# Rotary engines & vehicles using them

## Quasiturbine – WANKEL THEORITICAL DEFICIENCY

Bonjour,

The texte below is a clarification about the
WANKEL THEORITICAL SWEPT VOLUME DEFICIENCY
which has been added to the page
http://quasiturbine.promci.qc.ca/QTpasWankel.html

Meilleures salutations, Gilles
www.quasiturbine.com
*******************

WANKEL THEORITICAL SWEPT VOLUME DEFICIENCY

For the piston, it is obvious that the relaxation volume is generated
by the movement of the piston surface (because it is a one dimensional
engine : piston axis only),
which guaranty that the effect of applied pressure
opens an extra volume rigorously equal to the gas expansion volume.
However, this is not generally the case with other engines, and specially
with rotaries
(which are 2 dimensional engines : radially and tangentially, where it is
needed to defined 2 different volumes :
– one being the volume generated by the movement of the pushing tangential
surface.
– the other being the total observed volume.
Those two volumes are not generally equal.
For this reason, the Wankel total engine displacement volume does not give
the whole story about the engine.

WANKEL ANALOGIE 1 –
A PISTON WITH CYLINDER LATERAL CAVITIES

Lets imagine building a piston engine, and making thousands of deep cavities
every where in the cylinder wall,
such that the total volume of all cavities be twice the volume of the
cylinder.
This engine will run, but as the piston falls, hot gas will go into the
accessible cavities,
and there will be more and more cavities accessible as the piston gets near
The final volume will be 3 times the cylinder volume (2 will be the cavities
volume, 1 will be the cylinder itself).
Of course, the pressure during relaxation will fall due to this excess
volume cavities,
and the efficiency will not be good. That exactly what append in a Wankel !

This is not about chamber surface here, it is about an excess of volume.
The real volume measured at the end of the relaxation ìs three times the
volume
generated by the movement of the tangential surface of push !
It is like if the total piston volume at the end of relaxation
would be 3 time greater than the actual cylinder volume !

WANKEL ANALOGIE 2 –
A PISTON WITH CYLINDER HEAD MOVING UP !

Let looks at the Wankel excess swept volume intuitively.
Consider just one of the Walkel rotor surface, horizontal when at TDC and
vertical when at BTC.
By analogy to a piston, measured the distance (radius) from the engine
center
to this rotor surface when horizontally (TDC) and then when vertically
(BTC).
The difference correspond to the crankshaft drop (stroke) of the "piston
equivalent surface",
in relation to the engine center.
In piston engine, the "cylinder head" is fix and does not move during the
piston stroke.
Now, observe the distance from the engine center
to the contour wall vertically and horizontally just over those rotor
surfaces :
The "cylinder head equivalent" is closer inward when at TDC and further
outward when at BDC ?
During the 90 rotation, not only the "piston equivalent surface" of the
rotor has moved inward,
but the "cylinder head equivalent" contour wall has moved outward.
Worse, the cylinder head did move twice as much outward
compared to the rotor surface inward displacement !
It is also important to notice that the "piston equivalent surface"
component dropping (stroke) inward in the direction of the engine center
is at no time producing any contribution to the rotational force,
component is useless in the process,
even detrimental in making excess swept volume.
(the Quasitubine has no crankshaft and no inward net displacement).

WHAT DOES THAT MEANS ?

These two very simple intuitive look at the Wankel helps understanding and
accepting
the fact that there are 2 swept volumes,
one inward (making useful work on the crankshaft) and a larger one outward
not only unproductive,
but destroying the efficiency by reducing the internal pressure.
In the Wankel, the useful swept volume is not the actual chamber volume,
which make the Pressure-Volume P-V Diagram complex to use correctly !

Where does the lost of energy go ?
As explained in detail at http://quasiturbine.promci.qc.ca/QTpasWankel.html
compressing a gas increases its temperature, and allowing it to relax
(either by allowing the piston to move, or by allowing new cavities in the
chamber,
or by creating a geometric excess swept volume), the gas will cool down…
That is exactly what is happening in the Wankel,
where the excess swept volume lead to an excess cooling, and a poor cold
combustion !

In the Quasiturbine, there is no crankshaft,
and therefore no radial drop (stroke) of the equivalent ‘piston surface’,
and the radial distances of those blade surfaces to the centre are kept
unchanged from TDC and BDC.
The "cylinder head equivalent" moves outward by the exact amount of
"tangential surface of push" swept volume, which guaranty the Quasiturbine
is as efficient as the piston…
The Quasiturbine is very different from the Wankel !

HOW TO FIX THE WANKEL ?

Impossible, except by moving to the Quasiturbine concept, where this excess
swept volume is none.
The Quasiturbine does in fact correct the Wankel theoretical swept volume
deficiency,
and consequently, does not have its limitations…

SHOW ME THE WANKEL TANGENTIAL PUSHING SURFACE

The rotational force does not come from the volume but from the pressure
acting on the tangential surface.
Consider one pressurized Wankel chamber and find out the force generating
the rotation ?
This force is equal to the (pressure) time the (difference in radius of the
two contour seals contact),
time the (thickness of the rotor).

Draw the Wankel contour on a board, and cut the triangular rotor,
turn it slowly by pointing on the contour the difference between two
(this is proportional to the tangential surface of push).
Plot this difference of radius of the seals for different angle from TDC to
BDC
(ideally plot them on the radius, inward starting from the contour,
and at the chamber front seal for each rotor position).
These differential radius are proportional to the surfaces on which
the pressure is exercising useful work at each instant, and the movement of
this surface is
equivalent to the surface movement of the piston, so it does generate a
swept volume.

You find that the tangential surface evolution define a volume which is 1/3
of the total volume…
There is the internal lost in pressure in this two dimensional engine !

WHAT ABOUT THE ACTUAL ROTOR SURFACE SWEPT VOLUME ?

Are you more comfortable considering the rotor surface and the total off
center load on that rotor surface ?
You can precede that way as well.
In rotary engine there are generally 2 surfaces sweeping volume,
one is the rotor surface projected in the radial direction (unproductive
energy sweeping)
and the other is the tangential surface of push (the only useful tangential
pushing surface).
Failure to distinguish both component of the generating swept surfaces is a
common practice,
but propagate miss-comprehension of the Wankel.
Knowing at each time where the rotor surface is permit to calculate the
algebraic increment
of the surface swept volumes angular (radial and tangential) component for
each Wankel angle position,
and then make the total integration (addition).
Results are the same as the tangential surface of push radial method.

WANKEL ROTOR SWEPT VOLUME RESULTS

Starting with the same chamber maximum volume and with the same amount of
compress gas,
turning slowly both engine will show that the amount of work done
is much more with the piston and less with the Wankel.
This is due to the fact that the piston is a one dimensional engine (only
the piston axis affect the volume)
while the Walkel is a two dimensional engine (radius and tangential
coordinates both affect de volume)
that does not meet the Pressure-Volume P-V Diagram.
The Quasiturbine researchers did the calculation including the
correct Wankel swept volume integration and the results are shown in the
first graph at
http://quasiturbine.promci.qc.ca/QTpasWankel.html

The result show that the useful Wankel surface sweep volume
is only one third of the actual total chamber volume :
an extra geometric swept volume is not generating any power, but
catastrophically lowering the pressure…
which alone explain the poor efficiency of the Wankel !
No argument related to the shape of the combustion chamber explain correctly
the Wankel inefficiency,
but excess volume does… PV diagram correctly describes only engine
with no such excess volume, like the piston engines.
All Wankel PV diagram analysis are wrong… and high efficiency will never
be achieve…

This should help understand why in rotary engine, sweep volume are
generally different from the actual chamber volume. Sweeping volume need to
Better than showing the maths, we give 2 calculation methods for the swept
volumes.
Consequently, to properly analyze the Wankel,
it is important not to assume that its sweep volume is the actual internal
chamber volume.

