question for any 60' v6 engine

[evanesce69]

but mainly for our 3.0L

how many crank revs before all 6 cylinders fire?

am I right to assume every 60' degree we get a cylinder event, or would that be every 300 degree. a cylinder event being spark and ignition


I know there are at least 2 revs for a full ICPE cycle

What i dont know is the stagger of the cylinders and the total revs before all 6 cylinders have an event.




Gosh i feel so dumb for asking this question...........

 

[shadetree]

I assume you mean a 60 degree V6. Most modern V6's are even fire engines, and fire every 120 degrees of crank rotation. Many of the earlier V6's were of the odd fire design, and the as the name implies, the cylinders fired at differing degrees of rotation.

I have an odd fire Buick V6, never run since an overhaul about 4 years ago. I also have the transmission to go with it, also overhauled and never ran. I would sure like to make someone a gift of this POS. shady

 

[evanesce69]

Shady - here is what I thought after my post. I figure that 2 complete crank revolutions will allow all 6 cylinders to fire. 6 cylinders x 120 degree firing times = 720 degrees rotation, or 2 revolutions.

 

[Bob Ayers]

With DIS, and EDIS, ALL 6 plugs fire every revolution of the crank. They fire at near the top of the compression stroke, and again at near the top of the exhaust stroke. 2 plugs fire together, in series with each other.

 

[almostclueless]

With any four stroke engine it wil take two crank revolutions for all cylinders to fire.

 

[shadetree]

With DIS, and EDIS, ALL 6 plugs fire every revolution of the crank. They fire at near the top of the compression stroke, and again at near the top of the exhaust stroke. 2 plugs fire together, in series with each other.

He is talking about the power stroke. The waste spark at exhaust is of little consequence as it is very low voltage.

Shady - here is what I thought after my post. I figure that 2 complete crank revolutions will allow all 6 cylinders to fire. 6 cylinders x 120 degree firing times = 720 degrees rotation, or 2 revolutions.

That would be correct.

 

[Bob Ayers]

He is talking about the power stroke. The waste spark at exhaust is of little consequence as it is very low voltage.

WRONG!!!!! it is still many, many KILOVOLTS!!

 

[thegoat4]

The waste spark at exhaust is of little consequence as it is very low voltage.

What makes you think that the waste spark is low voltage?

 

[shadetree]

WRONG!!!!! it is still many, many KILOVOLTS!!

Ayers, you have about as much tact when dealing with people as Gross.
The following is from "Ford Fuel Injection & Electronic Engine Control," by Dr. Charles O. Probst, SAE;, page 126:

When A is beginning its power stroke with high compresson pressure, it will require high voltage to jump the spark gap, say 10kv. Power is proportional to voltage, so A gets the power spark.

B gets the low voltage waste spark, say less than 2-3 kv, because it is firing into the exhaust gasses at low cylinder pressure. The waste spark has no effect on emissions or power.

A = spark plug on power stroke
B = spark plug on exhaust stroke

shady

 

[Bob Ayers]

Ayers, you have about as much tact when dealing with people as Gross.
The following is from "Ford Fuel Injection & Electronic Engine Control," by Dr. Charles O. Probst, SAE;, page 126:

When A is beginning its power stroke with high compresson pressure, it will require high voltage to jump the spark gap, say 10kv. Power is proportional to voltage, so A gets the power spark.

B gets the low voltage waste spark, say less than 2-3 kv, because it is firing into the exhaust gasses at low cylinder pressure. The waste spark has no effect on emissions or power.

A = spark plug on power stroke
B = spark plug on exhaust stroke

shady

Shady, you have to do better than this!!! You lost a technical debate with me and MAKG with this guy's BS before!!! You need to learn from somebody else!!!

POWER IS NOT PROPOTIONAL TO VOLTAGE WITH COMBUSTION, ONLY WITH AC AND DC CIRCUITS, P=I*V !!!

The waste spark has no effect on emissions

Not true, if the plug misfired on the power stroke, the waste spark will ignite the unburned fuel/air mixture, reducing unburned hydrocarbon emissions!!

 

[shadetree]

Shady, you have to do better than this!!! You lost a technical debate with me and MAK with this guy's BS before!!! You need to learn from somebody else!!!

