hybrid car idea

If I get the money I am going to do a lot of research before building an electric or a hybrid, but I think that the best bet for the money is to buy your choice of used vehicle. The first hybrid that I read of was a small Subaru pickup truck. It got about 75 mph but only did about 45 mph. Who knows what someone could have done?

What I still haven't worked out is whether there is a viable scheme using three-phase 10 horsepower motors, and I don't even know if 10 hp three-phase is the same as 10 hp DC. Someone out there is putting 6v, 1000 amps through a DC motor and getting 90 mph on a VW Rabbit. That kit is something like $10,000 and most of that is the motor. A three-phase 10 hp motor is about $500. This might also make a good case for keeping the old transmission, which would make a lower torque motor more usable. You can replace half a dozen transmissions for $10,000.

 

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90mph in a Rabbit with only 8bhp? That seems a bit optimistic.

A polyphase motor in theory should be more efficient than a DC brushless design of the same specific output, but ultimately it will probably come down to which controller is more efficient. Synthesizing the three phases and regulating the phase angle from a DC supply would probably need a more complex controller. Something resembling three BLDC controllers, one for each phase, with each referencing the other to keep the right angle of seperation might work. Although the efficiency hit in the controller could be counteracted by having a motor with a much higher power/weight ratio.

 

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Horsepower isn't an easy thing to measure correctly. There is somehow a huge difference between electric motor horsepower and gasoline motor horsepower, and I think that it has something to do with the compressibility of the gases. Yes, you're transferring a certain measurable amount of momentum and all that, but exactly how that works is a riddle that I don't think the mathematicians have solved yet. There is no direct relationship between energy expended and momentum generated.

 

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Slow down - take time to think a little bit before posting.

Of course there is no direct relationship between momentum MV and even KE or 0.5MV^2.* Any high school physic student knows that.
For example a river barge full of coal has lots of MV but relatively little MV^2 whereas a bullet has relatively less MV but compared to its MV^2.

Certainly that is no "riddle" even to Mathematicians.

Horsepower is easy to measure. I worked one summer for Lion oil company.** They had a dynamometer - machine for measuring HP, torque etc at any speed the motor ran. (My only connection with it was to get free spark plugs, that had 100 or less hours of use, every Friday afternoon for my week-end trip home - very standardize drive, stopping at the one red light even it was green as I made very precise study of fuel economy vs. speed. (Filling tank to line scratched inside the filler tube always at same two pumps, at start and end of 100+ mile trip, parked in same position for air free tank etc.)

SUMMARY: Except for the spelling, there is absolutely nothing correct in this post of yours. - A new record even for you. Were you drunk?
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*even less relationship between MV and horsepower ass then with horsePOWER there is the time or rate of KE factor. I.e. even if there were a relationship between KE and MV there still would be none between MV and HP as the same KE in different interval is different HP.

**My summer job was to evaluate a new metal "packing" for a distillation tower they were building. First day on job I mixed 55gal each of benzene and toluene and spent remainer of the summer separating them again, several times. - That perforated metal chip packing was able to make 22 "theoretical plates" in the lab tower (about 3 meters tall) I was using. I wrote my report and went back to school - never did learn if they trusted my work enough buy that costly, but good, packing for the new tower.

 

Yes, exactly.

Horsepower is very easy to quantify. Using the horsepower equation we can see that it is a relationship of torque and RPM. HP is an expression of energy in the same way that wattage is a relationship of current and voltage. In that vein, one horsepower is an instantaneous energy measurement equivalent to 746 watts.

So the 6V/1000A motor previously mentioned is dissipating, via Ohm's Law, 6kW (or eight horsepower) but in reality the brake horsepower at the output shaft would be less than that. More like 6.5bhp assuming an average of 80% efficiency for an induction motor.

 

Billy, you seem unaware of something that I learned just recently. OK, maybe not, but the thing is, to push the piston and to get the piston to push the car through the drivetrain you have to develop momentum. Developing momentum favors moving larger masses using less energy, which is actually permissible in the momentum and kinetic energy equations.

 

Here's the deal about momentum versus kinetic energy: A 100 kilogram object travelling at 100 meters per second has a momentum of 10,000 (I don't know what unit) and a kinetic energy of 500,000 joules. A 1000 kilogram object moving at 10 meters per second has a momentum of 10,000 and a kinetic energy of 50,000 joules, the same momentum and 10 times less energy. Spinning up a small object to use to drive a large object is a very energy wasteful procedure. Now maybe someone has already written some really good papers on this subject, but it looks weird. Maybe the methods of transferring momentum do things that we aren't aware of.

