Bicycle riddle

On 5/28/2017 11:04 PM, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/28/2017 7:55 PM, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/28/2017 2:58 PM, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/27/2017 12:21 AM, Frank Krygowski wrote:
On Friday, May 26, 2017 at 11:48:13 PM UTC-4, Jeff Liebermann wrote:
On Fri, 26 May 2017 20:36:39 -0700 (PDT), Frank Krygowski wrote:

Bicycle riddle:
http://www.popularmechanics.com/scie...icycle-killer/
- Frank Krygowski

When you're done with that bicycle riddle, try this bicycle mechanical
problem:
http://www.popularmechanics.com/science/math/a26070/riddle-of-the-week-24/
(I guessed wrong, as usual).

I got it, but by a very different mental process than what they used.

I'll hold off discussing until others have had a chance to work on it.

- Frank Krygowski

So little interest in the riddle that was actually technical!


It was relatively obvious (by virtue of the fact that it was posed as a
riddle, there had to be some non-intuitiveness to the answer) and I deduced
the right answer. There just wasn't much left to talk about afterwards.

Just curious: How did you deduce the right answer?

I can see one very detailed way, requiring some simple math, and one
sort of shortcut way, no numbers required. I used the shortcut.


If you look at the system, the "gain" in torque
from pedals to wheels is less than one (for almost all bikes), therefore
the mechanical advantage is greater than one in the other direction. If you
had locked the drivetrain and pulled on the pedal with a string, there
would be equal and opposite forces on the pedal and the tire. Putting it
all together, if you apply some force to the tire and an equal force to the
pedal, the force applied to the tire will "win" and the bike will move
forwards.

Except you got it wrong!

For the pedal position shown (crank vertical, string pulling
horizontally backwards on the bottom pedal, and a bike with common
gearing) the bike moves _backwards_ when the string tries to rotate the
cranks in their normal direction.

See
http://www.popularmechanics.com/scie...f-the-week-24/

So the new question is: Why?

Or if people want to get more mathematical, we could ask under what
conditions would the bike move forward instead.


Yeah, that too... I actually did get it right, then botched it during the
explanation (and you can choose to believe that it or not, as you see fit).

I do believe you, because your explanation made sense right up to the
last two words.

I alluded to one possible situation where the bike would go forward, and
that was ultra low geared bikes. If the combination of long cranks, small
front chainring, large rear sprocket and small rear wheel combined to make
the bike move less than 2•pi•crank length for each rotation of the crank,
then the bike will move forwards (at first, at least).

Sounds good.

It also occurred to me that if the crank was not vertical, for some
crank angles a perpendicular string (as opposed to a horizontal one)
would still cause the bike to move forward instead of backward. The
limiting angle would depend on gear ratios, wheel diameter and crank
length. But I'm resisting the temptation to work out the formula.

--
- Frank Krygowski

Frank Krygowski wrote:
On 5/28/2017 11:04 PM, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/28/2017 7:55 PM, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/28/2017 2:58 PM, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/27/2017 12:21 AM, Frank Krygowski wrote:
On Friday, May 26, 2017 at 11:48:13 PM UTC-4, Jeff Liebermann wrote:
On Fri, 26 May 2017 20:36:39 -0700 (PDT), Frank Krygowski wrote:

Bicycle riddle:
http://www.popularmechanics.com/scie...icycle-killer/
- Frank Krygowski

When you're done with that bicycle riddle, try this bicycle mechanical
problem:
http://www.popularmechanics.com/science/math/a26070/riddle-of-the-week-24/
(I guessed wrong, as usual).

I got it, but by a very different mental process than what they used.

I'll hold off discussing until others have had a chance to work on it.

- Frank Krygowski

So little interest in the riddle that was actually technical!


It was relatively obvious (by virtue of the fact that it was posed as a
riddle, there had to be some non-intuitiveness to the answer) and I deduced
the right answer. There just wasn't much left to talk about afterwards.

Just curious: How did you deduce the right answer?

I can see one very detailed way, requiring some simple math, and one
sort of shortcut way, no numbers required. I used the shortcut.


