Re: Getting Started on Rollers (Calling Jobst)

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

Quoted message said:

Sure, quite simple. Not a physics major here, althought a few might be
lurking. The momentum of the wheels alone spinning at 40 mph is not
nearly
enough to move a 155 lb rider on a 20lb bike forward. Of course I stopped
pedalling when I hopped of of the rollers. Like I said, I gave a
demonstration to the mall managers. Howabout anchovies ? They really
stink.
OK, maybe I exageratted when I said "no forward motion whatsoever". What I
meant was no significant or perceptible forward motion. What that means is
that I did not travel.

Is the momentum of two lightweight wheels at 40 mph alone enough to move
175
lbs of bike a rider enough to crash into a wall.
If I am right then you can send me a six pack of the hoppiest beer that
you
can buy.
Where is Jobst Brandt when you need him ?

Steve aka Joe Pro

Joe Pro was the name of a East Coast Cat2 in the early 1980s.
"Mad Dog" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Joe Pro says...

Quoted message said:

Not that I do not believe you, but it is impossible. I once was involved

in

Quoted message said:

Quoted message said:

setting up some roller races at a large regional mall. We had to

demonstrate

Quoted message said:

Quoted message said:

to the mall manager that if someone fell off that he would not go

crashing

Quoted message said:

Quoted message said:

into the crowd or a plate glass window. I did the demonstration and

hopped

Quoted message said:

Quoted message said:

off of the rollers at speed. No forward motion whatsoever, and I did a

track

Quoted message said:

Quoted message said:

stand.

Dunno what to say about that, Joe. I only know what I've seen. Not only

have I

Quoted message said:

seen forward motion when others rode off, I've ridden it myself into the
basement wall with boxes flying all over the place.

Your "no forward motion whatsoever" claim doesn't make physical sense.
If
you're in a big gear and spinning, it's not unreasonable for the wheel to

be

Quoted message said:

going at the equivalent of 40 mph and most folks I know can easily

maintain 25

Quoted message said:

mph on rollers with no load unit strapped on. That's a fair bit of

angular

Quoted message said:

momentum on the average bike wheelset. Are you saying that the angular

momentum

Quoted message said:

of two spinning wheels just evaporates when you go off the rollers? It's

gotta

Quoted message said:

go somewhere. Or did you develop a process to convert it into pizza? If

so,

Quoted message said:

send me a veggie deluxe, light on the cheeze, very heavy on the garlic
and
jalapenos because it's gettin' deep here and I want to contribute to the

stench.

Quoted message said:

When you ride off of rollers, you fall sideways like a trackstand gone amiss
where you can't unclip in time. . . learned first-hand on more than one
occasion. . . If you are really good and TRY really hard, you can hit the
ground, maintain your balance and CRANK to get your mass moving forward. . .
like getting started from a full-stop track stand. But the rotation of
wheels, regardless of "speed" shown on your cyclo-compuker, has virtually
zero potential to move the total mass of rider and bike forward when it hits
the floor. . .

Actually, the "no forward motion" statement makes perfect sense. Wheels
weigh something like 5-6 pounds with less than half of that actually moving
and carrying momentum while riding on rollers. And ONLY the wheels are in
motion while on rollers . . . Rider plus bike weigh 175-200 lbs. (in my
case, more like 300 lbs. for the total package). 6 pounds cannot easily
cause 175 lbs. to be propelled forward. . . . An object at rest will stay
at rest unless acted upon by an external force. 6 lbs. doesn't pack
sufficient external force to alter the static physical condition of
175-200lbs. of rider and bike. . .

"Joe Pro" wrote

Quoted message said:

Sure, quite simple. Not a physics major here, althought a few might be
lurking. The momentum of the wheels alone spinning at 40 mph is not

nearly

Quoted message said:

enough to move a 155 lb rider on a 20lb bike forward. Of course I stopped
pedalling when I hopped of of the rollers. Like I said, I gave a
demonstration to the mall managers. Howabout anchovies ? They really

stink.

Quoted message said:

OK, maybe I exageratted when I said "no forward motion whatsoever". What I
meant was no significant or perceptible forward motion. What that means is
that I did not travel.

