Can an Electric Car Run Forever on a Slope with Rollover Power Generators?

[spikenigma]

Hi,

(Firstly, This is not a perpetual motion discussion. Nor am I suggesting anything of the sort. I'm just wondering where the energy losses enough to stall the system creep in.)

ok, I was having a discussion at lunchtime which I didn't have an answer to , so I'll post it here:

A Nissan leaf (http://en.wikipedia.org/wiki/Nissan_Leaf) is sitting at the top of a slope.

Along the length of the slope are rollover power generators (http://www.fahad.com/2006/03/electro-kinetic-road-ramp-clever-idea.html , http://news.bbc.co.uk/1/hi/england/somerset/4535408.stm)

The power generators are connected via superconducting material to an energy storage device at the bottom and charger which can recharge the car.

In short:

Variables

Efficiency of storage device and charger
Number of rollover generators
Gradient of slope (which will increase the constant energy consumption of the car by its percentage amount)
Length of slope

Constants

34kw per 100 miles for the car
10kw per rollover ramp generator


Now, the main point was that it is unclear when the system will stop working, because:

a) The car will always regain velocity (and thus any energy lost to heat, sound, light, friction) between ramps.
b) (Beyond "something like" 10% efficiency) it will also always generate more energy getting down the hill than it needs to drive back up it and thus will produce an excess - thermodynamically impossible.

Some quick back of the envelope calcs (don't crucify me):


Variables

Efficiency of storage device and charger - 30%
Number of rollover generators - 50
Gradient of slope - 20%
Length of slope - 100 miles


* Going down the slope, the car produces 500kw of power
* The Energy device can store and use 150kw
* The car uses 40.8kw to drive back up the hill

 

[Borek]

* The Energy device can store and use 150kw
* The car uses 40.8kw to drive back up the hill

Unless I am missing something it seems like you are using kW to measure energy. You can as well measure distance in pounds.

 

[spikenigma]

Unless I am missing something it seems like you are using kW to measure energy. You can as well measure distance in pounds.

...should be KwH

 

[JaredJames]

The energy the car has at the top of the hill is equal to the amount it has going down.

You can't have it use that amount of energy to produce XkW downwards and have it use YkW going up (where Y < X). At best it it would take all of the downwards energy exactly to get it back to the top.

Think of a hydroelectric dam. What you are proposing here is pumping the water straight back up into the dam with the power it produces on the downwards section and re-using it. Giving a net increase in energy produce each time (more produced than it takes to send it back up the hill).

This is not possible.

Didn't we have this exact discussion here before?

 

[spikenigma]

The energy the car has at the top of the hill is equal to the amount it has going down.

You can't have it use that amount of energy to produce XkW downwards and have it use YkW going up (where Y < X). At best it it would take all of the downwards energy exactly to get it back to the top.

this is my thinking also, but...

a) The car will always regain its velocity (and thus any energy lost to heat, sound, light, friction) between ramps.
b) (Beyond "something like" 10% efficiency) it will also always generate more energy getting down the hill than it needs to drive back up it and thus will produce an excess - thermodynamically impossible.

It will stop between ramps for a reason?

 

[JaredJames]

this is my thinking also, but...

There's no "but". You cannot produce more energy than you already have. If it has 1000 joules of energy at the top of the slope, that's all it can produce on the way down and it's the minimum it took to get it up there.

You can't get more out on the way down than you put in on the way up.

 

[spikenigma]

There's no "but". You cannot produce more energy than you already have. If it has 1000 joules of energy at the top of the slope, that's all it can produce on the way down and it's the minimum it took to get it up there.

You can't get more out on the way down than you put in on the way up.

Then (if you don't mind) talk me through what happens in the scenario generator by generator, and show some power calculations.

 

[xxChrisxx]

You appear to have plucked the figures from thin air (well taken them from a website and applied them blindly)

What about rolling resistance going down the hill, what about drag? Having the car constantly drive over something will slow it down.

So you can't say the textbook: gpe at top of slop = kinetic energy at the bottom.
As that's not the case here.

What this is describing is just an elaborate kinetic energy recovery system.

