Perpetual motion!

Quoted message said:

On Fri, 1 Feb 2008 03:27:17 -0800 (PST),

Quoted post said:

Hi All,

Suppose a Rube Goldberg bicycle of sorts was constructed with a gizmo
that raised some weights when the freewheel was freewheeling (down a
hill), and allowed these weight to lower when pedaling, using this
potential energy to assist in pedalling.

Would this contraption be faster over a hilly course with the system
engaged vs disengaged and just along for the ride?

Joseph

Dear Joseph,

Forget the weights--the more energy they store, the heavier they are
and the harder it will be for you to reach the top of the hill and
find out that they don't work.

You can in theory store a lot of energy in light weights by lifting them
very high.

Of course only in theory.

Quoted message said:

Just imagine storing force by winding a spring up by some brake gizmo
on the downhill. Or charging a battery.

Overall, you still lose.

You can't get more energy out than you put into closed system. You
can't even get as much energy out as you put in.

It's not a closed system though. You're losing energy to the wind. If
you can keep your speed more constant (so slower down and faster up
hills) you can have a higher average speed over the course for a given
maximum power and energy use. It's just like flattening the course out.

And then if you have to brake anyway for some reason, regenerative
braking is definitely a good idea in principle.

On Feb 1, 11:42 pm, Ben C <[email hidden]> wrote:
SNIP!

Quoted message said:

the hills when you take the stored energy back. You will have a
smaller deviation in your speed over the course and a higher average
speed for the same net energy expenditure. By avoiding such high speeds
you lose less energy in total to the wind. This is because wind
resistance power goes roughly as the cube of speed.

This is why in the real world when I am in a long ride/race (6+ hours)
that I am trying to make good time on, I don't pedal when over 45km/h
or so.

Joseph

Ben C said:

You could store a lot of energy in a carbon-fibre pea weighing 5g if you
winched it hundreds of miles into the air. That bike would probably work
quite well (this is a thought experiment of course...). On that machine
you'd winch the weight up at a much faster rate than that at which you
descended.

Dear Ben,

No, you can't store a lot of energy in a 5 gram pea by winching it
hundreds of miles into the air, at least not in the sense of useful
storage that allows the energy to be used for anything.

That's the problem with all perpetual motion schemes. They depend upon
ignoring frictional losses when storing and releasing energy, usually
by complicating matters to obscure what must happen or else by simply
ignoring the obvious.

Think about the practical energy losses involved in any scheme to
"winch" a pea a single mile into the air--that is, how much energy
will be needed to raise the pea with any mechanical contrivance?

Now think about the practical energy losses involved in releasing the
tiny amount of stored energy. The pea will not have enough energy to
pull any mile-long thread downward.

Consider a hailstone--the usefully available energy is demonstrated
when it strikes the ground. The same amount of energy reversed is
nowhere near enough to raise it back into the clouds. Even in free
fall, a pea reaches terminal velocity within a hundred feet.

Or consider the reverse. If you fire a typical .30-'06 bullet straight
up, it heads skyward at about 3,000 feet per second, roughly 2,000
mph. When it returns to the ground, the much denser and more
aerodynamic slug will be going far faster than the light, round pea,
but it won't be doing anywhere near 3,000 fps. The energy losses up
and down are not recoverable.

Joseph's original question about raising a weight using downhill power
and releasing it on the uphill could be simplified to having the rider
stand up on pedals on the downhill (raising the weight) and then
sitting down as he goes uphill.

The power to raise the rider's weight has to come from somewhere, some
of it will be lost in the raising, and more of it will be lost in the
lowering.

Overall, the center of mass will end up at exactly the same spot.
Shifting it up and down further will simply entail more energy losses.

Cheers,

Carl Fogel

Ben C said:

Quoted message said:

On Fri, 1 Feb 2008 03:27:17 -0800 (PST),

Quoted post said:

Hi All,

Suppose a Rube Goldberg bicycle of sorts was constructed with a gizmo
that raised some weights when the freewheel was freewheeling (down a
hill), and allowed these weight to lower when pedaling, using this
potential energy to assist in pedalling.