COMPARING ENGINES CONCEPTS

The useful Wankel surface sweep volume is only one third of the actual total
chamber volume.
Redoing the same calculation with the Quasiturbine show that the QT swept
volume is
exactly equal to the actual chamber volume, because it has been a contour
selection criteria.
Piston show no excess swept volume either, which simply explain their
superiority…

Almost every one does assume the energy is coming from the swept volume,
which is correct for the piston and the Quasiturbine engine, but not for the
Wankel.
Consequently, only "engine types" without any relaxation excess volume
can be successfully compared and match in results on the basis of
displacement volume and efficiency,
which means that piston and Wankel will not match, while piston and
Quasiturbine will
(both have zero excess relaxation volume).
The shaft RPM is the most interesting comparison basis

.

# 24 Responses to “Quasiturbine – WANKEL THEORITICAL DEFICIENCY”

Hi Saint-Hilaire,

I have conclusively proven this to be wrong in another thread but you didn’t
look at the evidence I supplied and didn’t respond to me. The spreadsheet
showed this, but as far as I could tell you did not download it. Here it is
again http://www.mikesdriveway.com/misc/rotaryworkoverexpansion.xls. (if you
get asked for a password push cancel). Have a look at it and tell me which
step is in error. It is quite simple, here is a basic outline of how it
works:

It calculates the work output from a rotary and piston motor over an
expansion from min vol to max vol assuming that the pressure starts at 1
unit and reduces as the volume gets bigger
Column A is the crank angle.
B is the volume in the chamber at the corresponding crank angle
C is the pressure in the volume
E is the tangential force on the rotor from the pressure
D is the level length over which the force acts to produce torque
F is the torque output, calculated from multiplying E and D
G is the work produced, calculated from multiplying torque by the shaft
rotation.

All of these bits of work are added to get the total work. This is done for
the piston and rotary engine and the result is the exact same amount of work

The whole concept under which you are selling your engine is false. This is
not to say that there is anything wrong with your engine, it just means your
sales technique are wrong. Potential buyers will most likely have an
engineering background and see this also.

> – one being the volume generated by the movement of the pushing tangential
> surface.
> – the other being the total observed volume.

These are equal!!! I proved this at the end of the other document you
ignored. http://www.mikesdriveway.com/misc/rotor.doc.

Michael Culley

"Michael Culley" <m…@nospam.com> wrote in message

news:408b3f4c\$0\$442\$afc38c87@news.optusnet.com.au…

- Hide quoted text — Show quoted text -

> Hi Saint-Hilaire,

> I have conclusively proven this to be wrong in another thread but you
didn’t
> look at the evidence I supplied and didn’t respond to me. The spreadsheet
> showed this, but as far as I could tell you did not download it. Here it
is
> again http://www.mikesdriveway.com/misc/rotaryworkoverexpansion.xls. (if
you
> get asked for a password push cancel). Have a look at it and tell me which
> step is in error. It is quite simple, here is a basic outline of how it
> works:

> It calculates the work output from a rotary and piston motor over an
> expansion from min vol to max vol assuming that the pressure starts at 1
> unit and reduces as the volume gets bigger
> Column A is the crank angle.
> B is the volume in the chamber at the corresponding crank angle
> C is the pressure in the volume
> E is the tangential force on the rotor from the pressure
> D is the level length over which the force acts to produce torque
> F is the torque output, calculated from multiplying E and D
> G is the work produced, calculated from multiplying torque by the shaft
> rotation.

> All of these bits of work are added to get the total work. This is done
for
> the piston and rotary engine and the result is the exact same amount of
work

> The whole concept under which you are selling your engine is false. This
is
> not to say that there is anything wrong with your engine, it just means
your
> sales technique are wrong. Potential buyers will most likely have an
> engineering background and see this also.

> > – one being the volume generated by the movement of the pushing
tangential
> > surface.
> > – the other being the total observed volume.

> These are equal!!! I proved this at the end of the other document you
> ignored. http://www.mikesdriveway.com/misc/rotor.doc.

> Michael Culley

Bonjour Michael,

Sure, you have shown that the piston is doing the same amout of work
than the Wankel when going though the "Micheal swept volume".
But this does not prove anything and this is not surprising,
because the swept volume you calculate
is about only 1/3 of the real total Wankel volume
swept by the rotor surface during the 270 degrees shaft rotation.
I told you, just compared the swept volume you calculated
with the real total Wankel BTC volume, and they are not equal.
This is the reason you match the amount of work calculated
for both piston and Wankel,
but if you use the real Wankel chamber volume
it will not work any more,
and your results will become similar to ours !
(your calculation has to many little round corners
and assumptions errors for me to systematically review it.
However, with time, I am sure you will fix them all
… in silence and quitly).

I notice your assertions have switch form
Wankel – Piston calculations objectives
to Quasiturbine wrong sale technique ?
(pure arbitrary and not substantiated destructive force,
just like others have fun to do… Fun is all what you get!)
You even support your destruction by saying
I did not answer you on rec.autos.rotary
where everyone can check I did answer
being generally forwarded to other groups as well…

If you want every one to know
that your understanding of marketing is better than physics,
there are other ways to go…

Avec mes respects et salutations, Gilles
http://www.quasituribne.com

"Saint-Hilaire" <saint-hila…@promci.qc.ca> wrote in message

news:RISic.67\$Ee6.13@charlie.risq.qc.ca…

- Hide quoted text — Show quoted text -

> "Michael Culley" <m…@nospam.com> wrote in message
> news:408b3f4c\$0\$442\$afc38c87@news.optusnet.com.au…
> > Hi Saint-Hilaire,

> > I have conclusively proven this to be wrong in another thread but you
> didn’t
> > look at the evidence I supplied and didn’t respond to me. The
> > showed this, but as far as I could tell you did not download it. Here it
> is
> > again http://www.mikesdriveway.com/misc/rotaryworkoverexpansion.xls. (if
> you
> > get asked for a password push cancel). Have a look at it and tell me
which
> > step is in error. It is quite simple, here is a basic outline of how it
> > works:

> > It calculates the work output from a rotary and piston motor over an
> > expansion from min vol to max vol assuming that the pressure starts at 1
> > unit and reduces as the volume gets bigger
> > Column A is the crank angle.
> > B is the volume in the chamber at the corresponding crank angle
> > C is the pressure in the volume
> > E is the tangential force on the rotor from the pressure
> > D is the level length over which the force acts to produce torque
> > F is the torque output, calculated from multiplying E and D
> > G is the work produced, calculated from multiplying torque by the shaft
> > rotation.

> > All of these bits of work are added to get the total work. This is done
> for
> > the piston and rotary engine and the result is the exact same amount of
> work

> > The whole concept under which you are selling your engine is false. This
> is
> > not to say that there is anything wrong with your engine, it just means
> your
> > sales technique are wrong. Potential buyers will most likely have an
> > engineering background and see this also.

> > > – one being the volume generated by the movement of the pushing
> tangential
> > > surface.
> > > – the other being the total observed volume.

> > These are equal!!! I proved this at the end of the other document you
> > ignored. http://www.mikesdriveway.com/misc/rotor.doc.

