Marco Acosta to drag race in a 4 Rotor!
 

Yep from the IDRC site

Marco Acosta will soon debute a 4-Rotor
Auto to compete against the V8's in the IDRC

damm
now were Talking major Power!
PP 4 rotor

hahaha, he is gonna rape those V8's.


the limiting factor is gonna be the eccentric shaft.
With too much power, its gonna flex like mother fu er

DAAAAAAAAAMN!!!!

It was only a matter of time, of course...

(Can you say 6-second "import"? :evilgrin: )

Rob at Pineapple builds em too! With his motors, e-shaft flex is'nt a problem. His 4 rotor is also shorter than the factory 4 rotor engines.

Someone, needs to come up with the $$ to buy a 4 rotor, and strap something like dual T88s on it, and throw it in a car like Abels, and go run 6s all day long. What class would a monster like that fit into??

Someone, with lots of money laying around...and too much time on there hands...*cou-Jimlab-gh*:D

CJ

unfortunatly, Jim hates rotaries ( :) ) and would rather have a h8ed v8

Re: Marco Acosta to drag race in a 4 Rotor!
 

why compete against the V8's? is the 4 rotor illegal in the 'regular' pro class? it still has less displacement than a supra 2JZ-gte motor....... bullish has 2 tube framed solaras that WILL run 6 seconds this year, and they dont have to run in the pro-V8 class.

a 2.6 liter 4 rotor is only slightly bigger than the 2.3 liter honda motors run in the pro calss. this is BS!!!!

HANDICAP THE GOD DAMNED SUPRAS!!!!!

The Supra engine is 3 liters... a 4-rotor based on 13B/20B parts would equate to 5.2l...

I always thought thought 4 rotors and V10s were banned from NHRA Sport Compact.

why wouldn't a 26B have double the displacement of a 13B?
13B=1.3 liter
26B=2.6 liter

i dont agree with those funny ways of calculating rotary displacement. the combustion chambers add up to 1.3 liters, mazda says it's 1.3 liters, i'm gonna call it 1.3 liters........

Re: Marco Acosta to drag race in a 4 Rotor!
 

IDRC has no Idea. I was @ MVA shop today, Marcos pick up truck has the old 3rotor he had in the MX3 but this time he has a new intake and exhaust manifold. No 4 rotor for now...

13B = 1.3l per revolution
R26B = 2.6l per revolution

2.6l piston engine = 1.3l per revolution
5.2l piston engine = 2.6l per revolution

Simple no?

Who said anythign about revolutions?? We are talking about total displacement, not the Displacement after revolutions, or fires, or whatever.

Seems simple to me, lets run through it once more in better terms:

Piston displacement = area of cylinder (witht the "height" beign calculated by piston stroke) X number of cylinders

Rotary displacement = Area of space around the rotor in rotor housing X the number of rotorhousings.

It doesnt get much simpler than that.

Therefor a 2.6l Engine = 2.6l of displacement ROTARY OR PISTON.

Now if you want to get into the number of times they fire per revolutions or turns or Whatever, then thats a diff topic.

?????????

Re: Re: Marco Acosta to drag race in a 4 Rotor!
 

he's using his 'old' 3 rotor? so does thids mean that he recovered his stolen MX-3?

NHRA and IDRC are 2 diffrent events

here is the link to the article

read the "IDRC PRO CLass"

http://www.urbanracer.com/headline/idrc_051402.html

Wrong peejay, every compression chamber on the rotor, is responsible for 654 cc's. Now is 654 x 2 = 1308cc's. For the 2 rotor engine to complete the 4 cycles(intake,compression,power,exhaust) on just 2 chambers it takes 2 eccentric shaft revolutions, which = 1.3 liter. in just one revolution it will never complete its 4 cycles to a 1.3liter. 1 revolution will just get the engine to te point of Intake compression, the other revolution will bring the engine towards Power and exhaust, completing it's 4 cycles. In all reality is a 1.3 liter 4 cycle engine, with a very decent power stroke...

Re: Re: Re: Marco Acosta to drag race in a 4 Rotor!
 

It was the old, old frankenstein engine he broke plates on. before they stoled the Mx3. When they stoled the Mx3 it had the new so fresh and so clean engine, and the old 3 rotor was not in the car, he had it in the shop.

Guys, rotaries are the equivalent of a 2 cycle 4 stoke.....

They are only 1.3lt in capacity in 13B form.

A 500cc 2 stoke GP bike is only 500cc as well, but they compete against 1000cc 4 strokes, as they are breathing 1000cc in the time it takes the 4 stroke to complete its cycle.

