Compression Testing Questions

The next point of confusion: The high voltage provided by the secondary circuit of the coil has absolutely nothing to do with whether the battery is POS or NEG ground. That is determined by the direction of current thru the primary of the coil and by the direction of the physical windings of the wire on the coil's core and the way the primary and secondary wires are connected. We can't change the last two items inside the sealed coil, but the manufacturer can. We can change the current direction by connecting BAT+ to coil + and BAT - to coil -, or BAT+ to coil - and BAT- to coil +, thus causing the High Voltage out to be either NEG or POS as we choose.

To run the test you merely connect the comression gauge to the appropriate spark plug hole and crank the engine until the pressure stops rising. Record the pressure. Repeat on all cylinders. If there are large variations (more than 10 psi)in compression, squirt a couple of ounces of motor oil in the cylinders and repeat the test. If the pressures increase, the rings are worn. If they don"t, the valves are leaking OR you may have a head gasket leak. This is especially true if the low pressures are on adjacent cylinders. But it"s not a sure thing. You can have a couple of burned balves on adjacent cylinders also.

The correct connection of the coil is that the primary wire to the distributor must be connected to the coil terminal with the same sign as the battery ground. So if you have a positve ground battery, the small wire to the distributor must be connected to the (+) terminal on the coil. Otherwise you"ll loose about half the spark energy.
HOWEVER, you don"t need to do anything to the coil to conduct a compression test. In a lot of cases, you don"t have to have the ignition switch on to run the starter.

HALF?

"you"ll loose about half the spark energy."

0.01

1%

typically 100:1 turns ratio

"The correct connection of the coil is that the primary wire to the distributor must be connected to the coil terminal with the same sign as the battery ground. So if you have a positve ground battery, the small wire to the distributor must be connected to the (+) terminal on the coil. Otherwise you"ll loose about half the spark energy.

"What are you talking about? I"m refering to spark energy not voltage ratios or turn ratios."

And I still say, not 1/2 energy lost.

No energy lost. Energy=1/2 x L x i x i.
L doesn't change with current direction nor does i.

Virtually all the front mounts converted to POS ground run around happy "reversed" anyway, because with the mechanical connections at square can coil, they can't be changed to obtain a neg HV output. So, for these thousands, this is purely academic.

I didn"t make this up, do some research and you"ll find this info in a lot of different resources (Motor Publications, etc).

And that is one of the problems with the front monts IF you change polarity. This forum is full of people have problems with front mounts. I don"t know how many are due to having the wrong polarity but there are a lot of folks having weak sparks with front mounts.

Because it is easier for the more highly mobile (because they are on a hotter surface)electrons on the center hotter electrode to jump to the side (cooler) electrode than the other way around, the voltage will have to rise higher to establish the arc.
So, yes there is a better & worse way. Reversed, it is harder to establish the arc, but the energy supplied/stored & released by the coil is the same.

You don't lose any energy, you just need a higher voltage to start the arc. That leaves you with a smaller safety margin. By all means hook it up right if you can.

I've read reams on the subject. The problem is that folks use terms loosely and misinformation is spread, misinterpreted and repeated. Pretty soon it becomes "fact?". Do you concern yourself with setting a lead acid battery on wet concrete or the damp earth? You would be surprised how many do.

Today I received an e-mail that said, " how to be happy: Live like a dog, if you can't eat it or hump it, pi$$ on it and walk away". I probably should take that advice.

The secondary circuit in an ideal situation will only provide the same power as the input power to the primary side. For a 6 V circuit @ 4amps that"s 24 watts ideal. Now lets say that it takes 10000 V to bridge the spark gap and there is no reistance in the circuit. That means the that for this ideal system, the current in the spark gap is 24w/10000 V = 2.4 ma. Now lets put some resistance in the circuit and remember we still have an ideal coil. It still takes 10000 V to jump the gap.The coil responds and puts out 20KV but, but we still only have 10000V at the spark plug gap itself. Soooo 24w/20000 =1.2 ma. The spark energy is now 1/2 of what it was (1.2 x 10000 versus 2.4 X 10000) The rest of the energy is dissipated as heat in the circuit. In reality the coil has some loss so the situation is even worse.
Regarding the dead headed coil, I suppose you could have some small air gap at the coil tower/secondary lead for the case where the high tension circuit is open but it you have an infinite resitance in the high tension lead, you have no current flow and no power in theat circuit. No current, no heat to dissipate. Your primary circuit will still. Just because you induce a high voltage in a inductor soesn"t mean to induce current in the high tension circuit. You only have the potential for current flow. It like a dead headed pump, all kinds of pressure but no flow.
I"ll end this discourse and we can agree to disagree now.

Best regards,
Jerry

A MAJOR flaw. Energy does not equal Power!

Energy=Power x Time.

In your example, the 10,000 volt discharge will last twice as long as if it were 20,000 volt discharge. Hence, SAME Energy.

The coil also doesn't respond with 20, 000 volt output. The coil's output voltage will rise only to that voltage which is necessary to breakdown the air gap and establish an arc. After the initial breakdown & establishment of the arc, the voltage across the gap reduces even further because it requires less voltage to sustain an arc in the plasma created by the arc.
If you look at the HV waveform on an ignition oscilloscope, this is very evident.

There are no facts here!

