The Carryfast engine design discussion

UK PROFESSIONAL DRIVERS FORUMS OLD TIME LORRIES, COMPANIES AND DRIVERS (INTERACT
1101

[zb]
anorak:

Carryfast:
Tell us did your bs ‘derivation’ match specific peak torque measured at the fly wheel x 2.464 yes or no.I actually saw no engines ( like TL12 or Eagle ) being referred to in that bs…

It’s all up there^^^. Not yes, no. 4.π is the constant. Don’t know where you get the 4.22644 from. Don’t tell me you worked it out yourself (I would actually have some respect for you , if you had). Why would I have to mention an actual engine? The equation applies to all engines.

PS are you now agreeing that BMEP is pressure, or is that understanding still the preserve of the older kids?

Which part of BMEP is purely theoretical it has nothing to do with actual cylinder pressures.By looking at the equations you can easily see that BMEP is simply torque per cubic inch of dispalcement ( specific torque ) didn’t you understand.

There are other constants which can be used I’ve posted them within the above statements.None of them include bleedin Pi.

So tell us why can’t an F1 engine produce its peak torque thereby its peak BMEP from 1,200 rpm given sufficient boost.You’re obviously saying that the required cylinder pressures would be the same.

As I said BMEP doesn’t even actually exist in the real world so I’m obviously not agreeing with your total bs bordering on trolling.

1102

[zb]
anorak:

Carryfast:
Tell us did your bs ‘derivation’ match specific peak torque measured at the fly wheel x 2.464 yes or no.I actually saw no engines ( like TL12 or Eagle ) being referred to in that bs…

It’s all up there^^^. Not yes, no. 4.π is the constant.

So you’re saying that BMEP doesn’t = specific torque x 2.464.
Pi is a constant applicable to exactly what figure.
So tell us what is your figure for the Eagle’s BMEP.( Bearing in mind it obviously isn’t a literal real cylinder pressure figure at all regardless ).
As trying to boost an F1 engine to 200 psi + BMEP at 1,200 rpm would prove.

1103

Carryfast:

[zb]
anorak:

Carryfast:
Tell us did your bs ‘derivation’ match specific peak torque measured at the fly wheel x 2.464 yes or no.I actually saw no engines ( like TL12 or Eagle ) being referred to in that bs…

It’s all up there^^^. Not yes, no. 4.π is the constant. Don’t know where you get the 4.22644 from. Don’t tell me you worked it out yourself (I would actually have some respect for you , if you had). Why would I have to mention an actual engine? The equation applies to all engines.

PS are you now agreeing that BMEP is pressure, or is that understanding still the preserve of the older kids?

Which part of BMEP is purely theoretical it has nothing to do with actual cylinder pressures.By looking at the equations you can easily see that BMEP is simply torque per cubic inch of dispalcement ( specific torque ) didn’t you understand.

There are other constants which can be used I’ve posted them within the above statements.None of them include bleedin Pi.

So tell us why can’t an F1 engine produce its peak torque thereby its peak BMEP from 1,200 rpm given sufficient boost.You’re obviously saying that the required cylinder pressures would be the same.

As I said BMEP doesn’t even actually exist in the real world so I’m obviously not agreeing with your total bs bordering on trolling.

You have not understood the derivation I posted. It proves that BMEP is pressure- gas in a hole which is too small for it.

1104

[zb]
anorak:

Carryfast:

[zb]
anorak:

Carryfast:
Tell us did your bs ‘derivation’ match specific peak torque measured at the fly wheel x 2.464 yes or no.I actually saw no engines ( like TL12 or Eagle ) being referred to in that bs…

It’s all up there^^^. Not yes, no. 4.π is the constant. Don’t know where you get the 4.22644 from. Don’t tell me you worked it out yourself (I would actually have some respect for you , if you had). Why would I have to mention an actual engine? The equation applies to all engines.

PS are you now agreeing that BMEP is pressure, or is that understanding still the preserve of the older kids?

Which part of BMEP is purely theoretical it has nothing to do with actual cylinder pressures.By looking at the equations you can easily see that BMEP is simply torque per cubic inch of dispalcement ( specific torque ) didn’t you understand.

There are other constants which can be used I’ve posted them within the above statements.None of them include bleedin Pi.

So tell us why can’t an F1 engine produce its peak torque thereby its peak BMEP from 1,200 rpm given sufficient boost.You’re obviously saying that the required cylinder pressures would be the same.

As I said BMEP doesn’t even actually exist in the real world so I’m obviously not agreeing with your total bs bordering on trolling.

You have not understood the derivation I posted. It proves that BMEP is pressure- gas in a hole which is too small for it.