> > Michael Culley

> Bonjour Michael,

> Sure, you have shown that the piston is doing the same amout of work
> than the Wankel when going though the "Micheal swept volume".
> But this does not prove anything and this is not surprising,
> because the swept volume you calculate
> is about only 1/3 of the real total Wankel volume
> swept by the rotor surface during the 270 degrees shaft rotation.
> I told you, just compared the swept volume you calculated
> with the real total Wankel BTC volume, and they are not equal.
> This is the reason you match the amount of work calculated
> for both piston and Wankel,
> but if you use the real Wankel chamber volume
> for your pressure drop calculation,
> it will not work any more,
> and your results will become similar to ours !
> (your calculation has to many little round corners
> and assumptions errors for me to systematically review it.
> However, with time, I am sure you will fix them all
> … in silence and quitly).

> I notice your assertions have switch form
> Wankel – Piston calculations objectives
> to Quasiturbine wrong sale technique ?
> (pure arbitrary and not substantiated destructive force,
> just like others have fun to do… Fun is all what you get!)
> You even support your destruction by saying
> I did not answer you on rec.autos.rotary
> where everyone can check I did answer
> being generally forwarded to other groups as well…

> If you want every one to know
> that your understanding of marketing is better than physics,
> there are other ways to go…

> Avec mes respects et salutations, Gilles
> http://www.quasituribne.com

"Saint-Hilaire" <saint-hila…@promci.qc.ca> wrote in message

news:RISic.67\$Ee6.13@charlie.risq.qc.ca…

- Hide quoted text — Show quoted text -

> "Michael Culley" <m…@nospam.com> wrote in message
> news:408b3f4c\$0\$442\$afc38c87@news.optusnet.com.au…
> > Hi Saint-Hilaire,

> > I have conclusively proven this to be wrong in another thread but you
> didn’t
> > look at the evidence I supplied and didn’t respond to me. The
> > showed this, but as far as I could tell you did not download it. Here it
> is
> > again http://www.mikesdriveway.com/misc/rotaryworkoverexpansion.xls. (if
> you
> > get asked for a password push cancel). Have a look at it and tell me
which
> > step is in error. It is quite simple, here is a basic outline of how it
> > works:

> > It calculates the work output from a rotary and piston motor over an
> > expansion from min vol to max vol assuming that the pressure starts at 1
> > unit and reduces as the volume gets bigger
> > Column A is the crank angle.
> > B is the volume in the chamber at the corresponding crank angle
> > C is the pressure in the volume
> > E is the tangential force on the rotor from the pressure
> > D is the level length over which the force acts to produce torque
> > F is the torque output, calculated from multiplying E and D
> > G is the work produced, calculated from multiplying torque by the shaft
> > rotation.

> > All of these bits of work are added to get the total work. This is done
> for
> > the piston and rotary engine and the result is the exact same amount of
> work

> > The whole concept under which you are selling your engine is false. This
> is
> > not to say that there is anything wrong with your engine, it just means
> your
> > sales technique are wrong. Potential buyers will most likely have an
> > engineering background and see this also.

> > > – one being the volume generated by the movement of the pushing
> tangential
> > > surface.
> > > – the other being the total observed volume.

> > These are equal!!! I proved this at the end of the other document you
> > ignored. http://www.mikesdriveway.com/misc/rotor.doc.

> > Michael Culley

> Bonjour Michael,

> Sure, you have shown that the piston is doing the same amout of work
> than the Wankel when going though the "Micheal swept volume".
> But this does not prove anything and this is not surprising,
> because the swept volume you calculate
> is about only 1/3 of the real total Wankel volume
> swept by the rotor surface during the 270 degrees shaft rotation.
> I told you, just compared the swept volume you calculated
> with the real total Wankel BTC volume, and they are not equal.

But if by ‘swept volume’ you just mean the movement of the rotor, that isn’t
the measure of work.  If the volume starts out as X and ends up as Y,
regardless of how the volume is ‘swept’ back/forth, then the expansion is
just X:Y.  If the rotor ‘sweeps’ through a large space (ala sideways), but
the volume only expands from X to Y, then the only work done is calculable
from X:Y.  The larger ‘sweeping’ is irrelevent.  (The two end points show
two states, regardless of the process path traversed getting from one end
point to the other).

I think that is what Michael is getting at.  Since the end points determine
the total available work, that is enough to show the theoretical possible
work.  You mechanism may be able to more closely reach the theoretical than
the standard pistion, but not exceed it.  And of course, the efficiency of
the combustion your photodetonation can achieve is another interesting
aspect of the mechanism.

daestrom

Saint-Hilaire,

> Sure, you have shown that the piston is doing the same amout of work
> than the Wankel when going though the "Micheal swept volume".

No, I have used the actual volume in the chamber. If the volume starts at 65cc and 1 unit of pressure and moves to 130cc I have
assumed the pressure is 0.5 units, what other magic volume should I use????

> But this does not prove anything and this is not surprising,
> because the swept volume you calculate
> is about only 1/3 of the real total Wankel volume
> swept by the rotor surface during the 270 degrees shaft rotation.

Actually, if you add up all the small tangential movements of the rotor face from 0 to 270 degrees they come to 654ccs. If you then
measure the difference between the volume at TDC and BDC you also get 654cc. So even if the magic volume you speak of does exist, it
is zero.

> but if you use the real Wankel chamber volume
> for your pressure drop calculation,

I have, at every point in the speadsheet I have used the actual volume in the chamber.

> (your calculation has to many little round corners
> and assumptions errors for me to systematically review it.

This is a just a dismissal of something you know will show you are in error.

> However, with time, I am sure you will fix them all
> … in silence and quitly).

I would suggest you should fix your website quietly. I am trying to save you some face when you try to sell your concept to
companies with some knowledge, they will quickly see through this volume you speak of.

> You even support your destruction by saying
> I did not answer you on rec.autos.rotary
> where everyone can check I did answer
> being generally forwarded to other groups as well…

Michael Culley

"daestrom" <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote in message news:YJVic.98738\$M3.1020@twister.nyroc.rr.com…
> If the volume starts out as X and ends up as Y,
> regardless of how the volume is ‘swept’ back/forth, then the expansion is
> just X:Y.  If the rotor ‘sweeps’ through a large space (ala sideways), but
> the volume only expands from X to Y, then the only work done is calculable
> from X:Y.  The larger ‘sweeping’ is irrelevent.  (The two end points show
> two states, regardless of the process path traversed getting from one end
> point to the other).

> I think that is what Michael is getting at.  Since the end points determine
> the total available work, that is enough to show the theoretical possible
> work.

That’s right. It doesn’t matter how the volume gets from X to Y the same amount of energy will be released to the shaft.

Michael Culley

"daestrom" <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote in message

news:YJVic.98738\$M3.1020@twister.nyroc.rr.com…

- Hide quoted text — Show quoted text -

> > Bonjour Michael,

> > Sure, you have shown that the piston is doing the same amout of work
> > than the Wankel when going though the "Micheal swept volume".
> > But this does not prove anything and this is not surprising,
> > because the swept volume you calculate
> > is about only 1/3 of the real total Wankel volume
> > swept by the rotor surface during the 270 degrees shaft rotation.
> > I told you, just compared the swept volume you calculated
> > with the real total Wankel BTC volume, and they are not equal.

> But if by ‘swept volume’ you just mean the movement of the rotor, that
isn’t
> the measure of work.  If the volume starts out as X and ends up as Y,
> regardless of how the volume is ‘swept’ back/forth, then the expansion is
> just X:Y.  If the rotor ‘sweeps’ through a large space (ala sideways), but
> the volume only expands from X to Y, then the only work done is calculable
> from X:Y.  The larger ‘sweeping’ is irrelevent.  (The two end points show
> two states, regardless of the process path traversed getting from one end
> point to the other).

