Turbo header design

[Clarence Wood]

Thanks Tom!

At 10:04 PM 1/7/99 EST, you wrote:
>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
>  >>
>
>