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How is displacement measured?

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Old 01-09-10 | 09:58 AM
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How is displacement measured?

I have never really thought about it but I have never known for sure how displacement is measured. on a piston motor it is the max volume of a cylinder x the number of cylinders (π(1/2 bore)[ sup ] 2 [/ sup] . so technically the engine would never have the labeled space available at a given time b/c the pistons are not all down at the same time.

so my question "How is displacement measured?" relating to a rotary, is it measured the displacement between the intake "stroke" and compression? where the volume is the greatest then that times the total number of faces? or the total volume of a rotor housing- the rotor times the number of rotors?
Old 01-09-10 | 10:00 AM
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i assume it is the first option measuring each face, since piston motors measure each cylinder
Old 01-09-10 | 12:04 PM
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You will find a definite split on this question with insightful thoughts on both sides.

My thought is that the displacement is equal to the full amount air that can be inhaled into the cylinder or rotor housing at bottom dead center. This would be considered 100% efficiency.

There are sanctioning bodies that handicap rotaries with anything from 1.7 to twice its displacement to level the field for competition.

Quote: “so my question "How is displacement measured?" relating to a rotary, is it measured the displacement between the intake "stroke" and compression? where the volume is the greatest then that times the total number of faces? or the total volume of a rotor housing- the rotor times the number of rotors?”

Your thought of doubling the displacement considering the exhaust side displacement would be like doubling a 4 stroke engine’s size by considering the second set of cycles (3&4) as part of its displacement.

My Thoughts,
Barry
Old 01-09-10 | 12:58 PM
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From: morton, il
but where does mazda get 1.3L (13B)?

Does mazda measure the displacement of a face at "bottom dead center" x 3 faces x 2 rotors?



or each face (3) with 1 being at "bottom dead center" x 2 rotors

Old 01-09-10 | 02:45 PM
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Old 01-09-10 | 03:32 PM
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Originally Posted by superdan50
but where does mazda get 1.3L (13B)?

Does mazda measure the displacement of a face at "bottom dead center" x 3 faces x 2 rotors?



or each face (3) with 1 being at "bottom dead center" x 2 rotors

#1= 654cc times two rotors, or 654cc X 2= 1308cc

Barry
Old 01-09-10 | 08:33 PM
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so the #1 in the 2nd illustration = 218cc, correct?
Old 01-09-10 | 09:06 PM
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some racing groups use the term SWEPT VOLUME, which for measuring they fill a space with a liquid, and calculate the difference between smallest to largest volume. in cubic centermeters
makes some sense.

but it raises a lot of argument, depending who you talk with.

also what is the volumetric effiency of a rotary, another diffecult question.
Old 01-10-10 | 08:36 AM
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Originally Posted by superdan50
so the #1 in the 2nd illustration = 218cc, correct?
No, it is 654cc.

Take some clay and seal the intake port in that #1 of the 2nd illustration, then fill the space with water........ 654cc !

Barry
Old 01-10-10 | 08:52 AM
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Originally Posted by superdan50
but where does mazda get 1.3L (13B)?

Does mazda measure the displacement of a face at "bottom dead center" x 3 faces x 2 rotors?



