tough question...

I know the formula for cubic displacement of a piston engine, but does anyone know what the forumla is for the rotary engine??

May find what your after here..alot of good info

http://www.rotaryengineillustrated.com/index.html

Damn it, I just did a report on the rotary engine"Wankel Engine" this week for my Algebra 2 class.

Went to that sight, doesn't help, and I'm not quite sure there really is a formula. Because in the rotor it has areas ground in on the face of it to increase displacement. Quite a bit of calculus and trig, stuff I'm not into yet. But with the eccentric shape or w/e it's gotta be a hard formula.

I'm also doing a report on it right now Scott, I'll post the report when I'm done with it, suppose to have somewhere in the vicinity of 1000-1200 words, after that we use the report to make a Powerpoint presentation.

btw, I get the emails sent to me when someone replys, and I got one that said ivymike replied, bowtiebandit and scott replied, but for some reason ivymike doesn't have a post in this thread, is this some mistake by AF, can anyone tell me?

i'll look around tomorrow and see what i can find.

well basically its BDC swepped area * width of rotor + volume of the indent * how many rotors.

So yeah, complicated calc required. Have fun:)


edit: p.s. as a hint, the combustion volume of one face of one rotor is 654cc's, *2 for the engine 1308 cc's, hence 13b engine ;)

I 'm a bit curious how the volume of the indent changes displacement - does the indent get bigger and smaller during the cycle?

the indent is like having a dished piston. it would lower compression... it's adding volume to the chamber

are you suggesting that a dished piston affects displacement?

Originally posted by v10_viper
btw, I get the emails sent to me when someone replys, and I got one that said ivymike replied, bowtiebandit and scott replied, but for some reason ivymike doesn't have a post in this thread, is this some mistake by AF, can anyone tell me?

I replied, but then decided that I didn't like my original response, so I deleted it.

well, I think I've got a formula figured out for the change in "cylinder" volume from "bdc" to "tdc" for one face of a rotor:

let t = crank throw
let a = radius from rotor center to an apex
let w = the distance between the sidewalls

Q is an angle
eqn1 = sqrt(a^2 + t^2 -2*a*t*cos(2/3*Q + 60deg))

Area1 = integral from 0deg to 360deg with respect to Q of eqn1
Area2 = integral from 270deg to 630deg with respect to Q of eqn1
Area3 = 1.732*(0.5*a^2 - a*t)
Area4 = 1.732*(0.5*a^2 + a*t)

then the volume change for one face from min vol to max vol is

D = w * ((Area1-Area3) - (Area2-Area4))

Unfortunately, I have been unable to perform the symbolic integral of eqn1. Eqn1 is easy enough to integrate numerically, however, so if you have some figures to punch in for t, a, and w, you can (hopefully) calculate displacement.

I'm tempted to say that since there are three faces per rotor, and each rotor rotates once for every three crank revs, total engine displacement is (number of rotors) * D. I haven't really given it much thought, though.

technically yes it does.

your displacement is the area inside the combustion chamber x # of cyls.

if you run a piston that's dished, it will give x cc's.

if you run a high comp piston that intrudes into the combustion chamber, you'll get a smaller cc #...

make sense?

to give a bad analogy...
a pie pan is a low comp piston... put something in it (mound shaped in the middle, and the volume drops...

it's miniscule, and not worth changing your 5.0 to a 4.5 or something like that, but it makes a difference.

Originally posted by flylwsi
technically yes it does.

your displacement is the area inside the combustion chamber x # of cyls.

if you run a piston that's dished, it will give x cc's.


You might want to research this a bit further. Hint: displacement is not related to the volume of the combustion chamber (assuming that when you say combustion chamber, you mean the volume above the piston at TDC).

and how would you measure your displacement then, if that's not the case?

(and yeah, i meant at full compression as you assumed... )

...by the amount of volume that is displaced by the pistons: the BDC volume minus the TDC volume times the number of cylinders.

It's calculated as follows:
pi * (bore diam)^2 / 4 * stroke * number of cylinders

The compression ratio, (BDC volume)/(TDC volume), does depend on the combustion chamber volume, as I believe you mentioned earlier...

Originally posted by ivymike1031

Q is an angle
eqn1 = sqrt(a^2 + t^2 -2*a*t*cos(2/3*Q + 60deg))

Unfortunately, I have been unable to perform the symbolic integral of eqn1.



Pfffft....pussy....

