flyrotary@lancaironline.net Mailing List Archive

Thank you; Leon.

That’s how I had always looked at it; but with all the math spinning about, I thought maybe I had overlooked something.

Subject: [FlyRotary] Last Word on Displacement was Re: [FlyRotary] Solved!!!! => Rotary AirFlow Equation? Help?

Hi Guys,

I have that feeling of "de ja vue" I've been here before. Here's how to understand it all without a million pages of marthematics.

The face of a rotary piston presents itself to the inlet port each revolution. (forget about the fact that the rotor has 3 faces - it only confuses the issue!!). Each piston of a 4 stroke piston engine go through an "Otto Cycle" every TWO revolutions.

With a rotary, you get ONE suck, per ROTOR, per REV. With a 4 stroke, you get ONE suck per PISTON, per (get this) TWO revs!!!

The same goes for Power Pulses. A rotary gets ONE power pulse per REV, per ROTOR. With a 4 stroke piston engine, you get ONE power pulse, per PISTON, per TWO revs.

1. The "SWEPT VOLUME" of a 6.5B/13B/20B rotary is 654 cc per rotor.

2. For EACH revolution of the E-Shaft, EACH rotor sucks 654 cc.

3. So, for a 6.5B, it sucks 654 cc, for a 13B, it sucks 1308 cc PER REV, and for a 20B it sucks 1962 cc.

I'll leave you to do the conversion into Cu.In.

Now, a 4 stroke engine completes it's Otto Cycle in TWO revolutions. So to equate a rotary to a 4 stroke "Otto Cycle" piston engine, we must compare how much a rotary sucks in TWO revolutions.

According to my limited mathematical ability, (I could never "run the numbers" with the alacrity of the blokes on ACRE! (}:>)),

A 6.5B is equivalent to a 1308 cc 4 stroke piston (Otto Cycle) engine. (two times 654 cc per rev)

A 13B is equivalent to a 2616 cc 4 stroke piston (Otto Cycle) engine. (two times 1308 cc per rev)

A 20B is equivalent to a 3924 cc 4 stroke piston (Otto Cycle) engine. (two times 1962 cc per rev)

It is THAT simple. Compare Granny Smith Apples with Granny Smith Apples (not mangoes or bananas!).

Now there are a couple of caveats.

1. The rotary, when ported, has the ability to easily run 125% to 130% volumetric efficiency. So it has a slightly higher ability to draw air than a piston engine. Runners, throttle bodies, and injectors must be sized accordingly.

2. This being so, it will put out slightly (up to 20%) more power than the equivalent swept volume of a 4 cycle piston engine of the same SWEPT capacity.

So there you have it. It's THAT simple!

Enjoy!

Leon

----- Original Message -----

From: Ed Anderson

To: Rotary motors in aircraft

Sent: Friday, February 11, 2005 8:18 AM

Subject: [FlyRotary] Solved!!!! => Rotary AirFlow Equation? Help?

> Ed, I think you've blinded yourself with science 8*)
>
> Each face of the rotor will cycle 40cid of air on each revolution. You
> can ignore the eshaft. Every time the rotor turns around, each face
> will have processed 40cid. If a rotor is turning at 2000 RPM, then three
> faces will each cycle 2000*40cid of air. You bring in referencing off
> the eshaft, and all of a sudden only 4 faces are processing air. I
> think that is the mistake.
>
> You reference 720 degrees of eshaft rotation. But all that means is
> that the rotors haven't completed their cycle. You're just ignoring
> every third rotor face.
>
>

Ernest, I think you are making my very point.

1. Each rotor has 3*40 = 120 cid of volume displace per 360 deg rotation and with two rotor that is 2*120 = 240 CID each revolution.

so we have 2000*240 cid/1728 = 277.77 CFM exactly what quantity the formula gives when treating the rotary as a 160 CID 4 stroke engine with e shaft at 6000 rpm - but that is for a FULL 360 deg rotation of the rotor. See part 2 next {:>).

2. Example

160*6000/(2*1728) = 277.77 CFM which agrees with the above but if the 160 CID figure is correct then that means that 160/4 = 4 , only 4 rotor faces worth of displacement have been considered in the formula. Which if that is correct then the rotor have only turned 240 degs and not 360 deg. But if the rotor has only turned 240 deg (240*3 = 720Deg Checks!) then that means that to make a valid comparison with 1 above, we need to reduce the amount of rotation in statement 1 to 240 deg instead of 360 which means that we only have 277.77*4/6 = 185 CFM which leaves me back where I started.

There is no question that if the two rotors have completed their 360 deg rotation - then all six faces have completed the cycle.

That gives 40*6*2000/1728 = 277.77 CFM, so lets not debate that point - we both agree! .

My problem has been the use of the 160 CID equivalent in the formula. I understand how it was arrived at ( I believe) its simply the amount of displacement occurring in the rotary at 720 deg of e shaft rotation which is the standard of comparison with other engines.

At 720 deg of e shaft rotation then indeed 160 CID of displace has occurred.  If I calculate using the 240 deg of rotor rotation I again get 240/360 * 6 = 4 rotor faces. But if that is indeed the actually amount of displacement of the rotor then according to our calculations in 1 above we need to use 240 deg not 360 and that gives 185 cfm not 277.77.

I mean I don't care if we use 240 CID (the total displacement across 1080) or 160 CID (the total displacement across 720 deg - the standard for comparison) but there should be some logical consistency between the two.

WAIT! Stop the presses I Finally Understand! The key is the 240 CID total displacement for 360 rotor degrees or for 1080 deg of e shaft rotation.

It was not a math problem or logic problem, my problem was a reference transformation problem!

             Using the rotor representation we have Air flow = 240 * 2000/1728 = 277.77 this is rotary based on 360 deg of rotation. Now if we  take that formula (which I think we agree on), I then want to convert it into a formula that conforms to the 720 deg standard. So I can compare apples and apples.

Here's what happens.

1. First the e shaft rpm is the e shaft rpm and is the standard reference point for rpm of engines. So we can simply multiply our 2000 rpm rotor rpm by 3 to reference it to the eshaft. So 3* 2000 = 6000. That "transformation" gives us the rpm figure in the "Standard" equation.

2.  However, to reference the 240 CID displacement to the 720 deg standard we must use the ratio of the transformation from 1080 to the 720 reference system. Because the amount of displacement IS affected by the choice of reference. This gives 720/1080 * 240 = 160 CID referenced to the standard.

So now the formula should migrate from rotor reference of 240*2000/1728 = 277.77 to e shaft reference of 160*6000/1728 = 277.77 by reference system transformation. To do that:

1.   We multiply the rotor rpm by 3 - the rpm does not care whether its 720 or 1080 referenced, one revolution/min is one revolution/min.

2.   The 720 deg standard DOES affect the amount of displacement so 720/1080 *240 = 160 CID.

Taking the rotor reference formula above we can directly transform it to the e shaft reference standard. Doing so gives us



     [720/1080*240]*[3*2000]/1728 = [0.666*240]*[6000]/1729, taking the first factor 0.6666* 240 = 159.9999 = 160 and we have

Airflow = 160CID*6000/1728 = 277.77 So the rotor reference is transformed to the e shaft reference of 720 deg.

Thanks all for helping me out of my problem area. I just knew there had to be a connection someway. If this is not correct - please refrain from informing me {:>).

Be heartened, Ernest, I have been faithfully using the 160 formula - just wanted to understand what appeared to me to be a difference when looking at it from two different perspectives. If the "reference" transformation is applied to the rotor equation then it translates into the 720 reference state.

Best Regards and thanks for your input

Ed