auricomXL

anybody know how many degrees ATDC any given rotor face will have a peak mechanical advantage onto the e-shaft?

as an example, a piston enine reaches it's peak mechanical advantage when the con rod is perpendicular to the crank throw. this means the peak advantage angle is a function of stroke and conrod length, usually being around 20 degrees i think.

anybody know the numbers for peritrochoidal geometries???

if it helps, i am asking this question to determine when i should hit peak combustion pressure so i can begin to form an ignition advance curve.

I feel dirty...I haven't thought about conrods in a LONG time

ericgrau

iTrader: (1)

In a rotary I think you're more worried about detonation than breaking the rotor. Unless you're running insane boost and already have detonation under control.

j9fd3s

iTrader: (3)

find a copy of the "rotary engine" book, its in pdf on the web. pretty sure that kinda of info would be in there.

imo you're prolly gonna end up with timing in the same range as everyone else...

auricomXL

yes, i most likely will, the problem is that noone seems to be willing to post their ignition advance curves, so i am going a more complicated route.

additionally, i am going to be running about 8psi on 9.7:1 rotors, which hardly anyone on the internet does.

i'll check that book out...
thanks,
max

Kyrasis6

I don't have numbers, but yes peak force is attained somewhere in the first 20 deg after TDC in a rotary as well, give or take a few.

auricomXL

so this point is where you want your combustion cycle pressure to max out as well right?

this is controled with the ignition advance table while taking into consideration the following factors:

surrounding material temp (hotter surfaces, less advance)
intake charge temp (hotter air, less advance
fuel octane rating (lower octane, less advance)
rotation velocity of the engine (lower RPM, less advance)
intake pressure (higher pressure (boost), less advance)

would everyone agree with that?

what sort of advance does one typically see under heavy boost? as low as 10*BTDC?

moving along the WOT curve, an NA motor will require lots of advance at low RPM. less advance around peak torque (because teak torque mean peak VE which means higher peak pressure in the rotor chambers), and then more advance up high after torque begins to fall off.
do i have that right?

rotary8

At approximately 45 degrees after top dead center, the rotor has the greatest mechanical advantage over the eccentric shaft. If you are running 9.7 rotors it is not really a big deal it just means you have to be a little bit more conservative with your timing curve . If you are going to turbo the engine you can start with 10 degrees leading and 0 on trailing and after you have dialed in your fuel mixture you can move your timing and see if the engine likes it but more timing sacrifices the life of the engine so keep that in mind.

j9fd3s

iTrader: (3)

yep thats about right.

RacerXtreme7

iTrader: (2)

As he stated ^^ above, 45* ATDC you get maximum torque transfer from the rotor to the eccentric shaft. If you really want to maximise your timming, get a pressure transducer and run it in your trailing plug hole while on a dyno using a precision encoder for crank angle. Its what I plan on doing eventually

~Mike.............

Speed of light

iTrader: (8)

The rather late occurrence of maximum leverage (45 deg) suggests this engine will do well with a slower burning fuel, e.g., race gas, etc..

On a related topic: What is the equivalent rod ratio of the rotary? Anyone know? I suspect it's going to be on the high side.