# Rookie introduction



## joris (Dec 17, 2007)

Hello,

what type of battery is it?
A 2 speed gearbox? Please, explain. will you use a gearbox from an other existing car? A gearbox doesn't have great efficiency (30% i thought).
You also wanne have a wide range so probably a great cost for batteries.
0-100 in 4 sec looks GREAT to me!
I've been reading for a while in "metricmind" his site and my toughts get stuck at the battery management system 

Maybe i first will test with an electric engine DC with the electronics from a fork truck on a little kart that i have. And then take 4 heavy duty car batteries.

Joris


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## blastergti (Dec 28, 2007)

Welcome to the forum joris! =D 
- I bet the guys in here will help you out =D. 
I have not converted a car yet, but i have been looking up the stuff that i want to use myself, which are somehow pricy, but surely gives the blast i need when the pedal hits the floor ^^, 
- I have links to all of these things, maybe it is something useful?
For example, the motorkit i want to use peaks 600Nm of torque! 

Lexus, that is going to be a monster! 
What motor kit will you use? and if possible, can you post a link of those batteries? They seem to have the right caliber for us three =D

/ /// Oscar


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## joris (Dec 17, 2007)

Hi fellas,

a 600NM engine  VERY nice! What engine is it? Is it AC also? What are the nominal rpm's of all these engines you all use?
Is it like normal 1500 (4 pole engine) of 3000rpm (2pole engine)?
If you have a 4-pole engine then your frequency has to be 10 times more than if you want 15000 rpm's...
I've learned that the bearings just can handle the double. I.e. 3000rpm with a normal asynchroon induction motor.

blastergti... i see gti, what car is it then?

joris


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## blastergti (Dec 28, 2007)

http://www.azuredynamics.com/pdf/AC90+DMOC645%20-%20June%202007.pdf
Here is the link =D and all the specifications you want to know!
Ac of of course! =D Only the best is good enough!

I dont know about the price! but here is a link to a complete AC conversion kit!
* edit: batteries are not included >.<
http://www.electroauto.com/catalog/price-pts.shtml#ackits (click on "ac kits" in the list) the site is a little bit wierd =P
"Heavy Vehicle Manual Transmission Kit"
"Light Vehicle Direct Drive Kit"
"Light Vehicle Manual Transmission Kit"

-Same manufacturer, but a little bit weaker motor, If you talk to them they may make a kit for you with the 600Nm (ac 90) motor!

The GTi? nah =) just for fun =D 

// / Blaster GTi!


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## joris (Dec 17, 2007)

Nice links!
Specially the first one!
Only, my question is: Is 50kW (+/-70hp) enough? I know that the torque is a whole other stroy. I'm very curious how this kit will response.
Will you use the original gearbox or how will you mount the engine to the wheels? Do you use an gearbox plus differential? Or is it only differential?

My first nice car (i still have it) has a 1.3 turbo engine (105hp and 150Nm)and it is in an lightweight car (800kg).
0-100km in about 8sec (factory numbers) and 8.2 with G-Tech
1/4mile time at low 17sec (internet numbers)

If i ever have an electric car with this performance i will be VERY GLAD!!!

At the moment i doesn't have a lot of time so can(t read a lot and make a lot of mistakes.

Greetz!!!


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## blastergti (Dec 28, 2007)

-That day i convert my car, is 3,5 years away, as earliest 
That´s because i am studying to a 100% Automobile electrician right now, and i am only 18 years old in february...

But that - believe my words - Does not stop me at all! 

_*--> Transmission:*_
-Lets say i would convert today, i would go rearwheeldrive, and finding a 6speed BMW M3 e36 gearbox, differential from a M535 e28, and a driveshaft from a E36 328. 
These goodies would be put into a Volvo 740,960,940,240, with a clutch so solid it will kill the asphalt before the clutch itself break 
What car i would use would be a Volvo 960,240,740,940. 
They are really easy to modify and find, they are also very light. 
There is tons of these for coffee prices on the swedish trading sites! The transmission should also be easy to find, it just to call to the scrapyards to find a smashed car, and get the stuff before anyone else does! The good thing is that the tranny parts i listed, takes 600 Nm easily! 
-Of course, this is the dream mix, and is not supercheap after all... 

