S65 engine weight - and some surprises!

[8k3]

Originally Posted by xfirer1guy

Ummmm... I'm confused. You realize you wrote an equation agreeing with my agrument, but you disagree?!

F=PA is all you need to understand to settle this argument. How do you not understand that A in the equation is BORE? Bore is indeed a parameter used to determine engine torque.

What he is saying is this, for a given displacement, say ~3.5L, simply changing the stroke and/or bore to KEEP you at 3.5L will not change the torque, here is an example of how you can change the bore/stroke multiple ways and stay at 3.5L

65.8mm stroke x 92mm bore (stock bore size) = 3.499 L
75.2mm stroke (stock stroke) x 86mm bore = 3.494 L
82.25mm stroke x 82.25 bore = 3.496 L

Now, if you change the stroke OR the bore to increase the overall dispalcement, THEN the torque increases (which is obvious, bigger engine say a 4L has more than a 3.5L)

[Billj747]

Originally Posted by CanAutM3

By increasing cylinder pressure with better overall efficiciency.

The GT3 engine flows better (volumetric efficiency), better utilises the intake charge (direct injection) and probably has less friction losses.

The GT3 engine example proves my point about the bore and stroke relationship. It has a much bigger bore vs stroke relationship (102/77.5) compared to the S65 (92/75.2). This bigger bore allows for a bigger head surface area and for bigger valves, which in turn improves volumetric efficiency. So oversquare engine don't produce less torque, myth busted .

I set it up for you to knock it down

Originally Posted by Obioban

Lolwut?
S54 stock redline: 8000 rpm
S65 stock redline: 8400 rpm
That's 5% higher, not >10% higher

Comparing modded to stock makes no sense. And, for that matter, my S54 had a 8700 rpm redline for a bit (albeit by mistake, but no damage was done to the engine).

Stock vs stock (aka, with stock redlines), the S54 has high piston speed and a longer stroke, too

S54 has a VERY long stroke for it's RPM. Same with the GT3 RS 4.0.

[CanAutM3]

Originally Posted by Billj747

I set it up for you to knock it down

Teamwork

[MilehighM3]

Originally Posted by regular guy

I thought you were going to say "74 pages of this BS...like the bearing Wiki thread?"

Edited for accuracy

[regular guy]

Originally Posted by regular guy

I thought you were going to say "75 pages of this BS...like the bearing Wiki thread?"

Edited for accuracy and syntax.

[speedaddictM3]

Originally Posted by US///M3

It's used on container ships...

http://jalopnik.com/the-largest-pist...loo-1480385814

Anyone know the rod bearing clearance for this engine?

[Taskmaster]

Those LS number aren't right though, as A Clutch and flywheel should be included in those weights, and they usually are about 40lbs combined. Are they included on the BMW and Ford's weight?

[Justin*]

Originally Posted by ibmike

Are you two^ serious? What the Hell? you going to have 10 pages of this BS?

I like it. Part of the point in being here is to learn. I studied volumetric efficiency at the tiniest level in college. Aircraft mechanic. Other than learning the concepts; honestly, there isn't much use after that. So, this stuff is interesting.

It was really only taught to me in laments terms. I've always known the outcome is actually the same whether the bore is increased to gain displacement or the stroke is increased to gain displacement. I do like over squared engines. I think they do allow for better fuel atomization.

[catpat8000]

Originally Posted by TheAxiom

Those LS number aren't right though, as A Clutch and flywheel should be included in those weights, and they usually are about 40lbs combined. Are they included on the BMW and Ford's weight?

You make a good point. I spent few minutes today trying to narrow in on weights - it is surprisingly difficult to get good information.

Just looking harder at the LS3, I've found a variety of numbers: 415, 418, 443, 468, none yet seemingly authoritative from GM. I thought I had seen 415 lbs from the GMPP perf catalog but on returning, I cannot confirm. That will teach me not to include links.

I will try and get better data.

Pat

P.S. I watched someone weigh an S65 engine block yesterday - it weighed exactly 86 lbs.

[MilehighM3]

Perhaps Benvo can get his removed S65 weighed to compare the actual number vs what is advertised?

[swamp2]

Originally Posted by catpat8000

You make a good point. I spent few minutes today trying to narrow in on weights - it is surprisingly difficult to get good information.