BMEP isn’t real.It’s a hypothetical comparator.
How about you answering the questions.
You know like does 100 lbft per litre x 2.464 match your silly calaculation.
Why can’t an F1 engine produce its peak torque at 1,200 rpm given sufficient boost.
Are you saying that the cylinder pressures would be the same even though though the BMEP figure wouldn’t have changed.
Which proves no link whatsoever between real cylinder pressure and the abstract hypothetical but not real BMEP figure.
The specific torque figure wouldn’t have changed but your engine would be toast.
Oh wait you obviously don’t beleive that BMEP just means specific torque nothing more.

1105

Good evening, welcome to the Carryfast Engine Design Discussion, in which we investigate all aspects of mechanical engineering, on a fundamental level, a bit like religion, but worse. Over to our titular representative:

Carryfast:
BMEP isn’t real.

1106

[zb]
anorak:
Good evening, welcome to the Carryfast Engine Design Discussion, in which we investigate all aspects of mechanical engineering, on a fundamental level, a bit like religion, but worse. Over to our titular representative:

Carryfast:
BMEP isn’t real.

I believe that CF has mangled up these letters and what he really means is “BNP” which he his obviously a founder member of the Leatherhead BNP faction ! With his warped views on politics I reckon this could be the reason for the mix up ! So stand by your beds Lads as when he realises his glaring error he will pile back in to explain why Jeremey Corbyn is the PM we never had !! :open_mouth: :unamused: :wink: :laughing: :laughing: :laughing: :laughing: :laughing: :laughing:

1107

[zb]
anorak:
Good evening, welcome to the Carryfast Engine Design Discussion, in which we investigate all aspects of mechanical engineering, on a fundamental level, a bit like religion, but worse. Over to our titular representative:

Carryfast:
BMEP isn’t real.

150.8 x 1,216 lbft = 183,372.8 / 742.65 ci = 246 psi or 16.9 Bar BMEP

1216 lbft /12.17 litres = 99.9 lbft per litre x 2.464 = 246 psi or 16.9 Bar BMEP.

Which part of those equations don’t just mean expressing specific torque.
As an abstract hypothetical pressure measurement which doesn’t actually exist anywhere at any time within the engine nor does 2 x that figure mean anything at all.

I’m also sure those figures beat an F1 engine.

Feel free to boost the F1 motor to match that specific torque/BMEP figure oh and we want it at less than 1,500 rpm not 15,000 rpm.

1108

Carryfast:
There are other constants which can be used I’ve posted them within the above statements.None of them include bleedin Pi…

You must be aware that, if you walk round in circles, the distance you cover is π x the diameter of the circle, multiplied by the number of times you go round?

1109

[zb]
anorak:

Carryfast:
There are other constants which can be used I’ve posted them within the above statements.None of them include bleedin Pi…

You must be aware that, if you walk round in circles, the distance you cover is π x the diameter of the circle, multiplied by the number of times you go round?

The force on the piston and con rod x leverage at the crankshaft makes the torque not the crankshaft.
The piston and rod are reciprocating they ain’t going round in circles.
In this case we’re talking specifically about the force x distance that the power stroke puts into the crankshaft.
That’s measured at the flwheel.What has the flywheel or crankshaft got to do with the resulting torque figure measured in lb force on the piston/rod x ft stroke divided by engine capacity.

Did you actually read the two different equations which reached the same result of BMEP no need for Pi in those equations.

1110

Edit
Watch this space as Anorak uses Pi to show how a turbo F1 engine can make exactly the same BMEP at 1,200 rpm as it can at 15,000 rpm obviously with the same cylinder pressures in both cases.BMEP means cylinder pressure right.

1111

Carryfast:

[zb]
anorak:

Carryfast:
There are other constants which can be used I’ve posted them within the above statements.None of them include bleedin Pi…

You must be aware that, if you walk round in circles, the distance you cover is π x the diameter of the circle, multiplied by the number of times you go round?

The force on the piston and con rod x leverage at the crankshaft makes the torque not the crankshaft.
The piston and rod are reciprocating they ain’t going round in circles.
In this case we’re talking specifically about the force x distance that the power stroke puts into the crankshaft.
That’s measured at the flwheel.What has the flywheel or crankshaft got to do with the resulting torque figure measured in lb force on the piston/rod x ft stroke divided by engine capacity.

Did you actually read the two different equations which reached the same result of BMEP no need for Pi in those equations.

“Torque T is applied to the crankshaft of a running engine, by a brake. It is reacted by a force at the big end journal, transmitted through the conn rod.”

Read that first line again, it contains the clue. The answer begins with N and there is a 3 in it.