The riddle is, how do we efficiently convert a certain amount of energy to momentum and get it to the wheels? Experiments with light and heavy flywheels would demonstrate that we can actually create these situations in the real world, where we have a flywheel with a lot of energy and little momentum, or one with a lot of momentum and little energy. I think that electric motors may show the way to optimum solutions of this riddle.

 

I'm trying to understand what happens to the kinetic energy that we are trying to convert to momentum.

In my example I was taking a lot of energy and converting it to much less energy at the same momentum. There must be a kinematic model that allows less energy input to produce the same momentum.

 

I gave you the barge vs bullet example earlier. Perhaps if I give another example with the same KE but different momentum (and conversely) you will understand this better:


With the same KE throw a golf ball vs roll a bowling ball. The bowling ball has much more momentum. MV and MV^2 are different functions, stop tying to find the proportionality between them. Or if you prefer you can give the same momentum to the bowling ball with less expendature of energy.

Work on the equation giving the relationship between apples and oranges instead.

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Hey. Smart guy. Your sugar-cane alcohol produces a certain amount of energy, exactly measurable and predictable. When you use it in an engine that engine converts it to a certain amount of momentum, but that amount of momentum is, AS I HAVE ALREADY SAID YOU TWIT, not going to be the same amount of momentum from every engine or even the same engine in a different car or different gearing system. I am not trying to find the kind of proportionality that I already spend a lot of time explaining doesn't exist, and its non-existence is basic to my thesis. WHAT I WANT TO KNOW IS WHAT CHANGES THE PROPORTION OF ENERGY TO MOMENTUM AND HOW TO EXPLOIT IT. Answer when sober.

 

For the third time:
Monemtum is mv, which to help you, I will call "M" but keep in mind "m" is for mass
Kinetic energyis mv^2, which to help you, I will call "K" (To keep it simple for you, I have omitted the customary factor of 0.5 - this is just a scale change, like then number of ounces vs number of Kilograms etc.)

Thus one can state: K =vM or in words, what changes the proportion of energy to momentum is v, the speed.

Recall from first post example: river barge has much less v than the bullet, (if both have the same M).
Recall from second post example: bowling ball has much less v than the golf ball, (if both have the same M).

You can not "exploit" it. The monemtum and kinetic energy are two differnt aspect of any moving object. Their functional relationship in this case is that K is a linear function of M and that function is very simple: just multiply M by v to get K. (or v/2 if you wish to return to the more conventional energy scale.)

This is exactly like surface area, A, and volume, V, are two different aspect of any object. In this case the relationship will depend upon the shape but the functional relationship is more complex that just multiplying A by a shape dependent constant to get V in the general case.

For example if the object is a cube, then A = 6L^2 and V=L^3 so V = (1/6)LA which is quite like K = vM case with (6/L) playing the role of v. In general the relationship between V and A is always of the form V = kLA where k is a different constant for each different shape, L is some characteristic linear dimension of the object, and A is still the surface area. In the special case of the cube, k=(1/6) and L is the edge length. (If I made no algebra mistake, interestingly for the sphere, k=(1/6) also as the "characteristic linear dimension of the object, L" is obviously the diameter,D. Or again, V = (1/6)LA = (1/6)DA. )

Your question is of the same mature as:

How can I exploit the surface area to change the volume? Answer is you can not unless you change the shape.

How can I exploit the momentum to change the Kinetic energy? Answer: you can not unless you change the mass (or the speed).*
Or if you prefer:
How can I exploit the Kinetic energy to change the momentum? Answer: you can not unless you change the mass (or the speed).*

If you still do not understand - I quit. you can re-read my three posts some more.
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*Normally one of these two is held constant. Or one of M vs K is held constant to make comparisions.

 

Billy, I understand that you took the long way around to tell me something that I already knew.

 

No - Your confusion is evident in your post 68 where you say:

"The riddle is, how do we efficiently convert a certain amount of {knietic} energy to momentum and get it to the wheels? Experiments with light and heavy flywheels would demonstrate that we can actually create these situations in the real world, where we have a flywheel with a lot of energy and little momentum, or one with a lot of momentum and little energy."

One does not convert energy to momentum as they are not the sort of thing as I explained to you three times. One kilojoule of energy can be assoicaited with a very large momentum (in massive object, like a river barge barely moving) or very tiny monentum in a fast moving grain of sand. Thus, to speak of converting energy to momentum just reflect lack of understanding that they are not at all related. One can equally well and equally stupidly speak of converting {electrical} energy into watts. Fact that your pair are both associated with motion is exactly paralleled in this example where both are quanties associated with electricy.