> I think that is what Michael is getting at.  Since the end points
determine
> the total available work, that is enough to show the theoretical possible
> work.  You mechanism may be able to more closely reach the theoretical
than
> the standard pistion, but not exceed it.  And of course, the efficiency of
> the combustion your photodetonation can achieve is another interesting
> aspect of the mechanism.

> daestrom

Bonjour Daestrom,

I have no problem with what you are saying,
but it is not what Michael is doing.
We all agree on the initial state,
but Micheal final state is an intermediary one,
not the absolute final one as experienced in the Wankel.

The Wankel rotor surface pivots arount its median point
in reference to the engine radius,
while rotating around the engine,
and while translating inward of the engine
(that make three different movements).
Micheal defined the swept volume
see http://www.mikesdriveway.com/misc/rotor.doc
as the volume swept only by the
radial inward perpendicular rotor surface movement,
and he assume that this swept volume
is the Wankel BDC volume
(at least he uses it as such for pressure drop calculation,
pretending he is using the total Wankel BDC).
According to our calculation.
this account for only about one third of the real volume ?
Understandibly, using his partial volume
inflates his calculated Wankel internal pressure
and allow to match the work done by the piston.

It would take him only a minute to draw a little rectangle
with a surface equal to his calculated swept volume,
and to superimposed it to the unit Wankel layout,
and see it is by far too small…
Rather than to verified this point,
Micheal keep assuming that his calculated
"Micheal swept volume" is the final Wankel BDC volume !

Unless I get his collaboration to check that,
not much else I can do for now…
But I still hope he will check it one day…

Meilleures salutations, Gilles
http://www.quasiturbine.com

"Saint-Hilaire" <saint-hila…@promci.qc.ca> wrote in message news:6RYic.84\$Ee6.45@charlie.risq.qc.ca…
> Micheal defined the swept volume
> see http://www.mikesdriveway.com/misc/rotor.doc
> as the volume swept only by the
> radial inward perpendicular rotor surface movement,
> and he assume that this swept volume
> is the Wankel BDC volume

My definition of swept volume is 654ccs. Is the difference between BDC and TDC volume of the wankel engine different to this?

> (at least he uses it as such for pressure drop calculation,
> pretending he is using the total Wankel BDC).

No, I am using the volume in the chamber! Have a look at the volumes used in the spreadsheet and tell me what you think they should
be.

> According to our calculation.
> this account for only about one third of the real volume ?

I have asked you to show me these calculations but you have not produced them. Instead you asked me to use "intuition"

> According to our calculation.
> this account for only about one third of the real volume ?
> Understandibly, using his partial volume
> inflates his calculated Wankel internal pressure
> and allow to match the work done by the piston.

No, I am using the *actual* volume in the chamber. The formulae for the volume is MaxVol (1-cos(angle/1.5)) / 2 + MinVol. Do you
disagree with this?

> It would take him only a minute to draw a little rectangle
> with a surface equal to his calculated swept volume,
> and to superimposed it to the unit Wankel layout,
> and see it is by far too small…
> Rather than to verified this point,
> Micheal keep assuming that his calculated
> "Micheal swept volume" is the final Wankel BDC volume !

I have checked it. If the volume starts at 65cc at TDC it will end up being 654+65=719cc at BDC. My swept volume is also 654cc,
which is derived from adding up all the little tangential movements of the rotor face in, say, 1 degree increments. Every way I look
at it I get 654ccs, what volume do you think it is?

> Unless I get his collaboration to check that,
> not much else I can do for now…
> But I still hope he will check it one day…

I really think you should do some rethinking.

Michael Culley

"Michael Culley" <mculleyNOS…@optushome.com.au> wrote in message

news:c6hn0u\$c30ej\$1@ID-180780.news.uni-berlin.de…

> The formulae for the volume is MaxVol (1-cos(angle/1.5)) / 2 + MinVol

> I have checked it. If the volume starts at 65cc at TDC
> it will end up being 654+65=719cc at BDC.
> My swept volume is also 654cc,

> Michael Culley

Bonjour Michael,

This is interesting me very much.
I can live with your volume formula
and assuming your check has been done OK
(which I think you did consienciously),
I am force to suspect a surface or lever
angular projection discrepency with my own results (?).

While you are in the midle of this fresh calculation,
our has been done more than 9 years ago,
and this is the reason it is not easy for my
to provide you details at this time.
However, engine physics did not change much meanwhile…

(distance of the apply load from the engine centrer)
to which the load pressure force is apply to create torque
is zero at TDC and TDB,
and increase and decrease progressively in between ?

2) Does the load pressure force applied to create torque
is calculated by using the rotor surface component
projected in that same tangential force direction ?
(not using the whole rotor surface to calculate the force,
but the projection of that surface in that tangential direction).

then I will try to find time this week
to dig into our original calculation archeives
in order to try to consiliate both calculations.
All this for my own interest
and the engine community information,
which deserve the truth on my website as well as elswere.

Merci et bonne nuit, Gilles
http://www.quasiturbine.com

"Saint-Hilaire" <saint-hila…@promci.qc.ca> wrote in message news:ls0jc.92\$Ee6.72@charlie.risq.qc.ca…
> (distance of the apply load from the engine centrer)
> to which the load pressure force is apply to create torque
> is zero at TDC and TDB,
> and increase and decrease progressively in between ?

This is correct. Lever length starts and ends at zero and peaks at 15mm. Formulae is     Lever = 15 sin(angle / 1.5)

> 2) Does the load pressure force applied to create torque
> is calculated by using the rotor surface component
> projected in that same tangential force direction ?
> (not using the whole rotor surface to calculate the force,
> but the projection of that surface in that tangential direction).

The entire surface is used because the tangential direction is assumed to be at 90degrees to the rotor face. You could do it
otherwise but then the lever length would be different. For example, if you assumed the lever was a constant 15mm then at TDC and
BDC the surface area perpendicular to the lever would be zero, giving the same results.

> then I will try to find time this week
> to dig into our original calculation archeives
> in order to try to consiliate both calculations.
> All this for my own interest
> and the engine community information,
> which deserve the truth on my website as well as elswere.

Thankyou.

Michael Culley

"Michael Culley" <mculleyNOS…@optushome.com.au> wrote in message

news:c6i5bl\$bnm5v\$1@ID-180780.news.uni-berlin.de…
> "Saint-Hilaire" <saint-hila…@promci.qc.ca> wrote in message

news:ls0jc.92\$Ee6.72@charlie.risq.qc.ca…

> This is correct. Lever length starts and ends at zero and peaks at 15mm.

Formulae is     Lever = 15 sin(angle / 1.5)

> The entire surface is used because the tangential direction is assumed to

be at 90degrees to the rotor face. You could do it
> otherwise but then the lever length would be different. For example, if

you assumed the lever was a constant 15mm then at TDC and

> BDC the surface area perpendicular to the lever would be zero, giving the
same results.

> > then I will try to find time this week
> > to dig into our original calculation archeives
> > in order to try to consiliate both calculations.
> > All this for my own interest
> > and the engine community information,
> > which deserve the truth on my website as well as elswere.

> Thankyou.

> Michael Culley

Bonjour Michael,

I did roughly sketch the TDC and BDC states
In order not to let you inactive this week,
I like you to look at this skech at
http://www.quasiturbine.com/QTImages/WankelStroke040426.gif

Meilleures salutations, Gilles
http://www.quasiturbine.com

In rec.autos.rotary daestrom <daestrom@no_spam_heretwcny.rr.com> wrote:

> from X:Y.  The larger ‘sweeping’ is irrelevent.  (The two end points show

Some of the available power is wasted in the rotary, used to push the rotor
sideways instead of applying power to the output shaft.

