>
> Ok, I hate to sound dumb here, but this chart is way over my
> head. I'm
> sure it provides everything I need to know, but I don't know what
> questions to ask of the chart. To start, is there a problem
> with the
> displacement shown on the turbo spreadsheet? The spreadsheet
> says that
> the 13B displacement is 159.6 cubic inches. That sounds way to
> high. I
> looked up the displacement in "cubic centimeters" and converted
> it:
> 13B = 1308 cc = 79.788 cubic inches
> 20B = 1962 cc = 119.682 cubic inches
>
> I'm in Dallas and do most of my flying in 90-100 degree F. My
> field
> altitude is 600ft. I usually fly under 15k altitude. I just
> want to
> know how to select a turbo that will help me maintain 350 hp up
> to about
> 18k altitude on those hotter that hell Texas Summer days. Can
> anyone
> give me the "Cliff Notes" or "Turbos for Dummies" version
> Don
>
> No Don it is indeed 159 c.i. equivalent. The three rotor is 240
> c.i.
> roughly. That is how much air they pump in one rev at 100% VE. I
> am not sure which spread sheet you are looking at but it is
> probably
> Dave Leonard's. Can you help Don out Dave?
>
> You will also need that Turbochargers book Don. I hope you save
> messages :)
>
> Paul Lamar
>
>
> Eh yeah, Im not going there... that is too close to the old "is
> it a
> 4-piston, 6-piston, or 2-piston equivalent?" :-)
>
> Lets just use 159 c.i. and leave it at that :-)
>
> (ok, each chamber is used once per rev instead of once every
> other rev
> like in a typical piston engine... so double the equivalent
> volume)
>
> --
>
> David Leonard
>
> Turbo Rotary RV-6 N4VY
> N4VY.RotaryRoster.net
>
www.RotaryRoster.net
> leonardiniraq.blogspot.com
>
> So you went there after all :)
>
> Paul Lamar ...No rotor no motor.
>
>
> Okay,
>
> I'm going to chip in here as well. I use the actual
> displacement in
> my spreadsheet, 80 ci (1300cc) for the 2 rotor, 120 ci for the
> 3-rotor etc.
>
> Of course, with the rotary there is one combustion event per rotor
> for each 360 degrees rotation of the engine shaft, which is double
> that of a piston engine. So the effective displacement is double.
>
> Regards,
>
> Gordon.
>
> I am going to have to add a section to my web site on this subject.
> It keeps turning up like a bad penny :)
>
> Here is what Ansdale said about it and he did not obviously put it
> to bed back in 1969 :)
>
> Paul Lamar
Yeah, I was just struck by a moment of melancholy when I realized
that Vance wasn't the one spearheading the objections..... He's missed.
Chris Cosman
Yes. Vance died unconvinced.
Here is my latest Ansdale patch work on the issue.
--
Paul Lamar ...No rotor no motor.
Paul,
I don't understand what you are trying to say.
The actual physical displacement of the Wankel combustion chamber --
that is, the volume of any of the three "flanks" of the rotor -- is
what it is. In the Ansdale illustration that is 500 cc.
In a 13B engine it is 1300 cc for the two rotors or 650 cc per
rotor. That is the actual physical volume and that is the amount of
air-fuel charge that is combusted in each power stroke. Is this not so?
That is why the naming conventions of all Wankel engines that I have
seen are based on actual displacement.
For each complete revolution of the three-sided rotor, there are
three power strokes. Since the engine shaft turns three times for
every turn of the rotor, that means one power stroke per turn of
engine shaft.
That means the "equivalent" or "effective displacement is twice that
of a piston engine. But what actually happens is that each
combustion chamber of half the size has two power strokes, in the
same time frame as the piston engine has one power stroke of a
combustion chamber that is twice as big.
So how does it make it right to pretend that the Wankel combustion
chamber is twice as big? In order to do that we then also have to
pretend that a power stroke only occurs once every two turns of the
engine shaft.
To me that's like calling a 650 cc 2-stroke a 1300 cc engine.
What is wrong with this picture? I don't get it.
Regards,
Gordon.
Ah well you have touched on the solution.
The rotary is not a two stroke. It is a four stroke.
Lets turn it around. How much air would a 2.6 liter two cylinder
TWO STROKE engine pump in one revolution?
As far as HP is concerned the ONLY thing that matters is how much
air a four stroke engine pumps in ONE REVOLUTION at 100% VE.
Not the physical displacement of x number of chambers.
The definition of displacement, as far as engines are concerned,
is only valid in context with the type of stroke.
Here is what Ansdale has to say about that.
13B rotaries are raced in the 2.6 liter class world wide
for a good reason. The only reason they generate more power
than a 2.6 liter four cylinder engine is they turn faster
and breath better. Lately that is less true as the piston
engine RPM has been raised by detail development and the
use of high strength to weight ratio materials like titanium.
Also four valves per cylinder has improved the breathing.
It is very rare for any 2 valve cylinder to have a VE over one.
The RX8 engine is just over one and P-ports are typically
1.2.
Formula One (2.5 liter as I recall) V8 engines are now
turning 19,000 RPM. At least four valves per cylinder with
a very short stroke so larger valves can be used for a given
physical cylinder displacement. You will not find the details
of a competitive Formula One engine anywhere. Its a well
kept secret. However I would not be surprised if the VE's were
well over one. Applying the RPM difference to a 2.6L p-port rotary
making 340 HP at 10,000 RPM would make the Formula One engine
about 19,000/10,000 times 300 HP or 655 HP which is about
right for a Formula One engine.
Paul Lamar ...No rotor no motor.
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