ivanson

i have a single zc vtec. and i just pulled of the head
today and found out its a dam ex motor...damit man.
but anywho.
anyone know how much i can shave the head down?
i heard it can go to .030.
and about with high compression pistons.
would it have valve to piston damage?
please any help is most apreciated

lessonsinspeed

I assume you're milling the head to get higher compression? I'd stick with a new head gasket or HC pistons... and heres why: if for any reason you need to go back, its as simple as using the stock parts. Milling the head is permanent. Plus, using an aftermarket head gasket or pistons from a good company means you know that you shouldn't have interference issues... with the head, you're guessing to a degree, and if you go to far, the head is worthless.

As for your question, I glanced in my B16 shop manual, and couldn't find anything.

BILLETGRIP

i milled .062 off of my z6 head(on a6 block) with no problems. i just retarded my cam timing 2 degrees. its been running that way for over 2 years now.

Mista Bone

.062 milled = 5 degrees cam retard.

I'll assume you installed the cam WRONG at 1 tooth advanced (9.75 degrees) then minus the seven degrees of retard from your adjustments and milling.

BILLETGRIP

I may have missed putting on the timing belt by a tooth but i did install the cam RIGHT. Either way i was sharing my experience with milled heads. You can go more than .030. I have went as far as .062 without problems. I do know you cant go much further than that without cutting into your valve seat.

jlicrx

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Mista Bone &raquo;</TD></TR><TR><TD CLASS="quote">.062 milled = 5 degrees cam retard.</TD></TR></TABLE>
on a D-series SOHC engine, 0.062" = about 1.53 degrees

BILLETGRIP

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by jlicrx &raquo;</TD></TR><TR><TD CLASS="quote">
on a D-series SOHC engine, 0.062" = about 1.53 degrees</TD></TR></TABLE>

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Mista Bone &raquo;</TD></TR><TR><TD CLASS="quote">
I'll assume you installed the cam WRONG at 1 tooth advanced (9.75 degrees) then minus the seven degrees of retard from your adjustments and milling.</TD></TR></TABLE>


And actually 1 tooth would be 9.4736842 degrees on a sohc.





Modified by BILLETGRIP at 5:25 PM 6/28/2007

Mista Bone

my reply was killed by HT lagging.............

but 9.5 (4.75 x 2 ) degrees is correct.

BILLETGRIP

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Mista Bone &raquo;</TD></TR><TR><TD CLASS="quote">

9.5 (4.75 x 2 ) degrees is correct.</TD></TR></TABLE>


Where does the 4.75 x 2 come from?
I just did 360 / 38 = 9.4736842. Im not knockin your equation, just wonderin how you got it like that?

jlicrx

You may see all kinds of different things on the internet about how many degrees the camshaft rotates due to milling the head. Unfortunately, most of them are not correct.

When you mill the head or block, the distance between the centerline of the crankshaft and the centerline of the camshaft(s) is reduced by the amount removed from the head or block. What effect does this actually have on the camshaft timing?

In order for the camshaft to be in time with the crankshaft, with zero degrees advance or retard, it must remain in the same position, relative to the crankshaft, with the milled head as it was with the stock head. If you are going to maintain this relationship, you will need an adjustable timing gear.

In order to determine the effects of the milling, you need to understand what is actually happening. The length of the timing belt between the crankshaft gear and the camshaft gear is a fixed distance, due to the belt having teeth which meshes with the gears. When you mill the head or the block, the actual distance between the crankshaft and camshafts is obviously reduced, but the belt length and number of teeth between the crank and cam gears remains constant. This means that in order for the belt to remain tight, without slack on the front side of the engine and remain meshed with the gears, the outside ring of the adjustable camshaft gear has to rotate clockwise, while at the same time keeping the camshaft in the same position relative to the crank. The distance the outside of the gear rotates is equal to the amount milled. The next question is “How much in degrees does it move?”

In order to calculate what this means in camshaft degrees, we use a simple math formula for the Properties of a Circle, where:

Angle A (in degrees) is = 180a / pi(r)

with “a” being the length of the arc or distance that gear rotates (equal to amount milled) and “r” being the radius from the center of the cam gear to the belt.