Is the momentum of two lightweight wheels at 40 mph alone enough to move

175

Quoted message said:

lbs of bike a rider enough to crash into a wall.
If I am right then you can send me a six pack of the hoppiest beer that

you

Quoted message said:

can buy.
Where is Jobst Brandt when you need him ?

Steve aka Joe Pro

Interesting problem. Here is what I came up with:

1884g Shimano R500 wheels without cassette or skewer

440 two Michelin Pro2 Race tires 23mm

200 two Continental tubes

50 two Zefal rim tapes

2574 TOTAL weight

453.6 grams is one pound.

5.67 pounds is total weight of wheels.

Assume the total energy of the wheels rotating at 40 mph is converted into
forward motion of the rider and bike. Assume the entire mass of the wheels
is at the tires. Assume the bike is 20 pounds and the rider is 155 pounds
giving a total of 175 pounds.

Kinetic energy is one half mass times velocity squared, if I remember my
grade school physics,

The velocity of the rider and bike after coming off of the rollers is V in
the following equation.

E=5.67x40x40/2=175xVxV/2

This gives V=7.2 mph

Actually 20 mph is more reasonable for a fairly fast ride on rollers. This
brings the exit speed down to 3.6 mph. And since not all of the mass of the
wheels is at the periphery as assumed above, recalculate taking the weight
of the rims instead of the wheels. Mavic Open Pro rims are 425g each so that
the rotating mass drops from 2574 to 1540 g. This drops the exit speed from
3.6 mph down to 2.8. This is 4.1 feet per second, which is much faster than
I would have guessed. But if you put on the brakes fairly quickly, you will
not have gone very far.

It seems to me that forgetting to stop peddling is far more important than
the energy in the rotating wheels.

Quoted message said:

Where is Jobst Brandt when you need him ?

He's probably riding to the beach, like I did today. Californians
don't need no steenking rollers... . Predicted 75F tomorrow. See you
in San Gregorio!

Alfred Ryder said:

Actually 20 mph is more reasonable for a fairly fast ride on rollers. This
brings the exit speed down to 3.6 mph. And since not all of the mass of the
wheels is at the periphery as assumed above, recalculate taking the weight
of the rims instead of the wheels. Mavic Open Pro rims are 425g each so that
the rotating mass drops from 2574 to 1540 g. This drops the exit speed from
3.6 mph down to 2.8. This is 4.1 feet per second, which is much faster than
I would have guessed. But if you put on the brakes fairly quickly, you will
not have gone very far.

I think you did a good job of showing what we are looking at here. To
simplify your equation a bit:

V= v*(m/M)^.5

... where large letters are the total bike and rider and small letters
are the wheels only. If you wanted to get picky I suppose you could put
the kinetic energy of all the rotating parts in the equation (including
the riders feet and legs), and also include the rotational energy after
the rider jumps off...

But, there is a bigger problem... how much energy gets dissipated by
the rear tire skidding on the floor? Could be a lot...

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

Quoted message said:

"Joe Pro" wrote

Quoted message said:

Sure, quite simple. Not a physics major here, althought a few might be
lurking. The momentum of the wheels alone spinning at 40 mph is not

nearly

Quoted message said:

enough to move a 155 lb rider on a 20lb bike forward. Of course I

stopped

Quoted message said:

Quoted message said:

pedalling when I hopped of of the rollers. Like I said, I gave a
demonstration to the mall managers. Howabout anchovies ? They really

stink.

Quoted message said:

OK, maybe I exageratted when I said "no forward motion whatsoever". What

I

Quoted message said:

Quoted message said:

meant was no significant or perceptible forward motion. What that means

is

Quoted message said:

Quoted message said:

that I did not travel.

Is the momentum of two lightweight wheels at 40 mph alone enough to move

175

Quoted message said:

lbs of bike a rider enough to crash into a wall.
If I am right then you can send me a six pack of the hoppiest beer that

you

Quoted message said:

can buy.
Where is Jobst Brandt when you need him ?

Steve aka Joe Pro

Interesting problem. Here is what I came up with:

1884g Shimano R500 wheels without cassette or skewer

440 two Michelin Pro2 Race tires 23mm

200 two Continental tubes

50 two Zefal rim tapes

2574 TOTAL weight

453.6 grams is one pound.

5.67 pounds is total weight of wheels.

