The Carryfast engine design discussion

UK PROFESSIONAL DRIVERS FORUMS OLD TIME LORRIES, COMPANIES AND DRIVERS (INTERACT
1121

[zb]
anorak:

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?

The brake as in dyno it’s generally bolted to the floor.The brake discs and wheels don’t produce any torque either.
Nor does the crankshaft other than the throw which provides the distance side of the equation.

The con rod doesn’t go ‘round and round’ like the wheels do.
It only applies force to the crankshaft during the power stroke x the leverage of the stroke.At all other points in each revolution of the engine it’s doing the square root of fa.
No answers to the questions yet I see who would have thought it.

1122

Carryfast:
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 4 - 7 makes the torque.Nothing else.

Edit.

1123

[zb]
anorak:

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?

Ah! You are making this overgrown non engineering school boy anorak think now!

1124

Carryfast:
…The brake discs and wheels don’t produce any torque either.

The brake caliper just hangs in space. No need for a bracket to attach it to the axle. No need for tyres either- why would you need friction, when there is no torque being transmitted through the wheels?

Carryfast:
Nor does the crankshaft other than the throw which provides the distance side of the equation.

Wtf?

Carryfast:
The con rod doesn’t go ‘round and round’ like the wheels do.

Yes it does- at the bottom. The top bit goes up and down. If you want to pretend that you can discuss engine design, you should take note of that.

Carryfast:
It only applies force to the crankshaft during the power stroke x the leverage of the stroke.At all other points in each revolution of the engine it’s doing the square root of fa.
No answers to the questions yet I see who would have thought it.

What questions?

1125

[zb]
anorak:

Carryfast:
…The brake discs and wheels don’t produce any torque either.

The brake caliper just hangs in space. No need for a bracket to attach it to the axle. No need for tyres either- why would you need friction, when there is no torque being transmitted through the wheels?

Carryfast:
Nor does the crankshaft other than the throw which provides the distance side of the equation.

Wtf?

Carryfast:
The con rod doesn’t go ‘round and round’ like the wheels do.

Yes it does- at the bottom. The top bit goes up and down. If you want to pretend that you can discuss engine design, you should take note of that.

Carryfast:
It only applies force to the crankshaft during the power stroke x the leverage of the stroke.At all other points in each revolution of the engine it’s doing the square root of fa.
No answers to the questions yet I see who would have thought it.

What questions?

The brake caliper doesn’t go round and round nor does the dyno.

It appplies a braking force to the flywheel, which thereby measures the force x distance applied by the piston on the con rod x the distance of the stroke during the power stroke.The piston and con rod only travels between TDC to BDC during that it doesn’t go round and round.

The CRANKSHAFT turns half a revolution.
The CON ROD doesn’t TURN anywhere that’s why it has a big end bearing in it to allow the CRANKSHAFT to TURN within it.NOT the bleedin CON ROD that’s going up and in this case only down.
WTF have the vehicle’s brakes got to do with the force applied to the crankshaft by the piston and the con rod and the multiplication of that force by the stroke.

The question you know like
Are the two results of the calculations I provided correct for the BMEP of the Eagle at 1,216 lbft compared to all your algebra bollox ?

If BMEP supposedly relates to cylinder pressures why not an F1 engine at 200 psi + BMEP at 1,200 rpm.
You’re obviously saying that it would be no different.

1126

essexpete:

[zb]
anorak:

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?

Ah! You are making this overgrown non engineering school boy anorak think now!

Until the realisation that it’s the CRANKSHAFT that’s going round in circles within the big end bearing NOT the bleedin CON ROD smacks you between the eyes.

The rod doesn’t ‘meander’ anywhere between TDC and BDC.The small end bearing allows the big end to move side ways with the throw of the crank shaft that’s all.It’s the crankshaft which is doing all the turning.

All I’m seeing here is nerds who’d like to think they know how an engine works. :unamused: :laughing: :laughing:

1127

Carryfast:

essexpete:

[zb]
anorak:

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?

Ah! You are making this overgrown non engineering school boy anorak think now!

Until the realisation that it’s the CRANKSHAFT that’s going round in circles within the big end bearing NOT the bleedin CON ROD smacks you between the eyes.

