Hour Record broken

Paul R said:

"steve" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

On 19-Jul-2005, smacked up and reeling, "Laz" <[email hidden]>

blindly

Quoted message said:

formulated
the following incoherence:

Quoted message said:

Thanx for the news. I've always maintained an interest in the Hour. But

I

Quoted message said:

Quoted message said:

wish they would go back to the Absolute Hour as it was more exciting to
see
technology play it's part.

I'll register disagreement on that one. Let the hour record be the test

of

Quoted message said:

the rider himself.

Personally Id prefer to see the technology taken substantially out of
cycling on the road, too. For example in the grand tours, Id make the

rule

Quoted message said:

that you have to ride the same bike (gearing changes allowed, but the same
wheels), handlebars, and helmet in the road stages and the TTs. Was it

last

Quoted message said:

year or the year before where one TT crested a tough climb in the middle,
and half the contenders switched bikes at the summit? That's not cycling.

From the article:

"In his attempt, Sosenka was using a 3.2 kg wheel and 190 mm cranks. The
reason for the heavy wheel was that although it was harder to get up to
speed, it was easy to maintain it. "

He basically used his rear wheel like a flywheel. The inertia of that heavy,
spinning wheel would keep him coasting at speed far longer than with regular
wheel.

So, another use of technology to reduce the hour record.

One of the things I like about mountain biking is that the racer must remain
self-sufficient. You can't change your wheel if you get a flat - you've
gotta fix it.

Not any more! UCI rules allow outside assistance in mountain bike races
(although non-UCI NORBA races still prevent it).

--
Steven L. Sheffield
stevens at veloworks dot com
bellum pax est libertas servitus est ignoratio vis est
ess ay ell tea ell ay kay ee sea eye tee why you ti ay aitch
aitch tee tea pea colon [for word] slash [four ward] slash double-you
double-yew double-ewe dot veloworks dot com [foreword] slash

In article
<[email hidden]>,

Quoted post said:

Paul R said:

He basically used his rear wheel like a flywheel. The inertia of that heavy,
spinning wheel would keep him coasting at speed far longer than with regular
wheel.

So, another use of technology to reduce the hour record.

You can't coast during an hour record attempt, or you won't
make it. Of course you aren't coasting anyway since it's a
fixed gear. The inertia of a bike and rider moving at 40+ km/h
keeps the pedals spinning. Suppose a regular rear weighs about
0.8 kg and his was 2.5 kg heavier. You get an additional
2.5 kg * speed of linear momentum for the wheel moving forward,
and the same amount for its additional rotational inertia.

Yes, where by "the same amount" we mean a bit less. Suppose a
1 kg mass is at 0.3 m from the axle of the wheel, and the wheel
roll-out is 2.15 m.
The linear momentum is 1 * speed.
The angular momentum is
0.3 * speed / 2.15 * 2pi = 0.877 * speed.

So the angular momentum is quantitatively less than the linear
momentum. Therefore it takes less than two masses hanging off the
frame to develop the same momentum as putting one mass in the
rotating wheel.

However angular momentum and linear momentum are measured in
different units and do not convert one to the other; even though
you and I are acting as if they are equivalent. (This is not a
quibble; the explanation can be expressed succinctly but not here
and now.) Anyway, momentum is probably not the defining concept
here. We are better off talking about kinetic energy.

The kinetic energy of one kilogram at 40 km/hr is
1/2 * (40000/ 3600)^2 joule = 61.7 joule

At 40 km/hr the angular spped is
(40000/3600)/2.15 * 2pi = 32.5 /sec
The kinetic energy in the rotation is
1/2 * (32.5)^2 * (0.3)^2 joule = 47.4 joule

Again, more energy stored hanging two masses from the frame
contrasted with putting one mass in the rotating wheel.

Quoted message said:

I think he could have got the same result by using a regular
wheel but putting a 5kg lead weight under the seat.

Merckx used as light a bike as possible for his attempt.
He could have put weights on the wheel if he wanted, so it's
legitimate Merckx-era technology.

--
Michael Press

First, on Sosenka's web page http://www.sosenka.cz/ is this
picture: http://www.sosenka.cz/uvodni.jpg
It doesn't tell us much about the wheel, but it could explain
how he was able to do such a good distance for the hour.
I wonder why the "derny" rider is riding a full-on motorcycle
but wearing a bike helmet. For god's sake don't tell anyone
in rec.bicycles.misc.

Michael Press said:

Quoted message said:

You get an additional
2.5 kg * speed of linear momentum for the wheel moving forward,
and the same amount for its additional rotational inertia.

