Perpetual motion!

<[email hidden]> wrote: You don't get 1.3 watts uphill.

Quoted message said:

You confused watts with watts per minute. (clip)

^^^^^^^^^^^^^^^^^^^^^
Dear Carl: Oy, do YOU have a wrong number! Watts is a rate of energy
production, flow or use. Watt-second, watt hour or kilowatt-hour, being
the product of energy and time, is a unit of energy. "Watts per minute"
would be the rate of change of power, which is possible, but hardly ever
used--and certainly not appropriate in this context.

Ben C said:

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.

I'm not convinced this pursuit makes any sense, but if the battery
charge rate is the limiting factor, could he use big capacitors?
--
Andrew Muzi
www.yellowjersey.org
Open every day since 1 April, 1971

A Muzi said:

Ben C wrote:

[...]

Quoted message said:

Quoted message said:

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.

I'm not convinced this pursuit makes any sense, but if the battery
charge rate is the limiting factor, could he use big capacitors?

Possibly one day. On that Wikipedia page, the EEStor (see also
http://en.wikipedia.org/wiki/EEstor) has a claimed energy density of
1MJ/kg. So still not much weight-- you could store up that 1000m descent
in 1kg rather than in 400g.

But such capacitors may not actually exist yet. All we know is they've
been patented.

Capacitors that do exist are about 1/100th that energy density, so would
weigh too much to be any good for this kind of thing.

On Sun, 03 Feb 2008 11:12:36 -0600, Ben C <[email hidden]> may

have said:

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?

Consult your local physics text. The faster the mass is accelerated,
the more energy is used - unrecoverably - in *changing* its speed. At
the extreme end of the several effects of this, if you tried to
recover the descending mass' energy in less time than it falls under
the effects of gravity, the resulting recovery rate is *zero* because
the mass will simply not move downward that fast. Conversely, if you
try to lift it at a rate that puts an initial acceleration of an
additional 10G on it to get it moving, you've just lost a bit of
additional energy in accelerating the mass suddenly because the force
that must be applied to it is greater than if it is moved more slowly,
but the potential energy which it will store is not increased.

Quoted message said:

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.

Only if the rate of acceleration is negligible. This is actually
fairly important. It's one of the reasons why mechanical regen
braking systems are so inherently inefficient; to be useful and
effective in real applications, they must react *fast*, and this is
always wasteful.

Quoted message said:

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.

However, for a dead-equivalent comparison of the two, the brake
systems must also be used *identically*. If you treat the gadget as
"not a brake" then the gadget-engaged bike is always going slower at
the bottom of the descent. Not much, but still slower. Trying to
treat the gadget as a voluntary brake, and requiring that the other
bike brake equally at the same time, eliminates the implied advantage
granted by one of the theoretical aspects of the original postulate,
so it isn't allowed; ergo, the original postulate seems to be asking
whether *with all other aspects equivalent* the device will produce a
faster run in total, and the math shows that in all cases given only a
single variable, it will not...because the gadget-engaged bike will
arrive at the end of the descent later than the gadget-disengaged one,
and the energy recovery will be insufficient to make up the deficit on
the climb.

Quoted message said:

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.

No, it doesn't work out that way when you try to build one. Really.
I'm not kidding about this.

Quoted message said:

Quoted message said:

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

I don't think necessarily always.

Yes, always.

Quoted message said:

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.

At a charge/discharge loss of at least 30 to 40%, or were you unaware
of the inefficiencies of batteries in this regard? (Looking ahead, I
see that you're aware...but haven't considered the implications fully)
Yes, they have impressive density (at phenominal cost and with some
significant fragility issues), but they also have typical chemical
cell energy recovery limitations.

Quoted message said:

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.

It's extraordinarily inefficient in general, which is why even though
the theory has been around since the 1920s and prior, no one has yet
built a functional, actual-vehicle-useful regen brake (mechanical or
electric) that recovers more than 15% of the energy of vehicle
braking, and that figure is achieved only under absolutely ideal
conditions. 5% is more typical of a "successful" design, and every
add-on system yet devised adds hardware which comes with a weight
penalty that eats all of the gain and more. If the drivetrain design
doesn't inherently incorporate a zero-penalty mode of energy recovery,
the penalty eats the gains every time.

Quoted message said:

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.

Which runs smack into Fogel's Objection; you can't get there from
here. No matter, though...

Quoted message said:

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.

