How much does it help?

[Terry, NJ]

How much does boring an engine .040 increase the compression with a stock head? I know it's slight, But I'm curious what the ratio is. My calculater can't be found and I'm not sure my pencils are sharp enough anymore. Let's see Pi R2 X 3.915= ahhh... I can't do it in my head anymore!
Terry

[ken ct]

Not even sure the comp would go up, as your making the chamber bigger not smaller as when milling head.It might come down slightly. I could be wrong on this thinking???? ken ct.

[ctlikon0712]

Yes, Ken that thinking is in error. The combustion chamber stay's the same, the bore volume increases so you are sucking in more volume and so are compressing slightly more volume into the same area thereby increasing the compression ratio.

[allen]

Terry every little bit helps, going 040 and run a H/C head from synders I could always tell the difference. Now for the math formula I guess that would be billfold x Desire = results

[MikeK]

Stock 4.22:1 becomes about 4.31:1 @ 0.040" overbore. A weenie 0.09 increase.

[Terry, NJ]

Sponsored Links (Register now to hide all advertisements)

Thanks Mike, I was on the same page with Craig. I got to thinking about this and it became a little obsession with me. Now for the next question! If I shaved .050 off the head with the bore also oversized what would the CR be.
Again, I realize that the increase will be small, But I was jest wonderin'.
Terry

Originally Posted by MikeK

Stock 4.22:1 becomes about 4.31:1 @ 0.040" overbore. A weenie 0.09 increase.

[ctlikon0712]

Terry, I think that would depend on the chamber volume and shape. Sorry I don't know the answer. But I am interested too!

[MikeK]

Originally Posted by Terry, NJ

Thanks Mike, I was on the same page with Craig. I got to thinking about this and it became a little obsession with me. Now for the next question! If I shaved .050 off the head with the bore also oversized what would the CR be.
Again, I realize that the increase will be small, But I was jest wonderin'.
Terry

Not as easy to calculate. The shape of an A combustion chamber is a geometric complex and includes the flycut. You need to start with the print dimensions and calculate the surface area, then figure volume change based on the 0.050.

[Flathead]

The A combustion chamber is huge, milling won't help much. Milling would have more effect with a smaller chamber size like a V8 Ford.

[ken ct]

Originally Posted by ctlikon0712

Yes, Ken that thinking is in error. The combustion chamber stay's the same, the bore volume increases so you are sucking in more volume and so are compressing slightly more volume into the same area thereby increasing the compression ratio.

Oh well i was wrong ,thanks for the correction. ken ct/

[Terry, NJ]

Actually I believe it is do-able without the shape of the C.C. But I'm having a hard time visualizing it. It goes something like this, 200 Cu ins. = 50 cu ins per cylinder. A 4 1/4 stroke gives a 50" displacement. Someone somewhere can properly interpolate how much more volume it takes to create 4.22. I just am not sure how to it. Then subtract the volume of the CC from the total. The real hard job is calculating the volume of the .050 cut. Once that is done, subtract that from the total and recalculate the CR. However, A simpler way would be Find the tables that someone once posted here giving the amount of CR increase with each increment of stock removed and see what .050 would give me.
Terry

Originally Posted by MikeK

Not as easy to calculate. The shape of an A combustion chamber is a geometric complex and includes the flycut. You need to start with the print dimensions and calculate the surface area, then figure volume change based on the 0.050.

[Fordors]

Got a head on the bench? Take some 1/4" graph paper, lay it on the surface of the head and trace the chamber. Now you can determine how many one inch squares there are and count the left over and irregular (use a bit of estimating when adding these) squares. Once you know how many square inches of area there are and convert to length X width X .050 to find volume. Example- 6 sq. in. = 2 X 3 X .050 = .30 sq. in. Seems like a simple way, or am I all wet?

[2manycars]

Originally Posted by allen

Terry every little bit helps, going 040 and run a H/C head from synders I could always tell the difference. Now for the math formula I guess that would be billfold x Desire = results

I like that. I am going to hang a sign in my shop "Billfold x Desire = results".