Yes, where by "the same amount" we mean a bit less. Suppose a
1 kg mass is at 0.3 m from the axle of the wheel, and the wheel
roll-out is 2.15 m.
The linear momentum is 1 * speed.
The angular momentum is
0.3 * speed / 2.15 * 2pi = 0.877 * speed.

If we're going to be pedantic, not quite right. The angular
momentum of the 1kg mass is 0.78 * speed/radius, because the
(0.3/0.34) ratio of where the mass is over the wheel radius
enters the angular momentum as the square: L = I*w, I=mass*(0.3)^2,
w=angular speed. See below.

Quoted message said:

So the angular momentum is quantitatively less than the linear
momentum. Therefore it takes less than two masses hanging off the
frame to develop the same momentum as putting one mass in the
rotating wheel.

Yes, I made the approximation that all the weight is at the rim.
Formally the angular momentum is I*w and the rotational kinetic
energy is I*w^2/2, where I is the moment of inertia and w the
angular speed. The linear speed is v = w*r. The moment of
inertia is I=K*m*r^2 where K is some factor less than one,
depending on how close you can get the weight to the rim.

Quoted message said:

However angular momentum and linear momentum are measured in
different units and do not convert one to the other; even though
you and I are acting as if they are equivalent. (This is not a
quibble; the explanation can be expressed succinctly but not here
and now.) Anyway, momentum is probably not the defining concept
here. We are better off talking about kinetic energy.

The force required to accelerate the rear wheel is
proportional to (1+K) where the 1 comes from linear momentum
and the K comes from angular momentum. The total kinetic
energy in the spinning wheel is also (1+K)*0.5*m*v^2.
Whether K is 0.8 or 1 is not very important. The point is that
its only effect is to slightly increase the inertia of the bike
plus rider, the same as a weight under the seat would.

On a fixed gear, the total inertia is what carries your pedal
stroke through the dead spot. However, slightly more weight
means more power lost to rolling resistance.

Quoted message said:

The kinetic energy of one kilogram at 40 km/hr is
1/2 * (40000/ 3600)^2 joule = 61.7 joule

At 40 km/hr the angular spped is
(40000/3600)/2.15 * 2pi = 32.5 /sec
The kinetic energy in the rotation is
1/2 * (32.5)^2 * (0.3)^2 joule = 47.4 joule

Again, more energy stored hanging two masses from the frame
contrasted with putting one mass in the rotating wheel.

In article
<[email hidden]>,

Quoted post said:

First, on Sosenka's web page http://www.sosenka.cz/ is this
picture: http://www.sosenka.cz/uvodni.jpg
It doesn't tell us much about the wheel, but it could explain
how he was able to do such a good distance for the hour.
I wonder why the "derny" rider is riding a full-on motorcycle
but wearing a bike helmet. For god's sake don't tell anyone
in rec.bicycles.misc.

Michael Press said:

Quoted message said:

You get an additional
2.5 kg * speed of linear momentum for the wheel moving forward,
and the same amount for its additional rotational inertia.

Yes, where by "the same amount" we mean a bit less. Suppose a
1 kg mass is at 0.3 m from the axle of the wheel, and the wheel
roll-out is 2.15 m.
The linear momentum is 1 * speed.
The angular momentum is
0.3 * speed / 2.15 * 2pi = 0.877 * speed.

If we're going to be pedantic, not quite right. The angular
momentum of the 1kg mass is 0.78 * speed/radius, because the
(0.3/0.34) ratio of where the mass is over the wheel radius
enters the angular momentum as the square: L = I*w, I=mass*(0.3)^2,
w=angular speed. See below.

Yes. So that is
angular momentum = (0.3)^2 * speed / 2.15 * 2pi =.263 * speed.
I was trying to be pedantic in the _good_ sense. <:^)

Quoted message said:

Quoted message said:

So the angular momentum is quantitatively less than the linear
momentum. Therefore it takes less than two masses hanging off the
frame to develop the same momentum as putting one mass in the
rotating wheel.

Yes, I made the approximation that all the weight is at the rim.
Formally the angular momentum is I*w and the rotational kinetic
energy is I*w^2/2, where I is the moment of inertia and w the
angular speed. The linear speed is v = w*r. The moment of
inertia is I=K*m*r^2 where K is some factor less than one,
depending on how close you can get the weight to the rim.

Quoted message said:

However angular momentum and linear momentum are measured in
different units and do not convert one to the other; even though
you and I are acting as if they are equivalent. (This is not a
quibble; the explanation can be expressed succinctly but not here
and now.) Anyway, momentum is probably not the defining concept
here. We are better off talking about kinetic energy.