First, you're trying to *store* the energy that he *used* in getting
there; do that, and he's coming off the bottom of the hill a lot
later, and slower if it's dragging him back for the whole run.
Second, electric generators have mechanical losses that are pretty
significant in themselves, and they're rather speed-sensitive; outside
their optimal output rpm range, they lose efficiency rapidly. Add in
a generous 30% loss for the battery cycle, and you've just discarded
at least half of the energy that you robbed from your cyclist in the
downhill run. And then he doesn't get it all back on the climb, and
he's behind both where he'd be with it disengaged and where he'd be
without it present at all. Ergo, leave the mess on the bench at the
shop, and he's even farther ahead.

Quoted message said:

Not sure if you can charge batteries that quickly though.

It doesn't matter. It still doesn't work to produce a gain in the
overall result. If you really want to understand why, there are lots
of theses available from engineering students, experimenters,
inventors and researchers who have tried to crack the problem in the
past. Toyota has the closest approach to success due to a quirk in
the design of the Prius drivetrain; it's inherently able to function
as a regen brake, and it still makes only a trivial difference in fuel
usage. (I have a relative and several friends who own them; they're
remarkable vehicles, though I dare say that I'm the only one among
that group aside from my brother...whose undergrad degree was in
EE...that really understands what's going on in that wonderful little
device Toyota cooked up.)

--
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 Sun, 03 Feb 2008 17:36:55 GMT, "Leo Lichtman"

may have said:

"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.

But you misunderstand the effects of gravity and acceleration in this
exact instance; because the mass is being moved against the effects of
gravity, to accelerate the mass significantly upward will require a
greater amount of force than if it is less suddenly put into
motion...but the potential energy it will store is unchanged. Thus
more energy is required - and some is wasted - when the weight is
moved rapidly vs when it is moved gradually. Conversely, when it is
lowered, the slower it moves, the more energy is recovered. The
faster it moves downward, the less force it can generate since some of
the accelerative force of gravity is being consumed in increasing the
velocity of the mass; at the extreme of the example, if the mechanism
tries to lower it faster than it would fall under gravity, *no*
recovery takes place at all, and all of the potential energy is lost
in accelerating the mass; it falls, it hits the stop at the bottom,
and nothing has happened except that its potential energy has changed
again without a singe joule getting recovered for the rider.

Quoted message said:

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.

You seem to be saying that it's faster to go slower. It may be less
"wasteful" to go slower if that can be done without shedding the
velocity in a manner that fails to make it available later, but in the
case of a descent under gravity, the rider's energy input is assumed
to be zero...so from the rider's standpoint, *nothing* is being wasted
with no recovery-brake in operation, and the maximum amount of
potential energy is being applied to making the rider move. The
problems which always make a storage system into a loser are: First,
the efficiency of storage of all potentially useful systems is
significantly below 100%, so that the recovery will never be high
enough to justify the gadget; second, the weight penalty that they
impose will always make the climb in the problem into an energy sink
for the rider; and third, if you try to bank any energy, you slow the
rider...resulting in a deficit at the midpoint which the inefficiency
guarantees cannot be regained. Though you might make an argument for
using a regen brake in place of a friction brake, experience has long
since proven that this is weight-intensive and ineffective even with
the best tech we can build or even theorize about; it always
underperforms the results obtained by same system without the extra
hardware. Discussions of "is it faster with ot without the device
engaged" are really silly since if the whole point is to improve
efficiency...and the greatest improvement is achieved by deleting the
device...then whether it's beneficial when it's present and engaged vs
disengaged is truly irrelevant. Fiurthermore, if you're talking about
using a regen system in an instance where a friction brake would not
have been applied at all, then you're definitely talking about a
loser, no question about it.

Quoted message said:

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.

Actually, as I read it, the reverse it the case; he seems to have
postulated that the only difference between the two runs should be the
engagement or disengagement of the device. Given that as the
operational parameter, the engagement of the device is a loser.

I've said enough on this. Continuation will be by others, if any.

--
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" wrote: (clip) I've said enough on this. Continuation will
be by others, if any.
^^^^^^^^^^^^^^^^^^^^
Good. It is totally frustrating to try to hold a discussion with someone
who fails or refuses to understand conventional (and elementary) physics.

Werehatrack said:

On Sun, 03 Feb 2008 11:12:36 -0600, Ben C <[email hidden]> may

have said:

Werehatrack said:

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

Consult your local physics text. The faster the mass is accelerated,
the more energy is used - unrecoverably - in *changing* its speed.

There's nothing about that in my local physics text.

Work = Force x distance.

Now if I lift the weight slowly against gravity, I apply its weight
constantly + a tiny initial amount to get it moving, over a distance. I
get all that back (assuming no air-resistance etc.).

If I want to shoot it up in the air, actually accelerating it as well as
overcoming its weight, I can apply more force over the same distance. So
more energy.

