
> >than the horizontal distance, it's 2/pi (0.6366), times it, so it looks
> >doubtful that the number we are looking for is going to come out to 20
> >or 40.
> >
> >The ratio of horizontal movement to vertical movement is 2piR to 4R, or
> >pi to 2. If you multiply 20 miles times pi/2, you get approximately
> >12.75. So a point on the circumference has to average 12.75 mph
> >vertically while averaging 20mph horizontally.
> >
> >Furthermore, the point on the circumference speeds up and slows down
> >twice per rotation, as opposed to once in the forward direction. So it
> >has to accelerate to 25.5mph and then decelerate to zero once on the
> >way up and then do it all again on the way down.
> >
> >Correct?
>
> Dear Doug,
>
> I think that the difference is that you may be thinking of averaging,
> curves, and angles, while I'm talking about simpler straight forward
> and straight up rates of motion.

Nope, I'm talking about straight-line motion.

> If the axle moves forward at 20 mph, then the rim must spin at 20 mph.

Not true. Each and every point on the rim is constantly changing
speed, in every direction, and in total.

> The exact top of the rim is always doing 20 + 20 = 40 mph straight
> forward.

Correct.

> The exact bottom of the rim is always doing 20 - 20 = 0 mph in every
> direction relative to the ground.

It is doing 0, but it's not 20-20. It's 0-0.

> (If the exact bottom of the rim did anything except 0 mph relative to
> the ground, it would leave a skid mark.)
Correct.

> The exact horizontal forward part of the rim is always moving 20 mph
> straight down at the same instant that it's doing 20 mph forward.

Incorrect. It it moving 25.5mph straight down.

> Same for the exact horizontal rearward part of the rim, except that it
> moves straight up at 20 mph.

Incorrect again. It is moving 25.5mph straight up.

> Viewed from the side, a pinpoint light on the rim proceeds in an
> endless series of half-circle leaps, slowing to 0 mph forward speed
> each time it "bounces" off the ground, accelerating to 40 mph forward
> speed at the top of the half-circle, and slowing back down to 0 mph.

Correct. But that movement is not symmetric. It involves more forward
movement that up-and-down movement. Therefore the total distance
travelled is not the same in the two directions.

> With a rim moving forward and rotating at 20 mph, the maxium straight
> upward or straight downward speed is obviously limited to 20 mph,
> since the axle does not move up or down.

Faulty logic. Do the math, Carl. Each point on the [circumference of
the] rim moves 20mph in the forward direction and 12.75mph in the
vertical direction when the bike moves 20mph. Therefore each point
must average 20mph forward and 12.75mph upanddown. Furthermore, being
a circle, changes in sine and cosine are symmetric, so the increases
and decreases in horizontal and vertical speeds are mirror images.
Therefore the wheel must be going twice the average, or 40mph at the
top of its run. And it must be going twice the average, or 25.5mph,
when at the midpoint of its upstroke, and again at the midpoint of its
downstroke.

The error in your logic is in your claim that any point on the rim is
moving at 20mph just because the wheel is rotating, and the bike is.
But you can arrive at the correct answer without thinking about that,
just do the math of the vertical movement and divide by time.

Do you agree that while the forward movement of one rotation is 2piR,
the vertical movement is 4R?

Cheers,
Doug

 

[[email protected]]

On 24 Aug 2006 15:53:23 -0700, [email protected] wrote:

>
>[email protected] wrote:
>> On 24 Aug 2006 13:56:08 -0700, [email protected] wrote:
>>
>> >> >> The ends of the horizontal spokes are going forward at 20 mph and also
>> >> >> going either up or down at 20 mph.
>> >> >
>> >> >Are you sure? Forward at 20, yes, but up and down too?
>> >> >
>> >> >Cheers,
>> >> >Doug
>> >>
>> >> Dear Doug,
>> >
>> >Dear Carl,
>> >
>> >> Let's leave the side wind for further down.
>> >OK
>> >> All figures are slightly high because the spokes don't reach the edge
>> >> of the tire--20 mph for the spokes is 20 mph minus a little bit.
>> >Okey Doke
>> >> Everything on the bicycle is moving forward relative to the ground at
>> >> 20 mph.
>> >
>> >Disagree. The wheel, where it touches the ground, is not moving
>> >forward at all.
>> >
>> >> The wheel is also rotating at enough rpm that its outer edge is doing
>> >> 20 mph.
>> >
>> >At two points [on the outer edge] but not at the rest of it.
>> >
>> >> So the top of the top spoke is going 20 + 20 = 40 mph forward.
>> >>
>> >> The bottom of the bottom spoke is going 20-20 = 0 mph forward.
>> >
>> >The bottom of the bottom spoke is not 0-0, not 20-20.
>> >
>> >> The entire front horizontal spoke is going 20 mph forward, while its
>> >> rim end is also moving 20 mph straight down.
>> >
>> >Really? I would have said that it is moving approximately 25mph down.
>> >Remember that during one rotation of the wheel, any point on the
>> >circumference moves forward by one circumference, so, 2piR, but up and
>> >down by two diameters, so, 4R. The vertical distance travelled is less
>> >than the horizontal distance, it's 2/pi (0.6366), times it, so it looks
>> >doubtful that the number we are looking for is going to come out to 20
>> >or 40.
>> >
>> >The ratio of horizontal movement to vertical movement is 2piR to 4R, or
>> >pi to 2. If you multiply 20 miles times pi/2, you get approximately
>> >12.75. So a point on the circumference has to average 12.75 mph
>> >vertically while averaging 20mph horizontally.
>> >
>> >Furthermore, the point on the circumference speeds up and slows down
>> >twice per rotation, as opposed to once in the forward direction. So it
>> >has to accelerate to 25.5mph and then decelerate to zero once on the
>> >way up and then do it all again on the way down.
>> >
>> >Correct?
>>
>> Dear Doug,
>>
>> I think that the difference is that you may be thinking of averaging,
>> curves, and angles, while I'm talking about simpler straight forward
>> and straight up rates of motion.
>
>Nope, I'm talking about straight-line motion.
>
>> If the axle moves forward at 20 mph, then the rim must spin at 20 mph.
>
>Not true. Each and every point on the rim is constantly changing
>speed, in every direction, and in total.
>
>> The exact top of the rim is always doing 20 + 20 = 40 mph straight
>> forward.
>
>Correct.
>
>> The exact bottom of the rim is always doing 20 - 20 = 0 mph in every
>> direction relative to the ground.
>
>It is doing 0, but it's not 20-20. It's 0-0.
>
>> (If the exact bottom of the rim did anything except 0 mph relative to
>> the ground, it would leave a skid mark.)
>Correct.
>
>> The exact horizontal forward part of the rim is always moving 20 mph
>> straight down at the same instant that it's doing 20 mph forward.
>
>Incorrect. It it moving 25.5mph straight down.
>
>> Same for the exact horizontal rearward part of the rim, except that it
>> moves straight up at 20 mph.
>
>Incorrect again. It is moving 25.5mph straight up.
>
>> Viewed from the side, a pinpoint light on the rim proceeds in an
>> endless series of half-circle leaps, slowing to 0 mph forward speed
>> each time it "bounces" off the ground, accelerating to 40 mph forward
>> speed at the top of the half-circle, and slowing back down to 0 mph.
>
>Correct. But that movement is not symmetric. It involves more forward
>movement that up-and-down movement. Therefore the total distance
>travelled is not the same in the two directions.
>
>> With a rim moving forward and rotating at 20 mph, the maxium straight
>> upward or straight downward speed is obviously limited to 20 mph,
>> since the axle does not move up or down.
>
>Faulty logic. Do the math, Carl. Each point on the [circumference of
>the] rim moves 20mph in the forward direction and 12.75mph in the
>vertical direction when the bike moves 20mph. Therefore each point
>must average 20mph forward and 12.75mph upanddown. Furthermore, being
>a circle, changes in sine and cosine are symmetric, so the increases
>and decreases in horizontal and vertical speeds are mirror images.
>Therefore the wheel must be going twice the average, or 40mph at the
>top of its run. And it must be going twice the average, or 25.5mph,
>when at the midpoint of its upstroke, and again at the midpoint of its
>downstroke.
>
>The error in your logic is in your claim that any point on the rim is
>moving at 20mph just because the wheel is rotating, and the bike is.
>But you can arrive at the correct answer without thinking about that,
>just do the math of the vertical movement and divide by time.
>
>Do you agree that while the forward movement of one rotation is 2piR,
>the vertical movement is 4R?
>
>Cheers,
>Doug

Dear Doug,

Put the bike upside down on a railroad flatcar.

Have the flatcar do 20 mph forward.

The wheel does not rotate.

Every point on the wheel moves forward, just like the bicycle frame at
20 mph.

Now spin the wheel so that the rim/tire/simplified-spoke-ends are
doing 20 mph.

***

Axle = 20 mph forward relative to ground.

Top of rim = 20 mph forward relative to axle.

Axle = 20 mph forward relative to ground.

20 + 20 = 40 mph for top of rim relative to ground.