The secondary circuit in an ideal situation will only provide the same power as the input power to the primary side. For a 6 V circuit @ 4amps that"s 24 watts ideal. Now lets say that it takes 10000 V to bridge the spark gap and there is no reistance in the circuit. That means the that for this ideal system, the current in the spark gap is 24w/10000 V = 2.4 ma. Now lets put some resistance in the circuit and remember we still have an ideal coil. It still takes 10000 V to jump the gap.The coil responds and puts out 20KV but, but we still only have 10000V at the spark plug gap itself. Soooo 24w/20000 =1.2 ma. The spark energy is now 1/2 of what it was (1.2 x 10000 versus 2.4 X 10000) The rest of the energy is dissipated as heat in the circuit. In reality the coil has some loss so the situation is even worse.

Power, then energy, then strike off in any direction!
Where ever you go....you can't evade physics! The energy stored in the coil is 1/2 L x i^2 AND it is ALL released when the points open! I is NOT lost in space, not even 10% of it!

Some people, as you appear to be, don't want to learn, their minds are like concrete, all mixed up and permanently set.

Don't listen to me, just go to the library, read a physics book, an engineering 101 text, you'll learn something and maybe not spread misinformation in the future!

Yep, we do disagree!
If you can't put out accurate info, don't put out anything at all.
To remain silent, you may be thought a fool, open your mouth and remove all doubt.

No temper lost her, just frustration at my inability to communicate. If you will re-read the thread, you will see that I never called you "wrong" or even "incorrect" as you did me.

You clearly show your loose footing as you dance about like a prize fighter, with ever changing subject argument, for example:
1) you lose half the energy.
2) you lose power
3) output of high tension circuit is zero
4) insert resistance in circuit by opening gap (gap is not a resistance) (a resistance exists in the conducting plasma after arc is established)
5) then a baseless calculation of ma in secondary circuit tied to fictional resistance
6) spark current is reduced

The obvious trend here is to not directly address the question, but rather jump somewhere else.

From early on, I stated that "I was here to learn, where does 1/2 energy go?" That got a response of "the spark will be weaker", not an answer as to where did the 1/2 energy go.
So, I asked again & got "because there is essentially more resistance in the high tension circuit "....well we didn't insert a resistor or re-wind the coil with finer wire. So, still no answer as to where did 1/2 energy go? "I didn"t make this up.."

"I"m sorry that you are unable grasp these arguments that address the fact that the spark current is reduced by reversing the polarity of the dc electrical system with out changing the "polarity" of the coil.

"I didn"t make this up" isn't much of an argument to grasp. Nor is "But it does..".

"You didn"t enter this discussion for any other reason than to tell me I was wrong."

Well, I never did tell you that, but I'm still looking for ananswer as to where did half the energy go? And I'm still sticking with the physics & electronics text books in that the coil stores AND releases 1/2 L i^2 every time. If an arc is established, it is all released into that arc and if the gap is so large as to prohibit an arc (dielectric strengh so large-not resistance), then that energy which is still stored in the coil, is still all released. Released either through a breakdown of insulation within the coil, or if that insulation is so good as to prevent that arc, then released by causing an oscillating exchange between the inductance of the primary and the capacitor, until all the energy is dissipated in those elements plus the small amount that is always lost in the arc at the points as they open.

Now if you would stop spreading "reverse polarity of the coil causes loss of half the spark energy" and speak of the primary issue with reverse polarity, that being that it is harder to establish the arc due to electron emission from hot/cold metal, then we could stop this exchange and get on with better things AND I won't even call you wrong, not correct, or incorrect.

If you have acceptable compression then theirs no need for a leak down test in most cases, never hurts to do a leak down test tho, it will tell ya more than a compression test.

On a compression test the first few hits on the gauge tell ya the ability of the rings to seal, the next 2 to 3 tell ya the ability of the valves to seal. Lets just say most of the time. When you run up on stuck/carbon-ed up rings then it may not B possible to bring up the compression so that's why you will hear to add oil to the cylinder and repeat. If oil brings the compression up then its a tell tell sign their is a ring problem. Added oil will have very little effect on valve seal. In this case you would STOP and go about possible ways to unstick the rings, if that does not help its time to open it up and go deep. A leak down test would confirm if the lost leakage was the rings.

It would take 2 much time for me to write it up step by step. First do a compression test if it shows a problem are low numbers move on to a leakage test, if still a problem go to a run'n compression test. Most folks don't go threw the steps to nail were the problem lies B-4 they tear into a engine. I read professional sites and even their compression/leak down/ vacuum test have become a lost art. Dunno but B-4 I go into a engine I like to know which cylinder is in question even if its a head gasket I like to know what cylinder to focus on.

Their is some tips on my web site if you can cipher threw it. Like Jethro Bodean I wanted to B a Brain surgeon / astronaut/ movie producer / 00 spy but on a allowance of 50 cents a week found out rite fast I would have to B a goober.

2. The only compression tests I'm familiar with on an N are wet & dry tests. Both call for a hot engine, choke & carb open, all 4 plugs removed. So, leaving the key off would address your coil question.

3. The threads in the head are 14mm straight threads. My compression gauge came w/ two adapters & one of them fit the N head.

4. The elbow is not required but sure is nice. It is not a standard plumbing item, so take your gauge w/ you to the hardware store.

• 50 Tips

Link

The real old stuff has standard tapered pipe threads.