Would this contraption be faster over a hilly course with the system
engaged vs disengaged and just along for the ride?

Joseph

Dear Joseph,

Forget the weights--the more energy they store, the heavier they are
and the harder it will be for you to reach the top of the hill and
find out that they don't work.

You can in theory store a lot of energy in light weights by lifting them
very high.

Of course only in theory.

Quoted message said:

Just imagine storing force by winding a spring up by some brake gizmo
on the downhill. Or charging a battery.

Overall, you still lose.

You can't get more energy out than you put into closed system. You
can't even get as much energy out as you put in.

It's not a closed system though. You're losing energy to the wind. If
you can keep your speed more constant (so slower down and faster up
hills) you can have a higher average speed over the course for a given
maximum power and energy use. It's just like flattening the course out.

And then if you have to brake anyway for some reason, regenerative
braking is definitely a good idea in principle.

Dear Ben,

Try to give some figures illustrating how you're going to store and
release energy on a bicycle so that the same power from the bicyclist
produces a faster time overall.

Let's say that you start at the top of a 10-mile hill and descend at
30 mph in 20 minutes. Then you pedal back up the same 10-mile hill at
10 mph, which takes 60 minutes.

Total time, 80 minutes.

Figure out how many watts you're going to store going downhill and how
much that _must_ slow you down, and you can figure out how many
watt-minutes of energy you'll have.

Then spread that energy out over the uphill climb and see whether it
lets you reach the top in less than 80 minutes.

Most perpetual motion schemes involve forgetting the times involved or
the absolute speed gains over time rather than the more dramatic
percentage speed gain with time ignored.

Cheers,

Carl Fogel

1. Is there an energy loss going up and down hills compared to flat
riding?

The answer to this is clearly yes, no one can average as high a speed
in rolling terrain as flat.

Much of that loss is the extreme speed going downhill losing potential
energy to wind losses.

2. With a no loss storage device, can some otherwise lost energy be
retained.

Again yes.

So it is just a matter of finding a storage device whose efficiency
loss is less that of the loss of rolling terrain. That of course is
quite hard to do. The one in vogue right now is electric regeneration.
For a number of reasons, that would not work well on a bicycle. The
weight of the batteries (or capacitive device) alone would knock it
out of contention. Some sort of air or liquid compression system could
work theoretically, however the pressures required to be low loss
would again not be practical for bicycles (not sure that anyone has
successfully created such a mechanism for any vehicle). A high speed
very light flywheel is probably the best bet and has had some success
in heavy vehicles.

Regular readers of Scientific American and Popular Science will have
encountered many such storage system discussions over the years. The
largest recovery of energy is in systems with frequent stops, such as
city traffic (which is why the Prius has a higher city mpg rating). It
is unlikely that any of the systems would work for such light
vehicles. And the energy loss in question is rather small
quantitatively. Not really worth pursuing.

Tom Nakashima said:

So at the race, all these fathers & mothers brought their exotic cars.
Painted
all shinny and decaled beautiful. All I had was a unpainted wedge, wasn't
even sanded and looked pretty crude, but I ended up winning the Adult race.

Did you also offset one front wheel so it didn't really carry any
weight? That can make enough difference in friction to give a Pinewood
an edge. I *think* it's within the rules.

velodancer said:

1. Is there an energy loss going up and down hills compared to flat
riding?

The answer to this is clearly yes, no one can average as high a speed
in rolling terrain as flat.

Much of that loss is the extreme speed going downhill losing potential
energy to wind losses.

2. With a no loss storage device, can some otherwise lost energy be
retained.

Again yes.

So it is just a matter of finding a storage device whose efficiency
loss is less that of the loss of rolling terrain. That of course is
quite hard to do. The one in vogue right now is electric regeneration.
For a number of reasons, that would not work well on a bicycle. The
weight of the batteries (or capacitive device) alone would knock it
out of contention. Some sort of air or liquid compression system could
work theoretically, however the pressures required to be low loss
would again not be practical for bicycles (not sure that anyone has
successfully created such a mechanism for any vehicle). A high speed
very light flywheel is probably the best bet and has had some success
in heavy vehicles.