I.e. to speak of converting “energy to momentum” or to speak of converting “energy into watts” are equally silly and only reflect the ignorance of the speaker.

 

More to the subject:

http://www.forbes.com/forbes/2008/0616/074.html?partner=globalnews_newsletter

Forbes gives some data on the new* Litium-ion battery designed for a car soon to be on the market at the lowest cost yet for full size car:

700 pound batter gives 120 mile range and currently costs, even made in China, $25,000 but they hope to bring price down to $6000 in five years with greater production etc. The maker, Tianjin Lishen Battery Joint-Stock, claims to have created a battery that won't overheat by replacing heat-generating cobalt oxide electrodes with cooler-running iron phosphate.*

Testers in China are driving the cars around the clock. Rubin aims to have 30 prototypes on the road in the U.S. and China this summer so contract testing companies can pinpoint problems. Rubin plans to warrant the battery, which is supposed to last ten years, for at least 125,000 miles to ease buyer doubts.
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*Not the earlier standard cobalt oxide type that catches fire, exploids, etc.

 

You truly cannot be serious. What you claim to have explained to me three times is what I explained to you first. It's exactly the same thing. I am the one who told you that there is no direct relationship between momentum and kinetic energy. The actual mathematical relationship is that the formula for kinetic energy is the first integral of the formula for the momentum.

If an object has a kinetic energy of 100 joules that implies that somehow or another 100 joules plus something to make up for inefficiencies can be used to make that object go at that particular speed. We know that an elastic collision with an object of just the right weight carrying that many joules of energy will actually do this, and I'm not going to open Maxwell's book right now, I have to get to bed (yes I have the book). A much lighter object will have to carry far more joules of energy to have the same effect, and a much heavier object will need far fewer joules of energy. If you use a lever, a weight that you drop on one side of the see-saw will flip the weight on the other side up into the air with the amount of kinetic energy that you choose. If the second weight is ten times lighter, it will have ten times the kinetic energy, assuming for convenient math that the see-saw weighs nothing.

If you efficiently transfer the momentum from a heavy weight to a light weight, and the see-saw lever is the easiest way, you actually multiply the kinetic energy that you start with.

Yes, whatever, you convert momentum to momentum, no shit, but we have systems that transform the equivalent of kinetic energy (measured in watt seconds which come 4.19 to the calorie of heat) to the equivalent of kinetic energy. One example that might tell you what I am talking about is the idea of converting fifty watts (as in joules per second) of energy to light to try to drive a motor. That will only turn the vanes of a radiometer. Use the same fifty watts to heat a working fluid for a Stirling engine and you will see a lot more action. The same amount of input energy, by known technology, produces a highly variable amount of momentum when you turn it into light,or moving fluids, or moving weights. You literally get more momentum, up to some point, the more you push with the energy. You have choices that can produce a little bit of momentum or a lot of momentum from the same energy input, and that is what I mean by converting kinetic energy (or equivalent) to momentum.

 

As this is both false and NONSENSE, I did not read further.

I will give you $1,000 for any machine, including a light weight see-saw, that can get more energy output that the energy input to the machine. Let me make it a more modest request:

Instead of your example of a 10 fold increase, just sell me a see-saw that only doubles my energy and the right to copy it. I will build more, hook them in series, and solve the world's energy problems forever.

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LOL

PS You should have quit when you only appeared to be confused, instead of dumb (in the sense of totally ignorant of the most basic of physics laws: conservation of energy).

 

Billy, kinetic energy is not a conserved quantity. Momentum is. This makes for a delightfully confusing universe, doesn't it?

Offer rejected about the money, though. If I come up with a practical device I know I can get a lot more money than that out of it.

 

the law of conservation of energy does not state that kinetic energy is conserved, nor did I. You stated ever falsely and foolishly that your see-saw (if light etc.) could create 10 times more output kinetic energy than the input kinetic energy!!!!

Your example was a weight falling on one end of the "perfect see-saw" flipping a weight only one tenth as massive, sitting on the other end of the see-saw up into the air.

You were thinking that the smaller weight would need to be going 10 times faster to conserve momentum, so its KE would be mV^2 vs the original Mv^2 KE as to conserve momentum Mv = mV. (Where the capital letters, of course are for the bigger quantities. I.e. M = 10m and V= 10v in your example.)

That is simply all wrong and NONSENSE. The sooner you admit your error the less lasting is the impression the your are just very ignorant.