The rod doesn’t ‘meander’ anywhere between TDC and BDC.The small end bearing allows the big end to move side ways with the throw of the crank shaft that’s all.It’s the crankshaft which is doing all the turning.

All I’m seeing here is nerds who’d like to think they know how an engine works. :unamused: :laughing: :laughing:

Hahaha!!! If the crankshaft goes round and round, but the conn rod does not, surely someone has forgotten two bolts and a cap. Maybe they did not realise that there were tensile forces involved, so specified too-small bolts, and the whole thing came apart?

“The small end bearing allows the big end to move side ways…” Plenty of end float, just in case?

Carryfast:
“All I’m seeing here is nerds who’d like to think they know how an engine works.”

1128

[zb]
anorak:
Hahaha!!! If the crankshaft goes round and round, but the conn rod does not, surely someone has forgotten two bolts and a cap. Maybe they did not realise that there were tensile forces involved, so specified too-small bolts, and the whole thing came apart?

“The small end bearing allows the big end to move side ways…” Plenty of end float, just in case?

Carryfast:
“All I’m seeing here is nerds who’d like to think they know how an engine works.”

How does the con rod go round and round when it’s attached to the piston.
The two bolts and the cap form part of the big end bearing.
You know the bearing that allows the crankshaft to revolve because the con rod doesn’t revolve with the crankshaft at any point.It just swings from side to side like a pendulum on its small end bearing pivot relative to the movement of the piston in the bore that it’s attached to.
Obviously only to one side during the power stroke but, unlike the induction and compression and exhaust strokes, both ends are under massive compressive load which is proportionate to the amount of leverage that you can provide to make the equivalent torque and that’s all about the relative stroke measurement.No need for Pi.
7% less leverage means 7% more force on the con rod required.Created by either more piston area and/or more cylinder pressure.
Which is a lot when multiplied by the smaller bearing areas v piston area.
Remind me how does supposed ‘BMEP’ figure account for all those variables.
The figure only knows specific torque it doesn’t know how that torque figure was made whether by more piston area or more cylinder pressure or more leverage.
It’s only meant to compare the specific torque output of different comparable engines.
Good luck with the 246 psi BMEP TL12 let alone that 200 psi BMEP at 1,200 rpm F1 engine.
BMEP doesn’t mean cylinder pressure.

1129

Carryfast:
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.

Absolutely correct if you apply the ‘Leatherhead Third Law of Motion’. But the passengers are going to be waiting a very long time at the bus stop. All together now: “The wheels on the bus go round and round, round and round, round and round, the wheels on the bus go round and round, round and round, round and round, all day long.”

Why do you think some vehicles have a diff lock? and why do vehicles lose traction on ice or mud? How much torque is being produced then?

Read that first line of Anorak’s derivation again.

“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.”

1130

Blimey- no reply. The last effort must have resulted in a stint in the corridor, outside the headmaster’s office.

This lull in proceedings allows us to catch up on the rapidly-developing narrative: so far, we have learned that the TL12 is a decidedly inferior engine to the RR Eagle Diesel, because its stroke is a bit shorter (142 vs. 152mm). There is no way that it could have been developed to make more than the torque it made in its 1980 version. The failure modes will be “hammering” of the bearings and “blowing” of the cylinder head gasket. To do the same torque as the later versions of the Eagle, which has the same capacity, the compressive force will be higher, as a direct result of that shorter stroke. In other words, that blowing, hammering force is proportional to the BMEP divided by stroke, or multiplied by piston area, if you can do kiddie algebra.

However, BMEP does not exist, and maths is for nerds, so we do not believe our own beliefs. Maybe the peak cylinder pressures are different, given that, for the same torque output, the engines have almost exactly the same capacity and, to produce the same torque, must use the same amount of fuel/air mix, and would both use the same injection timing, and injection pumps and turbochargers of contemporary design? That would explain it but, so far, we have no other information than bore, stroke and torque, so the comparison, with its emphatic conclusion, is based solely on the premise that Force= BMEP x Area. Like Pressure x Area, but BMEP is not pressure. It is expressed in the units of pressure for no reason at all, because the nerds don’t know how engines work.