Just looking harder at the LS3, I've found a variety of numbers: 415, 418, 443, 468, none yet seemingly authoritative from GM. I thought I had seen 415 lbs from the GMPP perf catalog but on returning, I cannot confirm. That will teach me not to include links.

I will try and get better data.

Pat

P.S. I watched someone weigh an S65 engine block yesterday - it weighed exactly 86 lbs.

There is no standard for engine weights! Some obvious variables include the presence/absence of:

Starter
Flywheel
Power steering pump
Oil
Coolant (prob most specs are dry even though that is largely irrelevant)
Belts
Fan(s)

Also:

How much of the intake system
How much of the exhaust system (headers only, probably nothing typically included)
How much of the electronics and wiring harness
How much if any of cooling/oil hoses

Would be nice if there was an SAE/JIS/EU standard for this. Based on this I think most engine weight comparisons are highly uncertain and almost for sure apples to oranges.

[swamp2]

Originally Posted by CanAutM3

If you study a bit longer, in the subsequent semesters of an engineering degree, you learn a bit more on the intricacies of a piston engine:
...
Tada , torque depends on cylinder pressure, displacement and crank angle only. No bore or stroke.

Your expressions are both incorrect, missing terms and approximate. That has led you to an incorrect conclusion. You shouldn't be quite so arrogant here... Your conclusion would be much closer to correct in the case of some hypothetical idealized engine with zero mass piston assemblies, rods and crank (or perhaps at such low rpms that the inertia of all components is insignificant).

Let's first talk about missing terms:

Torque on the crank comes from both the gas pressure and from inertia terms. You have completely set aside all inertial terms. There are two inertial terms, one from the crank inertia and one from the piston and rod inertias. Not surprisingly the latter is quite complex containing a full Fourier series in term of the crank angle, center of mass distances to rod bearings, mass of piston, mass of rod, etc. However, ignoring terms related to the crank angle position there are terms in both of these inertial terms that vary as:

Ti ~ m*ω^2*r^2

where m is both piston and part of the connecting rod mass, ω is rotational speed (SI units) and 2*r is the stroke. The inertial terms clearly depend on stroke. There are no "hidden terms" that would combine with the stroke to provide displacement (which we are fixing) since nothing in these terms depend on area nor gas pressure.

Approximations:

Your final derived expression only considers the gas pressure load. It contains sin(θ) dependence for the crank angle dependence. This is an approximation. Your trig/geometry or cross product is incorrect likely because the angle the connecting rod makes relative to the cylinder is not the same as the angle the crank makes with a reference plane. This angles have a complex relationship.

If we allow the approximation that the stroke/2 is << the connecting rod length the correct expression becomes:

Tp ≈ P*D/2*(sin(θ) + (r/2*l)cos(θ))

l = connecting rod length, D is displacement

We have basically dropped terms of order (r/l)^2. The r/2*l term for the S65 is about 1/8th as large (over 10% of) as the plain sin(θ) term. In contrast the (r/2*l)^2 term is only about 1-2%. Either way even the torque from the gas pressure term depends on stroke. In all of these terms an increase in stroke provides an increase in torque at constant displacement.

Any engine simulator with some numerical rigid body kinematics capability (which I'm sure Ricardo's system has) will capture all of these effects with no r/l approximations and all of the terms in any infinite series. Thus I'm confident at constant displacement stroke effects will easily show up in such simulation as the other poster alluded to.

Again only in the case of massless reciprocating parts, massless crank and l/r being infinite one can arrive at your conclusion.

[BPM]

Originally Posted by MilehighM3

Perhaps Benvo can get his removed S65 weighed to compare the actual number vs what is advertised?

Sure, it's in a box so I don't mind. I presume you want the weight with the oil drained? Or we can subtract whatever a quart weighs x 8.

The measurement will be with no throttle bodies or alternator. I do have an extra manifold I can put on. Clutch and flywheel is also attached.

[MilehighM3]

Originally Posted by BPMSport

Sure, it's in a box so I don't mind. I presume you want the weight with the oil drained? Or we can subtract whatever a quart weighs x 8.

The measurement will be with no throttle bodies or alternator. I do have an extra manifold I can put on. Clutch and flywheel is also attached.