...by the amount of volume that is displaced by the pistons: the BDC volume minus the TDC volume times the number of cylinders.

this means that high comp pistons will give you a different displacement than low comp, yes?

that's what i was saying...

sigh... no, it doesn't mean that. observe:

single cylinder engine
combustion chamber volume : V.cha =0.07L
stroke: 90*mm
bore: 90*mm

displaced volume per cylinder (doesn't matter what V.cha is):
V.disp = pi*(90*mm)^2/4 * 90*mm = 0.573L

TDC volume = V.cha = 0.7L
BDC volume = V.cha + V.disp = 0.643L

Compression ratio:
(V.cha+V.disp)/V.cha = 9.2:1

Use a deep-dish piston with a chamber volume of 0.08L and see what happens:
V.disp = pi*(90*mm)^2/4 * 90*mm = 0.573L
V.cha = 0.8L
(V.cha+V.disp)/V.cha = 8.2:1

Displacement is not changed by the dish in the piston. It is only affected by the bore and stroke. Compression ratio is changed by the dish, because it depends on both the displacement and the chamber volume.

If you want to reality-check your idea of what "displacement" is, look up the combustion chamber volume of any engine at all, and compare that to the displacement of the engine. Example:
Ford Mustang II 302 V8
http://www.mustangii.org/tech/heads.shtml
Chamber volume in 1980 D8OE-AB engine: 69cc
number of cylinders: 8
Displacement of this engine, per your method: 8*69cc = 552 cm^3 = 34 in^3
Displacement of this engine, in reality: 302 in^3 or 4949 cm^3

There's a big, big difference between your "displacement" and the real displacement of this example engine. Try any others, if you care to.

Cody Fassler
Computer; 5th period
Ms Hjelm
“How Rotary Engines Work”

In this report I will be explaining how rotary engines work and how they are different from normal 4-stroke piston engines work. A rotary engine is an internal combustion engine but works totally different from the way a conventional piston engine works. Dr. Felix Wankel conceived the original idea. This is why the first rotary engines were called Wankel Rotary engines as used in the Mazda RX-7 sports car.
In a regular piston engine the same space does all the 4 stages, intake, compression, combustion, and exhaust. A rotary engine does each of those, but each stage is done in it’s own part of the housing. The continuing motion of the rotor uses air and pressure to compress the mixture that is then ignited and keeps the rotor moving. The rotor is triangular shaped, and the sides are slightly curved outwards and have pockets ground in them to increase displacement.
The housing is something relative to what you would create with a Spiro graph. The path the tips of the rotor follows keeps the tips always touching the edge of the housing, this also works by offsetting the rotor from the center. It also uses metal rings on each side of the rotor to seal the side for combustion.
The shape of the housing almost looks like an oval but it is an epitrochoid. Describing how this shape is created is difficult but it is sort of like if you took two circles and overlapped just a little bit of the outside of them over one another, sort of like a MasterCard emblem.
The intake and exhaust ports are actually made into the housing. They use no valves just slots; one slot connects to the exhaust, the other to the throttle.
The output shaft is basically the crankshaft, just looks different. It has round lobes that are eccentrically mounted, which means they are offset from the center of the shaft. The amount of lobes relates to the amount of rotors mounted on it. With the lobe being offset it acts much like a crank handle on a which, it gives the rotor leverage to turn the output shaft. How the rotor turns inside of the housing pushes the lobe around in tight circles. The output shaft turns 3 times for every one revolution of the rotor.
Because of the rotor moving around in the way that it does changes the displacement of the chambers created by the triangular shape change to create each stage in the engine.
The intake phase starts as the tip of the rotor passes the intake port. As it passes the intake port it is at its smallest displacement, as it moves around it increases displacement, drawing in air like if a balloon were to expand itself and draw the air in. The air drawn in mixes with fuel, which then moves on into the next phase. When the peak of the rotor passes the intake port the chamber or phase becomes sealed off and compression starts. As it continues rotating the size or volume decreases which compresses the air/fuel mix inside. When it is at is minimum volume it meets the spark plugs and combustion starts. Most rotary engines have dual spark plugs because of the shape of the compression chamber. When it ignites it builds up pressure that makes the rotor move into its next phase. Once the peak of the rotor passes the exhaust port the combustion gases are free to leave out the exhaust. As it moves on the chamber contracts to move the rest of the exhaust out the port. One thing about the rotary engine is that the three faces of the rotor are constantly working on one part of the cycle. In one whole revolution there will be three combustion strokes, which also means the output shaft spins three times the speed of the rotors.
A rotary engine has many differences from a regular piston engine. It has many defining characteristics. The main one is it has fewer moving parts. It has a total of 3 moving parts, two rotors and the output shaft. Even the most simple convention engine has at least 40 moving parts. It includes pistons, connecting rods, crankshaft, timing gears and belts, and the valve train which consists of valves, camshafts, valve springs, rocker arms, and pushrods. More moving parts increases friction of the engine, which in turns takes away power from the engine. Rotary engines also have more reliability in turn for this.
The next difference is that it operates more smoothly. All the parts continue moving in one direction rather than changing directions violently. The rotary engines also use counterbalances that are used to get rid of any vibrations. The power delivery is also smoother because each combustion phase lasts 90º throughout the rotor’s rotation. Because the output shaft turns 3 times the speed of the rotor so the combustion phase lasts 270º of the output shaft’s rotation. This means that if you have a one-rotor engine that it delivers power for ¾ of each revolution. When you compare this to a single-cylinder piston engine combustion occurs only 180º out of every two revolutions, which in turn calculates to a quarter of each revolution of the crankshaft.
Motions made by the engine are also slower than that in a conventional piston engine. Because the rotor spins at one-third the speed of the output shaft parts move slower which make it more reliable.
There are some challenges to a rotary engine though. Manufacturing costs can be higher because the number of these engines is not produced as much as piston engines. It is more difficult to make them meet U.S. emissions regulations. They also use more fuel because of the thermodynamic efficiency is lowered by the long shaped combustion chamber and the low compression ratio. You can up the compression ratio thought by using forced induction instead of leaving it naturally aspirated. A form of this is typically a turbocharger, although a supercharger can also be used. Turbo- and superchargers work by propelling air into the engine to increase power. The normal amount of air that a normally aspirated engine gets at sea level is 14.7 psi or pounds per square inch. When you add a turbo, depending on size and type, you can have anywhere from 6-50 psi going into an engine. Diesels use turbocharger’s to get a high compression so the diesel fuel ignites. Any thing over twenty psi is dangerous and creates wear on your engine. Top fuel dragsters use 50 psi; their engines are also rebuilt after every run. Turbochargers are usually hooked up into your exhaust system and a turbine runs off the exhaust, which then turns a shaft that then spins yet another turbine or compressor that moves air through a series of ducts into your intake manifold so your engine gets more air in it. A supercharger works the same way, just that it gets propelled by having a pulley on it to run off of your crankshaft. The pulleys on a supercharger can be changed to smaller or bigger to allow for more or less airflow.
In the new Mazda RX-8 they have a new rotary engine renamed the Renesis. The most important effect is the new placement of the exhaust port. Instead of being opened by the tips of the rotors, it is now placed on the side to be opened by the sides of the rotor. This engine being only 80 cubic inches develops 250 horsepower without using forced induction. That is over 3 horsepower per cubic inch. Normal piston engines that are naturally aspirated are usually only around one horsepower per cubic inch.
That was my report on rotary engines and everything about them. I got most of my information from www.howstuffworks.com as well as self-knowledge.

