Turbo header design

[TMead17327 at aol.com]

In a message dated 1/7/99 12:36:28 PM Central Standard Time,
clarencewood at centuryinter.net writes:

I think it should look like this:

 		1	8	4	3	6	5	7	2
1	0	P		C		I		E	
	90		P		C		I		E
2	180	E		P		C		I	
	270		E		P		C		I
3	0	I		E		P		C	
	90		I		E		P		C
4	180	C		I		E		P	
	270		C		I		E		P

The first column counts the strokes, and the second column indicates degrees
from top dead center of the #1 piston.

Visualize the  #1 and #2 pistons, whose rods are mounted on the same crank
journal, firing.  #2 fires (the P at the bottom right corner of the chart).
The #1/#2 journal rotates (along with the rest of the crankshaft, hopefully!!)
90 degrees, and #1 is at TDC (the P at the top left of the chart).  After
another 90 degrees of crank rotation, the #7/#8 journal , which is 180 degrees
off from the #1/#2 journal, has brought #8 to TDC (P in the "8" column).  90
degrees later, #7 is at TDC, but at the beginning of the intake stroke (I in
the "7" column).  You'll get the idea if you picture the whole thing rotating.

I hope after all that, I got it right!?

Tommy
TMead17327 at aol.com

  

<<                 Firing order         
           1  8  4  3  6  5  7  2
 Stroke
   1.      P  C  I  E  P  C  I  E       (P=power,C=comp,I=intake,E=exht)
   2.      E  P  C  I  E  P  C  I
   3.      I  E  P  C  I  E  P  C
   4.      C  I  E  P  C  I  E  P
  >>