Assume the total energy of the wheels rotating at 40 mph is converted into
forward motion of the rider and bike. Assume the entire mass of the wheels
is at the tires. Assume the bike is 20 pounds and the rider is 155 pounds
giving a total of 175 pounds.

Kinetic energy is one half mass times velocity squared, if I remember my
grade school physics,

The velocity of the rider and bike after coming off of the rollers is V in
the following equation.

E=5.67x40x40/2=175xVxV/2

This gives V=7.2 mph

Actually 20 mph is more reasonable for a fairly fast ride on rollers. This
brings the exit speed down to 3.6 mph. And since not all of the mass of

the

Quoted message said:

wheels is at the periphery as assumed above, recalculate taking the weight
of the rims instead of the wheels. Mavic Open Pro rims are 425g each so

that

Quoted message said:

the rotating mass drops from 2574 to 1540 g. This drops the exit speed

from

Quoted message said:

3.6 mph down to 2.8. This is 4.1 feet per second, which is much faster

than

Quoted message said:

I would have guessed. But if you put on the brakes fairly quickly, you

will

Quoted message said:

not have gone very far.

It seems to me that forgetting to stop peddling is far more important than
the energy in the rotating wheels.

Science rules !!!!

Laz

When I rode off the side of the rollers that were on the tile floor of Ansel
John's shop in West Carrollton OH in 1968, it left two (yes, two) black skid
marks on the tile, each two or three inches long. The bike didn't progress
much, if at all.

Al Williams

Al Williams said:

When I rode off the side of the rollers that were on the tile floor of Ansel
John's shop in West Carrollton OH in 1968, it left two (yes, two) black skid
marks on the tile, each two or three inches long.

Of course! There would be two; not just the rear... doh!

Alfred Ryder said:

"Joe Pro" wrote

Quoted message said:

Sure, quite simple. Not a physics major here, althought a few might be
lurking. The momentum of the wheels alone spinning at 40 mph is not

nearly

Quoted message said:

enough to move a 155 lb rider on a 20lb bike forward.

Quoted message said:

Interesting problem. Here is what I came up with:

Quoted message said:

1884g Shimano R500 wheels without cassette or skewer
440 two Michelin Pro2 Race tires 23mm
200 two Continental tubes
50 two Zefal rim tapes
2574 TOTAL weight

Quoted message said:

453.6 grams is one pound.

Quoted message said:

5.67 pounds is total weight of wheels.

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.

First I'll assume that the spokes and hub don't contribute much to the
rotational inertia. Then I'll estimate the distance from axis of
rotation to the center of mass for the rim/tire/tube/tape combination to
be 622 mm. Finally, I'll use 500g as an estimate of the rim mass. Then

I = m r^2
= (500 + 220 + 100 + 25) g * (622 mm)^2
= 0.33 kg m^2

The rotational speed (w) is 2 pi * linear velocity / circumference, or

w = 2 * pi * 40 mph / (2 * pi * (622+46) mm)
= 27 rad/sec

(note the mixed units; conversion factors required!)

So the kinetic energy of one rolling wheel is

KE = (I w^2)/2
= 0.33 kg m^2 * (27 rad/sec)^2 / 2
= 117 J

Now translate that into forward velocity of the rider. Assuming the 175
lb total for bike and rider, and assuming that all of the KE of the
wheels is conserved (none lost to skid marks, brakes, etc)...

KE = (m v^2)/2
v = sqrt(2 * KE / m)
= sqrt(2 * (2*117J) / 175 lb)
= 2.4 m/s (5.4 mph)

(note the extra 2 in the equation for the KE of two wheels, and the
mixed units requiring conversion factors that I have left out)

Summary: I disagree with your physics, but my conclusion is the same.
The rotational inertia of the wheels *alone* won't propel the rider very
far or very fast. A little energy lost to skidding wheels, maybe a
slight touch of the brakes, and the rider might move a few feet before
stopping.

By the way, I have tested this experimentally, but at less than 40 mph.
The experiment was unintended 🙂.

--
Dave
dvt at psu dot edu

Wheels by BFWG said:

When you ride off of rollers, you fall sideways like a trackstand gone amiss
where you can't unclip in time. . . learned first-hand on more than one
occasion. . . If you are really good and TRY really hard, you can hit the
ground, maintain your balance and CRANK to get your mass moving forward. . .
like getting started from a full-stop track stand.