[Terry, NJ]

By George!, I think he's got it! Sounds good to me! Although....I could fill it with a measured (cu Cent.) amount. Convert the CCs to Cu. Ins. Then divide the amount into 50.250 cu. ins, or rather a corrected amount of cyl volume using 3.915 ins, and it should give me the CR. Actually, we have an almost perfect way to do this. Think about this, Each 1 1/16in of stroke (4.250) has a value of 1 in the CR. And the .22 equals the ratio of volume left. The volume of Combustion Chamber would be .22 of the volume of 1 1/16 in of stroke or 1/4 of the displacement of the cylinder (50.250 cubic inches) or 12.562 X .22= volume of the combustion chamber, I think! If I'm right, then the answer would be about 3 cubic inches (Estimated)
The problem with this is that if we visualize the c.c. as a series of .050 layers stacked on one another, we notice that the first layer to come off has the greatest amount of volume and it grow smaller with each layer. The last .050 layer is not equal in volume to the first layer due the sloping or rounded walls of the chamber. Back to the drawing board!
Terry

Originally Posted by Fordors

Got a head on the bench? Take some 1/4" graph paper, lay it on the surface of the head and trace the chamber. Now you can determine how many one inch squares there are and count the left over and irregular (use a bit of estimating when adding these) squares. Once you know how many square inches of area there are and convert to length X width X .050 to find volume. Example- 6 sq. in. = 2 X 3 X .050 = .30" Seems like a simple way, or am I all wet?

[Fordors]

Without setting the head dead level and filling it from a graduated burette until the liquid is .050 down (as measured with a depth micrometer) and noting the CC's then covering the chamber with plexiglass and completely filling the rest of the chamber I see no other way. The depth mic suggestion is dubious at best, but the graph paper will give the easiest method and the closest estimate as it will be exclusive of the angle and radii of the chamber walls.

[PC/SR]

Take a string or wire, form it along the edge of the combustion chamber to shape, then straighten the it and divide the length by 4, and it is then a square of with sides of that length. Multiply by 2 for the square inches, and that is the area. Multiply by the depth removed for the volume that is removed by the milling. I did that once by have long forgotten the result, but seem to recall that removing .050 was around 1 cubic inch. I cc'ed some heads too, and it seems one was about 168 cc and one was around174, but dont rely on it. I have heard that the early heads had a smaller combustion chamber and a slightly higher compression ratio.

[Fordors]

Ingenious, as always there is more than one way to skin a cat. Thanks PC/SR .

[Marco Tahtaras]

Originally Posted by PC/SR

Take a string or wire, form it along the edge of the combustion chamber to shape, then straighten the it and divide the length by 4, and it is then a square of with sides of that length. Multiply by 2 for the square inches, and that is the area. Multiply by the depth removed for the volume that is removed by the milling. I did that once by have long forgotten the result, but seem to recall that removing .050 was around 1 cubic inch. I cc'ed some heads too, and it seems one was about 168 cc and one was around174, but dont rely on it. I have heard that the early heads had a smaller combustion chamber and a slightly higher compression ratio.

Ok, to the first point I wish the geometry was that easy! Take that theory and apply it to five point star where the points align in a 4" diameter. Then apply the same to a simple 4" diameter circle. The star will have a significantly greater perimeter (string length) but the circle with have a far greater surface area.

I can't put my hands on my notes at the moment but for my compression ratio spreadsheet I included a stock "A" head gasket compressed to .050". I hope my memory is on track but believe I came up VERY close to one cubic inch or just over 16 cc. Assuming my memory is correct it's funny we have a match here!

I ran Terry's numbers of .040" over and the milled STOCK head and it's so insignificant that I'll leave the fun to others.

[BILL WILLIAMSON]

Terry,
Simple practical answer: Boring will help a "LITTLE" & milling the head will help a "LITTLE!" I have NO formula to back that up. (The dumb Dog's laughin' at me AGAIN!) Bill W.

[PC/SR]

Marco: Thats a brain teaser all right. Something to think about while doing the maintenance and fixing things over the winter. Will get back to you sometime......