***

Axle = 20 mph forward relative to ground.

Bottom of rim = 20 mph backward relative to axle.

20 - 20 = 0 mph for bottom of rim relative to ground.

***

Now view the wheel from dead ahead in the dark.

It doesn't matter whether the axle is moving toward you, standing
still, or moving away from you.

A light on the rim will appear to bounce back and forth either way
from the center of the tire, both up and down.

It will slow to 0 mph up/down at the top and bottom of the tire.

It will accelerate to 20 mph either straight up or straight down in
the exact middle of its travel, when it is in line with your eye and
the axle--that's the definition of a rim spinning at 20 mph.

***

Draw a circle with 4 tangent 2-inch lines at the top, bottom, and both
sides.

If the circle spins around a stationary axle at 20 mph, then each
2-inch tangent line is a 20 mph velocity, speed plus direction.

To make the axle move forward at 20 mph, add four horizontal 2-inch
lines at the appropriate spots.

The two lines cancel to 0 mph at the bottom of the circle.

The two lines add to each other for 40 mph at the top of the circle.

The horizontal lines do not affect the vertical travel, which remains
20 mph, either up or down.

Cheers,

Carl Fogel

 

[[email protected]]

> Dear Doug,
>
> Put the bike upside down on a railroad flatcar.
>
> Have the flatcar do 20 mph forward.
>
> The wheel does not rotate.
>
> Every point on the wheel moves forward, just like the bicycle frame at
> 20 mph.

Dear Carl,

Okay.

> Axle = 20 mph forward relative to ground.
> Top of rim = 20 mph forward relative to axle.
> Axle = 20 mph forward relative to ground.
> 20 + 20 = 40 mph for top of rim relative to ground.

Okay.

> Now view the wheel from dead ahead in the dark.
> It doesn't matter whether the axle is moving toward you, standing
> still, or moving away from you.
> A light on the rim will appear to bounce back and forth either way
> from the center of the tire, both up and down.
> It will slow to 0 mph up/down at the top and bottom of the tire.
>
> It will accelerate to 20 mph either straight up or straight down in
> the exact middle of its travel, when it is in line with your eye and
> the axle--that's the definition of a rim spinning at 20 mph.

Wrong. It will accelerate to 25.5mph both straight up and straight
down. Furthermore, there is no definition of a rim spinning at 20mph.
Rims spin, like everything else, at RPMs, not MPHs. It is undefined to
speak of a rim spinning at an MPH. This is critical.

> If the circle spins around a stationary axle at 20 mph, then each
> 2-inch tangent line is a 20 mph velocity, speed plus direction.

Carl, a circle DOES NOT "spin at 20mph". The If you -must- speak of
the circle as a whole, then the circle has forward speed of 20, and a
vertical speed of 0. The numbers 20 and 40 come into play when you
look at the forward speeds of various points on the rim. It really has
nothing to do with the vertical speeds at all - with the exception of
the relationship pi over 2.

You keep trying to claim that anything that is true about the forward
movement is also true of the vertical movement, but that's simply not
true.

Carl, you agree that a point on the rim will accelerate and decelerate
twice per rotation. Do you also agree that the distance travelled per
rotation is 2piR forward, 0 backwards, 2R up, and 2R down?

Cheers,
Doug

 

[[email protected]]

On 24 Aug 2006 16:54:09 -0700, [email protected] wrote:

>> Dear Doug,
>>
>> Put the bike upside down on a railroad flatcar.
>>
>> Have the flatcar do 20 mph forward.
>>
>> The wheel does not rotate.
>>
>> Every point on the wheel moves forward, just like the bicycle frame at
>> 20 mph.
>
>Dear Carl,
>
>Okay.
>
>> Axle = 20 mph forward relative to ground.
>> Top of rim = 20 mph forward relative to axle.
>> Axle = 20 mph forward relative to ground.
>> 20 + 20 = 40 mph for top of rim relative to ground.
>
>Okay.
>
>> Now view the wheel from dead ahead in the dark.
>> It doesn't matter whether the axle is moving toward you, standing
>> still, or moving away from you.
>> A light on the rim will appear to bounce back and forth either way
>> from the center of the tire, both up and down.
>> It will slow to 0 mph up/down at the top and bottom of the tire.
>>
>> It will accelerate to 20 mph either straight up or straight down in
>> the exact middle of its travel, when it is in line with your eye and
>> the axle--that's the definition of a rim spinning at 20 mph.
>
>Wrong. It will accelerate to 25.5mph both straight up and straight
>down. Furthermore, there is no definition of a rim spinning at 20mph.
>Rims spin, like everything else, at RPMs, not MPHs. It is undefined to
>speak of a rim spinning at an MPH. This is critical.
>
>> If the circle spins around a stationary axle at 20 mph, then each
>> 2-inch tangent line is a 20 mph velocity, speed plus direction.
>
>Carl, a circle DOES NOT "spin at 20mph". The If you -must- speak of
>the circle as a whole, then the circle has forward speed of 20, and a
>vertical speed of 0. The numbers 20 and 40 come into play when you
>look at the forward speeds of various points on the rim. It really has
>nothing to do with the vertical speeds at all - with the exception of
>the relationship pi over 2.
>
>You keep trying to claim that anything that is true about the forward
>movement is also true of the vertical movement, but that's simply not
>true.
>
>Carl, you agree that a point on the rim will accelerate and decelerate
>twice per rotation. Do you also agree that the distance travelled per
>rotation is 2piR forward, 0 backwards, 2R up, and 2R down?
>
>Cheers,
>Doug