Regular readers of Scientific American and Popular Science will have
encountered many such storage system discussions over the years. The
largest recovery of energy is in systems with frequent stops, such as
city traffic (which is why the Prius has a higher city mpg rating). It
is unlikely that any of the systems would work for such light
vehicles. And the energy loss in question is rather small
quantitatively. Not really worth pursuing.

Dear VD,

In this thread, there's a crucial and easily overlooked distinction
between Joseph's scenario and regenerative braking.

Joseph is asking about the overall speed difference between no braking
at all (just go downhill) and braking that somehow stores energy
efficiently (go slower down hill, store energy, and release it for the
uphill).

In other words, no-braking versus braking and trying to recover
energy.

Regenerative city braking is quite different.

In city driving, the vehicle must brake and slow down for turns and
stops, whether the energy is stored or not.

That's braking versus braking and trying to recover energy.

Cheers,

Carl Fogel

On Fri, 1 Feb 2008 11:52:06 -0800 (PST),
"[email hidden]" <[email hidden]> may have

Quoted message said:

How much energy is it to lift the 10kg 1m?

That's simple physics. 98 Joules will be stored in the weight, but
the amount used to get it there will be greater. The faster you lift
the weight, the more you lose to acceleration, and the less of the
consumed energy you can recover later. Given fairly typical losses
and a slow rate of lift, probably closer to 101J would be used to lift
the weight, of which a probable maximum of 95J can be recovered when
the weight is lowered by the same mechanism.

Quoted message said:

How much does that slow the descent?

Over a 1km 5% grade, bloody little, no matter how fast (within
practical limits) you lift the mass. 100J is just not all that much
energy. On a 5% 1km grade descent, the rider is going to be on the
brakes for a significant portion of the distance...or going much too
fast for safety at the bottom. Terminal velocity is probably going to
be reached.

Quoted message said:

How much faster is the 1km climb with this "additional"
energy?

Much slower; the exact figures depend upon the initial velocity of the
system at the start of the climb and the mechanism's descent rate for
the mass, but for any reasonable rate of application of the available
energy, the 100J is easily consumed in the first 80 to 100m of the
climb just by the effort needed to get the 10Kg itself up the grade.
However, assuming that it feeds force out at the same rate that force
was fed into it, and assuming that this rate was below the point at
which the storage was at a rate that slowed the descent, then due to
friction losses in the mechanism it will not even completely
compensate for the added effort required to carry the additional
weight up that section...and from there on, it's entirely and solely
additional dead weight that the rider must haul up by additional power
output. If the rate of storage and recovery is higher, the only net
effect is that the point at which it becomes dead weight is reached
sooner. The amount of energy that can be stored is limited by
gravity, and the amount that can be recovered is *always* less than
what went in.

This actually is very close to something that was available as an
extra-credit exercise in my Phys I class in college. I didn't need
the extra credit at that point, so I didn't go through the whole
thing, but the bottom line was that this is an energy sink when
real-world factors come into play.

--
My email address is antispammed; pull WEEDS if replying via e-mail.
Typoes are not a bug, they're a feature.
Words processed in a facility that contains nuts.

On Fri, 01 Feb 2008 21:12:58 -0700, [email hidden] may have

Quoted message said:

Regenerative city braking is quite different.

In city driving, the vehicle must brake and slow down for turns and
stops, whether the energy is stored or not.

That's braking versus braking and trying to recover energy.

And in any event, the other-poster-cited vehicle, the Toyota Prius,
derives most of its added efficiency not from regen braking but from
the fact that the hybrid drivetrain can make much more efficient use
of its IC engine than is done with a conventional drivetrain. Toyota
admitted somewhere that regen braking makes only about a 5% difference
under optimum conditions, which isn't much...and optimum conditions
are seldom obtained.

--
My email address is antispammed; pull WEEDS if replying via e-mail.
Typoes are not a bug, they're a feature.
Words processed in a facility that contains nuts.