Gradient of slope (which will increase the constant energy consumption of the car by its percentage amount)

Also I can assure you that this isn't the case. If you have a trip computer with an active MPG read out, look at it next time you go up a hill. The load change is no different for an electric motor.

When you are going in a straight line, you have to overcome RR and aero drag. Going up a hill you need to add the weight of the car into the mix.

 

[JaredJames]

There's no need for calculations.

You cannot produce more power than it takes to get the car up the slope. You can add all the little "tweaks" you like but tit doesn't change why it won't work.

At best you could have a 100% efficient system where the car would produce the exact amount of energy coming down the slope as it would need to get back up. You cannot have greater than 100% efficiency - that would be a PMM.

 

[spikenigma]

Apologies, my mistake. I seem to have wondered into the childrens forum where "nu-uh :tongue2: " with more words, or "didn't think of wind resistance didya buster - case closed! (sucks lollypop) " constitute the descriptive answers I've asked for.

A better answer would have been "at the first ramp the entire system has x KwH, the car travels back up to $ mph, at the second however only has...y kwH...at the bottom...the system has a state of...meaning it cannot make the journey back up due to x% of efficiency losses/comes to a halt at ramp number () ".

If it's trivilially easy show me with calcs and describe the state of the system with figures at each jucture!

 

[xxChrisxx]

There are simply too many assumptions that need to be made to do any meaningful calcualtions. All we ca say is what you have assumed is not really correct.

Read my post.

 

[spikenigma]

There are simply too many assumptions that need to be made to do any meaningful calcualtions. All we ca say is what you have assumed is not really correct.

Read my post.

Ah, I see then. So the equivilent of:

"didn't think of wind resistance didya buster? - case closed! (sucks lollypop) "

...an intellectually honest person would have said "assuming wind %/friction %/drag %, then...at the first ramp...at the second ramp..."

I'll wait for somebody else to reply with a more concise answer.

 

[xxChrisxx]

Gradient of slope (which will increase the constant energy consumption of the car by its percentage amount)

To show you simply just how wildly wrong this is.



Assuming constant speed (60mph) on a flat.

60mph = 26.8 m/s.
car = 500kg = 5000N (for ease of calc)
Crr Rolling resistance tyre to floor = 0.01

Car Cd = 0.3
Frontal area = 1m^3
air density = 1.22

Load requirement on a flat:
Rolling resistance = F = Crr*N
=0.01*5000
=50N

Aero Drag = 0.5pV^2CdA
= 0.5 * 1.22 * 1 *0.3 * 26.8^2
=131N.

Total load requiremnt on a flat = 181N.

Load from gonig up a hill.
F=mgsin (theta)

F=5000 * sin (20)
=1710N.

Your load requirement going up a 20% graduient are 9.5x that of driving on a flat.

So your power input figure is horrbly off to begin with. Note that this is LOAD not POWER useage, your power requirement to dive up a 20% gradent at the same speed as cruising on the flat will be higher than 9.5x that.


Now, it really is impossible to predict how much energy the leaf would acutally generate, and how much it would slow down after going over each generator bump.

 

[JaredJames]

spikenigma, put it like this:

Let's assume there are zero losses at all, no resistance, nothing.

The car has a mass of 1000kg, the hill is 1km high and gravity is 9.8m/s².

We take the bottom of the hill as 0 energy level.

At the top of the hill the car will have mgh Joules of energy = 1000*9.8*1000 = 9800000 Joules.

That is the maximum energy the car will have in this system.

At the top of the hill that is how much energy it will have and to get it there you must give it that much energy. You can only get out what is available (energy cannot be created or destroyed) so the most you could get out is that value. Which would be the exact amount to get you back to the top.

This of course assumes 100% efficiency in the system.

To get more energy out you would need greater than 100% efficiency - an example would be for every 1 Joule you put in you get 2 Joules out - this is impossible.

Values such as speed of the car are irrelevant. It will always accelerate with gravity. The only way to increase the velocity above this is to use an engine - which requires additional energy input.

It may only take X amount to get you up the hill, but you'll only get X back out coming down (coasting under ideal conditions). To use the engine to give more energy on the way down uses more energy than it will produce.

Now drop the attitude. Your numbers are ridiculous and not realistic (as per chris) and you can't expect anything meaningful from it.