Clarence A Dold – Hidden Valley (Lake County) CA USA  38.8-122.5

Hi Saint-Hilaire,

From the diagram it looks like you are assuming that the stroke is 2e (e being eccentricity of 15mm) but if you sum up all the
little tangential movements of the rotor face you get 3e. So the swept volume is 3 e a (where a is rotor face area), which is 3 x 15
x 182 x 80 = 655200 or 655ccs (I think the 182 should really be 181 point something so the end result is 654ccs).

Michael Culley

- Hide quoted text — Show quoted text -

"Saint-Hilaire" <saint-hila…@promci.qc.ca> wrote in message news:Mbajc.441\$Ee6.64@charlie.risq.qc.ca…
> "Michael Culley" <mculleyNOS…@optushome.com.au> wrote in message
> news:c6i5bl\$bnm5v\$1@ID-180780.news.uni-berlin.de…
> > "Saint-Hilaire" <saint-hila…@promci.qc.ca> wrote in message
> news:ls0jc.92\$Ee6.72@charlie.risq.qc.ca…

> > This is correct. Lever length starts and ends at zero and peaks at 15mm.
> Formulae is     Lever = 15 sin(angle / 1.5)

> > The entire surface is used because the tangential direction is assumed to
> be at 90degrees to the rotor face. You could do it
> > otherwise but then the lever length would be different. For example, if
> you assumed the lever was a constant 15mm then at TDC and
> > BDC the surface area perpendicular to the lever would be zero, giving the
> same results.

> > > then I will try to find time this week
> > > to dig into our original calculation archeives
> > > in order to try to consiliate both calculations.
> > > All this for my own interest
> > > and the engine community information,
> > > which deserve the truth on my website as well as elswere.

> > Thankyou.

> > Michael Culley

> Bonjour Michael,

> I did roughly sketch the TDC and BDC states
> In order not to let you inactive this week,
> I like you to look at this skech at
> http://www.quasiturbine.com/QTImages/WankelStroke040426.gif
> Your comment will be appréciated.

> Meilleures salutations, Gilles
> http://www.quasiturbine.com

"Michael Culley" <mculleyNOS…@optushome.com.au> wrote in message

news:c6kg30\$d247e\$1@ID-180780.news.uni-berlin.de…
> Hi Saint-Hilaire,

> From the diagram it looks like you are assuming that the stroke is 2e

(e being eccentricity of 15mm) but if you sum up all the
> little tangential movements of the rotor face you get 3e.

So the swept volume is 3 e a (where a is rotor face area), which is 3 x 15
> x 182 x 80 = 655200 or 655ccs

(I think the 182 should really be 181 point something so the end result is
654ccs).

> Michael Culley

Bonsoir Micheal,

My sketch at
http://www.quasiturbine.com/QTImages/WankelStroke040426.gif
is a reasonable observation, but it does not prove anything.
Only a carefull detail calculation answer it all,

I am not assuming that the stroke is 2e,
2e is the Wankel crankshaft diameter
and consequently the drop (stroke) from up to down.
Thruly, I do not yet understand your 2e and 3e comment ?
Basically, TDC state and BDC extrem energy states
should integrate pretty much all midway considerations (?)…
I do not see why 3e is shows-up ?
I will work on it when I have some time and will get back to you.

You probably understand that from my point of view,
the Quasiturbine has merit of it own,
and while I think it is interesting to make
a fair comparaison and comment about the Wankel,
this is not for us an obligation, neither a need,
and the Quasiturbine has its own interest and advantage anyway…
As I told you, our calculations was done about 9 years ago,
and our Wankel comment web page was written about 3 years ago,
part on best fidele souvenir of our calculation conclusion.

Nevertheless, I intend to give some time
to this conciliation problem which as become intriging to me…
Meanwhile, may be someone else
will come up with a useful remark ?

Bonne soirée, Gilles
http://www.quasiturbine.com

<d…@Quasiturbi.usenet.us.com> wrote in message news:c6jb95\$crt\$5@blue.rahul.net…
> Some of the available power is wasted in the rotary, used to push the rotor
> sideways instead of applying power to the output shaft.

This is not true, the same amount of power goes into the shaft as with the piston motor. More may be wasted by the large surface
area or inefficient combustion but not by pushing the rotor sideways. See the link to the spreadsheet in my other post.

Michael Culley

"Saint-Hilaire" <saint-hila…@promci.qc.ca> wrote in message

news:sMkic.22\$Ee6.11@charlie.risq.qc.ca…

> Bonjour,

> The texte is a clarification about the
> WANKEL THEORITICAL SWEPT VOLUME DEFICIENCY
> which has been added to the page
> http://quasiturbine.promci.qc.ca/QTpasWankel.html

> Meilleures salutations, Gilles
> http://www.quasiturbine.com
> *******************

Bonjour Michael,

I should have not find the time,
but the interest for this matter got me !
and I have not yet a final conclusion.

1)

> It calculates the work output from a rotary and piston motor over an
> expansion from min vol to max vol assuming that the pressure starts at 1
> unit and reduces as the volume gets bigger
> Column A is the crank angle.
> B is the volume in the chamber at the corresponding crank angle
> C is the pressure in the volume
> E is the tangential force on the rotor from the pressure
> D is the level length over which the force acts to produce torque
> F is the torque output, calculated from multiplying E and D
> G is the work produced, calculated from multiplying torque by the shaft
> rotation.

2)
Lever length starts and ends at zero and peaks at 15mm.
Formulae is     Lever = 15 sin(angle / 1.5)

3)
> Does the load pressure force applied to create torque
> is calculated by using the rotor surface component
> projected in that same tangential force direction ?
> (not using the whole rotor surface to calculate the force,
> but the projection of that surface in that tangential direction).

The entire surface is used because
the tangential direction is assumed to be
at 90 degrees to the rotor face.
You could do it otherwise but then the
lever length would be different.
For example, if you assumed the lever
was a constant 15mm then at TDC and¸
BDC the surface area perpendicular to the lever would be zero,
giving the same results.

4)

> The formulae for the volume is MaxVol (1-cos(angle/1.5)) / 2 + MinVol
> I have checked it. If the volume starts at 65cc at TDC
> it will end up being 654+65=719cc at BDC.
> My swept volume is also 654cc,

5)
> From the diagram it looks like you are assuming that the stroke is 2e

(e being eccentricity of 15mm) but if you sum up all the
> little tangential movements of the rotor face you get 3e.

So the swept volume is 3 e a (where a is rotor face area), which is 3 x 15
> x 182 x 80 = 655200 or 655ccs

(I think the 182 should really be 181 point something so the end result is
654ccs).

Refering to my rought draft at :
http://www.quasiturbine.com/QTImages/WankelStroke040426.gif

A)
I agree that all the volume
has to be swept one way or an other,
but you often mention your swept volume
without actually calculating it
(either the swept volume generated by
the inward or by the tangential movement).
You just take the total chamber volume
as your swept volume. Is this right ?

B)
Your formulae for the volume is
MaxVol (1-cos(angle/1.5)) / 2 + MinVol
which match perfectly the initial and final volumes,
and is certainly close to reality,
but quite emperical and certainly not
rigouriously correct to my opinion,
considering the compexity of the contour wall
and the rotor surface movement ?
Why dont you calculate the volume increment
for each angle. It is a bit of work,
but unavoidable to make the calculation unconstestable ?

