Does the VP44 completely shut off fueling when truck is coasting in gear on a downgrade (manual transmission)?

[Tractorman]

I can't seem to find a source that has a conclusive answer to the question, "Does the VP44 completely shut off fueling when truck is coasting in gear on a downgrade?"  It would seem to me that the VP44's high pressure solenoid valve would stop receiving signals under this condition if the APPS signal was at 0% (foot off throttle).  Does anyone know if this is true with certainty?

 

An example would be:  truck with a manual transmission coasting along on a down grade at 1700 rpm with foot off of throttle.  

 

- John

 

[Mopar1973Man]

Yes. When engine load is ZERO then the VP44 stops injecting and just bypasses to the return line. This applies to all engine and transmission combos.

[Silverwolf2691]

mine is zero with the can bus fuel going to like 3-13 every now and then.. but more or less I would call it shut off.

[Mopar1973Man]

Foot off throttle, my truck shows zero constant. EGTs drop to right about coolant temp on long down grades. Winchester Grade is quite long and typically see EGTs right at 180 to 200°F. Quadzilla reports zero engine load, back when I had the OBDLink LX it showed zero fuel injected (flow in GPH).

[Silverwolf2691]

   huh..   

 

wonder why the difference? engine load reads 0 all the through the hill unless I push the clutch in.. but my canbus does the bouncy bounce alternating between 0 and 3-13 (when coasting in gear)..

[Tractorman]

Actually, before I posed my question I reviewed some of @Me78569's data logs.  I looked at several of them and they all showed very similar recordings.  The one posted below is titled "Stock".  Note when the throttle position goes to zero (engine rpm 1977 and vehicle speed 51 mph), the CanBus fuel goes to zero followed by an occasional bump of fueling even though the throttle position remains at zero and engine rpm remains well above idle.

 

This kind of supports what @Silverwolf2691is saying.

 

- John

 

 

Edited April 27, 2022 by Tractorman

[Me78569]

load is based on canbus it's simply a 0-100% of 0-4096 fueling command.   so 3 or 13 canbus command = %0 load still %1 would be > 40 

[Tractorman]

Thanks for the quick response.  Just for clarification, are you saying that there is no fueling happening when canbus command is 3 or 13 with throttle position at 0% and engine rpm well above idle?  Just making sure I understand correctly.

 

- John 

Edited April 27, 2022 by Tractorman

[Me78569]

canbus is the fueling command, period,   If you see a reported canbus value of 3 then that means the vp44 is asking for ~.1 % of possible fueling duration.   

 

so yes it is fueling when canbus is 3, but fueling so little it is likely not enough to effect anything.

[Max Tune]

Wouldn't the lowest fueling be whatever idle is. Foot off pedal and coasting downhill should be idle volume no matter rpms, no? 

 

Or do the terms fueling and flow or volume of fuel injected mean different things?

Edited April 27, 2022 by Max Tune

[Silverwolf2691]

Lowest fueling would be zero..

 

The fueling command/canbus fuel is the 0-4095 number you see tossed around..

 

Engine load is that number converted to a 0-100% scale.

 

Canbus fuel is a signal only I think.. vp44 decodes that and converts that into a milliseconds of energized for the fuel solenoid.

 

Amount of fuel is the byproduct of this..

 

I think the system does this as a stall check or something. Possibly part of the idle validation system.. Rpms are too high for idle so it cuts fuel to bring them down, and its just checking at random intervals..

 

If you had idle levels of fuel injected at a coast the volume of fuel that could build up could potentially hydrolock the engine.. Maybe  ... Need to noodle this idea some more..

[Max Tune]

Should not cause a hydro lock because it should still fire on compression stroke just like when you start the engine. Even with throttle closed there has to be air in the cylinder. The effects of ignition with engine spinning 1700 and throttle closed wouldn't push on piston with much force. 