On D-series SOHC engines, the pitch or distance between teeth on the timing belt is 9.5 mm. This means that with 38 teeth on the gear, the circumference (distance around at the belt) is 38 x 9.5 = 361 mm. Using this, we can find the radius (r) of the cam gear from the center to the belt using the formula that says the radius = circumference divided by 2 x pi. This would then be r = 361 / 2 x 3.141592654 = 57.4549mm. To get this to inches, divide by 25.4, which then the radius equals 2.262”. Plugging this number into the formula above would result in Angle (in degrees) = 180 x a / 3.141592654 x 2.262 = 25.3297a. For head milled 0.010” the angle would be 25.3297 x 0.010 = .253297 degrees. For head milled 0.062” the angle would be 25.3297 x 0.062 = 1.57 degrees.

You can also get the same result by dividing the circumference of 361 mm by 360 degrees which gives you one degree on the gear equal to 1.0027778 mm or 0.039479 inches

For all practical purposes, you could say that milling 1 mm or 0.3937 inches would equal one degree

The numbers are different for a B-series since the B-series gears are smaller with only 34 teeth. For B-series the degrees would be 28.3096a.





Modified by jlicrx at 10:39 AM 6/29/2007

Mista Bone

Your math is flawed, esp. since you ASSUMED the cam gear has a 361 mm circumfrence.

Peak to peak = 114mm
valley to valley = 107.5mm

(114 + 107.5)/2 * pi = 347.93 MM (lets just call it 348mm)

already 3.6% error in your math.

according to your math 348 / 25.4 = 13.7

13.7 / 360 = .038" for one degree.......

"For all practical purposes, you could say that milling 1 mm or 0.3937 inches would equal one degree"

Thats quite a MAJOR error right there.


BTW, Tim wants to know "What is the unladed airspeed velocity of a swallow?

The answer to life, the universe, and everything is 42

scrooge

whoa math class

BILLETGRIP

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by jlicrx &raquo;</TD></TR><TR><TD CLASS="quote">


For all practical purposes, you could say that milling 1 mm or 0.3937 inches would equal one degree





</TD></TR></TABLE>

1mm = .03937 . .3937 = 10mm. Sorry but i had to call you out on it. The rest of that made absolutely no sense to me. I like drawings that have physical evidence rather than a bunch of math that can only be considered a theory

Mista Bone

I did the math years ago........and I could be wrong......but another person came up with the same numbers.

I'm not doubting his numbers QUITE yet, until I can prove him wrong........or prove myself wrong. Been too tired mentally to think it thru.

If I've ready found a few holes, but like Hawking I can admit I'm wrong also.

Firing up the old TI-30 III, damn thing still works!

I'm debating and thinking that the cam moves at 1/2 speed relative to the crank......so the .038 would be reduced to .019" per degree of milling.

Mista Bone

but something I just thougt about......

I know the "1/2 tooth cam error" when installing a D16 cam into a D15 setup. I'm the one who figured it out.

1/2 tooth = 4.75 degrees.

I've also measure the deck heights of D15 and D16 blocks, 207mm vs. 212mm.

5mm = 4.735 degrees

.197" = 4.735 degrees

.0416" = 1 degree

Which means I just proved myself wrong with my own numbers.

I've even figured out the registers on the head to check for milling, nonVTEC SOHC in the pic

BILLETGRIP

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Mista Bone &raquo;</TD></TR><TR><TD CLASS="quote">

I'm debating and thinking that the cam moves at 1/2 speed relative to the crank......so the .038 would be reduced to .019" per degree of milling.</TD></TR></TABLE>


If this is true wouldnt the cam gear have to be twice the diameter of the crank pulley?

jlicrx

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by BILLETGRIP &raquo;</TD></TR><TR><TD CLASS="quote">If this is true wouldnt the cam gear have to be twice the diameter of the crank pulley?</TD></TR></TABLE>true - the crank gear is 1/2 the diameter of the cam gear with 1/2 the number of teeth

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by BILLETGRIP &raquo;</TD></TR><TR><TD CLASS="quote">1mm = .03937 . .3937 = 10mm. Sorry but i had to call you out on it. The rest of that made absolutely no sense to me. I like drawings that have physical evidence rather than a bunch of math that can only be considered a theory</TD></TR></TABLE>typo when i edited the original post - it does say 0.03937 in the sentence just above

4drEF

I love when people reply to threads that are a decade old!