Dear Doug,

I think that you are confused about rpm and speed.

A point on the edge of the rim travels along a circular path at 20 mph
when your bicycle goes 20 mph,

That's how your bicycle odometer and speedometer work. RPM alone is
useless. You must also have the circumference to figure out the speed.

I also think that you need to look into the difference betwen a speed
and a velocity.

As I have repeatedly said, the pure forward speed of the leading
spoke's point on the rim (a horizontal velocity) is 20 mph straight
forward, and the pure downward speed of the same spot is 20 mph
straight downward.

Cheers,

Carl Fogel

 

[[email protected]]

> Dear Doug,
>
> I think that you are confused about rpm and speed.

No Carl, you are

> A point on the edge of the rim travels along a circular path at 20 mph
> when your bicycle goes 20 mph,

No, Carl, it does not. It -averages- 20mph. Only in two points is it
actually doing 20, the point at 90 degrees and the point at 270 degrees
where 0 is at the top. All other points are moving either less than or
greater than 20.

> That's how your bicycle odometer and speedometer work. RPM alone is
> useless. You must also have the circumference to figure out the speed.

No, Carl, it is not. They compute the speed of the bike and the
distance travelled from the known circumference of the wheel and a
count of the rotations. Not from the speed of a point on the rim.

> I also think that you need to look into the difference betwen a speed
> and a velocity.

And I think you need to run some numbers.

> As I have repeatedly said,

- incorrectly -

>...the pure forward speed of the leading
> spoke's point on the rim (a horizontal velocity) is 20 mph straight
> forward, and the pure downward speed of the same spot is 20 mph
> straight downward.

Incorrect. Carl, why do you not do the math? The pure downward speed
of the leading point of the rim is 25.5mph. Do the math and show us.
If your logic is correct, the numbers will verify it.

Cheers,
Doug

 

[[email protected]]

On 24 Aug 2006 18:46:23 -0700, [email protected] wrote:

>> Dear Doug,
>>
>> I think that you are confused about rpm and speed.
>
>No Carl, you are
>
>> A point on the edge of the rim travels along a circular path at 20 mph
>> when your bicycle goes 20 mph,
>
>No, Carl, it does not. It -averages- 20mph. Only in two points is it
>actually doing 20, the point at 90 degrees and the point at 270 degrees
>where 0 is at the top. All other points are moving either less than or
>greater than 20.
>
>> That's how your bicycle odometer and speedometer work. RPM alone is
>> useless. You must also have the circumference to figure out the speed.
>
>No, Carl, it is not. They compute the speed of the bike and the
>distance travelled from the known circumference of the wheel and a
>count of the rotations. Not from the speed of a point on the rim.
>
>> I also think that you need to look into the difference betwen a speed
>> and a velocity.
>
>And I think you need to run some numbers.
>
>> As I have repeatedly said,
>
>- incorrectly -
>
>>...the pure forward speed of the leading
>> spoke's point on the rim (a horizontal velocity) is 20 mph straight
>> forward, and the pure downward speed of the same spot is 20 mph
>> straight downward.
>
>Incorrect. Carl, why do you not do the math? The pure downward speed
>of the leading point of the rim is 25.5mph. Do the math and show us.
>If your logic is correct, the numbers will verify it.
>
>Cheers,
>Doug

Dear Doug,

Since you insist, I suggest that you might ponder an instantaneous
velocity (not speed) of 28.3 mph (not 25.5) at 45 degrees to the
ground (not forward and not vertically) on a bicycle rim of any size
whose axle is moving forward at 20 mph.

Cheers,

Carl Fogel