or each face (3) with 1 being at "bottom dead center" x 2 rotors

What you marked as "1" is actually chamber at its full displacement-geometric difference between TDC and BDC and in case of 13B, its 654ccm.
Old 01-10-10 | 09:27 AM
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Originally Posted by Barry Bordes
There are sanctioning bodies that handicap rotaries with anything from 1.7 to twice its displacement to level the field for competition. Barry
Handicap?? We can be happy that certain sanctioning bodies are allowing compete rotaries on uneven playing field. At ALMS when BK motorsport was campaigning Courage C65 with 20B, it was in class with N/A limit of 3,4 liters-as we know, 20B is equal to 3,9 and even was allowed to run larger inlet restrictor. Same in drag racing, rotaries are allowed with larger turbochargers... Nice example where sanctioning bodies take into account inefficiencies of wankel engine was R26B. Breathing ability like 5,2 liter but classed as 4708ccm. Its that easy, wankel engine will always have about 10-15% higher specific air consumption, so from equal capacity, assuming same MAX VE%, it will have 10-15% lower BMEP-torque and in the same time, assuming same AFR, 10-15% higher BSFC.
Old 01-11-10 | 08:13 AM
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Originally Posted by Liborek
Handicap?? We can be happy that certain sanctioning bodies are allowing compete rotaries on uneven playing field. At ALMS when BK motorsport was campaigning Courage C65 with 20B, it was in class with N/A limit of 3,4 liters-as we know, 20B is equal to 3,9 and even was allowed to run larger inlet restrictor. Same in drag racing, rotaries are allowed with larger turbochargers... Nice example where sanctioning bodies take into account inefficiencies of wankel engine was R26B. Breathing ability like 5,2 liter but classed as 4708ccm. Its that easy, wankel engine will always have about 10-15% higher specific air consumption, so from equal capacity, assuming same MAX VE%, it will have 10-15% lower BMEP-torque and in the same time, assuming same AFR, 10-15% higher BSFC.
Liborek, All true but;
13B = 1300cc
20B = 2000cc
R26B = 2600cc
A fair question would be, what size would Felix Wankel or Kenichi Yamamoto say that it is?
Barry
Old 01-11-10 | 10:51 AM
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Originally Posted by Liborek
What you marked as "1" is actually chamber at its full displacement-geometric difference between TDC and BDC and in case of 13B, its 654ccm.
so if #1 (equivalent of bottom dead center) is 654ccm shouldn't the 13b be a 39b?

because (correct me if im wrong) the displacement of a piston motor is measured as if all the pistons were at BDC at the same time (obviously not possible if the engine is working).

and realistically speaking each face is the equivalent of a piston (each face has a intake, compression, combustion, and exhaust "stroke).

so since a piston is measured with all pistons at BDC (unrealistic) then shouldnt a rotary be measured as if all the faces were at BDC (also unrealistic)?
Old 01-11-10 | 12:19 PM
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Originally Posted by superdan50
so if #1 (equivalent of bottom dead center) is 654ccm shouldn't the 13b be a 39b?

because (correct me if im wrong) the displacement of a piston motor is measured as if all the pistons were at BDC at the same time (obviously not possible if the engine is working).

and realistically speaking each face is the equivalent of a piston (each face has a intake, compression, combustion, and exhaust "stroke).

so since a piston is measured with all pistons at BDC (unrealistic) then shouldnt a rotary be measured as if all the faces were at BDC (also unrealistic)?
We could say that full engine displacement is when all pistons/rotor faces do the work, so you are right. Again in RE by Kenichi Yamamoto it is nicely described and clearly stated, that wankel rotary working cycle is finished after 1080°-in case of 13B, 3924ccm.

But if we want to compare, its like this:
1,3 2-stroke-all work done in 1 revolution
2,6 4-stroke-all work done in 2 revolutions
3,9 wankel cycle(13B)-all work done in 3 revolutions

notice-same work has been done...
Old 01-11-10 | 12:35 PM
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http://rx7.com/techarticles_displacement.html
Old 01-11-10 | 05:02 PM
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so the common handicap given to 13b giving it a displacement of 2.6l is based off the piston motor offering 1 full cycle in 720 deg. of crank rotation.

i feel that this common handicap is biased and doesnt "translate" well. yes in it takes 720 deg for 1 4stroke cycle, this is enough for each piston to complete an up/down trip twice. so shouldnt a 2L piston motor be 4L (2L x up/down 2x) of displacement?
Old 01-11-10 | 05:06 PM
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the most accurate solution would be if the displacement was 3.9L. o well, i guess i didnt fall in love with the rotary b/c of simplicity.
Old 01-12-10 | 12:27 AM
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A little off topic, since current formula one engines are N/A 2.4L v8's and are rev limited to 18,000 rpm and make 670+hp all the while having a MPG of two to four mile's....so technically there real displacement would be a lot higher??
Old 01-12-10 | 01:15 AM
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how do you figure? displacement has nothing to do with rev limits, mpg, or hp (at least in the context of your question). (granted displacement, revs, compression are direct factors of hp and effect mpg.)
Old 01-12-10 | 08:36 AM
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Originally Posted by superdan50
so the common handicap given to 13b giving it a displacement of 2.6l is based off the piston motor offering 1 full cycle in 720 deg. of crank rotation.

i feel that this common handicap is biased and doesnt "translate" well. yes in it takes 720 deg for 1 4stroke cycle, this is enough for each piston to complete an up/down trip twice. so shouldnt a 2L piston motor be 4L (2L x up/down 2x) of displacement?
Its not handicap, its equivalence.