 

[spikenigma]

Your load requirement going up a 20% graduient are 9.5x that of driving on a flat.

So your power input figure is horrbly off to begin with.

Congratulations, you've just proven it takes more Energy to go up a hill than down it!

Now, it really is impossible to predict how much energy the leaf would acutally generate, and how much it would slow down after going over each generator bump.

Then you're quite literally fudging the question.

If (as you say) it is impossible to predict "how much energy the leaf would generate" it could be argued that it could potentially generate a billion watts and fly back up the hill, or 10^10^10^10 watts and immediately collapse into a black hole and suck the hill in.

Why don't you try taking the car going down the hill once and calculate that?

 

[spikenigma]

spikenigma, put it like this:

Let's assume there are zero losses at all, no resistance, nothing.

The car has a mass of 1000kg, the hill is 1km high and gravity is 9.8m/s².

We take the bottom of the hill as 0 energy level.

At the top of the hill the car will have mgh Joules of energy = 1000*9.8*1000 = 9800000 Joules.

Thanks for quoting me basic thermodynamics rather than answering the specific scenario I outlined as requested.

 

[xxChrisxx]

spikenigma, put it like this:
Let's assume there are zero losses at all, no resistance, nothing.

You've simply restated the same thing in all your posts, as you've made the assumption that you can only get out what you put in.

Although correct it doesn't acutally prove anything. You've got to start from the point of assuming that it is possible. If we could get accurate figures, and not assumptions then it's trivial to prove that even assiung perpetual potion is possible the figures show it doesn't.



It can clearly be shows that he's vastly underestimated the power requirements to dive back up the hill. He's also overestimated how much power he'll get out. This is also ignoring the fact that 100miles on a 20% slope is 34 miles vertically.

 

[JaredJames]

Congratulations, you've just proven it takes more Energy to go up a hill than down it!

Seriously, lose the attitude or no one will bother helping you.

Your cars fuel converts to kinetic energy which, as you climb the hill becomes GPE. On the way down, that GPE becomes KE. That KE will equal the amount of fuel used to get up the slope (again under ideal conditions). So you don't gain anything. At best you'll break even.

 

[JaredJames]

You've simply restated the same thing in all your posts, as you've made the assumption that you can only get out what you put in.

It's not an assumption, it's a fact.

Although correct it doesn't acutally prove anything. You've got to start from the point of assuming that it is possible.

It isn't possible, period. If you want to entertain PMM threads that's up to you. I can throw numbers around but it doesn't mean anything.

 

[xxChrisxx]

Congratulations, you've just proven it takes more Energy to go up a hill than down it!
Then you're quite literally fudging the question.

If (as you say) it is impossible to predict "how much energy the leaf would generate" it could be argued that it could potentially generate a billion watts and fly back up the hill, or 10^10^10^10 watts and immediately collapse into a black hole and suck the hill in.

Why don't you try taking the car going down the hill once and calculate that?

You do realize that it's up to you to prove your initial assertion as part of the forum guidelines.

I've merely showed you've vastly underestimated your energy requirements to get up the hill. So it's no wonder you are showing a net gain.


If you carry this on, the mods will simply lock the thread. I'm trying my best to explain the error in your thinking.

 

[JaredJames]

Thanks for quoting me basic thermodynamics rather than answering the specific scenario I outlined as requested.

What I wrote applies to every scenario. Whether you like it or not.

 

[xxChrisxx]

It's not an assumption, it's a fact.
It isn't possible, period. If you want to entertain PMM threads that's up to you. I can throw numbers around but it doesn't mean anything.

For the purpose of explination, it doesn't matter either way. If you are trying to educate someone, you can't simply say 'the laws say so'. As they wil simply take that to mean 'lets assume you are wrong'.

Even though they are, they will just shut down at that point and stop listening.

Whether perpetual motion is or isn't possible is irrelevant as a 'law'. If we have real data we would find we get less out than we put in, which is another tick under the 1st law box. Leaving the PPM box rather bare.



Cliffs: You can't teach by assertion.

 

[spikenigma]

Seriously, lose the attitude or no one will bother helping you.

That's just it, you're not helping me.

I already know the scenario is incorrect and will come to a halt as I stated in the OP. You're simply restating obvious facts.