Edit- the post previous to cav551’s last post has just appeared after, I would speculate, some feverish editing. Do I sense a retreat from the TL12/Eagle slaughterhouse, because BMEP is not peak pressure? Has the Formula One engine been started up, to drown the laughter?

1131

cav551:
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."

The brake doesn’t ‘apply’ any torque at all only the piston and con rod acting on the crankshaft x the stroke/leverage ‘applies’ any torque.The brake just measures it by applying a counter force.Just like a 40 t truck climbing Mont Cenis.
Remind me why are transmissions only rated by the INPUT TORQUE of the engine never by the gross weight of the vehicle and/or the steepness of the hills they will be needed to climb.
You’ll know if it’s a 1,216 lbft Eagle powered Roadtrain v a TL12 powered one in that regard.
Not because of what the hill or the load says, that’s the same in both cases, but because of the torque produced by the engine.
Let me guess when you use a torque wrench you think that it’s the bolt that is applying the torque not the wrench.

Remind me again what is BMEP other than just a measure of specific torque to compare engines nothing more.It is too stupid to know exactly HOW that torque was produced by the engine whether by cylinder pressure or piston area or leverage.
No answer yet as to whether you can get 200 psi BMEP at 1,200 rpm from an F1 engine.
When you realise or understand why you can’t you’ll also know why you can’t get 246 psi BMEP from a TL12.

1132

Carryfast:
… you can’t get 246 psi BMEP from a TL12.

Is that because there is no such thing? What if you polish the rocker cover? Will a genie come out of the exhaust manifold?

Carryfast:
It is too stupid to know exactly HOW that torque was produced by the engine whether by cylinder pressure or piston area or leverage.

That is exactly what you are doing, by declaring that one engine, of similar capacity to another, but with a shorter stroke, will blow its head off, if both engines have the torque output increased. By saying that the TL12 will batter its bearings and destroy its head gasket, if it had been developed to produce a similar torque to the later Eagle engines, you are using the variable (BMEP/stroke). If you had other information, such as cyl pressure vs. crank angle for both engines, your declaration might have carried some weight, but you can only cite the parameter of which you deny the existence. That is the centre point of the stupidity.

1133

[zb]
anorak:

Carryfast:
… you can’t get 246 psi BMEP from a TL12.

Is that because there is no such thing? What if you polish the rocker cover? Will a genie come out of the exhaust manifold?

Carryfast:
It is too stupid to know exactly HOW that torque was produced by the engine whether by cylinder pressure or piston area or leverage.

That is exactly what you are doing, by declaring that one engine, of similar capacity to another, but with a shorter stroke, will blow its head off, if both engines have the torque output increased. By saying that the TL12 will batter its bearings and destroy its head gasket, if it had been developed to produce a similar torque to the later Eagle engines, you are using the variable (BMEP/stroke). If you had other information, such as cyl pressure vs. crank angle for both engines, your declaration might have carried some weight, but you can only cite the parameter of which you deny the existence. That is the centre point of the stupidity.

It’s only the TL12 that needs its BMEP/Specific torque increased.
The Eagle is already there at 1,216 ft that’s close enough to 100 lbft per litre to provide 246 psi BMEP.
Oh wait you’re in denial that that amount of specific torque output = 246 psi BMEP because you don’t accept the formula specific torque x 2.464.
So you need 43% more specific torque with a 7% leverage deficit.
Also a 2.5% piston area deficit v the Eagle’s leverage advantage.
Tell us how you’re going to make that torque reuirement.
The required extra cylinder pressure is going to wreck your head fastenings.
The required extra force required on the con rod is going to wreck your end bearings.
There is no stroke or piston area variable in the BMEP calculation.It’s is just based on specific torque at the flywheel regardless of how its produced.

Which makes BMEP an oxymoron in itself in creating the erroneous premise that more specific torque/BMEP by definition means more ‘cylinder pressure’.Which the two variables of piston area and leverage prove is bs.
Not to mention numerous documented statements confirming BMEP doesn’t mean cylinder pressure as I posted.It’s a non existent abstract hypothetical comparator that’s all.Might as well use the specific torque figure that it’s based on.

