snip again

    The forum seems to be open again?
    Anyway, I know, a lot has been said about the displacement of the
    rotary.
    My comments refer to engine calculations, or engine math, if you will.
    Mep in particular.

    As I see it, EVERYTHING IS WRONG with calling the Mazda Rotary a 1.3
    litre engine. Here is why:

    First, it is a four cycle engine and after all, in comparison to an RPE
    two shaft rotations must be counted to be equivalent.
    I think it is not correct to say any four-cycle engine performs like a
    two-cycle engine of half its displacement, regardless of Rotary or RPE.

    Secondly, one measure of engine performance and of its internal loading
    is the 'mean effective pressure' (mep). This is the reference piston
    pressure, the average between maximum and minimum gas pressure during
    the power stroke. (Work is calculated by using the stroke as the
    distance travelled.)

    The hp of an engine is influenced only by three things: displacement,
    mep and engine rpm, while torque is the product of displacement and mep
    only. Even the ratio of stroke-to-piston diameter has no impact on
    torque. Most people would dispute that, however.

    In four-stroke engines the hp is calculated thus:
    Hp = displacement * rpm / 120 * mep. (for 2-strokes the divisor is 60).
    In metric power is in kW, displacement is in litre, rpm in 1/min, mep in
    MPa. (No other factor is required.)

    Using the imperial system the displacement is in cu-inch and the mep in
    psi. A factor of 6600 is needed, since 6600 inlb/s equals one hp. (550
    ftlbs/s = 1 hp)

    In numbers: hp = displacement (cuin) * mep (psi) * n (rpm) / 792000.
    (6600 * 120 = 792000)
    mep in psi = hp * 792000 / ( displacement (cuin) * n (rpm) )

    Example, 2.6 litre, 4 cycle RPE, generating about 180 hp at 6000 rpm.
    (The number of cylinders is irrelevant). 2.6 litre equals ~ 160 cuin.

    Mep = 180 hp * 792000 / (160 * 6000) = ~ 150 psi. If 200 hp are
    generated, the mep is 165 psi.

    Today?s car engine average around 150 to 170 psi. 20 years ago it was
    about 10 to 30 psi lower. With charging, mep can be pushed up to as much
    as 200 psi. F1 engine are in the 230 psi range. Industrial engine are
    near 90 to 100 psi, even Diesels. However, they last ?forever? having
    10,000 hrs in life or more is not at all unusual. On the other hand F1
    engine with ~ 235 psi last for about 5 hours, that? all. However, their
    rpm is continuously very high.

    The mep is a measure of power output per displacement and also a good
    indicator of the life of the engine.

    When using 80 cuin or 1.3 litre displacement for the Mazda Rotary would
    result in near 300 psi mep. That number is unheard of for an ICE. That
    just can?t be right. The mep of the rotary is also near 150 psi in
    naturally aspirated use.

    Calculating with 2.6 litre (160 cuin) displacement, its mep is near 150
    psi, as it should be. Because of that, the Rotary can not be a 1.3 litre
    engine, but it must have a displacement of 2.6 litres.

    Of the three variables for power, mep is influenced mainly by the
    compression ratio, the porting and the resulting VE, as well as by the
    pressure of the intake charge, NA or charged.

    Regarding its displacement to overall size, the Rotary has an advantage
    there. An RPE of 2.6 litre displacement takes more space and is heavier.

    Rpm is limited in all engines by the inertia forces that are linear with
    the respective masses and proportional to the square of rpm. This is
    true not only for the RPE but also for the Rotary where there is no
    linear reciprocation, but a constant centrifugal force of rotor and
    eccentric. The direction of acceleration is pointing from the centre of
    the eccentric to centre of the eccentric shaft. A limit to the rpm
    applies to the Rotary just the same.

    Coming back to the displacement of the Rotary, calculating all three
    chambers would also be wrong as it takes three shaft revolutions for all
    three chambers to fire and that would also be incorrect as the speed is
    1.5 times higher compared to an RPE of 2.6 litres of same output. So the
    rotary is not a 3.9 litre engine either, it simply is an equivalent 2.6
    litre four-cycle engine. Two shaft rotations need be calculated. Period.

    On using 3.9 litre displacement its mep would also be incorrect at near
    100 psi which is lower than is being measured.

    What really counts is the displacement in volume per time at a certain
    engine rpm that must be compared. The Mazda Rotary compares to 2.6
    litres (160 cuin) displacement with regard to equivalent rpm to a four
    cycle RPE. The power output of the rotary also jives with that. Only 2.6
    litre can be right.

    Regards
    Rolf



    Rolf,

    The formula you used is designed specifically for piston engines.

    Hp = displacement * rpm / 120 * mep.

    The constant 120 is used because each cylinder has a power stroke
    only once for every TWO turns of the engine shaft. So because RPM is
    denominated in minutes, it is necessary to divide by 120 seconds to
    make the units come out right.

    If it was one power stroke per cylinder, per engine shaft turn -- as
    it is on the rotary, the formula would divide rpm by 60, the number
    of seconds in a minute.

    It's the same thing with calculations for volumetric flow. The
    rotary engine has one intake event for each engine shaft turn. (And
    one compression even, one combustion event and one exhaust event.
    The whole 4-stroke cycle happens in one turn of the engine shaft,
    for each rotor.)

    When you say the displacement of the engine is EQUAL to 2.6 liters,
    that is true, because it pumps the same amount of air per any given
    rpm as a piston engine of that size.

    However, that's like saying let's call  a Pound Sterling note a 2
    Dollar bill, because one Pound Sterling equals two Dollars.

    Does this make sense? Try walking into a store and buying something
    with a Pound note. You can try to convince the clerk that the two
    are completely identical and interchangeable...


    Here is what actually happens inside the rotary engine:

    Each intake event is 650 cc, not a molecule more. Each compression
    event is 650 cc (before compression), not a molecule more. Each
    combustion event is 650 cc (after expansion), not a molecule more.
    Each exhaust event is 650 cc not a molecule more.

    This is the simple fact. You and Paul and anyone who does not
    understand this can call it by its equivalent displacement.

    But people who are concerned with what is really going on inside the
    engine, not the EQUIVALENT compared to a piston engine, will stick
    to the REAL displacement.

    This is the real physical displacement of each rotor flank and it is
    650 cc. Period End of Story.


    Regards,


    Gordon.

    PS: I notice Jeff Spitzer's turbo spreadsheet also uses the real and
    correct displacement of 1300 cc.

    It is a ismple matter to switch between the two for calculation
    purposes and the result is exaclty the same. Both methods will
    calcualte absolutely correctly.

    This whole discussion is a tempest in a teacup. 1300 cc is the real
    displacement and 2600 cc is the equivalent displacment, when
    comparing ot a piston engine.



Jeff just said it is 2.6 L equavalent. He merely used 1300 cc to identify the engine.
You can't  have it both ways. If it is 1300 CC it won't pump 35 pounds
of air per minute. If it is not 35 pounds a minute it will not develop
the HP. This is a factual chart. Why don't you give it a rest for awhile.
I am sure it will dawn on you sooner or later.

Paul Lamar ...No rotor no motor.

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