'66 Deluxe

[Mike Kever Kombi]

Spectacular.

Almost time for another clunker bus.

[Six Volt]

This is what I needed to see. After watching all this, I'm sure I'll be at Harbor Freight looking for a cheap #$% blaster cabinet to sandblast small stuff. I'm amazed at how well stuff turns out when you can really clean, clean, clean. Really great stuff Bruce and thanks for sharing.

[SlashKirk]

Spectacular.

Almost time for another clunker bus.

I think he should buy that double cab off the same guy and fix it up next....


Awesome work Bruce!

[Dual Port]

I'm still using a 6v gas gauge on a 12v system, and Samba guys say you can use a 6v gauge forever on a 12v system, but I'm just not like that. I wanted to put a dropping resistor in the fuel gauge power feed. I read the amperage it pulled (76ma) on 12v and used Ohm's Law to figure the operating resistance at 12v, which was about 165 ohms. (12.6/.076=165.78) That meant if I put a 165 ohm resistor in the feed it should drop the voltage in half.

I bought a pack of 162 ohm resistors from Fleabay for a few bucks delivered.

Testing on the bench showed dead on- the resistor in series gave an operating voltage of 6.25v with a fully charged battery at 12.55v on a spare 6v gas gauge.

And used male/female protected spade connectors.


And some shrink tube.


This is what it looks like installed:

"Maybe" I can trust my gas gauge now......?

[Six Volt]

Great stuff. I always have trouble visualizing electrical issues when just reading. Showing is so much easier. I didn't fully appreciate the simplicity of resisters 'till seeing this. Come to think of it, I think LEAKOIL has a few resisters. Anyway, I always have trouble conceptualizing the difference between voltage and current. When reading, they seem to be the same thing to me. Pressure? Volume? Is there realy a difference? I guess so.

So what exactly makes a 12 volt battery 12 volts? Is it the number or size of the cells? I'll assume yes, and that the battery is our "storage tank". So here's where it gets tricky for us "readers." What defines current? I know current is measured in amps, but why are some 12 volt batteries different than others? It would seem the size of wires would effect the current? But wire size seems more related to resistence. So what's the magic in cranking amps that distinguishes 12 volt batteries? What the heck are amps in batteries and how do we conceptualize them?

So it seems you divided volts by amps to get ohms. I think I understand how you tested for amps at the 6V device, I just don't understand the amp concept. I'm guess I'm stuck on the amps part.

[Mike Kever Kombi]

Had a nice little response to your questions, but accidentally closed it. Will post again after work. It will make everything as clear as mud.

[Dual Port]

What defines current? I know current is measured in amps, but why are some 12 volt batteries different than others? It would seem the size of wires would effect the current? But wire size seems more related to resistence. So what's the magic in cranking amps that distinguishes 12 volt batteries? What the heck are amps in batteries and how do we conceptualize them?

Voltage is pressure, amperage is flow. Think of it like this- you have a garden hose on your house and a fire hydrant in front of your house. Both have (about) 70 PSI water pressure available in them, but the fire hydrant has much greater flow capacity. This is like comparing a large battery cable and a taillight wire. Both have 12v (pressure) available at them, but obviously the battery cable has the ability to pass much more current (think water flow, like a fire hose) to get more work done. The 18ga taillight wire is probably good for 10 amps or so, the 2ga battery cable is good for 300 or more amps. Trying to put out a house fire with a garden hose is a joke, and trying to run your starter with the 18ga taillight wire is a similar issue- the pressure is there, but the flow capability is not.

Your taillight wire has 12v available at it, and your starter cable also has 12v available at it, but the ability to flow current is dramatically different.

The measurement I took at the gas gauge for current flow is super tiny- 76ma (76 milliamps, which is 76- 1/1000s of an amp or .076 amp) is not enough to even make a taillight bulb glow dimly.

The reason some 12v batteries are different than others is the ability to flow current. Your VW battery only needs to be 300-400 CCA (Cold Cranking Amps). (I like overkill with batteries so mine is 650CCA I think.) If you put that in a heavy diesel truck that needs 3000CCA worth of (12v) batteries to get started, it obviously ain't gonna work.

Don't confuse CCA with CA or HCA. Some unscrupulous parts stores will advertise a 700CA (cranking amps) battery on sale to confuse you, because most people with think it's CCA. CCA is measured at 0F, CA is at 32F and HCA is at 80F. Ya gots to read the fine print.

A battery rated at 700 CA is probably only 500 CCA or so.

[Mike Kever Kombi]

lets see if we can muddy the waters even more. I will try to see if I can explain things, while sticking with the previously established water theme.

While current is measured in amps, and voltage is "pressure", amps would be more analogous to volume, more so than flow. flow would be a measure of resistance, which according to Ohm, would be would constant in a given medium. however the 2 have a direct correlation. That is however putting the cart before the horse, so we should start at the beginning with some basic laws and definitions.

electricity can be broken down (mostly) into three fundamental and separate aspects, and can be expressed in the following formula

V*A=W

Where V = volts, A = amps, and W = watts

in keeping with the water theme, volts would be the pressure, Amps would be the volume, and watts would be the potential for work (sorry no real water analogy here), which can be compared to HP. The more HP you have, the more potential for the work that can be performed.