If you look at the system, the "gain" in torque
from pedals to wheels is less than one (for almost all bikes), therefore
the mechanical advantage is greater than one in the other direction. If you
had locked the drivetrain and pulled on the pedal with a string, there
would be equal and opposite forces on the pedal and the tire. Putting it
all together, if you apply some force to the tire and an equal force to the
pedal, the force applied to the tire will "win" and the bike will move
forwards.

Except you got it wrong!

For the pedal position shown (crank vertical, string pulling
horizontally backwards on the bottom pedal, and a bike with common
gearing) the bike moves _backwards_ when the string tries to rotate the
cranks in their normal direction.

See
http://www.popularmechanics.com/scie...f-the-week-24/

So the new question is: Why?

Or if people want to get more mathematical, we could ask under what
conditions would the bike move forward instead.


Yeah, that too... I actually did get it right, then botched it during the
explanation (and you can choose to believe that it or not, as you see fit).

I do believe you, because your explanation made sense right up to the
last two words.

Thanks. This can be a pretty tough crowd some days.

I alluded to one possible situation where the bike would go forward, and
that was ultra low geared bikes. If the combination of long cranks, small
front chainring, large rear sprocket and small rear wheel combined to make
the bike move less than 2•pi•crank length for each rotation of the crank,
then the bike will move forwards (at first, at least).

Sounds good.

It also occurred to me that if the crank was not vertical, for some
crank angles a perpendicular string (as opposed to a horizontal one)
would still cause the bike to move forward instead of backward. The
limiting angle would depend on gear ratios, wheel diameter and crank
length. But I'm resisting the temptation to work out the formula.

Definitely, if the string is always perpendicular to the crank, the bike
can sometimes move forwards. The limiting case when the crank is horizontal
and the string is pulling straight down is intuitively obvious.

If the string is always pulling straight back, then I think the system has
the most "gain" in the forward direction when the cranks are vertical,
dropping to zero when the cranks are horizontal. For a system with a pedal
to tire torque gain of X, I would assume that the balance point would come
when sin(theta)•X = 1/X, but like you, I've got other stuff to do today
(and it's been a hell of a long time since my vector mechanics course).

On Monday, May 29, 2017 at 8:39:43 AM UTC-7, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/28/2017 11:04 PM, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/28/2017 7:55 PM, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/28/2017 2:58 PM, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/27/2017 12:21 AM, Frank Krygowski wrote:
On Friday, May 26, 2017 at 11:48:13 PM UTC-4, Jeff Liebermann wrote:
On Fri, 26 May 2017 20:36:39 -0700 (PDT), Frank Krygowski wrote:

Bicycle riddle:
http://www.popularmechanics.com/scie...icycle-killer/
- Frank Krygowski

When you're done with that bicycle riddle, try this bicycle mechanical
problem:
http://www.popularmechanics.com/science/math/a26070/riddle-of-the-week-24/
(I guessed wrong, as usual).

I got it, but by a very different mental process than what they used.

I'll hold off discussing until others have had a chance to work on it.

- Frank Krygowski

So little interest in the riddle that was actually technical!


It was relatively obvious (by virtue of the fact that it was posed as a
riddle, there had to be some non-intuitiveness to the answer) and I deduced
the right answer. There just wasn't much left to talk about afterwards.

Just curious: How did you deduce the right answer?

I can see one very detailed way, requiring some simple math, and one
sort of shortcut way, no numbers required. I used the shortcut.


If you look at the system, the "gain" in torque
from pedals to wheels is less than one (for almost all bikes), therefore
the mechanical advantage is greater than one in the other direction. If you
had locked the drivetrain and pulled on the pedal with a string, there
would be equal and opposite forces on the pedal and the tire. Putting it
all together, if you apply some force to the tire and an equal force to the
pedal, the force applied to the tire will "win" and the bike will move
forwards.

Except you got it wrong!

For the pedal position shown (crank vertical, string pulling
horizontally backwards on the bottom pedal, and a bike with common
gearing) the bike moves _backwards_ when the string tries to rotate the
cranks in their normal direction.

See
http://www.popularmechanics.com/scie...f-the-week-24/

So the new question is: Why?