A swedish guy built a 1200nm (insane huh??) Bmw who had these parts i listed included, plus stronger shafts to each wheel!
http://www.limmet.se/fakta.htm its in swedish, but check the numbers on the very top of the page!!

_*-->Possible Solution?:*_

The easiest way to make the car driveable is to lower the voltage a bit, it even will make your battery bill better, because 312 Volts do not just make themselfes, the batteries (lithiums) good and light (380kg) enough for 312V would cost 8 000 euros!!  so a voltage drop would be the best. 
You could also go with cheaper batteries to make 312V, 12V Optima yellowtop agm´s for example, but it would be very heavy, over 600 kg´s, or even 1 000kg!

_*--> Horsepower/Torque:*_ 
These are some of the trickiest things for me to explain! 
Torque is the raw power of the motor, shortly said, but is not enough for a good explanation: 
I want to say that it is the power creating cars to do massive burnouts or wheelspin very easy.
Example: A swedish BMW tuner have to carry 100 kg extra load in his cars trunk to prevent his 800Nm BMW to avoid wheelspin when driving normal, but a ferrari cant do that easy made wheelspin even if it has loads of horsepower.

CheerZ =D // - Blaster GTi


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## John (Sep 11, 2007)

Torque is force multiplied by the moment arm or the distance from the centre of rotation that the force is developed. Torque is what breaks things including the grip between the tyre and the road. Power is just torque multiplied by the revolutions per second. So torque can be amplified by gearing but power cannot. The Power at the input shaft to your gearbox will be the same at the driving wheels of the car minus some factor for the efficiency of the drive train. Torque on the other hand will be multiplied by the gear ratio minus losses. The force generated at the contact patch of the driving wheels will be the torque at the wheels divided by the rolling radius of the wheels. If that force exceeds the weight carried by the wheels multiplied by the coefficient of friction you'll burn rubber. Another thing you will notice from the derivation is that the faster you go the more power you will need to generate the same driving force at the back wheels as the torque stays the same but the revolutions climb.


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## blastergti (Dec 28, 2007)

Thanks a lot John! That was a great explanation!  

Joris!:

You maybe like to use the AC90 kit for your project, but if you are putting the voltage too low, for example 130V, your amp drag will be higher, causing shorter run-times, and making the cable thickness needed bigger. 

I also dont know if the voltage drop method will make the AC90 to decrease the power enough to keep your transmission from breaking. Transmissions are often built for the cars standard power and torque, and cant take much higher loads without either break or wear faster.

If this is the case, the weaker AC55kit -280Nm peak, -140Nm Continous model may be the better choice.

- I found all above out sunday night and had to inform you about this!

We have to find this out:
Wich motorkit and battery choice will give you the best, and most reliable performance for the money? 
- I think we have to discuss this and come up with something further on.

As i have learnt:

Higher voltage means less current, but maintained power, and the current is something we want as low as possible when it comes to electric vehicles of all kind. 
Higher current means that the batteries will not last as long as they would with higher voltage, and wires also need to get thicker and heavier.

- Correct me if i am wrong please!

// Best wishes! -Blaster GTi


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## joris (Dec 17, 2007)

Cool websites and VERY cool BMW!
In sweden they make a lot of powerfull cars! I saw many great movies from there.

It's great that you found it (last sunday  - over enthousiastic - bin there, done that, got the Tshirt)

They 've got nice kits but if it would be possible, i would like to take a normal AC motor with a normal Siemens invertor (or are there any other brands that make these things?). Even i was thinking to make an invertor myself but because i'm not a super-electronic myself... 
I want all these things to lower the costs because it will be very expensive if it's all finished.
Next problem is that with standard motor you have to take a battery pack around the 240V (delta) or 400V (star) and how far could you drive with batteries because after a few kilmeters of driving you battery pack will drop the voltage and then you haven't the power that you used to had. Sow you have to take a bit more batteries and so on. I know, i don't know a lot of batteries but i will learn it 

And indeed, how higher the voltage how lower the current and how less copper (wires, engine, ...) you need.