If somebody said that gravity is perpetual energy as the moon can tug on the Earth and not lose any velocity, I would state that it is indeed losing velocity and show that at year x the moon will either fly off or fall to Earth.

What I wouldn't do is the equivilent of what you've done, which is everything but answer the specific question I've asked.

 

[JaredJames]

That's just it, you're not helping me.

What I wouldn't do is the equivilent of what you've done, which is everything but answer the question.

If you understand you can't get more out than you put in, why do you believe you can get more energy out going down the hill than you put in going up?

 

[xxChrisxx]

That's just it, you're not helping me.

I already know the scenario is incorrect and will come to a halt as I stated in the OP. You're simply restating obvious facts.

If somebody said that gravity is perpetual energy as the moon can tug on the Earth and not lose any velocity, I would state that it is indeed losing velocity and show that at year x the moon will either fly off or fall to Earth.

What I wouldn't do is the equivilent of what you've done, which is everything but answer the question.

Ok, I'll have another pop at it.

We can think of this problem purely in terms of energy. We can also assume that the Leaf will give a full 10kW charge to each pad.

I'm also assuming that this is a constant 20% slope that is 100miles =161km long. So 55km vertically up.

Total energy available in the system = mgh.
The car will reach terminal velocity.
The car will ahve 0.5mV^2 KE at the end.
Assume frontal area = 1m^2
Cd = 0.3

We will assume the car starts off at terminal velocity.

Also assume the 'pad' for the energy recovery is 1m long.


Starting energy = mgh
= 500 *10 * 55000
275MJ.

Terminal velocity (eq is on wii page)
= 2 * 5000 / pACd
=165m/s.

10kW = J/S
at 1 m long each pad will activate in 1/165 of a second.

10000*165 = J
=1.65MJ

* 50 of them
=82.5MJ.


Final velocity = mv^2/2
=165^2*500/2
= 6.8MJ.

So if we add in the energy we started with:
275MJ

The energy at the end.
6.8 MJ KE + 82.5 MJ stored.
= 89.3 MJ.

So the total energy at the end =
89.3 - 273 = -183.7 MJ.


Basically it doesn't work ebcuase of drag and rolling resistance. Even if we used all the energy we had we would be able to get about 32% up the next hill.

However we've made a gain becuase if we didn't have that energy recovery we would only be able to get 2.5% up the next hill.

 

[spikenigma]

If you understand you can't get more out than you put in, why do you believe you can get more energy out going down the hill than you put in going up?

I understand it in a purely mathematical sense i.e. [Force to move an object up a gravity well >= Energy it loses moving down it], always.

We all understand basic thermodynamics which you seem to keep repeating, but yet, you can't seem to answer my question with any figures relevant to the scenario I have outlined.

Which is all I am asking!

 

[JaredJames]

We all understand basic thermodynamics which you seem to keep repeating, but yet, you can't seem to answer my question with any figures relevant to the scenario I have outlined.

Which is all I am asking!

If your own figures aren't relevant, how do I base any calculations off them?

Your best bet is to go with chris' above post showing energy.

 

[spikenigma]

Ok, I'll have another pop at it.

We can think of this problem purely in terms of energy. We can also assume that the Leaf will give a full 10kW charge to each pad.

I'm also assuming that this is a constant 20% slope that is 100miles =161km long. So 55km vertically up.

Total energy available in the system = mgh.
The car will reach terminal velocity.
The car will ahve 0.5mV^2 KE at the end.
Assume frontal area = 1m^2
Cd = 0.3

We will assume the car starts off at terminal velocity.

Also assume the 'pad' for the energy recovery is 1m long.


Starting energy = mgh
= 500 *10 * 55000
275MJ.

Terminal velocity (eq is on wii page)
= 2 * 5000 / pACd
=165m/s.

10kW = J/S
at 1 m long each pad will activate in 1/165 of a second.

10000*165 = J
=1.65MJ

* 50 of them
=82.5MJ.


Final velocity = mv^2/2
=165^2*500/2
= 6.8MJ.

So if we add in the energy we started with:
275MJ

The energy at the end.
6.8 MJ KE + 82.5 MJ stored.
= 89.3 MJ.