Or if people want to get more mathematical, we could ask under what
conditions would the bike move forward instead.


Yeah, that too... I actually did get it right, then botched it during the
explanation (and you can choose to believe that it or not, as you see fit).

I do believe you, because your explanation made sense right up to the
last two words.

Thanks. This can be a pretty tough crowd some days.

I alluded to one possible situation where the bike would go forward, and
that was ultra low geared bikes. If the combination of long cranks, small
front chainring, large rear sprocket and small rear wheel combined to make
the bike move less than 2•pi•crank length for each rotation of the crank,
then the bike will move forwards (at first, at least).

Sounds good.

It also occurred to me that if the crank was not vertical, for some
crank angles a perpendicular string (as opposed to a horizontal one)
would still cause the bike to move forward instead of backward. The
limiting angle would depend on gear ratios, wheel diameter and crank
length. But I'm resisting the temptation to work out the formula.

Definitely, if the string is always perpendicular to the crank, the bike
can sometimes move forwards. The limiting case when the crank is horizontal
and the string is pulling straight down is intuitively obvious.

If the string is always pulling straight back, then I think the system has
the most "gain" in the forward direction when the cranks are vertical,
dropping to zero when the cranks are horizontal. For a system with a pedal
to tire torque gain of X, I would assume that the balance point would come
when sin(theta)•X = 1/X, but like you, I've got other stuff to do today
(and it's been a hell of a long time since my vector mechanics course).

Please tell me you guys aren't talking about a cartoon picture of a bike.

wrote:
On Monday, May 29, 2017 at 8:39:43 AM UTC-7, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/28/2017 11:04 PM, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/28/2017 7:55 PM, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/28/2017 2:58 PM, Ralph Barone wrote:
Frank Krygowski wrote:
On 5/27/2017 12:21 AM, Frank Krygowski wrote:
On Friday, May 26, 2017 at 11:48:13 PM UTC-4, Jeff Liebermann wrote:
On Fri, 26 May 2017 20:36:39 -0700 (PDT), Frank Krygowski wrote:

Bicycle riddle:
http://www.popularmechanics.com/scie...icycle-killer/
- Frank Krygowski

When you're done with that bicycle riddle, try this bicycle mechanical
problem:
http://www.popularmechanics.com/science/math/a26070/riddle-of-the-week-24/
(I guessed wrong, as usual).

I got it, but by a very different mental process than what they used.

I'll hold off discussing until others have had a chance to work on it.

- Frank Krygowski

So little interest in the riddle that was actually technical!


It was relatively obvious (by virtue of the fact that it was posed as a
riddle, there had to be some non-intuitiveness to the answer) and I deduced
the right answer. There just wasn't much left to talk about afterwards.

Just curious: How did you deduce the right answer?

I can see one very detailed way, requiring some simple math, and one
sort of shortcut way, no numbers required. I used the shortcut.


If you look at the system, the "gain" in torque
from pedals to wheels is less than one (for almost all bikes), therefore
the mechanical advantage is greater than one in the other direction. If you
had locked the drivetrain and pulled on the pedal with a string, there
would be equal and opposite forces on the pedal and the tire. Putting it
all together, if you apply some force to the tire and an equal force to the
pedal, the force applied to the tire will "win" and the bike will move
forwards.

Except you got it wrong!

For the pedal position shown (crank vertical, string pulling
horizontally backwards on the bottom pedal, and a bike with common
gearing) the bike moves _backwards_ when the string tries to rotate the
cranks in their normal direction.

See
http://www.popularmechanics.com/scie...f-the-week-24/

So the new question is: Why?

Or if people want to get more mathematical, we could ask under what
conditions would the bike move forward instead.


Yeah, that too... I actually did get it right, then botched it during the
explanation (and you can choose to believe that it or not, as you see fit).

I do believe you, because your explanation made sense right up to the
last two words.

Thanks. This can be a pretty tough crowd some days.

I alluded to one possible situation where the bike would go forward, and
that was ultra low geared bikes. If the combination of long cranks, small
front chainring, large rear sprocket and small rear wheel combined to make
the bike move less than 2•pi•crank length for each rotation of the crank,
then the bike will move forwards (at first, at least).