If it would be possible i would like to take a differential of rearwheel driven car and put the electric motor put on the differential. Then i don't have losses from the gearbox and a better (how less parts, how less losses) efficiency. 
But then my problem is the motor (average ratio is 3.5:1) and if you count the 
diameter from my little fiat uno and punto it's like this:
13" wheels with 155/70 tyres becomes a diameter from (155*0.7*2+13*25.4 = 217+330 = 547mm) The contour is now 2*3.14*547= 1719mm
If we do the same with 165/65/14 then it will become 214.5+355.6 = 570.1 => 570*3.14=1791mm
Sow, lets take an average from 1800mm distance for one rotation.
When you drive about 100km/h you then is the wheel rotation about 925rpm from the wheel shaft. (100000m/h / 60minutes / 1,8meter = 925rpm).
Now the differential ratio is 3.5:1 sow when the wheel shaft makes a 1000rpm the motor schaft must take 3500rpm.

I would chose for a 4-pole engine that can take 1500rpm nominal you could take about the double of rpm out of it. What kind of engines sells the manufacter anyway? 
I learned (correct me if i learned it wrong ) that the nominal rpm from an electric engine could make the double off rpm (sow a 1500rpm motor could make up to 3000rpm with frequency regulator max; otherwise the winding will break due to centrifugal force and also the bearings won't last for long)

Because i'm a real rookie i probably will buy a kit.


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## blastergti (Dec 28, 2007)

Yes i think a kit for you is the best solution =)

 Motor to differential setup:

The motor on the differential style would be great, just as long as the motor doesn´t touch the ground. 
Azuredynamics motors are quite big, the AC90 is massive 14 inches, and the AC55 is 12 inches in diameter. But then, a long shaft from the motor in the engine bay to the diff is just as good. 
And then you dont have to worry about the groundclearence and performance  

Siemens and Brusas motors looks smaller than Azure´s, and would probably do the trick! 

Maybe i got this wrong, but did you want to build a rwd setup on your fwd?
If you want rwd, the best thing to do is to buy a car that is built for it since rwd cars are completely different in their construction for their purpose, an old Volvo would be great since they are really cheap, and easy to modify!

Price/performance:

I looked around and Siemens motors looks smaller than Azures but i dont have a spec PDF yet. 
And also, Brusa electronics from Switzerland looks smaller - but, no spec pdf.

But this turned everything upside down:

When i studied Brusas components on metricmindengineerings webpage, and saw that price/performance compared to Electroauto i almost thought i made the wrong calculation:

Brusas smallest controller alone would cost 6190 Euros ($9 522) + shipping,
A complete conversion kit from Electroauto.com with the AC55motor+controller would cost 6480 Euros! ($9 970) + shipping.
That really is a difference! 
Hopefully Siemens are cheaper than Brusa! I didnt find the price because it stood "inquire" at the price line...

Nominal RPM:
Dont know, but you are probably right! it sounds plausible though!  

// Best regards, wishes and everything! -Blaster GTi!


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## joris (Dec 17, 2007)

Therefor i would like to use an kitcar (rwd) and put the motor vertical on the pont (differential in french).
Do not have a lot of time; i'll try to post a figure


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## joris (Dec 17, 2007)

RWD


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## blastergti (Dec 28, 2007)

That diff setup is going to be good!
And i like the idea with the kit car, almost no wieght and lots of power!

Do you have a link to the kit car that looks interesting? If you got opportunity, could you post that link here? 
I would really like to see it!