The force required to accelerate the rear wheel is
proportional to (1+K) where the 1 comes from linear momentum
and the K comes from angular momentum. The total kinetic
energy in the spinning wheel is also (1+K)*0.5*m*v^2.
Whether K is 0.8 or 1 is not very important. The point is that
its only effect is to slightly increase the inertia of the bike
plus rider, the same as a weight under the seat would.

On a fixed gear, the total inertia is what carries your pedal
stroke through the dead spot. However, slightly more weight
means more power lost to rolling resistance.

Quoted message said:

The kinetic energy of one kilogram at 40 km/hr is
1/2 * (40000/ 3600)^2 joule = 61.7 joule

At 40 km/hr the angular spped is
(40000/3600)/2.15 * 2pi = 32.5 /sec
The kinetic energy in the rotation is
1/2 * (32.5)^2 * (0.3)^2 joule = 47.4 joule

Again, more energy stored hanging two masses from the frame
contrasted with putting one mass in the rotating wheel.

--
Michael Press

Jason Spaceman said:

According to CyclingNews a Czech rider with the Water & Soap team has set a new
hour rekkid ---> http://www.cyclingnews.com/news/?id=2005/jul05/jul19news6

There are more dopers on that team than anyone has seen since Gewiss.
I'm skeptical.

-Sonarrat.

In article
<[email hidden]>,

Quoted post said:

First, on Sosenka's web page http://www.sosenka.cz/ is this
picture: http://www.sosenka.cz/uvodni.jpg
It doesn't tell us much about the wheel, but it could explain
how he was able to do such a good distance for the hour.
I wonder why the "derny" rider is riding a full-on motorcycle
but wearing a bike helmet. For god's sake don't tell anyone
in rec.bicycles.misc.

Michael Press said:

Quoted message said:

You get an additional
2.5 kg * speed of linear momentum for the wheel moving forward,
and the same amount for its additional rotational inertia.

Yes, where by "the same amount" we mean a bit less. Suppose a
1 kg mass is at 0.3 m from the axle of the wheel, and the wheel
roll-out is 2.15 m.
The linear momentum is 1 * speed.
The angular momentum is
0.3 * speed / 2.15 * 2pi = 0.877 * speed.

If we're going to be pedantic, not quite right. The angular
momentum of the 1kg mass is 0.78 * speed/radius, because the
(0.3/0.34) ratio of where the mass is over the wheel radius
enters the angular momentum as the square: L = I*w, I=mass*(0.3)^2,
w=angular speed. See below.

Yes. So that is
angular momentum = (0.3)^2 * speed / 2.15 * 2pi =.263 * speed.
I was trying to be pedantic in the _good_ sense. <:^)

Quoted message said:

Quoted message said:

So the angular momentum is quantitatively less than the linear
momentum. Therefore it takes less than two masses hanging off the
frame to develop the same momentum as putting one mass in the
rotating wheel.

Yes, I made the approximation that all the weight is at the rim.
Formally the angular momentum is I*w and the rotational kinetic
energy is I*w^2/2, where I is the moment of inertia and w the
angular speed. The linear speed is v = w*r. The moment of
inertia is I=K*m*r^2 where K is some factor less than one,
depending on how close you can get the weight to the rim.

Quoted message said:

However angular momentum and linear momentum are measured in
different units and do not convert one to the other; even though
you and I are acting as if they are equivalent. (This is not a
quibble; the explanation can be expressed succinctly but not here
and now.) Anyway, momentum is probably not the defining concept
here. We are better off talking about kinetic energy.

The force required to accelerate the rear wheel is
proportional to (1+K) where the 1 comes from linear momentum
and the K comes from angular momentum. The total kinetic
energy in the spinning wheel is also (1+K)*0.5*m*v^2.
Whether K is 0.8 or 1 is not very important. The point is that
its only effect is to slightly increase the inertia of the bike
plus rider, the same as a weight under the seat would.

On a fixed gear, the total inertia is what carries your pedal
stroke through the dead spot. However, slightly more weight
means more power lost to rolling resistance.

Quoted message said:

The kinetic energy of one kilogram at 40 km/hr is
1/2 * (40000/ 3600)^2 joule = 61.7 joule

At 40 km/hr the angular spped is
(40000/3600)/2.15 * 2pi = 32.5 /sec
The kinetic energy in the rotation is
1/2 * (32.5)^2 * (0.3)^2 joule = 47.4 joule

Again, more energy stored hanging two masses from the frame
contrasted with putting one mass in the rotating wheel.

--
Michael Press