As I said, full engine displacement is determined when all parts of the engine do the work. 2-stroke do it in 1 rev, 4-stroke in 2 revs and wankel rotary in 3 revolutions. This is clear equivalence, nothing is biased.


.....so shouldnt a 2L piston motor be 4L......
no, because in these 4 strokes, only one is working...
Old 01-12-10 | 04:27 PM
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Originally Posted by superdan50
how do you figure? displacement has nothing to do with rev limits, mpg, or hp (at least in the context of your question). (granted displacement, revs, compression are direct factors of hp and effect mpg.)
Maybe my question is too simplified but still pretty straight forward..

Here's an example,let's say im racing and the racing body rule is 2.4L of displacement only and they are no rules regarding engine development.

The poor team has a honda N/A 2.4L I4 K24 engine making 380ish hp @9,500 rpm.

The rich team has a honda N/A 2.4L F1 style V8 engine making 670+hp @18,000 rpm.


Witch team you think is gonna win???
Old 01-13-10 | 11:55 AM
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Change your thread's title from...

"How is displacement measured?"

to

What do people think a rotary engine's displacement should be in relation to a 4 cycle engine?

Barry
Old 01-13-10 | 02:39 PM
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Originally Posted by Liborek
Its not handicap, its equivalence.

As I said, full engine displacement is determined when all parts of the engine do the work. 2-stroke do it in 1 rev, 4-stroke in 2 revs and wankel rotary in 3 revolutions. This is clear equivalence, nothing is biased.


.....so shouldnt a 2L piston motor be 4L......
no, because in these 4 strokes, only one is working...
so a 2L 2 stroke is the equivalent to a 4L.

Originally Posted by Chuck Norris FB
Maybe my question is too simplified but still pretty straight forward..

Here's an example,let's say im racing and the racing body rule is 2.4L of displacement only and they are no rules regarding engine development.

The poor team has a honda N/A 2.4L I4 K24 engine making 380ish hp @9,500 rpm.

The rich team has a honda N/A 2.4L F1 style V8 engine making 670+hp @18,000 rpm.


Witch team you think is gonna win???
i am sure there are rules in place for such a situation. how ever the answer to your question is in your question, $$$. the cost of a 2.4L v8 over that of a K24 eliminates the 2 competing. if you had a Ferrari would you race in a crowd of civics? or vice versa?

do you expect to see a F1 car at your local track day?

Originally Posted by Barry Bordes
Change your thread's title from...

"How is displacement measured?"

to

What do people think a rotary engine's displacement should be in relation to a 4 cycle engine?

Barry
initially it was about the measurement.
Old 01-13-10 | 07:22 PM
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If I may, here are my 2 c.

Long post, apologies and you don't have to read it...

The displacement, as some have underlined, is a different thing from attempting to quantify equivalence for engines that either have similar architecture with different cycles (2 and 4 stroke piston engines) or for engines that have an altogether different architecture (like a rotary).
The displacement of an engine is the total size of the combustion chambers, independently from how long it takes for firing cycles to happen or from how quickly you can reload and re-fire.
The measurement, therefore, is pretty straight forward. The 13B has 2 650 ccs combustion chambers, hence its size is 1.3. If you were to get two identical piston engines (identical cylinder bore, piston size and stroke) one 2 stroke and another 4 stroke, their displacement would be identical, so if one is 2 L, so is the other.

Different thing is to attempt to compare 2 engines and establish what the size of each should be for them to be considered comparable.
There are several issues with this, in fact, while it can be said that a 2 stroke is capable of twice as many power strokes over 2 revolutions as its hypothetical 4 stroke counterpart, the 2 stroke cycle engine architecture is such that its efficiency is not as high as the one of a 4 stroke due to the lack of valves and therefore tight seal/separation of intake/exhaust gasses.