So the total energy at the end =
89.3 - 273 = -183.7 MJ.


Basically it doesn't work ebcuase of drag and rolling resistance.

THANKYOU, Chris. This was the type of answer I was looking for.

Thinking in terms of energy then. How about if you had 4 times as many pads, (likely incorrect calcs) following:

Starting energy = mgh
= 500 *10 * 55000
275MJ.


Terminal velocity
=165m/s.

Per Pad
1.65MJ

200 of them (50 metres in total)
= 330MJ

So if we add in the energy we started with:
275MJ

The energy at the end.
6.8 MJ Kinetic Energy + 330 MJ stored.
= 336.8 MJ.


So the total energy at the end =
336.8 - 273 (to get back up the hill?) = 63.8 MJ excess


What have I done wrong? - the only thing I can think of would be 273 not being the starting energy, but being something else increasing in proportion with the number of pads?

 

[AlephZero]

The numbers in your "specific scenario" are in meaningless units, and/or bear no relation to reality.

Just to pick one of your numbers at random, ask yourself how fast a Leaf would be traveling downhill to generate 500 kW of power. The answer is about 1.5 times the speed of sound, down a 1 in 10 slope. Yeah, right...

In fairyland, your scenario is probably true (but maybe not on Wednesdays if somebody has annoyed the Wicked Witch). In the real universe, it isn't.

Edit: Even if your phyiscs is correct on your previous post (I didn't check), your "terminal velocity" of 165 m/s is about 370 miles per hour. That's impressive, for a Leaf.

 

[JaredJames]

What have I done wrong?

Did you allow for losses from the additional pads? You have to subtract the amount of energy gained from each pad from the initial energy provided (starting energy).

 

[spikenigma]

In fairyland, your scenario is probably true (but maybe not on Wednesdays if somebody has annoyed the Wicked Witch). In the real universe, it isn't.

Ha ha!, you what?, No scenario, super-engineering or no can break the laws of thermodynamics.

Just trying to find out where and how it breaks down mathematically.

 

[xxChrisxx]

THANKYOU, Chris. This was the type of answer I was looking for.

Thinking in terms of energy then. How about if you had 4 times as many pads, (likely incorrect calcs) following:

I knew this was coming.

We have the issue that we don't know how much the pads will slow the car down (it's clear they will as we are taking energy from the potential). It can't even be estimated at the speeds the car is going. So as you increase the number of pads you get an answer that is less and less accurate when you assume it won't slow down.


We also have the issue that we are dealing with an unrealistic scenarioto begin with. In reality as soom as the car hit the first ramp it would destroy the suspension and throw the car into the air.

 

[JaredJames]

Ha ha!, you what?, No scenario, super-engineering or no can break the laws of thermodynamics.

What's "super engineering"?

The laws of thermodynamics don't get broken, no matter how "super" the engineering happens to be.

Or, to keep chris happy: the laws of thermodynamics have never been observed to be broken, ever - even with an engineering feat as "super" as creating an artificial sun.

 

[xxChrisxx]

What's "super engineering"?

The laws of thermodynamics don't get broken, no matter how "super" the engineering happens to be.

Or, to keep chris happy: the laws of thermodynamics have never been observed to be broken, ever - even with an engineering feat as "super" as creating an artificial sun.

It's fine to state it's a law, as it is.
Just don't expect anyone to feel fulfilled if you assert the law as an answer.

 

[spikenigma]

I knew this was coming.

We have the issue that we don't know how much the pads will slow the car down (it's clear they will as we are taking energy from the potential). It can't even be estimated at the speeds the car is going. So as you increase the number of pads you get an answer that is less and less accurate when you assume it won't slow down.

I don't understand why the slowdown is even an issue, because between each ramp the car accellerates back up to terminal velocity - which is what prompted the entire scenario.

If we assume mathematically that each ramp slows the car down by x % of it's velocity, does that % - even an arbitrary one - make the car unable to return to the top of the hill?

We also have the issue that we are dealing with an unrealistic scenarioto begin with. In reality as soom as the car hit the first ramp it would destroy the suspension and throw the car into the air.

I realize it's infeasible in an engineering sense (a car traveling at terminal velocity on the ground).

But thermodynamically, you are proving to me what happens.