C)
The work element is the force time the displacement.
Your lever calculation could be fine (I do not know),
but in the Wankel, the undiscutable displacement
is the crankshaft 270 degrees movement
on a 2e diameter circumference.
You say that this method would give the same result,
but I have some doubt yet ?
Why dont you calculate the "tangential force"
on that 2e circumference (the only force that count),
where the displacement would be easily proportional
to the shaft angle rotation increment ?
This would remove an important concern ?

D)
In a piston, the surface on which the gas puses
do not rock back and forth,
but stay perpendicular to its movement at all time.
In the Wankel this surface is rocking
and is not always perpendicular to its movement
du to Wankel crankshaft rotation.
If you agree with this (?),
only the component of the rotor surface
perpendicular to the direction
of crankshaft movement is actually making work …
It may happend that some rocking effect
do cancel out, but it is difficult to conclude
before having actually done the detail calculations.

E)
To calculate the force, you say that
"the entire rotor face is used
because the tangential direction is
assumed to be at 90 degrees to the rotor face".
In fact, this is true of the tangential movement
in relation to the rotor center (on the crankshaft),
but it is not true in relation to the engine center,
which is of interest as engine torque production ?

F)
The crank shaft diameter is 2e.
Is it possible that the reason you get 3e
as the "effective crankshaft diameter"
is because you do not make proper projection
of the rotor surface in the
direction of the applied force ?
This 3e result of yours make still no sense to me yet.

As you know, I have no objective
related to the Wankel technologie,
except the curriosity of a fair understanding.
As you see, I have several potential interogations
(I do not say you are wrong yet),
and that is the reason
I have not reach a final conclusion yet,
and maybe you should not either (?).

Meilleures salutations, Gilles
http://www.quasiturbine.com

"Saint-Hilaire" <saint-hila…@promci.qc.ca> wrote in message news:O%ujc.2672\$Ee6.19@charlie.risq.qc.ca…
> A)
> I agree that all the volume
> has to be swept one way or an other,
> but you often mention your swept volume
> without actually calculating it
> (either the swept volume generated by
> the inward or by the tangential movement).
> You just take the total chamber volume
> as your swept volume. Is this right ?

There are 2 (or more) ways to calculate it. You can measure the minimum volume and maximum volume and find the difference or, you
can add up all the tiny tangential movements from 0 to 270 degrees. Both of these give me 654cc.

> B)
> Your formulae for the volume is
> MaxVol (1-cos(angle/1.5)) / 2 + MinVol
> which match perfectly the initial and final volumes,
> and is certainly close to reality,
> but quite emperical and certainly not
> rigouriously correct to my opinion,
> considering the compexity of the contour wall
> and the rotor surface movement ?
> Why dont you calculate the volume increment
> for each angle. It is a bit of work,
> but unavoidable to make the calculation unconstestable ?

It would be possible to calculate the volume at each point in time using some of the maths at the end of the document that I posted,
which I believe will come to the above formulae. The volume in the chamber is a simple sine wave. Considering that the volume is
made up of 2 rotations (centre of rotor rotates at crank speed and rotor rotates at 1/3 crank speed) it makes sense that the volume
formula would come out to be a sine wave.

> C
> The work element is the force time the displacement.
> Your lever calculation could be fine (I do not know),
> but in the Wankel, the undiscutable displacement
> is the crankshaft 270 degrees movement
> on a 2e diameter circumference.
> You say that this method would give the same result,
> but I have some doubt yet ?
> Why dont you calculate the "tangential force"
> on that 2e circumference (the only force that count),
> where the displacement would be easily proportional
> to the shaft angle rotation increment ?
> This would remove an important concern ?

The rotor moves from +e from the centre point to -e, a difference of 2e. But if you add up all the little tangential movements of
the rotor face you actually get 3e.

The tangential force always applies to a lever somewhere between 0 and e. I need to draw a diagram to explain what I mean properly,
I will do this tonight.

> D)
> In a piston, the surface on which the gas puses
> do not rock back and forth,
> but stay perpendicular to its movement at all time.
> In the Wankel this surface is rocking
> and is not always perpendicular to its movement
> du to Wankel crankshaft rotation.
> If you agree with this (?),
> only the component of the rotor surface
> perpendicular to the direction
> of crankshaft movement is actually making work …
> It may happend that some rocking effect
> do cancel out, but it is difficult to conclude
> before having actually done the detail calculations.

All of the sideways movement does not contribute to work being done. For example, at the very tip of TDC the rotor is only moving
sideway and the volume is not expanding, so no work is lost.

> E)
> To calculate the force, you say that
> "the entire rotor face is used
> because the tangential direction is
> assumed to be at 90 degrees to the rotor face".
> In fact, this is true of the tangential movement
> in relation to the rotor center (on the crankshaft),
> but it is not true in relation to the engine center,
> which is of interest as engine torque production ?

I will show this in my diagram which make it clearer as to what I mean, it is difficult to describe these things in words.

> F)
> The crank shaft diameter is 2e.
> Is it possible that the reason you get 3e
> as the "effective crankshaft diameter"
> is because you do not make proper projection
> of the rotor surface in the
> direction of the applied force ?
> This 3e result of yours make still no sense to me yet.

Because the rotor rotates 90degrees from TDC to BDC the expansion takes 270 instead of 180 for a piston motor. Because of this the
rotor face has made a larger amount of tangential movements.

> You probably understand that from my point of view,
> the Quasiturbine has merit of it own,
> and while I think it is interesting to make
> a fair comparaison and comment about the Wankel,
> this is not for us an obligation, neither a need,
> and the Quasiturbine has its own interest and advantage anyway…
> As I told you, our calculations was done about 9 years ago,
> and our Wankel comment web page was written about 3 years ago,
> part on best fidele souvenir of our calculation conclusion.

Of course, I am in no way disputing the quality of your engine. I can describe some advantages that I believe are not listed on your
website.

My rotary engine software may be of use to you, you can use it to measure the lever length etc. I plan to add a window to it to show
the tangential forces and lever length used to calculate torque.

Michael Culley

Hi Saint-Hilaire,

This is the diagram I promised. This shows how I calculate the torque from
the pressure and lever length. Because the pressure is pretty much the same
everywhere in the chamber the force will be even on the rotor and not put
rotational force on it. So the force can be shown as a line through the
centre of the rotor. It is then just a matter of determining the lever
length and multiplying by the force to get the torque.

It is easy to imagine how the lever length varies acording to 15
sin(angle/1.5). Start by imagining that the rotor does not rotate but has a
face always pointing the the right, the lever length would be 15sin(angle).
If you then rotate the rotor at 1/3 crank speed then it takes more crank
rotation to change the lever length, so you get 15 sin(angle – angle/3),
which is 15 sin(angle/1.5).

Having the volume follow a sine wave is a little more difficult to imagine,
but it will vary with the lever length. When the lever length is zero the
volume will change zero, when the lever length is greatest the volume will
change the most.

Regards,
Michael Culley

"Michael Culley" <mculleyNOS…@optushome.com.au> wrote in message

news:c6n9b4\$e3iv2\$1@ID-180780.news.uni-berlin.de…
> "Saint-Hilaire" <saint-hila…@promci.qc.ca> wrote in message

news:O%ujc.2672\$Ee6.19@charlie.risq.qc.ca…

> > A)
> > I agree that all the volume
> > has to be swept one way or an other,
> > but you often mention your swept volume
> > without actually calculating it
> > (either the swept volume generated by
> > the inward or by the tangential movement).
> > You just take the total chamber volume
> > as your swept volume. Is this right ?