1112

Carryfast:

[zb]
anorak:

Carryfast:
There are other constants which can be used I’ve posted them within the above statements.None of them include bleedin Pi…

You must be aware that, if you walk round in circles, the distance you cover is π x the diameter of the circle, multiplied by the number of times you go round?

The force on the piston and con rod x leverage at the crankshaft makes the torque not the crankshaft.
The piston and rod are reciprocating they ain’t going round in circles.
In this case we’re talking specifically about the force x distance that the power stroke puts into the crankshaft.
That’s measured at the flwheel.What has the flywheel or crankshaft got to do with the resulting torque figure measured in lb force on the piston/rod x ft stroke divided by engine capacity.

Did you actually read the two different equations which reached the same result of BMEP no need for Pi in those equations.

If you start mixing psi and lbft or even worse, psi and bar, as you appear to have been doing, you will end up with odd little numbers like 4.236, or whatever it was. Whatever you do, π is in there somewhere. Those “equations” you trot out, from some dusty old man’s Victorian training manual, are throw-away rubbish. What I am trying to show you is that you can work the equations out, yourself, from simple fundamentals, without having to remember anything.

Do you believe that the circumference of a circle is discovered by multiplying the diameter by π- not 2.464?

Good. Now, consider the following two other facts:

  1. The pistons in the engine go up and down.
  2. The wheels on the bus go round and round.

Is it not reasonable to assume that π will be involved, as soon as you start to compare the force above the piston to the torque on the crank?

1113

cav551:

Carryfast:
The force on the piston and con rod x leverage at the crankshaft makes the torque not the crankshaft.
The piston and rod are reciprocating they ain’t going round in circles.
In this case we’re talking specifically about the force x distance that the power stroke puts into the crankshaft.
That’s measured at the flwheel.What has the flywheel or crankshaft got to do with the resulting torque figure measured in lb force on the piston/rod x ft stroke divided by engine capacity.

Did you actually read the two different equations which reached the same result of BMEP no need for Pi in those equations.

“Torque T is applied to the crankshaft of a running engine, by a brake. It is reacted by a force at the big end journal, transmitted through the conn rod.”

Read that first line again, it contains the clue. The answer begins with N and there is a 3 in it.

The ‘brake’ isn’t going round in circles either just like the con rod isn’t.
The figure is only a measure of force x leverage by the con rod on the crankshaft nothing to do with the brake or the flywheel.
Just like transmissions are only rated by input torque.
The clue is contained in the two agreeing equations which I posted.
No Pi in that.
Even if there was you still end up with nothing more than a re stated version of specific torque.

Which just leaves the variable of the leverage side of the force x distance equation at the con rod which also proves that BMEP figure has nothing to do with cylinder pressures.
Because more leverage = less force required for the equivalent torque and the amount of leverage isn’t a part of the specific torque equation.
Good luck with boosting that F1 motor to 200 psi BMEP at 1,200 rpm.

1114

[zb]
anorak:
Whatever you do, π is in there somewhere. Those “equations” you trot out, from some dusty old man’s Victorian training manual, are throw-away rubbish. What I am trying to show you is that you can work the equations out, yourself, from simple fundamentals, without having to remember anything.

Do you believe that the circumference of a circle is discovered by multiplying the diameter by π- not 2.464?

Good. Now, consider the following two other facts:

  1. The pistons in the engine go up and down.
  2. The wheels on the bus go round and round.

Is it not reasonable to assume that π will be involved, as soon as you start to compare the force above the piston to the torque on the crank?

I didn’t mix it up at all you was the one who was moaning about imperial measurements then you start moaning again about me simply converting the offending figures to whatever you like.
Because it’s more inconvenient evidence which wrecks your silly arguments.

1 The pistons and con rods go up and down.
2 During half a revolution of the engine they draw in the charge proportionate to and according to Atmospheric pressure OR they let in a COMPRESSED FORCED INDUCTION charge.
3 During the next half a revolution they then compress that charge or already compressed charge.
4 During the next half a revolution the compressed charge is ignited and massively expands under combustion.
5 The resulting EXPANSION OF THE COMPRESSED, or COMPRESSED COMPRESSED, CHARGE pushes the piston down with a force proportionate to the piston area.That’s why it’s called the ‘power’ stroke’.
6 The same force is applied to the cylinder head.
7 The distance of the con rod acting on the crankshaft acts as a force multiplier/lever on that force proportionate to that distance.

The wheels on the AEC or any other bus produce no torque at all.
Only the combination of 1 - 7 makes the torque.Nothing else.

So tell us how will you put the equivalent 100 lbft per litre torque into the wheels of the 32-38t truck with the TL12’s 7% leverage deficit, 2.5 % piston area v leverage deficit and starting from a 43% deficit.
Then how much extra stress that will mean on your end bearings and head fastenings.
Here’s a clue cylinder pressure isn’t the same thing as BMEP and all the force applied to the con rod is multiplied by the difference between end bearing area v piston area at the rod ends.
The same applies to head fastenings area v piston area.

