Mailing List flyrotary@lancaironline.net Message #16881
From: Ed Anderson <eanderson@carolina.rr.com>
Subject: Re: [FlyRotary] Re: : Same HP = Same Air Mass <> same air Velocity II [FlyRotary] Re: Ellison, the missing piece
Date: Thu, 10 Feb 2005 10:12:00 -0500
To: Rotary motors in aircraft <flyrotary@lancaironline.net>
Tom,
 
I don't really think anyone can accurately make a generalization like that.  Certainly a PP 13B (such as power sports) which reportedly  produces 215 HP produces more power than a carb 360 Lycoming.  A street ported well tuned 13B will certainly produce 180HP.  But, given any two specific engines and depending on how well their induction/exhaust, etc. are set up, you could have one or the other producing the greater HP.
 
But, if someone has specific data that shows the 360 produces more power, then I would certainly like to know about it.
 
Ed A
----- Original Message -----
From: Tom
Sent: Thursday, February 10, 2005 9:14 AM
Subject: [FlyRotary] Re: : Same HP = Same Air Mass <> same air Velocity II [FlyRotary] Re: Ellison, the missing piece

It's my understanding that NA non-renesis rotary installations produce less power than 360s, Perry Mick might have a word on this.


Eric Ruttan <ericruttan@chartermi.net> wrote:
Warning top poster, who cuts the post size down.

A hopothises for your examination.

A 360 Lyc does not produce the same power as a rotary.

If true, then the Ellison card may not get enough air.

If not true, then there is no real reason why the Ellison cannot feed a
rotary.

Ed, I understand your math, but even if the local inlet velocity is much
higher, we dont care. the velocities adverage out to the same, as the
volume of air = velocity * carb area.

If the velocities are higher, the rotary consumes more air, and makes more
power.

Eric

----- Original Message -----
From: "Ed Anderson"
To: "Rotary motors in aircraft"
Sent: Thursday, February 10, 2005 8:31 AM
Subject: [FlyRotary] : Same HP = Same Air Mass <> same air Velocity II
[FlyRotary] Re: Ellison, the missing piece


Good question, Tom.

That interpretation did occur to me. I think the answer depends on your
assumptions, IF using commonly accepted formulas for calculating air flow vs
rpm and displacement (and considering both are positive displacement
pumps) - then the 360 CID lycoming turning 2800 rpm and the rotors in the
rotary turning 2100 rpm (6300 rpm E shaft) ingest the same total quantity of
air in one minute - approx 291 CFM. In comparing the two engines, its
accepted that you compare them over the standard 720deg 4 stroke cycle -
that means that 4 of the rotary faces have gone through their cycle in the
same 720 deg of rotation.

But, assuming the formulas are correct, then they both end up with the same
amount of air in the engine to create the same HP. I think my math is
correct on the smaller/unit displacement and longer period of rotation for
the rotary for the same intake of air. However, in both cases the air flow
is pulsating and pulsating differently. So if the total displacement for
the rotary over that 720 deg is less than the Lycoming and the time it takes
to complete that rotation is slower AND you still ingest the same amount of
total Air then the only way I can see that happening is the velocity of the
air in the rotary's intake has to be considerably higher than in the
Lycoming.

The only other alternative answer I see if that the commonly accepted
formula for comparing the rotary to the reciprocating 4 stroke is incorrect
(I got beat about the head mercilessly by a number of respected rotary
experts challenging that formula , so I wont' go there again (at least not
now {:>)).

Air Flow = Total Displacement * RPM/(2 - accounting for only every other
cylinder sucking on each rev * 1728 (conversion of cubic inches to cubic
feet) = TD*RPM/(2*1728)

For the 360 CID Lycoming at 2800 rpm, Air Flow = 360*2800/(2*1728) = 291.66
CFM

Using the commonly accepted notion that a rotary is equivalent to a 160 CID
4 stroke reciprocating engine because of the 4 faces of 40 CID that complete
there cycle in 720 deg.

For the 160 CID Rotary at 6000 rpm, Air Flow = 160 * 6300/(2*1728) = 291.66
CFM

So if both ingest the 291 CFM and the rotary has less total displacement
(over 720 deg) then disregarding any of my math on rotation period
differences you still have to account for why the rotary can ingest the same
amount of air with less displacement. (Now I must admit I have my
suspicions about the commonly accepted (racing approved) formula for the
rotary. However, if my suspicions about the rotary formula are correct, it
would make the rotary even more efficient at ingesting air - so I won't go
there {:>)).

If my logic and calculations are correct then this implies the Ve of the
rotary is considerably better than the Lycoming and is great than 100%. I
mentioned a few of the reasons why the Ve of the rotary may indeed be better
in the previous message.

Now, its possible that the stories about the Ellison not working well on the
rotary is just that - a story OR there could be a plausible physical reason
as I have poorly attempted to present.


Ed


----- Original Message -----
From: Tom
To: Rotary motors in aircraft
Sent: Wednesday, February 09, 2005 11:54 PM
Subject: [FlyRotary] Re: Same HP = Same Air Mass <> same air Velocity
[FlyRotary] Re: Ellison, the missing piece


Ed,

>The rotary has 40 CID displacement per face and 2 facesx 2 rotors = 4*40
or 160 CID for one rev. So the rotary has 22% less displacement per
revolution and the longer rotation period.<

and

>So if the rotary has less displacement of the sucking component and must
take 25% longer for each revolution. Therefore the only way it can obtain
an equal amount of air is for the intake air to have a higher velocity than
the Lycoming does.<

Isn't 'displacement' equal to the amount of air needing to be ingested?
So 22% less displacement equates to 22% less air and the rotarys longer
rotation period gives it more time for air to push in? And then the
intake air velocity should be lower?



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