Although the rotary is a different type of engine and perhaps the "sealing" of the gasses is not as much of an issue as it it might be in the 2 stroke piston engine, there are design characteristics that don't allow to say that its 4 stroke piston counterpart is is simply a double-sized piston engine.

This type of discussion is primarily academic, since each type of engine has peculiar characteristics that make it preferable for certain things and not so preferable for others. The bottom line being, if you are on the street and you are only worried about what your fuel efficiency or power or torque or whatever are, then it really doesn't matter what you drive.
If your needs are those of a more refined palate, you might perhaps say that you would rather have a sophisticated multi-valve, variable valve timing, super power/displacement, exotic materials built engine, which moves the conversation to a different area.

In racing, there are several ways of "estimating equality". As far as I am concerned, the only really good and interesting one is: here's the type of engine you use (piston, 2 stroke, or piston, 4 stroke, or rotary) and here's the displacement you're allowed. The rest is up to you.
This obviously has good and bad sides. It is exclusive, since it does not allow you to use any different type of engine, but it is honest and it keeps the heart of the competition in the engineering field (you can use all your creativity to make more power out of your 2, 3, 4, 5 L that the governing body has given you). It certainly has some downsides, in fact, this system fosters research (which is actually good), but it also promotes unbelievable expenses, because the potential for development is only limited by the amount of money you can pour into your program. I believe that ALMS has an interesting (however arguable) take on this matter: they tell you how much HP you can have in a given category and that pretty much sums it all. How you get it, is up to you. This is very good in curbing spending, however, it does have the opposite effect of displacement limitation, as it becomes utterly useless to develop the engine, all you need to do is to get some more displacement and away you go.

All this ranting simply to state what follows: it is difficult to compare engines with different architectures when it comes to have them pitted against each other in a competition. This is because their individual characteristics are not just and only about what their comparable displacements are, but rather the way they deliver the "goods", and also their ability to withstand certain types of stresses at a certain level of reliability. It doesn't make sense, for instance, to attempt to compare a rotary to a piston engine simply by balancing their displacements (whichever way this would be theoretically acceptable), because of their different power delivery and therefore, their superior adaptability to different uses and different situations within one use. For instance, in races where the engine is running at high rpms constantly, a rotary would probably be your friend over a piston engine, not only because of the nature of its power band, but also because of its superior reliability under those circumstances. When you would be exiting slow corners with the "wrong" gear, a piston engine would rather be your friend, because of its different power delivery. This is true even within one type of engine. It is an old story that F1 cars have had engines that have ranged from v8 to v12. What's best? Well, it all depends on where you race. At Monza, you'll be better off with a 12, hands down, but at Monaco... well, not quite the same story.
Which brings me to the original point, that is: there are formulas to compare different engines, however, due to their unique characteristics, it is simply impossible to draw a correspondence chart.
Old 01-13-10 | 11:37 PM
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Originally Posted by superdan50
so a 2L 2 stroke is the equivalent to a 4L.



i am sure there are rules in place for such a situation. how ever the answer to your question is in your question, $$$. the cost of a 2.4L v8 over that of a K24 eliminates the 2 competing. if you had a Ferrari would you race in a crowd of civics? or vice versa?

do you expect to see a F1 car at your local track day?



initially it was about the measurement.


First i clearly stated on my previous post that it was an example and not a question. in witch you have the nerve to state a non-answer from my example that is completely unrelated from my first post. Although i get what you mean in terms of money and engine development but even that statement is invalid,have you seen honda civics or even our car's kicking massive *** against exotics on youtube or what not? it's ******* hilarious it's a massive up yours to high society...


(With all due respect)Let me restated my example again and hopefully you can understand this time.


Engine one: Honda N/A 2.4L I4 K24 engine with Max R.P.M @9,500 H.P @ 380+


Engine two: Honda N/A 2.4L I4 K24 engine with Max R.P.M @ 18,000 H.P @ 670+



Which engine has a higher displacement??

Let the flaming begin..



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