The hose analogy from earlier is still valid, with a few caveats. lets assume you have a 100 gallon water tank, and the same said 2 hoses. if you try to pass 100 gallons per hour, through the garden hose, you will encounter more resistance than you would with the fire hose, in fact you could get to a point where you are passing too much water through in a given time and cause the hose to burst. The same thing happens in wiring, but for different reasons.

The reason that the thicker (smaller gauge) wire could produce the 300 amps, is simply because it does not have the resistance that the thinner (bigger gauge) wire does, before failure. Again, referring back to Ohm, resistance is a function of the medium involved, so copper, being less restrictive than say aluminum, could carry more amperage, in the same gauge wire. An Ohm, is a unit of resistance in a conductor. All conductive mediums have different resistance, as do different gauges within the same medium. That is, a 6awg wire can handle more current than a 10awg wire can. the flip side is, that a 6awg also inherently has more resistance, so trying to push low currents through is not going to happen any more than high current on too small a wire, though for different reasons.

too small a current on too big a wire, is like trying to pull a tractor trailer uphill with a goat. Sure, the goat may make it for a little while, but eventually he will just wear out and never make it to the top.

Too big a current with too small a wire is like trying to pass a baseball through a garden hose. It might make it a little bit, but sooner rather than later the hose is going to split.

so, back to V*A=W, and Ohms law, which states that the current through a conductor is directly proportional to the volts (Potential Difference) which can be expressed: I=V/R, where I = amps, V = volts, and R =resistance, we can work out the resistance of a known gauge wire, and calculate the wattage of the appliance or fixture we are trying to power, with our given current and voltage. or in other words, we know we have a 12v system, with a 40 watt bulb, we can use some algebra from our formula V*A=W or W/V=A. 40(w)/12(v)=3.33(a). Then on to Ohm I=V/R or V/I=R. 12(v)/3.33(a)=3.6, so we would size our wire accordingly.

so, back to Sean's pressure/volume difference, when reading current, you are reading the amps that are travelling across the conductor, are you measuring pressure or volume? The answer is yes. You may start off with the same volume (amps) upstream, but because of the resistance found in different conductors (big vs small, dissimilar conductors) the pressure downstream can, and will be different. The inverse is also true, you can start with the same voltage upstream, but end up with different current downstream, for the same reasons.

Batteries.

electricity is conducted 2 ways (mostly), through electrons, and through ions. when conducted through wire, electricity is transferred through electrons, this is partly where Ohms law, and thermodynamics come in to play with regards to wire sizes. a By product of transferring electricity through electrons (that little elementary particle that zips around a proton) is heat, excessive amounts of heat. Trying to push too high a current through too small a wire will produce more heat than the wire can dissipate, causing failure.

electricity in batteries is transferred via ions in an electrolytic solution. a battery is comprised of an array of cells. cells are (generally) 2 dissimilar materials, with an imbalance of electrons separated by an electrolyte. the imbalance of electrons will always try to right itself, and get as close to equal as possible (more or less), 1 electron to 1 proton. the ions in the electrolyte are imbalanced and "transfer" their energy from one pole to another (cathode and anode) and spit out the excess electron at one end (electricity). [This is not exactly how it happens, but it is close enough for this discussion]

The more cells (array of cells) a battery has, the more power (amps) that battery has available to it. What (mostly) determines (not always) the voltage available to a battery, is the chemicals used, and the dissimilar mediums. the interaction, and exchange of electrons within, and between the mediums will produce a certain voltage.

to get back to Sean, and why are some 12v different? that is the number of cells a battery has (mostly).

how do we conceptualize them?

try this analogy. in a room, you have 5 thirty year olds. In another room, you have 2 twenty nine year olds, 2 thirty one year olds, and 1 thirty year old. no matter how you slice it, the average age is thirty, in either room. thirty would be the volts (12v). and 5 would be the amps. try as they might, they can neither make themselves a year older, nor make their total more.

now in another room, you have 15 thirty year olds. they are still all thirty (12v) but now they have tripled their quantity (amps)

the same holds true for a battery. you can add more array, but not necessarily increase the voltage (series vs. parallel)

you understand how he tested for amps at the device. you know that he tested for current, a measure of amps, and you know he went from a 6v system, to a 12v system, doubling his previous voltage. so you can use your formula for I=V/R, insert known parameters (v and amps) V/I=R or 12.6/.076=165.78, or 165(166) ohms resistance. since he doubled his voltage, and we want to half it for the fuel gauge (6v), we need to double our resistance, or our Ohms. We already know, from our equation, that we have 165 ohms, and we are looking to double the total resistance to 330 Ohms. 165 are already present, so we need to add 165 Ohm resistor to get a total of 330 Ohms. Basically you are using algebra to solve for 1/2 V (to reduce 12v back down to 6v). V=RI or R=V/I or I=V/R

Hopefully I have succeeded in providing more questions than answers, and you will never get a firm grasp of electricity

just like I promised, clear as mud.

[Ken]

After reading that I feel like I have a PhD in Electrical Engineering. As we say in the mechanical/structural field - "It's only wires"

[WideFive]

A solution to a problem that never existed.