Werehatrack said:

On Fri, 01 Feb 2008 21:12:58 -0700, [email hidden] may have

Quoted message said:

Regenerative city braking is quite different.

In city driving, the vehicle must brake and slow down for turns and
stops, whether the energy is stored or not.

That's braking versus braking and trying to recover energy.

And in any event, the other-poster-cited vehicle, the Toyota Prius,
derives most of its added efficiency not from regen braking but from
the fact that the hybrid drivetrain can make much more efficient use
of its IC engine than is done with a conventional drivetrain. Toyota
admitted somewhere that regen braking makes only about a 5% difference
under optimum conditions, which isn't much...and optimum conditions
are seldom obtained.

Dear Werehatrack,

Yes, the regenerative claims are usually wishful thinking.

My father loved his little Honda hybrid.

I didn't point out that on a typical drive, he touched his brakes once
at the end of his driveway in Beulah, once again at a stop sign at the
top of a hill two miles away, and once again when he pulled into the
WalMart parking lot 25 miles later in Pueblo.

Touching the brakes gently three times in 27 miles didn't provide much
power to recover.

His high gas mileage was the product of a very small, very light, and
very aerodynamic car driving on a nearly empty 2-lane 65 mph highway.

If he hadn't been hauling the hybrid's heavy batteries, he'd have
gotten even better mileage.

Cheers,

Carl Fogel

Werehatrack said:

On Fri, 1 Feb 2008 11:52:06 -0800 (PST),
"[email hidden]" <[email hidden]> may have

Quoted message said:

How much energy is it to lift the 10kg 1m?

That's simple physics. 98 Joules will be stored in the weight, but
the amount used to get it there will be greater. The faster you lift
the weight, the more you lose to acceleration, and the less of the
consumed energy you can recover later. Given fairly typical losses
and a slow rate of lift, probably closer to 101J would be used to lift
the weight, of which a probable maximum of 95J can be recovered when
the weight is lowered by the same mechanism.

Quoted message said:

How much does that slow the descent?

Over a 1km 5% grade, bloody little, no matter how fast (within
practical limits) you lift the mass. 100J is just not all that much
energy. On a 5% 1km grade descent, the rider is going to be on the
brakes for a significant portion of the distance...or going much too
fast for safety at the bottom. Terminal velocity is probably going to
be reached.

Quoted message said:

How much faster is the 1km climb with this "additional"
energy?

Much slower; the exact figures depend upon the initial velocity of the
system at the start of the climb and the mechanism's descent rate for
the mass, but for any reasonable rate of application of the available
energy, the 100J is easily consumed in the first 80 to 100m of the
climb just by the effort needed to get the 10Kg itself up the grade.
However, assuming that it feeds force out at the same rate that force
was fed into it, and assuming that this rate was below the point at
which the storage was at a rate that slowed the descent, then due to
friction losses in the mechanism it will not even completely
compensate for the added effort required to carry the additional
weight up that section...and from there on, it's entirely and solely
additional dead weight that the rider must haul up by additional power
output. If the rate of storage and recovery is higher, the only net
effect is that the point at which it becomes dead weight is reached
sooner. The amount of energy that can be stored is limited by
gravity, and the amount that can be recovered is *always* less than
what went in.

This actually is very close to something that was available as an
extra-credit exercise in my Phys I class in college. I didn't need
the extra credit at that point, so I didn't go through the whole
thing, but the bottom line was that this is an energy sink when
real-world factors come into play.

--
My email address is antispammed; pull WEEDS if replying via e-mail.
Typoes are not a bug, they're a feature.
Words processed in a facility that contains nuts.

No question carrying the weight of the contraption makes the whole
enterprise a slow one (except on the descent!) but is it faster with
the system engaged, or the system disenganged? Seems to be. The effect
would be super small, but taking 100J out of the descent and putting
it into the ascent would make a difference. How much?

Joseph

Ben C said:

We'll say the bike and rider weigh 100kg. At the top of a 10 mile (or
16093 metre) hill, we've got 1609kJ of gravitational potential energy.