Congratulations on enough forward motion to get the rear wheel to
clear the front roller(s). Might be easier to do on the old small
single front roller folding Cinellis, but would defintiely be a
control issue on any larger roller, single or double. I'm guessing a
cyclocrosser with really quick reflexes to do it, accident or
otherwise. Me, I'd probably hit the roller with a pedal, then fold the
rear wheel...

Curtis L. Russell
Odenton, MD (USA)
Just someone on two wheels...

On Tue, 15 Nov 2005 05:17:10 GMT, "Alfred Ryder" <[email hidden]>

Quoted message said:

Quoted message said:

Where is Jobst Brandt when you need him ?

Steve aka Joe Pro

Interesting problem. Here is what I came up with:

1884g Shimano R500 wheels without cassette or skewer

440 two Michelin Pro2 Race tires 23mm

200 two Continental tubes

50 two Zefal rim tapes

2574 TOTAL weight
and other stuff...

Are you guys riding with rollers with platforms between the front and
rear rollers that is keeping the rear wheel up and off of both the
ground and smacking the [censored] out of the front roller in all this front
motion stuff? Seems like you have to figure the effort to do the hop
to clear all this stuff.

OTOH, I've never had enough spin to actually generate lift, so what do
I know?

Curtis L. Russell
Odenton, MD (USA)
Just someone on two wheels...

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.

First I'll assume that the spokes and hub don't contribute much to the
rotational inertia. Then I'll estimate the distance from axis of
rotation to the center of mass for the rim/tire/tube/tape combination to
be 622 mm. Finally, I'll use 500g as an estimate of the rim mass. Then

622mm would be the diameter hence r = 311mm making the kinetic energy
1/4 of that you calculated.

Quoted message said:

I = m r^2
= (500 + 220 + 100 + 25) g * (622 mm)^2
= 0.33 kg m^2

The rotational speed (w) is 2 pi * linear velocity / circumference, or

w = 2 * pi * 40 mph / (2 * pi * (622+46) mm)
= 27 rad/sec

(note the mixed units; conversion factors required!)

So the kinetic energy of one rolling wheel is

KE = (I w^2)/2
= 0.33 kg m^2 * (27 rad/sec)^2 / 2
= 117 J

29.25 J

Quoted message said:

Now translate that into forward velocity of the rider. Assuming the 175
lb total for bike and rider, and assuming that all of the KE of the
wheels is conserved (none lost to skid marks, brakes, etc)...

KE = (m v^2)/2
v = sqrt(2 * KE / m)
= sqrt(2 * (2*117J) / 175 lb)
= 2.4 m/s (5.4 mph)

2.7 mph then.

Quoted message said:

Summary: I disagree with your physics, but my conclusion is the same.
The rotational inertia of the wheels *alone* won't propel the rider very
far or very fast. A little energy lost to skidding wheels, maybe a
slight touch of the brakes, and the rider might move a few feet before
stopping.

By the way, I have tested this experimentally, but at less than 40 mph.
The experiment was unintended 🙂.

--
The email address is only valid until the end of next month. If you need to
contact me after that then please use
http://cgi.mellis.force9.co.uk/mailme.php

Mike Ellis (news) said:

622mm would be the diameter hence r = 311mm making the kinetic energy
1/4 of that you calculated.

Good catch. Thanks.

--
Dave
dvt at psu dot edu

George Mueller said:

Quoted message said:

Where is Jobst Brandt when you need him?

Quoted message said:

He's probably riding to the beach, like I did today. Californians
don't need no steenking rollers... Predicted 75F tomorrow. See you
in San Gregorio!

Well that was last week via Big Basin redwoods and Gazos Creek in
wonderful fall weather. This Sunday it was Mt. Hamilton where
beautiful fall foliage was augmented by a herd of a dozen adult wild
pigs (shoot then varmints) in Grant Park and scores of turkeys from
the San Jose hills to Livermore. Among other wildlife, on two
occasions, a Roadrunner crossed in front of us and stood there long
enough for us to appreciate its beautiful plumage. They did not say
"mbeep beep". A flock of about 100 migrating Canada Geese were
waiting for sunset to continue their trek as they ate grass on the
lawns of the large office center with the geyser like fountain at the
west end of Livermore.