POWER IS NOT PROPOTIONAL TO VOLTAGE WITH COMBUSTION, ONLY WITH AC AND DC CIRCUITS, P=I*V !!!

Not true, if the plug misfired on the power stroke, the waste spark will ignite the unburned fuel/air mixture, reducing unburned hydrocarbon emissions!!

I didn't lose any debate. I just quit arguing, because with you two, it is fruitless and boring.

When you and Gross can match the qualifications of Dr. Probst, I might pay a little attention to what you say. "This guy," as you put it, has written several books concerning automobiles. He consulted with Ford engineers and trainers on this particular book to train Ford technicians. It was given a Ford SVO part number M-1832-Z1. This is how I obtained my copy, thru a Ford dealership as it wasn't available anywhere else at the time.

Here are a couple more of his books:
Ford Fuel Injection and Engine Control: 1988-1993
Same title : 1980-1987
Bosch Fuel Injection and Engine Management

You are the same guy who told me one time that there was no such thing as a KAM circuit in Ford computers, so I pay very little attention to what you say technically. shady

 

[Bob Ayers]

I didn't lose any debate. I just quit arguing, because with you two, it is fruitless and boring.

When you and Gross can match the qualifications of Dr. Probst, I might pay a little attention to what you say. "This guy," as you put it, has written several books concerning automobiles. He consulted with Ford engineers and trainers on this particular book to train Ford technicians. It was given a Ford SVO part number M-1832-Z1. This is how I obtained my copy, thru a Ford dealership as it wasn't available anywhere else at the time.

Here are a couple more of his books:
Ford Fuel Injection and Engine Control: 1988-1993
Same title : 1980-1987
Bosch Fuel Injection and Engine Management

You are the same guy who told me one time that there was no such thing as a KAM circuit in Ford computers, so I pay very little attention to what you say technically. shady

I'm sure you read his book on how an alternator and voltage regulator works also!!!

 

[almostclueless]

Voltage is not the entire consideration.

The "waste" cylinder's gap MIGHT have a different voltage across it, as the contents of it are very different than the cylinder about to make power, and cylinder pressure will be very different. To jump a plug gap will still require kilovolts.

Either way, the current will be the same thru both gaps; it's a series circuit.

 

[shadetree]

I'm sure you read his book on how an alternator and voltage regulator works also!!!

Is this what you are talking about????

Q. What happens when a voltage regulator fails with a short circuit?

A. Assuming that this is not a catastrophic internal failure which would cause the alternator field circuit breaker to trip) the field windings would be able to draw the maximum possible current and would effectively turn the alternator full on. In this condition the alternator is producing its maximum rated power. You might think that this is not a problem, but due to rules governing inductors and capacitors an even bigger problem will develop: runaway output voltage. If there is not sufficient electrical load to use all of the available power from the alternator the voltage in the alternator stator windings will rise uncontrollably.

Both of these failures are a problem. The second failure, however, could do potential damage to the radios and other appliances in the electrical system.
shady

 

[Bob Ayers]

Is this what you are talking about????

Q. What happens when a voltage regulator fails with a short circuit?

A. Assuming that this is not a catastrophic internal failure which would cause the alternator field circuit breaker to trip) the field windings would be able to draw the maximum possible current and would effectively turn the alternator full on. In this condition the alternator is producing its maximum rated power. You might think that this is not a problem, but due to rules governing inductors and capacitors an even bigger problem will develop: runaway output voltage. If there is not sufficient electrical load to use all of the available power from the alternator the voltage in the alternator stator windings will rise uncontrollably.

Both of these failures are a problem. The second failure, however, could do potential damage to the radios and other appliances in the electrical system.
shady

No, this is what I'm talking about....

The problem in removing the battery cable with the truck running is if the regulator in the alternator is no longer regulating, when you disconnect from the battery, it can allow the alternator to go to full amperage output, and burn up a lot of stuff. If the alternator is dead, as it apparently is in your case, the truck simply dies same as turning off the ignition key when you remove the battery cable.

Next time, invest in a $5 Radio Shack volt meter, and check voltages.shady

My response!

It's the di/dt that creates the huge voltage spike when you remove a battery cable. This can happen just with the current load on the battery.
The load isn't sufficient, for the alternator to go "full amperage", but large voltage spikes will be created. The voltage output of the alternator
is controlled with the "voltage" regulator, not the current.