Sounds good.

It also occurred to me that if the crank was not vertical, for some
crank angles a perpendicular string (as opposed to a horizontal one)
would still cause the bike to move forward instead of backward. The
limiting angle would depend on gear ratios, wheel diameter and crank
length. But I'm resisting the temptation to work out the formula.

Definitely, if the string is always perpendicular to the crank, the bike
can sometimes move forwards. The limiting case when the crank is horizontal
and the string is pulling straight down is intuitively obvious.

If the string is always pulling straight back, then I think the system has
the most "gain" in the forward direction when the cranks are vertical,
dropping to zero when the cranks are horizontal. For a system with a pedal
to tire torque gain of X, I would assume that the balance point would come
when sin(theta)•X = 1/X, but like you, I've got other stuff to do today
(and it's been a hell of a long time since my vector mechanics course).

Please tell me you guys aren't talking about a cartoon picture of a bike.


There were two riddles, Tom. One was about card cheats and the other was
actually a physics problem.

On Sunday, May 28, 2017 at 2:34:03 PM UTC-7, Jeff Liebermann wrote:
On Sun, 28 May 2017 14:07:40 -0400, Frank Krygowski
wrote:

Actually, 53 bicycles in the room (as stated in the original riddle)
would mean something like 26.5 cards. Each "Bicycle" brand playing card
shows two bicycles, not just one.

Yep, although the alleged bicycle appears rather painful to ride
without a saddle and is missing the pedals:
https://cdn3.volusion.com/artgw.hyvvw/v/vspfiles/photos/CB00741-3.jpg

At least, that's what I assume, based on the name. On the cards,
they're illustrated as a head on view, ridden by angels. If not for the
name, one might suppose the angels are riding unicycles.

I would guess(tm) that a unicycle could be considered half of a
bicycle, thus returning the total count to 53.

At one time, Bicycle playing cards had recognizeable bicycles on the
back:
http://www.bicyclecards.com/wp-content/uploads/2015/06/18921.jpg

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

What kind of seat is that?

On 5/30/2017 4:48 PM, Doug Landau wrote:
On Sunday, May 28, 2017 at 2:34:03 PM UTC-7, Jeff Liebermann wrote:
On Sun, 28 May 2017 14:07:40 -0400, Frank Krygowski
wrote:

Actually, 53 bicycles in the room (as stated in the original riddle)
would mean something like 26.5 cards. Each "Bicycle" brand playing card
shows two bicycles, not just one.

Yep, although the alleged bicycle appears rather painful to ride
without a saddle and is missing the pedals:
https://cdn3.volusion.com/artgw.hyvvw/v/vspfiles/photos/CB00741-3.jpg

At least, that's what I assume, based on the name. On the cards,
they're illustrated as a head on view, ridden by angels. If not for the
name, one might suppose the angels are riding unicycles.

I would guess(tm) that a unicycle could be considered half of a
bicycle, thus returning the total count to 53.

At one time, Bicycle playing cards had recognizeable bicycles on the
back:
http://www.bicyclecards.com/wp-content/uploads/2015/06/18921.jpg

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

What kind of seat is that?


Brooks B.135 with some artistic license added.

--
Andrew Muzi
www.yellowjersey.org/
Open every day since 1 April, 1971

On Tue, 30 May 2017 14:48:20 -0700 (PDT), Doug Landau
wrote:

What kind of seat is that?

Masochists special bicycle saddle:
https://cdn3.volusion.com/artgw.hyvvw/v/vspfiles/photos/CB00741-3.jpg
If I hadn't been told that it was a bicycle wheel, I would swear that
the angel is sitting on the pointed end of a radio lattice tower.

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

On Tue, 30 May 2017 15:25:08 -0700, Jeff Liebermann
wrote:

On Tue, 30 May 2017 14:48:20 -0700 (PDT), Doug Landau
wrote:

What kind of seat is that?

Masochists special bicycle saddle:
https://cdn3.volusion.com/artgw.hyvvw/v/vspfiles/photos/CB00741-3.jpg
If I hadn't been told that it was a bicycle wheel, I would swear that
the angel is sitting on the pointed end of a radio lattice tower.