 

If you're coasting downhill, 1700 rpm, off throttle, you have engine braking holding you back from freewheeling coast. I wonder if it would make a noticeable difference if you were to shut off the key, completely cutting all fuel?

[Great work!]

I can confirm the fuel cuts to zero when coasting and even when quickly revving in neutral and instantly letting off. The 0 to 4095 is not milliseconds it is degrees scaled, basically every 512 of 4096 is roughly 6 degrees (crank) of injection. At 1000 rpm 1 millisecond is 6 degrees. At 3000 rpm 1 millisecond is 18 degrees.

However fuel delivery in ml or cc is not linear 0 thru 4095 because of the cam shape. the first few degrees and the last few degrees are moving the plungers very little. The bulk of injection is in the middle of the the cam lift. Just like valve lift specs there is advertised duration and at .050 the valve is flowing very little under .050 lift.

That's why you can see the injection as high as 12 to 15 degrees just to idle the engine . the first 5 degrees or so is doing almost nothing. However  doubling that 15 to 30 degrees is injecting a lot more than double the fuel in ml or cc.

[Tractorman]

3 hours ago, Silverwolf2691 said:

If you had idle levels of fuel injected at a coast the volume of fuel that could build up could potentially hydrolock the engine..

 

No hydro-lock would occur.   The small amounts of fuel would be injected at regular commanded timing.  

 

3 hours ago, Max Tune said:

Even with throttle closed there has to be air in the cylinder. The effects of ignition with engine spinning 1700 and throttle closed wouldn't push on piston with much force. 

 

Since you mentioned the phrase "throttle closed", that could be inferring there is a throttle plate - which there is not on a 5.9 liter Cummins diesel engine.  Just mentioning this for clarity.  Diesel engines without a throttle plate will not hold back the truck on downgrades with foot removed from throttle.  The energy generated on the compression stroke without fuel is immediately released on the following power stroke , so any engine braking is cancelled.

 

 

@Great work!, thank you for that explanation!  That is the answer I was looking for.

 

- John

 

-

[wil440]

7 hours ago, Tractorman said:

Diesel engines without a throttle plate will not hold back the truck on downgrades with foot removed from throttle.

Diesels without a throttle plate will hold back on a downgrade because the cylinder that is on compression does not have the previous firing cylinder powering the compression cylinder, this is linear across the firing order so constant, the vehicle weight and downgrade pushes all cylinders over compression

If a diesel with manual box is put in gear on a small slope it will not roll away this force is increased many times by the transmission ratios so the lower the gear the more retarding force there is

Diesels are better than petrol because of the higher compression ratios

[Tractorman]

7 hours ago, wil440 said:

Diesels without a throttle plate will hold back on a downgrade because the cylinder that is on compression does not have the previous firing cylinder powering the compression cylinder, this is linear across the firing order so constant, the vehicle weight and downgrade pushes all cylinders over compression

 

All four strokes of the four stroke cycle engine must be considered for engine braking effectiveness - not just the compression stroke

 

Since all engines have internal friction of varying degrees, this friction is not considered in the examples below for holding a vehicle back on a downgrade.  For simplicity, consider the following operation on a single cylinder engine:

 

Diesel engine without throttle plate

 

  intake stroke - intake valve open, no holdback

  compression stroke - all valves closed, air is compressed, holdback

  power stroke - all valves closed, compressed air expands, cancels compression stroke holdback

  exhaust stroke - exhaust valve open, no holdback

 

Net results - no engine holdback

 

Gasoline engine with throttle plate

  

  intake stroke - intake valve open, throttle plate closed, negative pressure produced, holdback

  compression stroke - all valves closed, throttle plate closed, air is compressed, holdback

  power stroke - all valves closed, throttle plate closed, compressed air expands, cancels compression stroke holdback

  exhaust stroke - exhaust valve open, no holdback

 

Net results - engine holdback on intake stroke.  If a closed throttle plate introduced a negative pressure of 25 inches of mercury (approximately 12 psi), then 12 psi would be working on each piston surface area to aid in holdback.  Available atmospheric pressure is the limiting factor for engine braking.