Where does the extra energy go? Into kinetic energy, but after I stop
applying the force, the mass continues to rise upwards, like a pancake
tossed from a pan.

The height it eventually reaches will give it a gravitational potential
energy corresponding to the work I put in.

No energy is unrecoverably lost just because I accelerated the object.

Quoted message said:

At the extreme end of the several effects of this, if you tried to
recover the descending mass' energy in less time than it falls under
the effects of gravity, the resulting recovery rate is *zero* because
the mass will simply not move downward that fast.

I don't really see how I even could try-- I could forcibly pull the mass
down, but that would just be putting more energy in, which I'd get back,
but I wouldn't get my original stake back any quicker.

Quoted message said:

Conversely, if you try to lift it at a rate that puts an initial
acceleration of an additional 10G on it to get it moving, you've just
lost a bit of additional energy in accelerating the mass suddenly
because the force that must be applied to it is greater than if it is
moved more slowly, but the potential energy which it will store is not
increased.

It continues to rise upwards after you let go of it, and does store more
potential energy.

Quoted message said:

Quoted message said:

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.

Only if the rate of acceleration is negligible. This is actually
fairly important. It's one of the reasons why mechanical regen
braking systems are so inherently inefficient; to be useful and
effective in real applications, they must react *fast*, and this is
always wasteful.

Speed of reaction in general is a significant consideration I'm sure--
like how to charge batteries quickly.

If we're saying "assume the mechanism is 100% efficient" then the rate
at which energy is stored and retrieved from the device (or the rate at
which a mass is accelerated) doesn't matter.

But in practice, I'm sure many possible mechanisms that might be
reasonably efficient at lower powers, become unworkable or less
efficient when you need to transfer energy in and out of them quickly.

Quoted message said:

Anything is possible, given the initial false assumption that we could
somehow store and release energy without significant losses.

That's why the first rule of formal logic is that you cannot reach a
valid conclusion if you start with a false premise.

I'd be more than happy with a validly false conclusion.

Anyway, on today's 90km fixed ride, where the wind turned at almost
the halfway point such that I had a headwind the whole time, I had
plenty of time to ponder the nature of realsesing energy with
significant losses.

So I have finalized the design of the contaption, and I think it would
do a fine job of maintaining a reasonable level of efficiency, though
perhaps at the expense of many other desireable characteristics.

A 2m or so long pipe is attached with hose-clamps to a bike's down-
tube, such that the pipe projects forward of the front wheel with the
end at some elevation more than 1m above the ground. A fine pulley is
attached to the forward end of the pipe. A high-flange hub is used on
the front wheel. A spool-like pulley is bolted to the flange of the
hub, and a length of high test fishing line is tied around this spool,
and fed trough the pulley, and tied to a suitable weight. A cinder-
block comes to mind.

Neverminding the fact that having a cinder-block dangling on a thin
line right in front of the front wheel might be somewhat
disconcerting, and have an adverse effect upon the bike's stability,
our intrepid test pilot commences riding down our hill. The line gets
wound around the spool much like a yo-yo. When the rider gets to the
bottom of the hill (which by happenstance is exactly when the cinder-
block is hoisted to the maximum!), he casually flips the steering
around as a BMX free-styler would do, and the block starts to descend
slowly unwinding the line from the spool and assisting the rider in
his ascent by applying the stored energy toward driving the front
wheel.

If nothing else, I maintain that this design is more efficent than a
5km tall massless evacuated tube with a pea inside.

Joseph

Quoted message said:

On Feb 4, 4:52 pm, Ben C <[email hidden]> wrote:

[...]

Quoted message said:

Quoted message said:

He's a tiny bit slower down the hill, but completes the whole ride 0.39
seconds quicker! So better to engage the device than not, if he has to
carry it anyway (but only if it's 100% efficient).

[...]

Quoted message said:

Quoted message said:

For a 150W device that stores about 14kJ (and weighs 10kg) he needs at
least around 80% storage and retrieval efficiency for it to be as good
as saving the 10kg by carrying no device at all, on this particular
course.

[...]

Quoted message said:

Excellent!

How inefficient can the 10kg weight gizmo be before the .39 seconds is
used up?

To my surprise it can be as bad as 65% efficient before you're better to
disengage it.

It may just be a bug in the program or some kind of numerical error, but
I think the reason is that if the storage device runs at a lower power,
the absolute amount of energy lost due to inefficiency is smaller.

In all these simulations, there's a 150W rider pedalling continuously,
both down and up the hill. If he has a 150W storage device that's only
50% efficient, half of his power is just being dissipated. That doesn't
make for a good journey time.

But if the storage device is only 1.3W and 50% efficient he's only
losing 0.65W.