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Hi Tom,
As I say, my comment isn't directed at the calculations at all. There
has been much discussion about what the displacement is, and how much of
the rotation to count. Mazda claiming 1.3 L for example.
I stand by my statement about the displacement. That seems to be
exactly the way Ed has used it.
Quote from Ed's post:
"Taken the rotor reference we have 6 faces of 40 CID displacement
in one revolution of the rotor. So at 6000 e shaft rpm (2000 rotor rpm)
we have
Air Flow (rotor Ref) = 6 * 40 * 2000/1728 = 277.77 CFM"
The 6 * 40 is the 240 cu. in. displacement, 2000 RPM, and 1728 to
convert to CFM. What am I missing?
Bob White
On Thu, 10 Feb 2005 20:33:43 -0800 (PST)
Tom <tomtugan@yahoo.com> wrote:
Sorry, Additonally, Bob you say "the rotary (uses it's displacement ) in 3
revolutions of the E-shaft or one revolution of the rotors." While
all of the displacement of all the rotors is disturbed during one
revolution of all of the rotors, Ed's work attempts to determine how
much combustion charge is consumed (factoring in displacement and
other factors) per revolution of the rotors and then determine CFM
based on consumed combustion charges and rpm. Apparently it isn't as
simple as what you said about 'uses it's displacement' Or so I think.
Tom
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