Just the long block without any intake or exhaust components, clutch or flywheel would be great. I'd be surprised if your tech didn't drain the oil prior to removal and the amount of coolant left in the block would be minimal as well. Just for the sake of comparison of course, to see if the advertised weight is remotely close to yours.
Thanks in advance!

[happos2]

1+1 = 3. Therefore, the S65 is awesome!

For real though...Good discussions!

[catpat8000]

Originally Posted by swamp2

There is no standard for engine weights! Some obvious variables include the presence/absence of:

Starter
Flywheel
Power steering pump
Oil
Coolant (prob most specs are dry even though that is largely irrelevant)
Belts
Fan(s)

Yes, obviously. I thought I had collected weights which included flywheel, no liquids, no starter, no power steering pump, intake and exhaust manifolds and no electronics for all engines.

But I cannot confirm some of my numbers. I'll re-post should I get better numbers.

Pat

[CanAutM3]

Originally Posted by swamp2

Your expressions are both incorrect, missing terms and approximate. That has led you to an incorrect conclusion. You shouldn't be quite so arrogant here... Your conclusion would be much closer to correct in the case of some hypothetical idealized engine with zero mass piston assemblies, rods and crank (or perhaps at such low rpms that the inertia of all components is insignificant).

Let's first talk about missing terms:

Torque on the crank comes from both the gas pressure and from inertia terms. You have completely set aside all inertial terms. There are two inertial terms, one from the crank inertia and one from the piston and rod inertias. Not surprisingly the latter is quite complex containing a full Fourier series in term of the crank angle, center of mass distances to rod bearings, mass of piston, mass of rod, etc. However, ignoring terms related to the crank angle position there are terms in both of these inertial terms that vary as:

Ti ~ m*ω^2*r^2

where m is both piston and part of the connecting rod mass, ω is rotational speed (SI units) and 2*r is the stroke. The inertial terms clearly depend on stroke. There are no "hidden terms" that would combine with the stroke to provide displacement (which we are fixing) since nothing in these terms depend on area nor gas pressure.

Approximations:

Your final derived expression only considers the gas pressure load. It contains sin(θ) dependence for the crank angle dependence. This is an approximation. Your trig/geometry or cross product is incorrect likely because the angle the connecting rod makes relative to the cylinder is not the same as the angle the crank makes with a reference plane. This angles have a complex relationship.

If we allow the approximation that the stroke/2 is << the connecting rod length the correct expression becomes:

Tp ≈ P*D/2*(sin(θ) + (r/2*l)cos(θ))

l = connecting rod length, D is displacement

We have basically dropped terms of order (r/l)^2. The r/2*l term for the S65 is about 1/8th as large (over 10% of) as the plain sin(θ) term. In contrast the (r/2*l)^2 term is only about 1-2%. Either way even the torque from the gas pressure term depends on stroke. In all of these terms an increase in stroke provides an increase in torque at constant displacement.

Any engine simulator with some numerical rigid body kinematics capability (which I'm sure Ricardo's system has) will capture all of these effects with no r/l approximations and all of the terms in any infinite series. Thus I'm confident at constant displacement stroke effects will easily show up in such simulation as the other poster alluded to.

Again only in the case of massless reciprocating parts, massless crank and l/r being infinite one can arrive at your conclusion.

I agree, the entire mathematical model required to establish instantaneous torque at the flywheel is extremely complex and inertia terms are important components to this.

The expression I quoted simply wanted to reflect the contribution of the cylinder pressure of one piston to the overall engine. I limited it to that for the sake of simplicity. The expression also does not take into account any of the frictional losses, so it is far from complete. It was simply intended to dispel the common myth that longer stroke engines are capable of more torque than bigger bore engines due to the longer "torque arm" provided by the longer stroke.

Where it also becomes much more complex, is when considering the impact of the variation of the volume caused by the piston travel on the cylinder pressure. This is where piston position relative to crank angle comes into play and where bore, stroke and connecting rod length all have an impact. That is also where Fourier series you mention become handy.

However, I am not sure I follow your expression Tp ≈ P*D/2*(sin(θ) + (r/2*l)cos(θ)). This equation implies that torque is generated from cylinder pressure at TDC and negative torque at BDC, which is geometrically not possible. How do you explain this?