Only if the figures for the incline (considering that there are no
16093 metre hills on Earth when measured vertically) and the gravity
acceleration cancel each other out...

In article <[email hidden]>,

Tom Nakashima said:

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

Quoted message said:

Many a student hour has been used in trying to overcome these
limitations in enginering schools around the world. Oddly enough, not
a single one has yielded a "better mousetrap".

Got that right.
Years ago when my son was young and in the cub scouts, I remember
the pinewood derby races. We actually had adult races to keep us out of the
kids projects. I asked quite a few physicist and engineers on weight
placement, of course everyone had different ideas, none on which I used.
You would think the scouts would have this down to perfection by now after
all these years of pinewood derby racing, but they still argue on where to
place the weight.

I ended up doing my own test with a sliding weight on a wedge for a curved
sloped track, and using an electronic light source to record time in
milliseconds. As it turned out, placing the weight above the rear wheels
(equally divided) and slightly forward (kind of about where one sits on a
bicycle slightly forward of the rear wheel) made for the fastest times.

So at the race, all these fathers & mothers brought their exotic cars.
Painted
all shinny and decaled beautiful. All I had was a unpainted wedge, wasn't
even sanded and looked pretty crude, but I ended up winning the Adult race.

The weight back falls further than the weight forward.

--
Michael Press

On Sat, 2 Feb 2008 02:00:03 -0800 (PST),
"[email hidden]" <[email hidden]> may have

Quoted message said:

No question carrying the weight of the contraption makes the whole
enterprise a slow one (except on the descent!) but is it faster with
the system engaged, or the system disenganged? Seems to be. The effect
would be super small, but taking 100J out of the descent and putting
it into the ascent would make a difference. How much?

You have provided insufficient parameters to answer that question
precisely, but I found no solution which made it faster for the broad
range of possible ones that you gave. It *always* came out as a
loser. 100J is a *tiny* amount of energy by comparison to the amount
lost to dragging the extra 10Kg up the climb.

--
My email address is antispammed; pull WEEDS if replying via e-mail.
Typoes are not a bug, they're a feature.
Words processed in a facility that contains nuts.

Werehatrack said:

On Sat, 2 Feb 2008 02:00:03 -0800 (PST),
"[email hidden]" <[email hidden]> may have

Quoted message said:

No question carrying the weight of the contraption makes the whole
enterprise a slow one (except on the descent!) but is it faster with
the system engaged, or the system disenganged? Seems to be. The effect
would be super small, but taking 100J out of the descent and putting
it into the ascent would make a difference. How much?

You have provided insufficient parameters to answer that question
precisely, but I found no solution which made it faster for the broad
range of possible ones that you gave. It *always* came out as a
loser. 100J is a *tiny* amount of energy by comparison to the amount
lost to dragging the extra 10Kg up the climb.

But how could it be worse than carrying the contraption anyway but not
engaging it? Assuming the mechanism itself is 100% efficient, which of
course it wouldn't be.

On Sun, 03 Feb 2008 03:20:43 -0600, Ben C <[email hidden]> may

have said:

Werehatrack said:

On Sat, 2 Feb 2008 02:00:03 -0800 (PST),
"[email hidden]" <[email hidden]> may have

Quoted message said:

No question carrying the weight of the contraption makes the whole
enterprise a slow one (except on the descent!) but is it faster with
the system engaged, or the system disenganged? Seems to be. The effect
would be super small, but taking 100J out of the descent and putting
it into the ascent would make a difference. How much?

You have provided insufficient parameters to answer that question
precisely, but I found no solution which made it faster for the broad
range of possible ones that you gave. It *always* came out as a
loser. 100J is a *tiny* amount of energy by comparison to the amount
lost to dragging the extra 10Kg up the climb.

But how could it be worse than carrying the contraption anyway but not
engaging it? Assuming the mechanism itself is 100% efficient, which of
course it wouldn't be.

