Getting Started on Rollers

"Mad Dog" <[email hidden]> wrote in message

Quoted message said:

Not for me, especially back at the age I was at the time. My rollers had
no
load fan or mag and they were large diameter rollers.

Rolling resistance is inversely proportional to roller size.

"Mad Dog" <[email hidden]> wrote in message

Quoted message said:

If
I could find my training log from those years, I could be more exact with
dates
and such, but it's buried down in the basement chaos and it simply ain't
worth
the effort.

Sounds like the perfect opportunity to do another 180 rpm roller dismount.

"Mad Dog" <[email hidden]> wrote in message

Quoted message said:

You won't shoot off since there is no forward momentum except in the
rotation of the wheels. If you do fall off, it is more like you move
about three feet forward and then fall over in slow motion.

Quoted message said:

Bzzzt! Wrongo! I used to ride rollers in the basement and twice rode off
of
them. Fortunately, we had a bunch of boxes stacked up against the wall I
ran
into that was a good 15 feet away. The rubber streak on the floor was at
least
7' long, so the rear wheel had enough momentum to be spinning hard for
that
distance.

Apparently you did not understand the previous post. It said
_no_forward_momentum_. Nothing about continuing to apply power.

You admitted "I don't recall any significant rubber being laid down by the
front tire. You could see an effect, like a short polish mark."

Case closed.

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

Quoted message said:

Carl Sundquist says...

Quoted message said:

As I mentioned in my original response about you wieghing about 30 pounds,
I
couldn't imagine how you could spin a rear wheel for that distance.

Considering the antics I witnessed at the mall fund raiser, I concluded
that the
paint on the basement floor was a key component. It was slick. The
cement
surface on the mall floor was relatively smooth but still considerably
coarser
than the painted basement floor. The typical skid mark at the mall was in
the
1' range. We got to clean them up so I got to know roller-dismount skid
marks
real well.

The grip on the smoother surface will be superior to that of the coarser
surface, as long as the surfaces are clean and dry. The smoother surface
will provide greater surface area to contact the tire. If the surface is wet
or dirty, the coarser surface will have greater grip. The smooth, wet
surface will cause hydroplaning, the smooth dirty surface will have the dirt
act as a lubricant as it is dislodged.

Of course, in the case of either wet or dirty, whatever contaminant is on
the smooth surface would minimize or prevent skid marks on the surface
itself.

In article <[email hidden]>, [email hidden] says...

Quoted message said:

Alex Rodriguez says...

Quoted message said:

You must be guisness book of records material.

Correct.

Quoted message said:

Unless you kept pedaling after you came off the rollers, there is no way
you wet 15 feet forward.

Dammit, pay attention! And don't confuse threads from other newsgroups. The
15' horizontal [censored] claim was on rec.climbing.

Ha, ha, ha. I seem to have difficulty hitting the 'N' key.
---------------
Alex

If four men are riding rollers in the same room, will they
spontaneously begin pedalling in synchronicity?

-RJ

Donald Munro said:

Kurgan Gringioni said:

Quoted message said:

I can't really argue with any of that, but what I'd like to know is:
how many grams does it take to save 2 minutes?

Quoted message said:

About 30 milligrams of methamphetamine hydrochloride USP.
You're welcome.

I always thought the [censored] that will kill them was more sophisticated.

PS Thanks for bringing us back on topic.

He didn't ask how many grams it would take to save two
minutes _and_ not test positive at the top of the hill.
For that kind of advice, you gotta pay the big bucks, or
at least get a membership to 53x12.com.

ronaldo_jeremiah said:

If four men are riding rollers in the same room, will they
spontaneously begin pedalling in synchronicity?

-RJ

Yes. And they will all gradually increase their cadences until they are
pedaling so fast that when they fall off, they will shoot forward into each
other at relativistic speeds, forming a supercollider, with the collisional
energy being so great it will tear a rip in spacetime, destroying the
entire universe. When you get them drunk at a party, physicists will admit
they refer to this scenario as the big gang bang.

You know, it struck me that the OP is probably thinking "screw cycling, I'm
buying a treadmill."