All the best! - Blaster GTI


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## bucket (Jan 8, 2008)

Hey Joris,

Als ik even zo vrij mag zijn om je in 't nederlands aan te spreken 
Velen op dit Forum zullen denken: wat ik godsnaam staat hier nu ???
Maar terloops wil ik je toch eens op de hoogte brengen van mijn eigen plannen ... Ik ben namelijk bezig met de restauratie van een Datsun 100A (1.0 liter benzine) en daarvoor heb ik al heel wat spullen bijeen gezocht (en gekocht ) Op een bepaald moment kon ik zelfs een volledige 100A redden van de schroothoop voor de prijs van het oud ijzer. Geen papieren, geen sleutels maar wel een hoop onderdelen !!!! Nadat ik die tweede wagen helemaal gestript had en op het punt stond om die toch maar als oud ijzer weg te voeren, kreeg ik dit niet over mijn hart. Al bij al is er wel veel werk aan de carrosserie maar volgens mij valt het nog te redden. En toen dacht ik ook aan een EV convertie !! Het is een licht autootje (geen 700 kg), geen stuurbekrachtiging, zelfs geen rembekrachtiger maar er bestaan genoeg voorbeelden van remconverties (zelfs met geventileerde schijven). Om mijn verhaal wat in te korten: mijn eerste datsun blijft ICE maar helemaal gerestaureerd en die tweede wordt een projectje. Ik heb ondertussen ook al een elektrische heftruck uit elkaar gesmeten maar daar staken 3 kleinere DC-motors in en ik heb eigelijk 1 redelijk grote nodig (blijf zoeken). Zo zie je maar, ook in dit Belgenlandje zitten zo van die zotten die willen scheuren op elektriek !!!

For all the other forum-members: above is some dutch between a couple of belgian guys 

greetz, Pieter


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## blastergti (Dec 28, 2007)

Hi pieter!
Welcome to the forum =D


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## Sandy (Jan 22, 2008)

Hi, I'm new to all of this, I've begun working on my first conversion about a month ago. So take my questions and comments for what they are, novice at best.

I'm confussed by the "low current" goal. Here's why, power is measured in Watts and Watts are Volts times Amps. I had the impression that if you had a battery that had 100 amps at 12 volts that, if the battery could be fully discarged down to zero amps (of course it can't) you would be able to get about 1200 watts of power. Now if you put two of these batteries in series you would now be able to get 100 amps at 24 volts, or 2400 watts. If you put the two batteries in parallel you would be able to get 200 amps at 12 volts, still 2400 watts.

I understand that with higher voltage you can use thinner wire, so I can see there would be a weight savings, but beyound that I don't understand the drive toward higher voltage. For example if a 3 phase induction motor is rated to run at 230 or 460. When it is run at 230V the amp's consummed are twice that of when it is run at 460V, but the kilowatts are exactly the same. So if that motor is set to run at 460V and you string 40 batteries each with a 50 amp capacity, won't you get just about the same result as if you set the motor to 230V, and run a string of 20 batteries with 100 amp capacity?

Also I've been told, by a local battery dealer, that the more batteries you string together the more problems you run into with keeping them "balanced".

Any clarifiction would be greatly appreciated.

Sandy


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## blastergti (Dec 28, 2007)

Hi and welcome to the forum Sandy!

The high voltage is the best choice. 
You are right about what you are saying. This will also give you much better run times. 
Your equipment will run cooler without compromising with performance. 
And your car won´t be that heavy, due to thinner cables.

The lower the amp draw, the more power will remain in the batteries, and your runtime will be increased. 

If one would go low voltage you would get a massive current - the power would be the same - that´s right, but - the run time would be shorter. 
This is because the power - rated in amps, comes from the batteries, and the more amps you use, the less power will remain. 
Think of a battery cell like a water canister put under more or less pressure! 
Think of electricity in general like water, it is good advice, it has helped me out many times. My teacher compares to water and hydraulics when we talk electricity, even my books i am working with! 


Battery balancing:
He is right. Now lets make an example for you: If you got a lead-acid battery with 6 cells where five of these keep the right voltage, but the sixth not, you will have a noticable decrease in power. It dont have to be much to make a lead-acid go weaker than it should, or any multicell battery.

That is because they are connected in series, it means that if one cell stops functioning - but the other five work, the current can´t go through the broken one, and the battery will be unusable. 