In the 1km descent/ascent on a 5% grade, the reduction in speed from
the device's engagement lengthens the time of descent there, and may
slow the exit velocity as well, depending upon when it is engaged. If
it is engaged with considerable skill in lieu of brakes at a point at
which the rider must hit the brakes, then it potentially incurs no
descent penalty, but that's an unquantifiable variable which is hard
to justify as a predictable result, (indeed, it makes the two runs
inherently different because braking application is already defined as
not the same) so the safer conclusion is that the descent time will be
slowed by a small amount that would not have occurred without it.
Without the device engaged, then, the rider would reach terminal
velocity sooner and/or come off the descent going faster. On the
climb, the storage/retrieval mechanical losses mean that you don't get
back all of the energy used in storing the reduced forward momentum,
so the rider does not gain as much as was lost on the decent. The
descent penalty has the rider starting the climb later and/or slower,
needing 100% of the lost energy to make up the gap, but he doesn't get
it. Device engaged = loser.

A side note: One major consideration is that the faster the device
soaks up the forward momentum, the more energy is lost via
acceleration of the storage mass; the closest approach to optimal
efficeincy is with the device very slowly lifting the weight and very
slowly lowering it, so that almost none of the energy is consumed in
accelerations. Ergo, it's really not possible for it to be used
efficiently as a manually-engaged substitute for brakes.

OTOH, even if the whole system is 100% efficient and all other factors
are the same, then here comes the paradox: Under ideal conditions,
it's still a loser. It's faster for the heavier object to roll down
one grade and up the other under the normal operation of gravity than
it is for the device to try to transfer energy from one segment of the
run to the other, because the rider's instantaneous speed with the
device disengaged will, at any given distance point on the run, be
equivalent to or higher than his speed with the device engaged. Ergo,
it always loses.

The more practical real-world problem is that given the effects of
drag and system friction, by its very presence it becomes a loser vs
not having it.

As stated elsewhere, this is essentially a regen braking problem;
engineers have been working on it for a long time, and the results
bear out the prediction that it only provides a positive result when
it can be achieved with zero additional equipment, operating in a
scenario where the regen system is substituting for the normal braking
system in a reasonably effective manner.

The whole thing's a bad move anyway since "no device" will always be
faster.

--
My email address is antispammed; pull WEEDS if replying via e-mail.
Typoes are not a bug, they're a feature.
Words processed in a facility that contains nuts.

Werehatrack said:

On Sun, 03 Feb 2008 03:20:43 -0600, Ben C <[email hidden]> may

have said:

Werehatrack said:

On Sat, 2 Feb 2008 02:00:03 -0800 (PST),
"[email hidden]" <[email hidden]> may have
said:

Quoted message said:

Quoted message said:

>No question carrying the weight of the contraption makes the whole
>enterprise a slow one (except on the descent!) but is it faster with
>the system engaged, or the system disenganged? Seems to be. The effect
>would be super small, but taking 100J out of the descent and putting
>it into the ascent would make a difference. How much?

Quoted message said:

Quoted message said:

You have provided insufficient parameters to answer that question
precisely, but I found no solution which made it faster for the broad
range of possible ones that you gave.  It *always* came out as a
loser.  100J is a *tiny* amount of energy by comparison to the amount
lost to dragging the extra 10Kg up the climb.

Quoted message said:

But how could it be worse than carrying the contraption anyway but not
engaging it? Assuming the mechanism itself is 100% efficient, which of
course it wouldn't be.

In the 1km descent/ascent on a 5% grade, the reduction in speed from
the device's engagement lengthens the time of descent there, and may
slow the exit velocity as well, depending upon when it is engaged.  If
it is engaged with considerable skill in lieu of brakes at a point at
which the rider must hit the brakes, then it potentially incurs no
descent penalty, but that's an unquantifiable variable which is hard
to justify as a predictable result, (indeed, it makes the two runs
inherently different because braking application is already defined as
not the same) so the safer conclusion is that the descent time will be
slowed by a small amount that would not have occurred without it.
Without the device engaged, then, the rider would reach terminal
velocity sooner and/or come off the descent going faster.  On the
climb, the storage/retrieval mechanical losses mean that you don't get
back all of the energy used in storing the reduced forward momentum,
so the rider does not gain as much as was lost on the decent.  The
descent penalty has the rider starting the climb later and/or slower,
needing 100% of the lost energy to make up the gap, but he doesn't get
it.  Device engaged = loser.