> There are 2 (or more) ways to calculate it.
> You can measure the minimum volume and
> maximum volume and find the difference or, you
> can add up all the tiny tangential movements from 0 to 270 degrees.
> Both of these give me 654cc.

From this answer, I can not find out if you do actually
algebraicaly calculate it, or if you are just making a statement ?
This is fundamental, because if you do not actually
calculate it, taking into account positive and negative swept volume,
there is no way you will ever match our calcutation…

- Hide quoted text — Show quoted text -

> > B)
> > Your formulae for the volume is
> > MaxVol (1-cos(angle/1.5)) / 2 + MinVol
> > which match perfectly the initial and final volumes,
> > and is certainly close to reality,
> > but quite emperical and certainly not
> > rigouriously correct to my opinion,
> > considering the compexity of the contour wall
> > and the rotor surface movement ?
> > Why dont you calculate the volume increment
> > for each angle. It is a bit of work,
> > but unavoidable to make the calculation unconstestable ?

> It would be possible to calculate the volume at
> each point in time using some of the maths
> at the end of the document that I posted,
> which I believe will come to the above formulae.
> The volume in the chamber is a simple sine wave.
> Considering that the volume is made up of 2 rotations
> (centre of rotor rotates at crank speed and
> rotor rotates at 1/3 crank speed) it makes sense that the volume
> formula would come out to be a sine wave.

There are here 2 rotations, composed with variable
relative centers, plus one substantial translation.
Overall, I can not catch your symplification and conclusion…

- Hide quoted text — Show quoted text -

> > C
> > The work element is the force time the displacement.
> > Your lever calculation could be fine (I do not know),
> > but in the Wankel, the undiscutable displacement
> > is the crankshaft 270 degrees movement
> > on a 2e diameter circumference.
> > You say that this method would give the same result,
> > but I have some doubt yet ?
> > Why dont you calculate the "tangential force"
> > on that 2e circumference (the only force that count),
> > where the displacement would be easily proportional
> > to the shaft angle rotation increment ?
> > This would remove an important concern ?

> The rotor moves from +e from the centre point
> to -e, a difference of 2e.
> But if you add up all the little tangential movements of
> the rotor face you actually get 3e.
> The tangential force always applies to a lever
> somewhere between 0 and e.
> I need to draw a diagram to explain what I mean properly,
> I will do this tonight.

but just could not figure out your 3e result.
In order to mach our calculation,
you shoud rather come down to 1e !

- Hide quoted text — Show quoted text -

> > D)
> > In a piston, the surface on which the gas puses
> > do not rock back and forth,
> > but stay perpendicular to its movement at all time.
> > In the Wankel this surface is rocking
> > and is not always perpendicular to its movement
> > du to Wankel crankshaft rotation.
> > If you agree with this (?),
> > only the component of the rotor surface
> > perpendicular to the direction
> > of crankshaft movement is actually making work …
> > It may happend that some rocking effect
> > do cancel out, but it is difficult to conclude
> > before having actually done the detail calculations.

> All of the sideways movement
> does not contribute to work being done.
> For example, at the very tip of TDC the rotor is only moving
> sideway and the volume is not expanding, so no work is lost.

I can not desagree more.
Not only you have to take into account
the projection of the surface in direction of mouvement,
but at the tip of the rotor face, the pressure
makes the pushing surface to pivot backward
in relation to the rotor gear theets,
and the work done by the pressure at the tip
does in fact cancel some of the work done
in other area of the rotor surface…

> > E)
> > To calculate the force, you say that
> > "the entire rotor face is used
> > because the tangential direction is
> > assumed to be at 90 degrees to the rotor face".
> > In fact, this is true of the tangential movement
> > in relation to the rotor center (on the crankshaft),
> > but it is not true in relation to the engine center,
> > which is of interest as engine torque production ?

> I will show this in my diagram which make it
> clearer as to what I mean,
> it is difficult to describe these things in words.

but this is an obvious argument hard to circumvent…

> > F)
> > The crank shaft diameter is 2e.
> > Is it possible that the reason you get 3e
> > as the "effective crankshaft diameter"
> > is because you do not make proper projection
> > of the rotor surface in the
> > direction of the applied force ?
> > This 3e result of yours make still no sense to me yet.

> Because the rotor rotates 90degrees from
> TDC to BDC the expansion takes 270 instead
> of 180 for a piston motor. Because of this the
> rotor face has made a larger amount of tangential movements.

True for the larger tangential movement,
but still the effective diameter should be 2e anyway ?
In fact, to consiliate with our calculation,
it shoud rather be 2e !
I do not see your point for a larger diameter ?

- Hide quoted text — Show quoted text -

> > You probably understand that from my point of view,
> > the Quasiturbine has merit of it own,
> > and while I think it is interesting to make
> > a fair comparaison and comment about the Wankel,
> > this is not for us an obligation, neither a need,
> > and the Quasiturbine has its own interest and advantage anyway…
> > As I told you, our calculations was done about 9 years ago,
> > and our Wankel comment web page was written about 3 years ago,
> > part on best fidele souvenir of our calculation conclusion.

> Of course, I am in no way disputing the quality of your engine.
> I can describe some advantages
> that I believe are not listed on your website.
> My rotary engine software may be of use to you,
> you can use it to measure the lever length etc.
> I plan to add a window to it to show
> the tangential forces and lever length used to calculate torque.
> http://www.mikesdriveway.com/engineapp

I agree, it can be very useful if I can end-up agreeing with it !
Lets try more…
Amicalement, Gilles  www.quasiturbine.com

- Hide quoted text — Show quoted text -

> Michael Culley

In rec.autos.rotary Michael Culley <m…@nospam.com> wrote:

> http://www.mikesdriveway.com/misc/rotortorque.jpg
> This is the diagram I promised. This shows how I calculate the torque from
> the pressure and lever length. Because the pressure is pretty much the same
> everywhere in the chamber the force will be even on the rotor and not put
> rotational force on it. So the force can be shown as a line through the
> centre of the rotor. It is then just a matter of determining the lever
> length and multiplying by the force to get the torque.

But the rotor does move sideways.  Some force vector is required for that.

This picture immediately conjurs the thought that the force applied to the
trailing side of the rotor face is to the right of the pivot point, and
would want to drive the rotor in that direction, countered by force applied
on the leading side of the rotor, which is a waste of those force elements.

You are also assuming that the "pressure is pretty much the same
caused by the long and narrow combustion chamber.

Clarence A Dold – Hidden Valley (Lake County) CA USA  38.8-122.5

"Michael Culley" <m…@nospam.com> wrote in message

news:408f9e61\$0\$436\$afc38c87@news.optusnet.com.au…

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> Hi Saint-Hilaire,

> This is the diagram I promised.
> This shows how I calculate the torque from the pressure
> and lever length. Because the pressure is pretty much the same
> everywhere in the chamber the force will be even on the rotor and not put
> rotational force on it. So the force can be shown as a line through the
> centre of the rotor. It is then just a matter of determining the lever
> length and multiplying by the force to get the torque.

> It is easy to imagine how the lever length varies
> acording to 15 sin(angle/1.5).
> Start by imagining that the rotor does not rotate but has a
> face always pointing the the right,
> the lever length would be 15sin(angle).
> If you then rotate the rotor at 1/3 crank speed then it takes more crank
> rotation to change the lever length, so you get 15 sin(angle – angle/3),
> which is 15 sin(angle/1.5).