--
Bill Asher

Jenko said:

Michael Press said:

Kinetic energy in two wheels:
2 * I * w^2 / 2 = 450 J

Does the front wheel spin as fast as the rear wheel?
Jenko, never used a roller

Yes, there's a rubber belt that connects the front and rear
drums, so the front wheel is driven by the drum. I never
tried it without the belt. I suspect it would be a lot harder
to balance and you would fall down.

Carl Sundquist says...

Quoted message said:

Apparently you did not understand the previous post. It said
_no_forward_momentum_. Nothing about continuing to apply power.

It also said nothing about NOT continuing to apply power.

Quoted message said:

Case closed.

Decades ago.

Quoted message said:

Jenko said:

Michael Press said:

Kinetic energy in two wheels:
2 * I * w^2 / 2 = 450 J

Does the front wheel spin as fast as the rear wheel?
Jenko, never used a roller

Yes, there's a rubber belt that connects the front and rear
drums, so the front wheel is driven by the drum. I never
tried it without the belt. I suspect it would be a lot harder
to balance and you would fall down.

After which you'd shoot across the room.

Carl Sundquist says...

Quoted message said:

The grip on the smoother surface will be superior to that of the coarser
surface, as long as the surfaces are clean and dry.

Carl, your basement floor may be dry and provide superior traction, but I assure
you that the basement I speak of, which was in Michigan, was never fully dry.
It's apparently the 11th commandment that MI basements will never dry. Humid,
dank, moldy - with a slicker'n'snot painted floor. I put the snot in there as a
thank-you to H2 for the petting, it really wasn't as slick as snot, but then
again some parts may have been, like the accumulation zones off to the sides of
the rollers.

Quoted message said:

Of course, in the case of either wet or dirty, whatever contaminant is on
the smooth surface would minimize or prevent skid marks on the surface
itself.

I'm sure there are plenty of other permutations.

William Asher said:

You know, it struck me that the OP is probably thinking "screw cycling,
I'm buying a treadmill."

http://www.bikeforest.com/tread/index.php

Carl Sundquist says...

Quoted message said:

Sounds like the perfect opportunity to do another 180 rpm roller dismount.

I've been expecting that. I'm certain that I can't get to 180 these days. I'd
venture a guess that by 160, I'd look like a heavy metal drummer riding a pogo
stick on crack. Besides, I no longer have the rollers nor that basement floor
and it's useless trying to repeat an experiment when the conditions are so
different. And since I no longer do alcohol, the vital mental attitude required
to do something even remotely that stupid is long gone. Plus, I've grown from
those anorexic days and the added mass might really ruin the fun. And, but,
also, you know you can never go back.

Michael Press said:

"[email hidden]" <[email hidden]> wrote:
Angular momentum is conserved.
Linear momentum is conserved.
Energy is conserved.

You are trying to argue that angular momentum converts to
linear momentum. In the world of physics it does not.

Look at the calculation I presented. I assume half the
energy of the turning wheels is converted into heating of
the floor and tires. The other half is converted into
kinetic energy of linear motion. I get a resultant speed
of ~ 5 km/hr for two 0.5 kg wheels. Angular momentum and
linear momentum are transferred to the motion of matter
associated with the earth to conserve these quantities.

Where did you get that factor of one-half of the energy going
into heat? Hand waving. When friction plays a large part in
a problem - as it must here, to decelerate the tire on contact
with the floor - arguing from energy considerations can easily
lead you astray. If you try figuring out how much energy goes
into friction in your solution, you'll see it is inconsistent.
There is a physical model behind my approach to the problem,
which I will now explain in tedious detail.

Before I get even more boring, let me say: I'm an observer,
not a theorist. I didn't try bunny hopping off rollers (I have
downstairs neighbors), but anyone can try this experiment
without rollers. Lift up the rear wheel of your bike. Put it in
high gear and crank the pedal as fast as you possibly can,
so the rear wheel is really going. Now firmly set the rear wheel
down on the floor (use an old rug), putting your weight on the
bike. You might get a little skid mark, but the forward motion
is trivial. It's not like the bike breaks out of your hand and tries
to run across the room.