A side note:  One major consideration is that the faster the device
soaks up the forward momentum, the more energy is lost via
acceleration of the storage mass; the closest approach to optimal
efficeincy is with the device very slowly lifting the weight and very
slowly lowering it, so that almost none of the energy is consumed in
accelerations.  Ergo, it's really not possible for it to be used
efficiently as a manually-engaged  substitute for brakes.

OTOH, even if the whole system is 100% efficient and all other factors
are the same, then here comes the paradox:  Under ideal conditions,
it's still a loser.  It's faster for the heavier object to roll down
one grade and up the other under the normal operation of gravity than
it is for the device to try to transfer energy from one segment of the
run to the other, because the rider's instantaneous speed with the
device disengaged will, at any given distance point on the run, be
equivalent to or higher than his speed with the device engaged.  Ergo,
it always loses.

The more practical real-world problem is that given the effects of
drag and system friction, by its very presence it becomes a loser vs
not having it.

As stated elsewhere, this is essentially a regen braking problem;
engineers have been working on it for a long time, and the results
bear out the prediction that it only provides a positive result when
it can be achieved with zero additional equipment, operating in a
scenario where the regen system is substituting for the normal braking
system in a reasonably effective manner.

The whole thing's a bad move anyway since "no device" will always be
faster.

--
My email address is antispammed; pull WEEDS if replying via e-mail.
Typoes are not a bug, they're a feature.
Words processed in a facility that contains nuts.

No doubt it is a bad idea ;-)

But your description ignores wind resistance and the relative
differences in resistance at low speed on the ascent vs high speed on
the descent. The descent time is slowed, but this is a good thing,
because it is slowed by saving some of the PE from the top of the hill
by not "wasting" it overcoming additional wind resistance.

Joseph

Werehatrack said:

On Sun, 03 Feb 2008 03:20:43 -0600, Ben C <[email hidden]> may
have said:

[...]

Quoted message said:

Quoted message said:

But how could it be worse than carrying the contraption anyway but not
engaging it? Assuming the mechanism itself is 100% efficient, which of
course it wouldn't be.

[...]

Quoted message said:

A side note: One major consideration is that the faster the device
soaks up the forward momentum, the more energy is lost via
acceleration of the storage mass;

Why does accelerating the storage mass lose energy? Assuming for now the
mechanism is 100% efficient, you get back any energy you put into
raising the mass, however quickly or slowly you raise it.

Quoted message said:

the closest approach to optimal
efficeincy is with the device very slowly lifting the weight and very
slowly lowering it, so that almost none of the energy is consumed in
accelerations. Ergo, it's really not possible for it to be used
efficiently as a manually-engaged substitute for brakes.

OTOH, even if the whole system is 100% efficient and all other factors
are the same, then here comes the paradox: Under ideal conditions,
it's still a loser. It's faster for the heavier object to roll down
one grade and up the other under the normal operation of gravity than
it is for the device to try to transfer energy from one segment of the
run to the other, because the rider's instantaneous speed with the
device disengaged will, at any given distance point on the run, be
equivalent to or higher than his speed with the device engaged.

I think I see what you mean, but I'm not sure it's right.

If there were no air-resistance, and the course doesn't require braking
for safety reasons anyway, the weight is a pointless herring. The rider
might as well store his descent energy as kinetic energy-- i.e. by not
raising the weight he's going faster at the bottom of the hill and that
energy helps carry him up the next one.

But with air-resistance in the equation, and with a 100% efficient
mechanism, the rider can store energy more efficiently by raising the
weight and reducing his speed on the descent. This is because, with
air-resistance, storing descent energy as kinetic energy is not 100%
efficient.

He will therefore go faster up the next climb, even though he's starting
the climb with a little bit less speed.