As for riding rollers, I haven't, but I have enough friends who did
and reported crashes in which some bent forks when hitting the wall
when the bicycle came off at speed. Jumping off the bicycle while at
speed, holding onto the saddle so it doesn't launch itself, also
works, but the rotating energy of the wheels must dissipate somewhere
so they coast down for a while.

Jobst Brandt

In rec.bicycles.tech Curtis L. Russell said:

Quoted message said:

When you ride off of rollers, you fall sideways like a trackstand
gone amiss where you can't unclip in time... learned first-hand on
more than one occasion... If you are really good and TRY really
hard, you can hit the ground, maintain your balance and CRANK to
get your mass moving forward... like getting started from a
full-stop track stand.

Quoted message said:

Congratulations on enough forward motion to get the rear wheel to
clear the front roller(s). Might be easier to do on the old small
single front roller folding Cinellis, but would definitely be a
control issue on any larger roller, single or double. I'm guessing a
cyclocrosser with really quick reflexes to do it, accident or
otherwise. Me, I'd probably hit the roller with a pedal, then fold
the rear wheel...

Coming off the rollers is generally a riding-off-the-side event and
that doesn't require riding over the forward roller.

Jobst Brandt

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.

Ron Ruff said:

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.

I hadn't thought of it that way. Now that you point it out, the mass of
rim/tyre/tubes/tape are all traveling at 40 mph or slightly less in the
scenario presented. Yes, that would have been easier to calculate.

Quoted message said:

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

He used the mass of the entire wheel (including hubs) traveling at 40
mph. That's why he came up with a much higher estimate than I.

Quoted message said:

Quoted message said:

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

Quoted message said:

Guess again.

If Alfred and I used the same physics, how do you figure that we came up
with different answers? We can't both be right.

--
Dave
dvt at psu dot edu

Ron Ruff said:

I think you did a good job of showing what we are looking at here. To
simplify your equation a bit:

V= v*(m/M)^.5

... where large letters are the total bike and rider and small letters
are the wheels only. If you wanted to get picky I suppose you could put
the kinetic energy of all the rotating parts in the equation (including
the riders feet and legs), and also include the rotational energy after
the rider jumps off...

But, there is a bigger problem... how much energy gets dissipated by
the rear tire skidding on the floor? Could be a lot...

It is a lot. The friction between the floor and the tire is what slows
the wheel down, so doing this problem in terms of conservation
of energy is somewhat futile. It's better to think about momentum and
the force the floor exerts on the tire during the skid. If you look at
this thread in rec.bicycles.racing, I posted a tediously-worked
out derivation there. The answer, in your notation above, is
final V = v/(1+M/m) (no square root, because momentum is being
conserved, not energy). If the rider stops pedaling when he hops off,
the forward speed generated when tires hit floor is near zero.

dvt said:

He used the mass of the entire wheel (including hubs) traveling at 40
mph. That's why he came up with a much higher estimate than I.

If Alfred and I used the same physics, how do you figure that we came up
with different answers? We can't both be right.

His final calculation was rims and tires, etc weight of 1540g (3.38lb),
and 20mph speed, 175lb rider:

20*(3.38/175)^.5= 2.78mph

At 40mph it would be double... 5.56mph

Using your numbers, 1000g (2.2lb, too low BTW), 40mph, 175lb rider:

40*(2.2/175)^.5= 4.48mph ... which is different than what you got. I
don't know why. Maybe a conversion factor was wrong, or some other
error? That's why it is good to simplify and use dimensionless ratios
before calculating... if possible.

Quoted message said:

It is a lot. The friction between the floor and the tire is what slows
the wheel down, so doing this problem in terms of conservation
of energy is somewhat futile. It's better to think about momentum and
the force the floor exerts on the tire during the skid. If you look at
this thread in rec.bicycles.racing, I posted a tediously-worked
out derivation there. The answer, in your notation above, is
final V = v/(1+M/m) (no square root, because momentum is being
conserved, not energy). If the rider stops pedaling when he hops off,
the forward speed generated when tires hit floor is near zero.

Great! The definitive answer... now we know why those skidmarks are
only a few inches long and the speed is nil.

But that still doesn't solve the mystery of why all the posts from this
thread are getting stuck in a similar (and very huge) one at RBR...