The design of Bicycle playing cards seem to date back to 1885. The
first primitive radio transmitters (called Hertzian oscillators) were
built by German physicist Heinrich Hertz in 1887... :-)
--
Cheers,

John B.

On Wed, 31 May 2017 08:15:16 +0700, John B.
wrote:

On Tue, 30 May 2017 15:25:08 -0700, Jeff Liebermann
wrote:

On Tue, 30 May 2017 14:48:20 -0700 (PDT), Doug Landau
wrote:

What kind of seat is that?

Masochists special bicycle saddle:
https://cdn3.volusion.com/artgw.hyvvw/v/vspfiles/photos/CB00741-3.jpg
If I hadn't been told that it was a bicycle wheel, I would swear that
the angel is sitting on the pointed end of a radio lattice tower.

The design of Bicycle playing cards seem to date back to 1885.

So it is written, so it must be:
http://www.bicyclecards.com/article/our-history/

The
first primitive radio transmitters (called Hertzian oscillators) were
built by German physicist Heinrich Hertz in 1887... :-)

Good point. That would make the bicycle on the card a Penny Farthing
style bicycle where the saddle is directly behind the head tube and
therefore invisible in a head-on view, as on the card.
https://www.google.com/search?q=penny+farthing+bicycle&tbm=isch
The proportions aren't quite right but everything else seems to fit.

There were lattice towers long before Hertz, Marconi, and the radio
pioneers. They were used for lighting, observation, warships, wind
power, and Eiffel's Tower. However, I will concede that the designer
of the bicycle playing card had a bicycle in mind, rather than a radio
tower.

Perception is everything and it still looks like a radio tower to me.

Is it my imagination, or is the angel on the card wearing a helmet?
There are larger angel drawings near the 4 corners of the card which
show the helmet better.
http://802.11junk.com/jeffl/crud/angel-helmet.jpg
It even has ventilation slits.

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

On Tuesday, May 30, 2017 at 11:17:24 PM UTC-4, Jeff Liebermann wrote:
Snipped
Is it my imagination, or is the angel on the card wearing a helmet?
There are larger angel drawings near the 4 corners of the card which
show the helmet better.
http://802.11junk.com/jeffl/crud/angel-helmet.jpg
It even has ventilation slits.

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

Most likely it's ust the common hair parted in the middle hairstyle worn by menn at that time.

Playing cards can be dangerous too! LOL One prisoner in 1930 commited suicide by making a pipe bomb from parts of 20 decks of cards.

http://i-p-c-s.org/faq/WilliamKogut.php

"He decided to commit suicide using only the rudimentary tools available to him in his prison cell. First of all, Kogut procured several packs of playing cards – a fairly innocuous possession, even in a prison. He began by tearing up several packs of playing cards, giving particular focus to obtaining pieces with red ink (at the time, the ink in red playing cards contained nitrocellulose, which is flammable and when wet can create an explosive mixture), and stuffed them into a pipe. He probably scraped the red ink off more than 20 decks of cards to get enough nitrocellulose. He then plugged one end of the pipe firmly with a broom handle and poured water into the other end to soak the card pieces. He then placed the pipe on a kerosene heater next to his bed and placed the open end firmly against his head. The heater turned the water into steam and eventually enough pressure built up inside the pipe to make it explode, killing him. Alternavively, he might have ignited the pipe bomb by striking the metal leg cap, as the glycerin would ignite if shaken enough. Someone reported that he languished for 2-3 days in the infirmary before dying.

Snipped

Explanation

In the 1930s, a substance called nitrocellulose was cropping up in all sorts of places, and still does now. Nitrocellulose is unstable, and decomposes easily, releasing nitric acid. This nitric acid further decomposes nitrocellulose, leading to a self-catalysing reaction. Nitrocellulose is also quite flammable, and when wet forms an explosive mixture. As Kogut lay his head down to rest, the warmth from the heater accelerated the reactions taking place within his improvised pipe bomb. Soon the concoction reached a critical state and exploded, killing Kogut instantly."

Cheers