Standard Definition of Displacement. Is there one?

[KevinK2]

I searched and could not find an ASTM, ASME, SAE, or DIN standard defining displacement.

My research lead me back to early "piston pumps" (predating steam engines). There displacement was defined as # of pistons x piston displacement, as these pumps worked on a 1 rev cycle.

I think this method was copied by the otto cycle engines, even though they needed 2 revs to complete a cycle.

When 2 cycle engines appeared, the same "piston pump" approach to displacement was used, but these engines need just one rev to complete all cycles. This advantage was handled in motorcycle racing by allowing about 50-70% more displacement for the 4 stroke bikes, that competed against the 2 stroke bikes.

With the 2 and 4 stroke engines, the "piston pump" rule could always be substantiated by considering each engine as a 1 rev pump to verify displacement.

Enter the Wankel.

The "1.3L" engine found today in Mazdas was very different, with dynamic combustion chambers, and 3 revs to complete a single chambers's 4 strokes. The developers apparently used a different rule to get displacement. They said you have 2 rotors, and each rotor fires once PER REV, a fired chamber is .64L, so displacement is 2 rotors x .64L per rotor = 1.3L displacement ... PER REV.

With this logic, and rotor = piston, a 2 stroke would have the correct displacement (per rev) for 100% efficency. The 4 stroke 5.0L engine would be 2.5L, based on Wankel's "1 rev" method. This is actually a good way to determine displacement, as it is consistent for all engines; 100% VE ingestion per rev.

I don't think anyone is prepared to call their 5.0 V8 a 2.50. So, if one were to apply the "piston pump" rule to the Wankel:

Each rotor has 3 distinct combustion chambers per rotor, that take 3 output shaft revs to fire. With 2 rotors, that's 3.8L fired in 3 revs.

As a wankel pump, each of the sets of 4-otto-cycles of a combustion chamber cycle, provide 2 pumping cycles for a pump (note that the valving and porting are ignored in the basic "piston pump" displacemnt calc). So you take 2 x the fired chambers (pumping strokes) and divide by 3 to get one rev, and you have 2.6L displacement rating, per the "piston pump" method. This is actually the number of chambers that fire in 2 revs, and is an equal basis for rating the extremely common 4 stroke piston engine.

Other than the "piston pump" method, is there any official standard for displacement?

Kevin

 

[KevinK2]

Limited here to NA, 2 and 4 stroke otto engines.

 

[btrueblood]

"just looking for a standard for rated displacement, and found none"

Probably because displacement is a fairly useless measurement, other than for "bragging rights" at the local drinking establishment? See Steve's post.

 

[Lou Scannon]

What about a Hemi??

Yeah, what about a Hemi?

 

[patprimmer]

The reason we cannot truly use revs of the output shaft to rate a Wankle engine is that there is a gear ratio between the output shaft and the rotor. It should be rated by each induction stroke at 100% VE times the number of chambers that can be filled in one full cycle.

Racing bodies put on correction factors to keep competition even.

Apart from that it does not really matter unless trying to do original estimate of fuel flow requirements

Regards
Pat
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[KevinK2]

From Steve

This whole issue was really stirred up by the likes of Norton and Mazda, competing in limited capacity racing, running a n engine whose capacity could never really be compared to the conventional engines.

The wankel is easily compared to a 4 stroke otto engine, which it is, with rotor faces vs piston tops. It's a clever version that uses "shared" combustion chambers. See my post at 14:06 for a direct piston engine equal to the 13B. To take it a step further, the stroke would only be 1.38" and the piston dia 3.0" .

 

[KevinK2]

Can anyone check my equivalent displacement 3.9L 6 cyl piston engine, for the Mazda 13B, assuming 100% VE. Data:

(6) .65L cylinders ( same as 6 .65L "1/3 moon" shaped pockets in 13B)

and, a 1.5 planetary-gear speed increaser on the output shaft of the 6 cyl engine, presenting the functional output shaft to the flywheel at 1.5X crank speed.

This creates the same duration of the otto cycle strokes.

Kevin

 

[BrianGar]

''Wankel engines can be classified by their geometric size in terms of radius (rotor center to tip distance, also the median stator radius) and depth (rotor thickness), and offset (crank throw, eccentricity, also 1/4 the difference between stator's major and minor axes). These metrics function similarly to the bore and stroke measurements of a piston engine. Displacement is 3√3radius·offset·depth, multiplied with the number of rotors (note that this only counts a single face of each rotor as the entire rotor's displacement, and is of course incorrect as there are three faces, equivalent to three piston faces, per rotor, i.e. equivalent to a three cylinder radial piston motor per rotor). Nearly all Mazda production Wankel engines share a single rotor radius, 105 mm (4.1 in), with a 15 mm (0.6 in) crankshaft offset. The only engine to diverge from this formula was the rare 13A, which used a 120 mm (4.7 in) rotor radius and 17.5 mm (0.7 in) crankshaft offset.''