Hope this may bring clarity! =D

All the best and the rest! // Blaster GTI


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## John (Sep 11, 2007)

Sandy the voltage drop caused by a component in a circuit is equal to the current conducted multiplied by the resistance of that component. So if that circuit was to conduct a higher voltage at a lower current the same wattage would be conducted with a lower voltage drop across the component. Say double the voltage at half the current this would equal half the voltage drop. That voltage drop multiplied by the current conducted equals the watts lost by that component. Half the voltage drop multiplied by half the current equals one quarter of the losses. So you can see that the higher voltage circuit can conduct similar wattage with substantially reduced losses due to resistance. We may be able to alter the resistance of our conductors to compensate by increasing the cross sectional area and shortening their length but we cannot alter the internal resistance of our batteries. Double the number of batteries in a string to double the voltage and you would have double the collective internal resistance but halved the current required to produce similar watts so the voltage drop would be similar multiplied by half the current for half the internal losses in the batteries. The usable charge recovered from batteries deteriorates significantly as the current draw increases and can easily dip below 50% for high current draw downs on lead acid batteries. I'm not quite sure what it means in terms of induction motor efficiency though. Magnetic field strength is proportional to ampere turns so the 460v would require twice the number of turns to produce similar field strength on half the amps. This is a bit outside my area of expertise though.


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## Sandy (Jan 22, 2008)

John, (and Blastergti) thanks for the replys, please check my understanding.

Even though we can use thicker or thiner wire in the motor to control the resistance we CAN'T do anything about what's going on inside the battery.

"Double the number of batteries in a string to double the voltage and you would have double the collective internal resistance but halved the current required to produce similar watts so the voltage drop would be similar multiplied by half the current for half the internal losses in the batteries."

So the batteries in series builds resistance as you add batteries, hence higher Volts and lower amps.

"The usable charge recovered from batteries deteriorates significantly as the current draw increases and can easily dip below 50% for high current draw downs on lead acid batteries."

So a big factor is that batterys will deliver more total usable power when discharged at a lower amps. So by putting them in series, you get the same total power (watts) initially, but by drawing the power out of the batteries at a ratio of higher volts lower amps the batterys perform better.

Is this right?

Sandy


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## Mastiff (Jan 11, 2008)

> So a big factor is that batterys will deliver more total usable power when discharged at a lower amps. So by putting them in series, you get the same total power (watts) initially, but by drawing the power out of the batteries at a ratio of higher volts lower amps the batterys perform better.
> 
> Is this right?


Yes, this usually is the case.

If you've ever noticed, batteries are sometimes rated at their 1-hour Amp Hour rating and their 20-hour Amp Hours rating.

Basically it means that if you draw less amps from the battery over a longer time the battery will produce more watt-hours of "total" energy.

It's due to the chemistry of some batteries each battery chemistry has it's ideal drain rate, draining them faster reduces the total energy the battery will produce.


Here's an example:
http://www.batteryspace.com/prod-specs/LA-12V10.pdf

20 hour rate F.V.(1.75V/cell)( 500mA to 10.50volts) ............................................................................10.0A.H.
10 hour rate F.V.(1.75V/cell)( 950mA to 10.50volts).............................................................................9.50A.H.
5 hour rate F.V.(1.75V/cell)( 1700mA to 10.50volts) ..........................................................................8.50A.H.
1 hour rate F.V.(1.55V/cell)( 6000mA to 9.30volts).............................................................................6.00A.H.


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## Sandy (Jan 22, 2008)

Mastiff, thanks for the reply.

I'm imagining that you get more out of the battery by discharging it over a longer time, because the battery will heat up less (lower internal resistance) and because there's more time for replenishment of the charge by chemical reactions within the battery.

Except for the time period you allow to get the power, if your after 100 watts, does it matter if you take it at 100 V 1 amp verses 1 V 100 amp?

I mean I can understand that if you take the 100 watts fast, it will be "harder" on the battery. But does the ratio of volts to amps actually matter or does it all come down to time?