Did those ‘Victorian’ constants provide the correct concurring answers or not ?.
Obviously not if you think the wheels of the bus also produce the power not just the result of the expansion of the ignited inlet charge acting on the pistons/con rods during the power stroke MULTIPLIED BY THE LEVERAGE provided by the STROKE measurement.

I won’t hold my breath waiting for an answer as to why can’t an F1 engine deliver its same BMEP but at 1,200 rpm.
Are you saying that the boost and resulting cylinder pressures required to do it would be the same as they are at 15,000 rpm ?.

1115

Carryfast:
… BMEP figure has nothing to do with cylinder pressures…

You will eat those silly words.

1116

[zb]
anorak:

Carryfast:
… BMEP figure has nothing to do with cylinder pressures…

You will eat those silly words.

You mean like I made you eat 2 x BMEP as meaning anything at all.

So tell us how would you make 200 let alone 246 psi BMEP with an F1 engine at 1,200 rpm.
By your reckoning the peak cylinder pressures should obviously be the same in all cases for the same BMEP.
BMEP is supposedly connected with cylinder pressures, right.

Which part of this didn’t you understand.
epi-eng.com/piston_engine_te … dstick.htm

Do you agree with the two results for the Eagle at 1216 lbft.Yes or no will do.

1117

Carryfast:

[zb]
anorak:

Carryfast:
… BMEP figure has nothing to do with cylinder pressures…

You will eat those silly words.

Blah blah.

By your reckoning the peak cylinder pressures should obviously be the same in all cases for the same BMEP…

No. You should just read what I said, without arranging your own flowers around it.

If you want to learn anything at all, just take the principle that the work done is the force times the distance. Surely you cannot add a falsehood to that?

1118

[zb]
anorak:

Carryfast:
By your reckoning the peak cylinder pressures should obviously be the same in all cases for the same BMEP…

No. You should just read what I said, without arranging your own flowers around it.

If you want to learn anything at all, just take the principle that the work done is the force times the distance. Surely you cannot add a falsehood to that?

You said BMEP is linked to actual/peak cylinder pressures.

You’re also the one who’s denying force x distance.

You clearly said that 2 x BMEP is all it would take in peak cylinder pressure for the TL12 to match the force x distance advantage of the Eagle.How when BMEP has no connection with actual/real cylinder pressure but which you’re still in denial of.

Yep the principle of force x distance is exactly my case in favour of Eagle over TL12.
How are you going to compensate for your 7% distance force multiplier deficit.Also your 2.5 % piston area deficit v distance multiplier.

Do you still deny 43% more specific torque required x 7% less distance multiplication with a 2.5% greater piston area.

So tell us if BMEP means cylinder pressure how can cylinder presssure be any different for engines with the same BMEP you know like extracting the same 200 + psi BMEP from an F1 engine but at 1,200 rpm.

So what does force on the piston/con rod x distance/stroke have to do with the wheels or the engine brake bolted to the floor.
As opposed to the more distance you’ve got to multiply the force, the less force you’ll need for the equivalent torque output.

Good luck with getting 200 + ‘psi’ ‘BMEP’ at 1,200 rpm from the F1 engine or 246 ‘psi’ ‘BMEP’ from the TL12 without them breaking or at least wearing out their rod end bearings too quick.

Now awaits you telling us that actual peak cylinder pressure and force required on the piston/con rod isn’t multiplied massively when it’s applied to end bearing area and cylinder head fasteners area.

Oh and you didn’t answer the question did those two concurring results match the BMEP calculation for the Eagle ?. Yes or no will do.
If yes how isn’t that just a different way of expressing specific torque just as the article correctly states and I’ve been saying for the previous pages.

1119

No answer.

The laws of physics, as understood by generations of children, on their way to becoming productive members of society, fly so far overhead that we can only wonder what they are.

1120

Carryfast:
…

The ‘brake’ isn’t going round in circles either just like the con rod isn’t…

Just spotted this little gem. We’re back in the nursery now.

If the brake doesn’t go round in circles, how come the discs and drums are made in that nice roundy shape? Round brake discs? Pffttt. I want mine shaped like the Shield of Leatherhead, and I don’t want them going round in circles.

As for the conn rod, the argument central to the function of the engine is that the bottom of the conn rod goes round in- guess what? Yes, a circle. Now, a question for the new boy in the class: if the top of the conn rod travels a distance S, from TDC to BDC, how far does the bottom of it travel, as it meanders its way around the crankcase?