How did I do?? any mistakes??

lets see if this kid did some plagerizing. not sure if thats spelled right, but you all know what I mean.;)

Originally posted by ivymike1031
well, I think I've got a formula figured out for the change in "cylinder" volume from "bdc" to "tdc" for one face of a rotor:
let t = crank throw
let a = radius from rotor center to an apex
let w = the distance between the sidewalls
Q is an angle
eqn1 = sqrt(a^2 + t^2 -2*a*t*cos(2/3*Q + 60deg))
Area1 = integral from 0deg to 360deg with respect to Q of eqn1
Area2 = integral from 270deg to 630deg with respect to Q of eqn1
Area3 = 1.732*(0.5*a^2 - a*t)
Area4 = 1.732*(0.5*a^2 + a*t)
then the volume change for one face from min vol to max vol is
D = w * ((Area1-Area3) - (Area2-Area4))


I got some dimensions for a particular rotary engine, and tried to check my earlier equation. The numbers didn't match, so I took a closer look and I found at least three errors in the formula above. The biggest error was that I was saying that the swept area was the integral of r with respect to angle, when it should have been (r^2)/2 with respect to angle. When you square eqn1 (radius), the radical drops out, and it becomes much easier to integrate. Shows what happens when you stay up too late staring at a monitor. Nonetheless, rather than try again to come up with a correct formula, I searched around online and found one.

http://members.rogers.com/sofronov/Cars/Mazda/Rotary.html

e=eccentricity. The amount of offset between the
eccentric shaft centerline and the rotor centerline

R=radius. Generating radius is the distance between
the centerline of the rotor and the apex

b=width. The trochoid chamber width

Vh=working chamber volume or single camber capacity
( displacement ) Vh=3 x ( square root of 3 ) x R x e x b

Note- the formula above does not take into account any cutout which may exist in the rotor (and I believe it shouldn't).

Regardless of error, it's great you spent some time doing that stuff. If something gets too complicated I usually stop doing it. :)

yeah, well, I'm combatting my compulsion by taking courses that will eventually lead to a non-technical degree... :)