Sandy


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## Mastiff (Jan 11, 2008)

Theoretically the power should be the same if you take it as 100v/1amp or 100amp/1v according to the wiki:
http://www.diyelectriccar.com/forums/showthread.php?t=6535
Read the bottom battery wiring in series and parallel.

One thing that does matter though is resistance.

Pretend I've got standard household electrical wiring(USA wiring, other people are crazy ), for now lets say the good stuff, it can carry 110Volts and 20 Amps.

Now, that same wire could also carry 220 Volts at 10 amps but it could NOT carry 55 Volts at 40 Amps.

Wires must be thicker to carry more amps otherwise they heat up and build resistance and in the case of the 55v 40 amps going though the household wiring, that causes the house to burn down.

Now think about the electrical grid, it runs in the 10,000+ volts I think and I think those are copper cored STEEL wound cables.(I could be mistaken)

But they're thin, the only way they they can get that kind of power though them is to use voltage because amps would require larger wires. (more copper, copper = $$$$)

But in an efficiency stand point, this is awful, since you need to step all of that voltage down.

What it comes down to is that in an EV application it's all about what voltage your motor runs at.

If you've got a 144volt DC motor, build a battery pack that supplies 144 volts, this would mean your controller would be doing less work converting the power to the required voltage for at-speed driving.

I'm not so sure about how it works in an AC system but I think it's relatively the same, the pack should have a nominal voltage near the motors voltage rating.


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## joris (Dec 17, 2007)

wow guys,

i went on a short vacation and i got stuck on the work and we got 3 pages... a lot to think about!

@ pieter:
van waar ben je? ik ben zelf van sint-niklaas. lijkt me wel eens leuk om af te spreken...

i've visited the national car salon (?) and we saw the "westfield" car... a LOT of money and the difference between build it your own and let it build is 3000€... 
i also know that poly cars aren't alloud anymore in belgium. only if you are certifieted... yup, thats belgium 
if we have a lot of luck we could get a test project car from italy..  we know the right person for papers en documents

grtz


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## bucket (Jan 8, 2008)

@ Joris

I'm from Izegem (W-Vl) and i'm just starting my way up into EV's. I'm restoring a datsun 100A at the moment (yes, its ICE) (also see: http://www.diyelectriccar.com/forums/showthread.php/just-short-introduction-myself-7709.html) and I have a spare donorcar which could be used for convertion. To get an idea of what is needed I also have a big go-cart that's going to drive electral. For now I'm just gathering parts and ideas and of course, restoring my first datsun 100A.
Will keep you posted on what I'm doing here ...


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## mattW (Sep 14, 2007)

Just to take a look at your example above; to get 100A at 1V you need a very low resistance circuit V=IR so R=V/I meaning R=1/100 or 0.01Ohms. The same wire at 100V could carry I=V/R=100/0.01=1000A and P=VI=100x1000=100kW of power compared to just 100W at 1V (1000x more power). 

To get 100V to put out 1amp your total resistance can be 100Ohms so if you use the same wire in the above situation at 0.01Ohms you can have 99.99 Ohm device (eg a light bulb) that does useful work so if P=VI=IxIxR then you have 1x1x99.99=99.99W of useful power dissipated in the light bulb and just 0.01W wasted in the wires compared to having all your power wasted in the wires in the 1V version.

Both circuits use 100W of power but at 100V you can use the power where you want it not lose it in the wires.


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## bucket (Jan 8, 2008)

@ MattW

I think you didn't get the picture in my little project above. The datsun 100A is a little oldtimer and the "100A" is just a type of datsun (just like they made the 120A, the 100A FII, the 120Y, and so on ...) In some countries they were called Datsun Cherry and I even think they never sold those cars in Australia. So, I did not refer to 100 amps !! Although I'm planning on converting one to electric, I'm not aming for a drag-racer. The car weighs about 700 kg (1500 pounds) so a light convertion should do it (thinking of a 96 V system and a 10 Kw DC motor) From my home to my work is only 2.5 miles (another 2.5 to get back home ) on really flat roads so range isn't a big deal for me. The only problem I have is that you can't convert a car in Belgium !!!! They won't allow any changes except the things that are off factory (not even wide rims or another steering wheel !!!) That's why I'm working on an oldtimer. Regulations on oldtimers over here are a little looser but then you have all the other problems of getting parts and rust (a lot of rust !!!) In my thread where I introduced myself I'm showing two of the same cars (one is a donor) and for now I'm restoring the first one (with the petrol engine) The second one (for parts) can be saved from salvage but it will need al lot of welding too. A last remark: all my projects are on a tight budget, it's a hobby for me and I'm also a lucky guy with a wife and two children so you can imagine that i'm not spending all of my free time "playing with my toys" 
Oh, and another thing: I'm also making an electric go-cart for my son. I already have a go-cart and a DC motor (from a forklift) and now looking on how to get those two things together.