BMEP couldn’t possibly be a measure of cylinder pressures no matter how much those with your erroneous ideas try to say it is.
It’s a measure of specific torque.Which is made up of the combination of cylinder pressure, piston area and leverage.Leverage having the advantage of multiplying the effort of both of the former and you’ve got a 7% deficit of it.

I’m still waiting for your answer to the question can you make an F1 engine produce its around same 200 psi BMEP but at 1,200 rpm.If not why not.

1134

Carryfast:
No answer yet as to whether you can get 200 psi BMEP at 1,200 rpm from an F1 engine.

Probably miles out with my answer, at low revs of 1200rpm an F1 engine at 200psi would I think stall the engine,
They have such a very short stroke & large piston area and can only create torque at high revs, the secret in an F1
car engine I believe is getting the gearing correctly set up for the track.

1135

Carryfast:

cav551:
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."

The brake doesn’t ‘apply’ any torque at all only the piston and con rod acting on the crankshaft x the stroke/leverage ‘applies’ any torque.The brake just measures it by applying a counter force.Just like a 40 t truck climbing Mont Cenis.
Remind me why are transmissions only rated by the INPUT TORQUE of the engine never by the gross weight of the vehicle and/or the steepness of the hills they will be needed to climb.
You’ll know if it’s a 1,216 lbft Eagle powered Roadtrain v a TL12 powered one in that regard.
Not because of what the hill or the load says, that’s the same in both cases, but because of the torque produced by the engine.
Let me guess when you use a torque wrench you think that it’s the bolt that is applying the torque not the wrench.

Remind me again what is BMEP other than just a measure of specific torque to compare engines nothing more.It is too stupid to know exactly HOW that torque was produced by the engine whether by cylinder pressure or piston area or leverage.
No answer yet as to whether you can get 200 psi BMEP at 1,200 rpm from an F1 engine.
When you realise or understand why you can’t you’ll also know why you can’t get 246 psi BMEP from a TL12.

oh dear never mind, stand up close behind the breech block of one of these then to prove that you are right.

youtube.com/watch?v=IUQpSiG0yuw

1136

dave docwra:

Carryfast:
No answer yet as to whether you can get 200 psi BMEP at 1,200 rpm from an F1 engine.

Probably miles out with my answer, at low revs of 1200rpm an F1 engine at 200psi would I think stall the engine,
They have such a very short stroke & large piston area and can only create torque at high revs

So proof that 200 psi BMEP in an F1 engine isn’t the same thing as 200 psi BMEP in a truck engine.
Which obviously means that it isn’t a measure of cylinder pressure.
It is just a measure of specific torque nothing more.
While a stroke deficit in a truck engine is obviously a very bad thing in that regard.

1137

cav551:

Carryfast:

cav551:
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."

The brake doesn’t ‘apply’ any torque at all only the piston and con rod acting on the crankshaft x the stroke/leverage ‘applies’ any torque.The brake just measures it by applying a counter force.Just like a 40 t truck climbing Mont Cenis.
Remind me why are transmissions only rated by the INPUT TORQUE of the engine never by the gross weight of the vehicle and/or the steepness of the hills they will be needed to climb.
You’ll know if it’s a 1,216 lbft Eagle powered Roadtrain v a TL12 powered one in that regard.
Not because of what the hill or the load says, that’s the same in both cases, but because of the torque produced by the engine.
Let me guess when you use a torque wrench you think that it’s the bolt that is applying the torque not the wrench.

Remind me again what is BMEP other than just a measure of specific torque to compare engines nothing more.It is too stupid to know exactly HOW that torque was produced by the engine whether by cylinder pressure or piston area or leverage.
No answer yet as to whether you can get 200 psi BMEP at 1,200 rpm from an F1 engine.
When you realise or understand why you can’t you’ll also know why you can’t get 246 psi BMEP from a TL12.

oh dear never mind, stand up close behind the breech block of one of these then to prove that you are right.

youtube.com/watch?v=IUQpSiG0yuw

No that’s all about head fastener loads.
The breech block is the cylinder head.No recoil with an engine though the head fasteners have to take all the of ■■■■■■■■■■ pressure and hold it.Why would you want more cylinder pressure instead of more leverage.