''In auto racing, the displacement of a Wankel engine is usually doubled for classing purposes. This is of course a marketing ploy and wrong. Using only a single face per rotor instead of three results in the nominal displacement being a third of actual. For calculating taxes in Japan, the displacement of Wankel engines is defined as the equivalent of 1.5 times the nominal displacement, so the 1300 cc 13B engines are taxed as 1950 cc, whereas the actual displacement is 3900 cc''

Pulled off the internet to save time.

See Pats last sentence in his last post.

Brian,

 

[KevinK2]

"... The only engine to diverge from this formula was the rare 13A, which used a 120 mm (4.7 in) rotor radius and 17.5 mm (0.7 in) crankshaft offset."

This reference was pre-13B, Mazda's version with 17.5mm offset.

 

[globi5]

The rotor of the Wankel does only spin at 1/3 of the speed of the crankshaft. Each rotor has 3 chambers with 0.65 l maximum volume each. Therefore each rotor displaces the volume of only one chamber (0.65 l) per crankshaft revolution .

Thus, the Mazda RX-7, RX-8 Wankel engines with 2 rotors displace 1.3 l per crankshaft revolution, which is why it's named 13B and not 39B.

1.3 l is the same volume a 2.6 l piston 4-stroke engine would pump (assuming same VE), which is why it is taxed, raced, compared to a 2.6 l conventional engine.

 

[globi5]

This sentence in the English written wikipedia page regarding the Wankel engine is actually false: "In auto racing, the displacement of a Wankel engine is usually doubled for classing purposes. This is of course a marketing ploy and wrong."

Naturally, the 1.3 l displacement has to be doubled for classing purposes unless the Wankel engine would race against 2-stroke piston engines. If one doesn't believe it - here's an animation showing that the rotor does only 1/3 of the revolution of the crankshaft revolution:

http://de.wikipedia.org/wiki/Wankelmotor

Piston as well as Wankel engines are essentially positive displacement pumps, which can easily be compared and classified (unlike centrifugal and axial flow pumps).

 

[KevinK2]

The rotor of the Wankel does only spin at 1/3 of the speed of the crankshaft. Each rotor has 3 chambers with 0.65 l maximum volume each. Therefore each rotor displaces the volume of only one chamber (0.65 l) per crankshaft revolution .

THUS, the Mazda RX-7, RX-8 Wankel engines with 2 rotors displace 1.3 l per crankshaft revolution, which is why it's named 13B and not 39B.

Using that logic, basically injested air per rev, a mustang 5.0 should be rebadged a mustang 2.5 ... ain't gonna happen. That was the clever logic that fooled people into thinking the 13B had the highest specific power of all NA automotive engines. The 5.0 is based an equivelant pump, swept volume in one rev.

1.3 l is the same volume a 2.6 l piston 4-stroke engine would pump (assuming same VE), which is why it is taxed, raced, compared to a 2.6 l conventional engine.

Globi5, are you speaking of two different volumes, in your 1st and 3rd paragraph?

Kevin

 

[KevinK2]

Globi5, let me clarify.

1) I found here that there is no SAE standard for displacement.

2) My research showed the first dicussion of volumetric "displacement" for machines occured with ancient piston pumps, predating the steam engine.

3) Otto cycle engines adopted this "piston pump" method for displacement rating, # pistons x area x stroke, or swept volume. This had a flaw in that it under-rated 2 stroke power potential. It also was independent of the number or revs to complete an engine cycle.

4) When the wankel was developed, they took advantage of the lack of a standard. The swept volume would have put it at 3.8L. So unlike the other otto's they assumed a "per rev" basis for displacement, as 2 chambers were swept per rev.

But, using the old "piston pump" method, you see how much is pumped per rev, as a pump where the 4 strokes for a single otto cycle yield )2) 2-stoke pump cycles. So:

(3.8L x 2) total pumped in 3 revs / 3 = 2.6L pumped per rev.

Kevin

 

[globi5]

1. The rotary engine still pumps only 1.3 l per revolution and not 3.9 l, because for the third time: The rotor only moves 120 degrees when the crankshaft which delivers the actual engine revolution moves 360 degrees.