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## joris (Dec 17, 2007)

Hi guys,

i've bought a buggy kart 150cc.
The licence plate will be from a 2-wheeler motorisezed bike and in belgium we don't have to go to the yearly inspection...
Now we can transform it into something we like.

I am still searching for an old elektro truck as a donor (batteries, engine, electronics,...)

When its arrived i will post some pics.

@Pieter: ZALIG BAKSKE!!!!


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## Sandy (Jan 22, 2008)

Can you help me understand this?

When I charge a battery, I use V = IR to decide what resistor I want to use to limit the current from the power supply to the battery. 

So for example with a 24 volt charger that rated for 1.5 amps, if I want to limit the charge current to say .9 amps, then I divide 24 by .9 and that gives me the ohms of resistance I need roughly 26.6 ohms. Then I read that Ni-Cad's have about 155 milliohms of resistance for each 1.2 volt cell. So my 21.6 volt pack has 18 cells. So the battery should have about 2.8 ohms of resistance. So that's a total of 29.4 ohms of resistance. So I'm expecting to get 24Volts / 29.2 ohms = .822 amps.

When I meter the amps in the circuit from the power supply through the resister to the meter and back to the supply the amps are just what I expect them to be nearly .9 amps. But when I put the battery in the circuit the amps drop way down, way way down, like form .9 Amps to .05 amps. 

Is it that the current goes into the battery and doesn't come out? I thought in any circuit the current would be the same across any two points.

Should I instead be using the difference in voltage between the battery and the charge circuit to figure the ohms? For example if the batteries fully charged will be 20.25, and my charge voltage is 21.15, do I take the difference ( .90 Volts ) and use that to determine the current and resistance? For example .9 Volts / 26.6 olms = 0.034 amps. This is much closer to what I'm actually seeing.

Please help!!!


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## Sandy (Jan 22, 2008)

I think I got it... Please check me... Thanks!

Purpose: Find out if I should have been using the difference between the charging voltage and the battery voltage to determine the resistance and the current.

Procedure: I will measure the charging voltage and the battery voltage take the difference and apply ohms law to determine what resistance is needed to achieve a specific charging current.

Given's: the resistance in a Ni-Cd 1.2 V cell is .155 ohms
battery has 15 cells
external resistor value is 10 ohms and can dissipate 10 Watts 

Data:

Voltage measured in battery 19.03 Volts
Voltage measured from power supply 24.32 Volts
Ohms measured across external resistor = 10.2

Calculations:

15 cells * .155 ohms = 2.325 ohms in battery.
Vsup - Vbat = Vdiff
24.32 - 19.03 = 5.29 
V=IR
5.29 = I * 2.325 ohms
5.29/2.325 = 2.275 amps (this amperage is to high for the battery
and above the capacity of the power supply)

Total ohms in series = 10.2 + 2.325 = 12.525 ohms
V/R = I 
5.29 / 12.525 = .422 (this in an amperage that the both battery and supply like)

V * I = Watts
5.29 * .422 = 2.266 (this is the Watts the battery and the resistor will dissipate)

Observations:

Actual measured current in circuit is .417 this is within 2% tolerance 

Conclusions:

Ohms laws applies to the relationship between to potentials, the potential difference.
In the case of charging a battery, because the battery has it's own potential the voltage or potential difference between the battery and the power supply is what ohms law applies to.


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