I no way way did I intend to be arrogant, my comment was just a wink at the reference to first semester physics . I did say it was more than 20 years ago, my memory might be rusty , I'll go back and check my old textbooks for the trigonometry.

EDIT:

Found the old textbooks; the geometry/trig to determine torque from crank angle is much more complex than I remembered:

T = P*(B/2)^2*pi * (S/2) * sin (θ + asin((S/2)/L*sin(θ))) / cos(asin((S/2)/L*sin(θ)))
or
T = P*D/2 * sin (θ + asin((S/2)/L*sin(θ))) / cos(asin((S/2)/L*sin(θ)))

Where
T = Torque
P = Cylinder pressure
B = Bore
S = Stroke
D = Displacement
L = Connecting rod length
θ = Crank angle with θ=0 at TDC

While the instantaneous torque value does change by increasing the stroke/bore ratio; from 0 to 180 degrees of crank angle, the increased bore/stroke ratio sometimes provides a greater leverage and sometimes a lesser one. What is interesting is that integrating this leverage from θ=0 to 180 yields the same result regardless of the bore and stroke relationship. Now I just need to add the varying cylinder pressure with cylinder volume; I remember the teacher demonstrating this to us, I just cannot (yet) find the mathematical demonstration in my old textbooks...

[swamp2]

Originally Posted by CanAutM3

The expression I quoted simply wanted to reflect the contribution of the cylinder pressure of one piston to the overall engine. I limited it to that for the sake of simplicity. ... It was simply intended to dispel the common myth that longer stroke engines are capable of more torque than bigger bore engines due to the longer "torque arm" provided by the longer stroke.

That still incorrect. There is more torque even not counting intertial terms.

Originally Posted by CanAutM3

However, I am not sure I follow your expression Tp ≈ P*D/2*(sin(θ) + (r/2*l)cos(θ)). This equation implies that torque is generated from cylinder pressure at TDC and negative torque at BDC, which is geometrically not possible. How do you explain this?

Error in that formula. I was also counting too much on a text rather than doing the full derivation manually. The correct expression is (again dropping terms of order (r/l)^2

Tp ≈ P*D/2*(sin(θ) + (S/4*l)*sin(2θ))

None of those complex arc functions are required. The exact expression including (r/l)^2 is not too much more complex.

Originally Posted by CanAutM3

While the instantaneous torque value does change by increasing the stroke/bore ratio; from 0 to 180 degrees of crank angle, the increased bore/stroke ratio sometimes provides a greater leverage and sometimes a lesser one. What is interesting is that integrating this leverage from θ=0 to 180 yields the same result regardless of the bore and stroke relationship.

This is only true if the combustion pressure is not a function of the crank angle yet in reality it is strong function of crank angle (i.e. P = P(θ)). It is much larger in the first half of the power stroke. As an aside I'm pretty sure it would also be true for constant combustion pressure with the additional (r/l)^2 terms.

Edit: In messing around numerically integrating the exact formula (with (r/l)^2 term) using a Gaussian approximation for the combustion pressure (which is qualitatively extremely close) I found that a 25% change in stroke only produces a 1-2% change in the integrated torque over the entire power stroke (θ:0-180°). Thus, not counting the inertial terms, "more stroke at a constant displacement does not make more torque" is a pretty decent engineering approximation. However, the inertial terms can be a very significant contributor to the torque, making this good engineering approximation nothing more than a point of mild curiosity (again for those engines with massless components or running at very low rpm). For a paper showing this have a look at this post (part of the great S65 bearing clearance debate...) and the paper I reference in it.

[swamp2]

Originally Posted by catpat8000

Yes, obviously. I thought I had collected weights which included flywheel, no liquids, no starter, no power steering pump, intake and exhaust manifolds and no electronics for all engines.

These manufacturers may have some natural consistency which is much greater than one might assume. Maybe your (their) numbers are pretty well "apples to apples". However, those pesky marketing guys always like to have their input which one can safely assume would contribute to more variation in what each manufacturers number actually means.