> Having the volume follow a sine wave
> is a little more difficult to imagine,
> but it will vary with the lever length. When the lever length is zero the
> volume will change zero, when the lever length is greatest the volume will
> change the most.

> Regards,
> Michael Culley

Bonjour Michael,

Thanks for your diagramm at :
http://www.mikesdriveway.com/misc/rotortorque.jpg
Truly, you are not doing much to help me.
More I look deeply in your calculation,
further away I stand…

It is obvious that there is a Wankel excess swept volume.
The question is what is the impact of
this excess volume on the shaft power output.
You pretend that this excess swept volume
while our calculation shows that this excess swept volume
is internally counter-productive and
distroy the efficiency of the Wankel engine.

In under to help understanding my point,
I have completed my rought draft at :
http://www.quasiturbine.com/QTImages/WankelStroke040426.gif
showing the positive and negative sweep volume force vectors.
My understanding is that your spreed sheet calculation
ignore this moment of forces arount the gear teeth ?

What we found is that the Wankel excess sweept volume
correspond to a pure rotational (tangential) force
which is apply to the tip of the rotor face,
and cancel other useful forces on other area of the same blade.
Because of the teeth rotor equilibrium point
and this cancellation of forces, there is
a strong net rotational forces apply on the the rotor face.
The net result could be stated like this :
‘The total Wankel chamber swept volume is composed
of 2/3 from positive forces plus 1/3 from negative forces,
for a net output equivalent of 1/3 of the total chamber volume’.

Even if not proprely stated, I suspect the earlyer
comment from d…@Quasiturbi.usenet.us.com
was in conformity with our own observation :
"Some of the available power is wasted in the rotary,
used to push the rotor sideways
instead of applying power to the output shaft"

In the Quasiturbine however,
there is not crankshaft, and no inward stroke,
since the blade distance to the center
stay constant during rotation.
All the Quasiturbine swept volume is
pure rotational (tangential) force.
In this case however, there is no competition
since all force vertors are positive
and adding up to produce torque and energy…

The Wankel was invented at the end of the 18 century
and Felix Wankel did work on it from 1926.
Still today, we are arguing about force vectors
… because too few did apparently make
an exhaustive détail analysis, or kept their result secret…
Your work merit to be continue,
and it could be useful to many, not only ourself…

Meilleures salutations, Gilles
http://www.quasiturbine.com

Hi, Saint-Hilaire

> From this answer, I can not find out if you do actually
> algebraicaly calculate it, or if you are just making a statement ?
> This is fundamental, because if you do not actually
> calculate it, taking into account positive and negative swept volume,
> there is no way you will ever match our calcutation…

I calculate it. I have never gone out and actually measured the swept volume but I would be *very* suprised if it turned out to be
different to 654ccs.

> I wait for your diagram,
> but just could not figure out your 3e result.
> In order to mach our calculation,
> you shoud rather come down to 1e !

Have a look at this spreadsheet, this shows 2 methods of calculating swept volume which both match. A description is in the
http://www.mikesdriveway.com/misc/calcrotarysweptvolume.xls

> I can not desagree more.
> Not only you have to take into account
> the projection of the surface in direction of mouvement,

Moving the rotor sideways does not require work (aside from minor friction losses), only the movement in the direction of the arrow
in my diagram requires work.

> but at the tip of the rotor face, the pressure
> makes the pushing surface to pivot backward
> in relation to the rotor gear theets,
> and the work done by the pressure at the tip
> does in fact cancel some of the work done
> in other area of the rotor surface…

Force is free, it does not matter that the top of the rotor is moving against the force. You could say that these parts of the rotor
face are putting work back into the gas to be used later.

> Thanks for your diagramm at :
> http://www.mikesdriveway.com/misc/rotortorque.jpg
> Truly, you are not doing much to help me.
> More I look deeply in your calculation,
> further away I stand…

I think we are getting off the track a little, what I suggest you do is, first verify my formulae for lever length. You could do
this using the software and a ruler and plotting a few points from 0 to 270 degrees. Verify my formulae for the volume of the
chamber. I’m not sure how you can do this, might require an actual engine. Once you have done that, reproduce my original

> It is obvious that there is a Wankel excess swept volume.
> The question is what is the impact of
> this excess volume on the shaft power output.
> You pretend that this excess swept volume
> while our calculation shows that this excess swept volume
> is internally counter-productive and
> distroy the efficiency of the Wankel engine.

I don’t say it produces energy, I say that there is zero excess volume.

> In under to help understanding my point,
> I have completed my rought draft at :
> http://www.quasiturbine.com/QTImages/WankelStroke040426.gif
> showing the positive and negative sweep volume force vectors.
> My understanding is that your spreed sheet calculation
> ignore this moment of forces arount the gear teeth ?

You are correct, I assume there is zero force on the gears. The eccentric shaft is at the centre of the rotor and the entire rotor
will be pushing on the eccentric shaft at the centre of the rotor. So the forces are evenly distributed and will not twist the rotor
in any way.

Even if the forces do act as you’ve shown them in your diagram, the effective lever length has been increased, which would result in
the same torque.

Michael Culley

<d…@Quasiturbi.usenet.us.com> wrote in message news:c6peod\$sg2\$1@blue.rahul.net…
> But the rotor does move sideways.  Some force vector is required for that.

Only friction forces will need to be overcome, which are fairly minor.

> This picture immediately conjurs the thought that the force applied to the
> trailing side of the rotor face is to the right of the pivot point, and
> would want to drive the rotor in that direction, countered by force applied
> on the leading side of the rotor, which is a waste of those force elements.

The pivot point is the centre of the eccentric shaft lobe and hence the centre of the rotor face. The pivot point is not where the
teeth of the gears meet.  If you imagine the motor in this diagram http://www.mikesdriveway.com/misc/rotortorque.jpg with no gears
at all, what would be the tendancy of the rotor to twist? Because the rotor is being held by the eccentric shaft at the rotors
centre the force is applied equally to either side of this pivot point and so it will not twist.

You could consider that the parts of the face moving towards the force are putting work back into the gas, and this work is used by
the parts of the face moving away.

> You are also assuming that the "pressure is pretty much the same
> everywhere" which is contrary to comments about the poor flame propogation
> caused by the long and narrow combustion chamber.

Pressure travels at the speed of sound and although one area might be burning poorly it will very quickly reach the same pressure as
other areas. Even so, it doesn’t affect the issue of excess volume.

Here’s another way to look at it. Imagine you have gas at a pressure and you want to release that pressure *without* gaining the
energy from it in some form. No matter how you release that gas the energy will be released through heat, sound, speed, lift, or
work through an output shaft. In the rotary this spare energy would have to go somewhere if it doesn’t go to the shaft. As the
rotary expands there is nothing to suggest that any more of this energy would go to anywhere else except the output shaft (besides
friction).

Michael Culley

Saint-Hilaire wrote:

> In under to help understanding my point,
> I have completed my rought draft at :
> http://www.quasiturbine.com/QTImages/WankelStroke040426.gif
> showing the positive and negative sweep volume force vectors.
> My understanding is that your spreed sheet calculation
> ignore this moment of forces arount the gear teeth ?

See, this is where it goes wrong. You assume the force vectors define
the resulting torque. But if you consider that there is little movement
at the ‘negative’ end of the rotor, this force is little more than a
pivot point. It would be proper to integrate the change of movement of
the rotor face by that force over time to determine the real torque.

I think this is the point that both Michael and daestrom are trying to
make when they say it doesn’t matter how you get from X to Y.

(By the way Mr. Dold, I lived in Clearlake for twenty years!)

Best, Dan.

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