2. For the reason above the rotary engine pumps 3 x 1.3 l = 3.9 l in 3 revolutions and definitely not 2 x 3.9 l and definitely neither 2.6 l per revolution.

3. A valveless 2 stroke piston engine pumps as much volume as it displacement indicates. And so does this (also valveless) rotary engine.

4. A four stroke engine only displaces half of its volume, because it only pumps its displacement every second revolution. (Thanks to its inlet valves which only open every second revolution.)
If you were to badge the displacement of a 4 stroke piston engine to half of its displacement, people would then ask: Why is a 4 stroke piston so much bigger than a 2 stroke or a piston compressor or steam engine, even though displacement is defined as

the volume displaced by a piston (as in a pump or an engine) in a single stroke

http://www.merriam-webster.com/dictionary/displacement

5. I don't recall any Mazda official ever trying to fool the public regarding the classification of its rotary engine, because nobody can fool simple geometry and nobody really cares anyway.

6. I don't even care about the displacement definition, I just tried to help regarding your question: "Can anyone check my equivalent displacement 3.9L"
And I also wanted to let you know that this wiki sentence above is false. But if you don't want to know the facts, don't ask for it and put whatever badges on rotary engines you prefer.

 

[KevinK2]

From Globi5

2. For the reason above the rotary engine pumps 3 x 1.3 l = 3.9 l in 3 revolutions and definitely not 2 x 3.9 l and definitely neither 2.6 l per revolution.

I said, configured as a true pump, where the four strokes for an IC engine cycle would be used as 2 cycles in a pump, the pumped fluid would be 2 x 3.9L in 3 revs. The fluid pumped, configured as a true pump, in one rev would be 2.6L. I believe these are facts, based on engineering interpretation.

6. I don't even care about the displacement definition, I just tried to help regarding your question: "Can anyone check my equivalent displacement 3.9L"

Read the title of the thread, it's all about displacemnt ratings, sorry you took a wrong turn. And I saw nothing related to my hypothetical 3.9L 6 cyl piston engine.

And I also wanted to let you know that this wiki sentence above is false. But if you don't want to know the facts, don't ask for it and put whatever badges on rotary engines you prefer.

Huh?

First, the "wiki" definition is consistent with the "swept volume" defination for piston engines, which was based on the historic "piston Pump" displacement rating.

Also, you are consitently using the air pumped by an IC engine in your explanations, which would put the 13B at 3.9L, if number of engine cycles doesn't matter ( ref 2-stroke, 4-stroke).

On the subject of displacement, I suggested using a simple piston pump analogy (predated the current concept of swept volume), which puts the 2-cycle and 4 cycle at same disp't, as it is now. It also puts the 13B at 2.6L,. All are considered to be true fluid pumps, 100% VE, and displacement is rated a 1 rev.

Kevin

 

[SomptingGuy]

Can anyone else here smell some kind of agenda? Like someone is hoping for validation?

- Steve

 

[ivymike]

I decided to help Greg fry his grandmother's eggs if he ever sends an invite.. until then I'm checking my emails instead of this thread.

 

[patprimmer]

Yep.

I've had enough.

Kevin now has 2 threads full of garbage.

Regards
Pat
See FAQ731-376 for tips on use of eng-tips by professional engineers &

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[GregLocock]

Smells? reeks is more like it. In the not too distant future, on an intertube near you, someone on a Wankel fanboi site will write 'the engineers at eng-tips said....'. And the dispute will rage on for another hundred posts.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[patprimmer]

Greg, but it won't be here, so who cares.

Hopefully this will all disappear at least from here tonight Aus time or tomorrow morning USA time.

Oh. The other one was not a thread, but simply a post in the hand built engines thread. Unfortunately I answered before I saw just how stupid it was getting here.

Regards
Pat
See FAQ731-376 for tips on use of eng-tips by professional engineers &

http://eng-tips.com/market.cfm

for site rules

 

[dicer]

Pretty easy stuff. Displacement

You were looking in the wrong place for a definition.
A good old dictionary would have been just fine.

Simply put, the removal of something and then something else takes its place. Thus a piston is removed and the volume that is left is what displaced it. And since that is all pretty much set by the Bore and the stroke, its a fairly simple thing to calculate it. Oh and there is another word that is used along with the word displacement, swept volume.
And another thing, the number of cycles and or rpm has nothing to do with the displacement calculation. If you enter that factor then you are concerning yourself with CFM or capacity.