13B CAS sensor wiring, timing and marks ?

[357supermagnum]

It looks like the Mazda 13B Wankel engine CAS sensor is wired like this:

CAS sensor wiring ( internal wires )
G+ Green ( the wheel with the 2 teeth or "spikes" ) This is the "cam wheel" as MS3 sees it.
G- White with black stripe
NE+ Red ( crank wheel,24 teeth)
NE- White ( crank wheel )

How to find TDC on a 13B engine:
This somewhat crude drawing is the front pulley with its key-way. When the key way is facing the intake, TDC is 90 degrees straight up as on the drawing.


The two "spikes" on the top wheel should line up with the purple line in this picture:


Originally, this sensor is set to 5 degrees timing, it looks like it needs to be set at exactly TDC so MS3 can calulate the timing correctly.


Can anyone confirm that i am right ?

[Matt Cramer]

MegaSquirt can calculate the timing with the sensor set wherever. You just adjust Tooth #1 Angle until you get the right timing.

[elturbonitroso]

if your front cover and your pulleys are a match just use the -5 mark on the pulley and sync your cas with that mark and you all set.

[357supermagnum]

MegaSquirt can calculate the timing with the sensor set wherever. You just adjust Tooth #1 Angle until you get the right timing.

Actually,the Trigger wizard should be used to set a correct offset.
The problem is that in my instance it is grayed out and cannot be accessed.
I am using the ini file supplied with the flash-able program for the MS3 cpu.





It is impossible to connect to the actual MS3 controller as this will be ordered in the autumn season..
Maybe this is a bug in the ini file or something ?

*edit* Looks like it is the #1 tooth angle ( deg btcd ) that is the setting i am looking for, should be 5 degrees.

[elturbonitroso]

your engine size should be 1308cc and not 2616,also your fix advance should be -5 if you want to use the first mark on your pulley.

[muythaibxr]

I always use a 5 deg tooth #1 angle, set timing to fixed -5, put a timing light on the leading wire for rotor 1 (it matters even if it is wasted spark) and turn the CAS to get it to line up. There is no need to use the trigger wizard.

[357supermagnum]

your engine size should be 1308cc and not 2616,also your fix advance should be -5 if you want to use the first mark on your pulley.

You are right about the timing, but wrong about the engine size.
The size rating on a Wankel Rotary Engine is a little different because its internals are nothing like a piston engine.
The actual *full* displacement of these tiny engines from Mazda is 2616 (really!), based on one complete revolution of the rotors. But one full revolution of the rotor creates 3 full rotations of the crankshaft. Since a piston engines rating is based on full displacement during 1 crank rotation they do that with the Rotary as well. So 1 crankshaft rotation is 1/3 of the 2616cc displacement, or 1.3L.
The reason the 1.3L Wankel gets rated as 2.6L for power output is that the combustion stroke is twice as long as in a piston engine. During a pistons travels during one crank rotation it is supplying power for 90 degrees of a 360 degree movement. A Wankel Rotary supplies almost 180 degrees of movement under the combustion phase of its rotation.

[dontz125]

Ummm ... 1.3l is 1/2 of 2.6l; 1/3 of 2616cc is 872cc. Don't know that it makes any difference to the discussion at hand.

[357supermagnum]

Ummm ... 1.3l is 1/2 of 2.6l; 1/3 of 2616cc is 872cc. Don't know that it makes any difference to the discussion at hand.

The Mazda 13B wankel engine has two rotors shaped roughly like a triangle. This makes for three combustion faces per rotor and a total of six for a standard two-rotor engine. Each face has a “swept volume” or displacement of 40ci(654cc) and there are a total of six faces. With this known, the engine displacement should be 40ci(654cc) times six to equal 240ci(3.9L), right? In a way, yes, but that would not be a comparable displacement to the 4-cycle engine.
Each rotation of the engine (360°) will bring two faces through the combustion cycle (the torque input to the eccentric shaft). This said, it takes 1080° or three complete revolutions of the crankshaft to complete the entire thermodynamic cycle. Obviously, we have a disparity. How can we get a number to compare to a 4-stroke engine? The best way is to study 720° of rotation of the two-rotor engine. Every 360° of rotation, two faces of the engine complete a combustion cycle. 720° will have a total of four faces completing their cycle. 40ci(654cc) per face times four faces equals 160ci or 2.6L. Mazda lists this at exactly 2616cc for the Rx3/rx4 engines.
So,2616cc is the "effective" displacement if it is compared to a piston engine.
Also, Aaron Cake has a page with information of how to set up Megasquirt for a 13B engine, the 2600 number is also mentioned in the Magamanual.
Aarons page is here :
http://www.aaroncake.net/rx-7/Megasquir ... amMS3X.asp

[elturbonitroso]

your engine size should be 1308cc and not 2616,also your fix advance should be -5 if you want to use the first mark on your pulley.