[catpat8000]

Originally Posted by swamp2

These manufacturers may have some natural consistency which is much greater than one might assume. Maybe your (their) numbers are pretty well "apples to apples". However, those pesky marketing guys always like to have their input which one can safely assume would contribute to more variation in what each manufacturers number actually means.

I did learn one interesting thing, which will probably elicit more comments welcoming me into 1997: The 6.2 liter LS3 has a claimed weight by GM of 385 lbs!

http://www.chevrolet.com/content/dam...ce-catalog.pdf

The catalog claims 415 lbs total which includes 30 lbs of crate and packaging material - 385 lbs!

This engine appears to include:
- intake manifold
- exhaust manifolds
- water pump
- no power steering
- no fluids
- no electronics
- flywheel

Chevrolet sure invests a lot in keeping the aftermarket perf guys happy. That catalog is chock full of information and all manner of engines. Chevrolet will sell you an LS3 with ~100hp over the stock LS3 in a crate with a 2 year warranty, for example.

[CanAutM3]

Originally Posted by swamp2

That still incorrect. There is more torque even not counting intertial terms.

I disagree; I still believe the previous conclusion is accurate where an increased torque arm is offset by a smaller piston area. I thought I could demonstrate this through instantaneous torque, but I was mistaken. The math is more complex and my previous demonstration was oversimplified and inaccurate. I now realize that to complete the demonstration, the full cycle of the engine needs to be considered.

Originally Posted by swamp2

Error in that formula. I was also counting too much on a text rather than doing the full derivation manually. The correct expression is (again dropping terms of order (r/l)^2

Tp ≈ P*D/2*(sin(θ) + (S/4*l)*sin(2θ))

I struggled a bit to reconcile the expression above, maybe due to misinterpretation on my part. By interpreting it the following way, I was able to come decently close to the longer expression I quoted in my previous post:

Tp ≈ P*D/2*(sin(θ) + (S/(4*l))*sin(2θ))

The latter however remains an approximation. Probably due the omission of the (r/l)^2 term you mention.

Originally Posted by swamp2

None of those complex arc functions are required. The exact expression including (r/l)^2 is not too much more complex.

The expression from my previous post was derived manually and most likely can be simplified to a shorter expression through some algebraic gymnastics; I was just too lazy to do so . But it remains mathematically valid and exact.

Originally Posted by swamp2

However, the inertial terms can be a very significant contributor to the torque…

I agree that when evaluating instantaneous torque values, the inertial terms do have an important impact. A piston engine relies in “storing” a part of the energy from the compression stroke in momentum for the other three strokes (exhaust, intake and compression). As for the translation momentum of the piston and rods (and other components), if I am not mistaken, the forces used to accelerate them in the first portion of the stroke are recuperated when they slow down in the second portion.

So for an engine operating at steady state, the inertial terms should cancel out through an entire 4 stroke cycle, wouldn’t they ?

Originally Posted by swamp2

Thus, "more stroke at a constant displacement does not make more torque" is a pretty decent engineering approximation. However, the inertial terms can be a very significant contributor to the torque, making this good engineering approximation nothing more than a point of mild curiosity (again for those engines with massless components or running at very low rpm).

I think it is much more than just that.

I now realize that trying to demonstrate this through force equations is way too complex. So I’ll try to present it from a different perspective using the conservation of energy principles instead. This gets rid of all the geometric and inertial factors.

Assume two isentropic engines operating at the same constant RPM for which all parameters are equal (cylinder count, displacement, compression ratio, volumetric efficiency, thermal efficiency, con-rod length, etc...) except for the bore and stroke.

Both engines are sucking-in the same amount of intake charge (same displacement, RPM and VE). So, for a given amount of rotations, the exact same amount of energy (chemical here) is input to the system. The output energy is measured by the torque produced over the given amount of rotations (work). If the engine with the longer stroke generates more torque, where is the other engine loosing its energy to? I know, this is still very simplified, but provides food for thought.

From a design perspective, varying the bore and stroke ratio allows other parameters to be altered such as volumetric efficiency, valve size, blow-by losses, friction losses and many more that will vary the torque output depending on operating conditions (low vs high RPM for example). But IMHO, the conclusion remains, for a given displacement, the mechanical benefit of a longer torque arm is offset by a smaller piston area.

[US///M3]

Now you guys are just showing off with your fancy formulas.