A wankel is no different, in the time it takes a 2.6lt 4 cycle piston engine to complete it cycle (2 engine revs) the wankel has completed two cycles, hence you double the capacity to "compare it to a 4 cycle piston engine" easy as that.

They are only a 1.3lt, but they do the same amount of work as a 2.6lt 4 stroke piston engine.

Ito, you are wrong, the wankel needs just ONE crank rev to complete a full cycle, Intake, Compression, Power, Exhaust.

1 cycle in a piston engine is exactly the same as 1 cycle on a rotary. First cycle is Intake. second is compression, third power 4th exhaust. is just that simple. we could get into power stroke degrees(equivalent to a 6cil. power stroke) and other factors that give advantage to a rotary vs a 4cil. piston engines. But a cylce is a cycle when we are talking about the "OTTO Cycles" 1,2,3,4 is the same for both engines, nothing more nothing less.

In two crank revs, 4 chambers have completed cycles. 4 X 654cc = 2616cc.

To verify this all you need to see is how many times the spark plug fires, it fires 4 times over the two rotors.

Another cross check for this is when you fit an Autometer tacho to a 2 rotor, to get it to read right you set it for a 4 cylinder engine, this is cause a 4 cylinder has 4 cylinder ignition pulses for every 2 engine revs, A rotary has the same except that it only has 2 "cylinders" but over 2 engine revs it sees 4 individual pulses !

This means that it does double the work and to compare it to a 4 stroke piston engine you should double it's capacity. If you are comparing it to a 2 stroke piston engine then yeah it is a 1.3lt engine in 13B form.

Ito it is an easy mistake to make, but you are wrong.

A 4 cycle engine in reciprocating form takes 2 engine revs to fire its spark plug to generate ONE POWER PULSE.

A 4 cycle wankel takes one engine rev to fire its spark plug to generate ONE POWER PULSE.

hmm...
 

for every one rotation of the eccentric shaft, the rotor turns 1/3 of a rotation. From intake to compression.

Because every one rotation of the E-shaft the rotor turns once. 1 divided by 3 is 1/3.

It takes 3 full e-shaft rotations to complete 1 full cycle from intake back to intake, but the other sides follow around too, so it takes 1 e-shaft rotation per side to reach combustion stage.

While running this would mean that each side is combusting at 1 full revolution of the e-shaft, now if each rotor is calculated 654cc total than that means that it is only running 218cc per rotor per revolution, however I do not believe this is the case.

I am not sure on this, but I think that the calculation is 654cc per rotor per rotation, which makes 654cc per side, making 654cc per combustion per rotation x2 rotors is still 1.3 L...

It has been argued that the engine is actually a 2.6 L of ACTUAL displacement, I don't know how they calculate that, but if you wanted to calculate it funny you could say that a 2 rotor is actually 2.0L for 1 rotation of the rotor x2 rotors and you have a 4.0 L engine. But that is stupid if we are taking about per rotation, where it stands 1.3L.

But also don't the rotors combust at different times? So its really only making 654cc at a time?


Yarg...I am confused now. no more math for me.

Rice you have just amazed me. never in the history of 10A,12A or 13B or 20B will the engine complete 4 cycles in just one eccentric shaft revolution. Rice you need to touch up more on this subject. It takes no more then "2" 360 degrees of eccentic shaft revs to complete it's 4 cycles. period. been there done it myself, I havent read this info. I personally took a front side housing with the front rotor and E shaft on, plus the timing pulley on. 1 360 degree rev. will only get the rotor to intake and compression, the 2nd rev. will get the engine pass power and exhaust, and it beggins all over. Rice I have read some of your post before, it seems like you know some of your stuff, but with this statement your way offffffffff...

Listen, you guys are confusing yourselfs on this.

We all agree that to make power the plugs need to fire...right?

So all you need to do is see haw many times the plug fires simple huh

I do not give a fuck if it's a 2 stroke, 4 stroke, wankel 1 cylinder 10 cylinder or 4 rotor ! ALL YOU NEED TO DO IS COUNT HOW MANY SPARKS YOU HAVE OVER AN ENGINE CYCLE.

For a 4 stroke 4 cylinder piston engine you have 4, count them 4 !!! (in two engine revolutions, cause it needs 2 revolutions to complete a cycle)

For a two rotor wankel you have 4, yes 4 mother fucking ignition pulses. THIS MEANS 4 POWER PULSES in the same two engine revolutions. THIS MEANS 654cc X 4 = 2616cc.

Which means that to compare it for what "work" it actually does then a wankel is the SAME, YES THE SAME as a 2616cc 4 stroke piston engine.