You didn’t answer the question why are transmissions rated by input torque.Not gross weight of the truck being moved.

Since when did the bolt apply any torque to a torque wrench.

1138

cav551:

Carryfast:

cav551:
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."

The brake doesn’t ‘apply’ any torque at all only the piston and con rod acting on the crankshaft x the stroke/leverage ‘applies’ any torque.The brake just measures it by applying a counter force.Just like a 40 t truck climbing Mont Cenis.
Remind me why are transmissions only rated by the INPUT TORQUE of the engine never by the gross weight of the vehicle and/or the steepness of the hills they will be needed to climb.
You’ll know if it’s a 1,216 lbft Eagle powered Roadtrain v a TL12 powered one in that regard.
Not because of what the hill or the load says, that’s the same in both cases, but because of the torque produced by the engine.
Let me guess when you use a torque wrench you think that it’s the bolt that is applying the torque not the wrench.

Remind me again what is BMEP other than just a measure of specific torque to compare engines nothing more.It is too stupid to know exactly HOW that torque was produced by the engine whether by cylinder pressure or piston area or leverage.
No answer yet as to whether you can get 200 psi BMEP at 1,200 rpm from an F1 engine.
When you realise or understand why you can’t you’ll also know why you can’t get 246 psi BMEP from a TL12.

oh dear never mind, stand up close behind the breech block of one of these then to prove that you are right.

youtube.com/watch?v=IUQpSiG0yuw

Well if the RR Roadtrain and the TL12 Roadtrain were both putting out 270 bhp it would be difficult to tell, well maybe the driver of the TL12 could has he passed the RR up the hill but all this as already been said you just ignore it. The TL12 was a very good engine giving few problems in service.
How long would this aforementioned torque wrench be would it be of a standard length or a 8 to 10 mm longer one that would give 900% more leverage and the rest would be history

1139

ramone:
Well if the RR Roadtrain and the TL12 Roadtrain were both putting out 270 bhp it would be difficult to tell, well maybe the driver of the TL12 could has he passed the RR up the hill but all this as already been said you just ignore it. The TL12 was a very good engine giving few problems in service.
How long would this aforementioned torque wrench be would it be of a standard length or a 8 to 10 mm longer one that would give 900% more leverage and the rest would be history

Which part of the Eagle was putting out more than 273 hp at 1,200 rpm didn’t you understand.
Good luck with 7% less leverage when you need to apply 1,216 lbft.
The rest was history for the TL12 from 1983 the only surprising thing being why bother with it at all.Unless you want the firm to fail.

1140

Carryfast:

cav551:

Carryfast:

cav551:
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."

The brake doesn’t ‘apply’ any torque at all only the piston and con rod acting on the crankshaft x the stroke/leverage ‘applies’ any torque.The brake just measures it by applying a counter force.Just like a 40 t truck climbing Mont Cenis.
Remind me why are transmissions only rated by the INPUT TORQUE of the engine never by the gross weight of the vehicle and/or the steepness of the hills they will be needed to climb.

oh dear never mind, stand up close behind the breech block of one of these then to prove that you are right.

youtube.com/watch?v=IUQpSiG0yuw

No that’s all about head fastener loads.
The breech block is the cylinder head.No recoil with an engine though the head fasteners have to take all the of ■■■■■■■■■■ pressure and hold it.Why would you want more cylinder pressure instead of more leverage.

You didn’t answer the question why are transmissions rated by input torque.Not gross weight of the truck being moved.

Oh what a clever one you are managing to introduce the subject of leverage agian. Yes a gearbox is a leverage device so it is the input torque which matters… Buy yourself a bag of gobstoppers. Better still buy a nice big bag.

Still denying equal and opposite reaction? and of course as you mentioned earlier the wheels on the bus don’t produce torque.

So this below is what? make believe ? Well it can’t be can it, because there is leverage involved ? You’d better make that an extra, extra large bag of super size gobstoppers.
x-engineer.org/automotive-engin … ue-engine/