  

Diesel engine without throttle plate, with exhaust brake actuated

 

  intake stroke - intake valve open, exhaust brake closed, no holdback

  compression stroke - all valves closed, exhaust brake closed, air is compressed, holdback

  power stroke - all valves closed, exhaust brake closed, compressed air expands, cancels compression stroke holdback

  exhaust stroke - exhaust valve open, exhaust brake closed, holdback

 

Net results - engine holdback on exhaust stroke.

 

Diesel engine without throttle plate, with engine brake actuated (a real Jake brake)

 

  intake stroke - intake valve open, engine brake turned on, no holdback

  compression stroke - all valves closed, engine brake turned on, air is compressed, holdback (major)

  power stroke - all valves closed, engine brake pops exhaust valves open briefly at TDC, compressed air suddenly released, negative air pressure during stroke,  holdback (minor)

  exhaust stroke - exhaust valve open, engine brake turned on, no holdback

 

Net results - major engine holdback on compression stroke, minor engine holdback on power stroke.

 

Please let me know if there are any errors or if something doesn't make sense in the above examples.

 

- John

Edited April 28, 2022 by Tractorman

[wil440]

sorry I disagree

In a 6 cylinder engine there is always one cylinder on compression or very close to it so compression one after the other with no power stroke from the previous firing cylinder so the endless cycle of having to compress air gives engine braking

 

 

Look at any firing order thats the compression events, doesn't matter what the cylinders that are not on compression are doing, doesn't matter what the other valves are doing 

 

With the foot off the pedal it is just a glorified compressor sucking air in the air cleaner and pumping it out the exhaust and we all know compressors take energy to work, the energy here is the wheels driving the motor through the transmission, they don't put " engage low gear " at the top of hills for nothing, the lower the gear the more energy it takes to drive the compressor as once a low gear is driven from the wheels it actually becomes the equivelant of a high gear at the flywheel trying to drive the motor faster

3 hours ago, Tractorman said:

Diesel engine without throttle plate

 

  intake stroke - intake valve open, no holdback

  compression stroke - all valves closed, air is compressed, holdback

  power stroke - all valves closed, compressed air expands, cancels compression stroke holdback

  exhaust stroke - exhaust valve open, no holdback

 

Net results - no engine holdback

compression stroke - all valves closed, air is compressed, holdback         Don't forget.... this just moves onto the next cylinder in the firing order, so always a cylinder compressing

 

[Tractorman]

1 hour ago, wil440 said:

compression stroke - all valves closed, air is compressed, holdback         Don't forget.... this just moves onto the next cylinder in the firing order, so always a cylinder compressing

 

Not arguing this point. but it seems that your focus is on the compression stroke only with no consideration of what happens on the next stroke (the power stroke) which is key.  At the beginning of the power stroke the cylinder is filled with stored energy in the from of highly compressed air just generated by the previous compression stroke.  That stored energy will now drive the piston downward with the same force that was used to compress the air in the first place.

 

  Compression stroke - stores energy in the form of highly compressed air by resisting and slowing the upward speed of the piston - results in momentary engine braking .

 

  Power stroke - releases energy of same parcel of highly compressed air driving the piston downward and accelerating the downward speed of the piston - results in no engine braking and piston acceleration is momentary. 

 

The actions of the compression stroke and power stroke cancel each other which nets zero engine braking.

 

The number of cylinders in an engine is not relevant as each cylinder operates independently in the four stroke cycle process.  This is why I simplified my explanation using a single cylinder engine.  The only thing individual cylinders have in common are air intake, exhaust outlet, and crankshaft connection.  

 

2 hours ago, wil440 said:

doesn't matter what the cylinders that are not on compression are doing, doesn't matter what the other valves are doing 

 

If this is true, then can you offer an explanation of what is happening with energy from the highly compressed air during the power stroke?