[...]

Quoted message said:

The more practical real-world problem is that given the effects of
drag and system friction, by its very presence it becomes a loser vs
not having it.

As stated elsewhere, this is essentially a regen braking problem;
engineers have been working on it for a long time, and the results
bear out the prediction that it only provides a positive result when
it can be achieved with zero additional equipment, operating in a
scenario where the regen system is substituting for the normal braking
system in a reasonably effective manner.

I don't see why it has to be zero additional equipment. A small amount
of extra weight could still be worth it.

Quoted message said:

The whole thing's a bad move anyway since "no device" will always be
faster.

I don't think necessarily always. There's a good table of energy
densities (energy per unit volume and mass) on this page:

http://en.wikipedia.org/wiki/Energy_density

That claims 2.5MJ/kg for a "Lithium Thionyl Chloride Battery". In one of
those you could store the energy of a 1000 metre descent of a 100kg
bike+rider in just 400g of batteries. About half a water bottle.

Then it becomes a matter of comparing charging and discharging
efficiency with the efficiency of storing the energy as kinetic energy
instead which is very inefficient at higher speeds because of air
resistance.

Consider the 100kg bike at the bottom of a 1000m descent. Potential
energy at the top is 1000 * 100 * 9.8 = 980kJ.

Now suppose all that were transferred to kinetic energy at the bottom.
0.5mv^2 = 980kJ, which, solving for v, gives a speed of 504kph.

Since the rider is actually probably doing only 50kph, energy storage as
k.e. is only about 10% efficient. A 400g battery, which you can charge
and discharge at maybe 60% efficiency (rough estimate) could well be a
better bet.

Not sure if you can charge batteries that quickly though.

"Werehatrack" (clip) One major consideration is that the faster the device

Quoted message said:

soaks up the forward momentum, the more energy is lost via
acceleration of the storage mass; the closest approach to optimal
efficeincy is with the device very slowly lifting the weight and very
slowly lowering it, so that almost none of the energy is consumed in
accelerations. (clip)

^^^^^^^^^^^^^^^^^^^^^
You seem to have an incorrect understanding of acceleration. It does not,
in itself, waste energy. It converts some other form (in this case,
potential energy) into kinetic energy. Conversely, if a bicycle arrives at
the foot of a hill going fast, the deceleration as it coasts uphill is the
conversion of kinetic energy back to potential energy. This conversion,
back and forth, is part of conservation of energy. If any energy is wasted,
it is due to friction, not acceleration.

That is the point of this debate. By storing energy in some form other than
velocity, the friction due to air drag is reduced, so less is wasted. Put
it this way: If you have a choice of arriving at the bottom of a hill with
a certain amount of energy, you are better off having it in the form of
potential energy than kinetic. This is because power lost to friction goes
up as the cube of velocity. It's even better than that. Coming down the
hill, when you are trying to gather energy for the coming climb, you lose
more if you get up more speed. Better to "bank it" in the form of potential
energy (by lifting a weight.)

Again, let's not argue about the wisdom of adding weight to a bike. The OP
was smart enough to pose the question in a form which cancels that out.

Quoted message said:

On Sun, 3 Feb 2008 09:04:38 -0800 (PST),

Quoted message said:

Consider a 10 mile hill at 10 mph up, 30 mph down.

For example, saving 9 watts per minute for 20 minutes (plus the extra
braking time) downhill does not provide 10 watts for 60 minutes uphill
(less the slight speed increase).

You get 180 watt-minutes downhill at 10 watts per minute for 20

^^^^^^^^^^^^^^^^^^^^^^
carelessly forgot to cut

Quoted message said:

minutes. That gives you only 3 watts per minute for 60 minutes uphill,
not the 9 watts per minute at which you stored the energy.

Both times and their wind drag also change from the no-storage
scenario, with the downhill taking longer and the uphill increasing in
time and wind drag.

Cheers,

Carl Fogel

Aaargh! Failed to clean up above. Started with 10 watts per minute,
found 9 watts divided more simply into the example.

Cheers,

Carl Fogel