Okay, here is the full model of the problem. This may seem
overly mathematical for rbr, but it's just multiplication. I'm going
to assume the rider, unlike Mad "Cuddles" Dog, stops pedaling.
Consider one wheel, with mass m1 = 0.8 kg at the rim, rotating at
speed v1=14 m/s. The instant it hits the floor, half the bike+rider
weight is on it, call this m2=40 kg. (Double the quantities to
do two wheels).

As the wheel hits the floor, the contact patch is moving at
14 m/s backward relative to the floor. The friction force of the floor
on the tire is F_fric = mu*m2*g, where g=9.8 m/s^2 and mu=0.7 is the
coefficient of friction (0.7 is about right for rubber on concrete).
Now what happens is the frictional force pushes forward on the
tire, both decelerating the wheel and pushing the bike forward.
This happens for a very short time (the skid) until the rotation
speed is decelerated to match the forward speed of the bike.
Once the speeds match, the skidding ends and the bike rolls
forward normally.

The skid time is dt and the impulse (change in momentum)
delivered to the wheel is P=F_fric * dt = mu*m2*g*dt. This
actually puts a torque on the wheel to change its angular momentum,
but since the torque and the wheel speed are both measured at
the rim, the factor of radius is the same for both. The
wheel is decelerated by P/m1, so the final wheel velocity
is v1_final = v1 - P/m1. The force puts the same change in
momentum P into the (half-weight) bike+rider, so it accelerates
the bike from 0 to v2_final = P/m2.

The skidding stops when the bike speed and wheel speed match,
so v2_final = v1_final. This means mu*g*dt = v1 - mu*g*dt*(m2/m1).
Simplifying, v1 = (1 + m2/m1) * v2_final. This is almost the
same as what I derived earlier from conservation of momentum
(it's different becase earlier I neglected the small angular momentum
remaining in the rotating wheels at the end of the skid). When I
substitute in the numbers:
final speed of bike, v2_final = 0.275 m/sec (a whopping 0.6 mph)
time of skid dt = 0.04 seconds.

Most of the rotating wheel's energy has been dissipated during the
skid as friction slows the wheel. During the skid the bike speed goes
only from 0 to 0.275 m/s, so the skid mark is very short. This
calculation gives 0.5 cm, but a real tire's contact patch is longer
than that - the skid mark has to be at least as long as the contact
patch. BTW, notice that the final speed doesn't depend on the
coefficient of friction, but the time of skid does.

My work here is done. By now you should all be bored enough
to be looking forward to Laff@me's reports from the libel trial.

Robert Chung says...

Quoted message said:

Nope. I'd say, however, that if riders can accelerate in the corners of a
velodrome with decreasing power, they can also accelerate with constant
power.

Now that you're on a roll, go ahead. It's a loaded question from the start.

Regardless, the power output was specified as constant in the original question
and the data you show does not match that spec, so is the next step to start
making unprovable assumptions?

Quoted message said:

My work here is done. By now you should all be bored

not at all. (although i'll admit (thanks to bill "sex maniac" asher)
that i did start to fantasize about getting you drunk at a party. 😉

heather

dvt said:

Kinetic energy for linear motion is (mv^2)/2. For rotational energy, as
in the wheels spinning, the energy is (Iw^2)/2, where I is the
rotational inertia and w is the angular velocity.

Do you realize that this is exactly the same? Iw^2/2 is just the
kinetic energy equation written in polar coordinates. But we already
know that the only velocity that matters is at the rim/tire... and we
know that speed. You've gone to a lot of extra trouble to calculate the
same thing.

Alfred Rider had the right idea, and the equation reduces to:
V= v*(m/M)^.5

Quoted message said:

Summary: I disagree with your physics, but my conclusion is the same.

Guess again.

Robert Chung said:

William Asher said:

You know, it struck me that the OP is probably thinking "screw cycling,
I'm buying a treadmill."

http://www.bikeforest.com/tread/index.php

Oh. My. God.

The antichrist!

I wish I could read a review of this that went like this:

http://www.theonion.com/content/node/40311

--
Bill Asher

William Asher says...

Quoted message said:

I wish I could read a review of this that went like this:

http://www.theonion.com/content/node/40311

Million laughs, Mr. Bill. I especially liked the part where he talked about
force feeding them their own balls. That's a stretch.