 

Just want you to know that I have always respected your point of view and continue to do so - you have clearly shown that you are a knowledgeable person through many posts on this forum.  In the end we may have to agree to disagree, which is okay with me

 

- John

 

 

 

 

 

[wil440]

10 hours ago, Tractorman said:

If this is true, then can you offer an explanation of what is happening with energy from the highly compressed air during the power stroke?

I really cannot see the downward force on the top of the piston on the power stroke adding the same amount of energy that was used to compress the air on the compression stroke as for one a portion of that energy will be converted to heat, more energy will be used to compress air than will be released on the power stroke and then there is the next firing cylinder to compress.

Do a test on your truck as yours is manual, 1st gear idling on a slope versus neutral idling on a slope, i'll bet 1st gear is slower, this is engine braking, get 20 people to push your truck while in 1st gear idling  clutch in down a slight slope then let the clutch out and see what happens, just tell them to keep their faces away from the tailgate

 

I used to compete in off road trials in a diesel landrover and have gone down some really serious banks, always low 1st gear engine braking holding the vehicle to a nice descent speed and have also seen some people decend at a rapid rate after the crap landrover gearbox has jumped out of low 1st.

 

 Take a Cat777 dump truck slightly older one without the auto retarder with the Cat C32 V12 32ltr diesel, put that in 1st gear and that will go down some grade empty at walking pace usually with the rear wheels hopping and skidding, loaded at 100 ton payload yes you will need to use the retarder also due to the overall weight

 

We'll have to agree to disagree as while I also respect your point of view diesels have the best engine braking of all engines

 

Edited April 29, 2022 by wil440

[Tractorman]

On 4/28/2022 at 3:44 PM, wil440 said:

I really cannot see the downward force on the top of the piston on the power stroke adding the same amount of energy that was used to compress the air on the compression stroke as for one a portion of that energy will be converted to heat, more energy will be used to compress air than will be released on the power stroke

 

I agree with your statement, but I think the amount of energy lost would be insignificant at higher engine rpm's. 

 

If a vehicle is decelerating with an engine speed of 1800 rpm, that would mean a camshaft speed of 900 rpm.  Dividing 900 rpm by 60 seconds equals 15 revolutions per second.  Dividing one second by 15 revolutions equals .066 seconds to complete one revolution.  This means that four strokes of the four stroke cycle are completed in .066 seconds.  This would also mean the time to complete the compression and power stroke (two strokes) would be .033 seconds, or the time to complete the power stoke only, would be about .017 seconds.  I just can't visualize that much energy being lost in 17 micro seconds.  

 

Feel free to double check my math.

 

On 4/28/2022 at 3:44 PM, wil440 said:

Do a test on your truck as yours is manual, 1st gear idling on a slope versus neutral idling on a slope, i'll bet 1st gear is slower,

 

Not necessary to perform the test because you and I would agree with the results - yes, first gear would be slower.  But, I am saying it would be slower for a different reason.  That reason would be resistance from friction (overcoming engine internal resistance by forcing the engine to rotate) which applies to all engines.  I did perform a test that I think demonstrates this.  I have a level tool that allows me to measure percent of a grade.  I found a down grade that averaged just over 3 percent and performed the following tests.

 

  Test #1.  With truck stopped, engine off, transmission in 3rd gear, park brake released - truck did not move (observed this condition for over a minute).  This indicates to me that engine friction in combination with gear reduction is holding the truck from moving.  Any compressed air in a cylinder on compression stroke would have long been bled off.

 

  Test #2.  Started engine and brought truck up to 20 mph in 3rd gear, then shut off engine.  Truck speed very slowly began to increase.  This indicates to me that engine compression is not doing much to hold the truck back.

 

This testing may not be very scientific, but it supports how I think it should work.

 

I think I hijacked my own thread....,

 

- John

 

 

P.S.  I posted this late so you would have something to look forward to in the morning.

 

Edited April 30, 2022 by Tractorman