You are right about the timing, but wrong about the engine size.
The size rating on a Wankel Rotary Engine is a little different because its internals are nothing like a piston engine.
The actual *full* displacement of these tiny engines from Mazda is 2616 (really!), based on one complete revolution of the rotors. But one full revolution of the rotor creates 3 full rotations of the crankshaft. Since a piston engines rating is based on full displacement during 1 crank rotation they do that with the Rotary as well. So 1 crankshaft rotation is 1/3 of the 2616cc displacement, or 1.3L.
The reason the 1.3L Wankel gets rated as 2.6L for power output is that the combustion stroke is twice as long as in a piston engine. During a pistons travels during one crank rotation it is supplying power for 90 degrees of a 360 degree movement. A Wankel Rotary supplies almost 180 degrees of movement under the combustion phase of its rotation.

i know how a rotary engine works i built and tune them but the setting that i told you its the ones that you have to use if you want your ms to work with the recent codes the info in the manual its for the old code(s its outdated).

[muythaibxr]

your engine size should be 1308cc and not 2616,also your fix advance should be -5 if you want to use the first mark on your pulley.

You are right about the timing, but wrong about the engine size.
The size rating on a Wankel Rotary Engine is a little different because its internals are nothing like a piston engine.
The actual *full* displacement of these tiny engines from Mazda is 2616 (really!), based on one complete revolution of the rotors. But one full revolution of the rotor creates 3 full rotations of the crankshaft. Since a piston engines rating is based on full displacement during 1 crank rotation they do that with the Rotary as well. So 1 crankshaft rotation is 1/3 of the 2616cc displacement, or 1.3L.
The reason the 1.3L Wankel gets rated as 2.6L for power output is that the combustion stroke is twice as long as in a piston engine. During a pistons travels during one crank rotation it is supplying power for 90 degrees of a 360 degree movement. A Wankel Rotary supplies almost 180 degrees of movement under the combustion phase of its rotation.

This is so wrong I don't know where to start.

Mazda rated the displacement based on 1 full e-shaft rotation, or 1.3L. On the MS, if you select "rotary" stroke, you should set displacement to 1308cc, and 2 rotors.

It actually takes 3 e-shaft rotations to get all three rotor faces on both rotors through a combustion event. The 2.6L comes from a comparison to piston engines, which take 2 revs to burn their whole rated displacement. In 2 revs the 2-rotor 1.3L engine burns 2.6L (assuming 100% VE).

Ken

[muythaibxr]

Ummm ... 1.3l is 1/2 of 2.6l; 1/3 of 2616cc is 872cc. Don't know that it makes any difference to the discussion at hand.

The Mazda 13B wankel engine has two rotors shaped roughly like a triangle. This makes for three combustion faces per rotor and a total of six for a standard two-rotor engine. Each face has a “swept volume” or displacement of 40ci(654cc) and there are a total of six faces. With this known, the engine displacement should be 40ci(654cc) times six to equal 240ci(3.9L), right? In a way, yes, but that would not be a comparable displacement to the 4-cycle engine.
Each rotation of the engine (360°) will bring two faces through the combustion cycle (the torque input to the eccentric shaft). This said, it takes 1080° or three complete revolutions of the crankshaft to complete the entire thermodynamic cycle. Obviously, we have a disparity. How can we get a number to compare to a 4-stroke engine? The best way is to study 720° of rotation of the two-rotor engine. Every 360° of rotation, two faces of the engine complete a combustion cycle. 720° will have a total of four faces completing their cycle. 40ci(654cc) per face times four faces equals 160ci or 2.6L. Mazda lists this at exactly 2616cc for the Rx3/rx4 engines.
So,2616cc is the "effective" displacement if it is compared to a piston engine.
Also, Aaron Cake has a page with information of how to set up Megasquirt for a 13B engine, the 2600 number is also mentioned in the Magamanual.
Aarons page is here :
http://www.aaroncake.net/rx-7/Megasquir ... amMS3X.asp

The megamanual should not be used for the ms3. The ms3 manual should. If Aaron Cake's site says use 2600cc, his instructions are for an old firmware. The proper settings are "Rotary" stroke and 1308cc, and 2 rotors.