What's my displacement? Help!

[Bearcuda]

Alright you motor gurus, help me out here. I'm trying to determine what my 440s displacement is since it had a .040 over bore. According to the 1969 specs, the 440 had three slightly different bore and strokes:
1969 Cu. In. 340 383 440 Bore (in.) 4.04 4.25 4.32 Stroke (in.) 3.31 3.38 3.75 Compression 10.50:1 10.00:1 10.10:1 Induction 1x4 bbl. 1x4 bbl. 1x4 bbl. Horsepower/RPM 275/5000 330/4400 375/4600 Torque (lbs.-ft.)/RPM 340/3200 425/3400 480/3200
I know the formula is (bore area x stroke)x # of cylinders right? If the bore is one of those above numbers, what is the bore area? What's my cid?

[matt63]

The 440 only had one bore and stroke.  I believe the stroke is 3.75 and the bore is 4.32.  Boring an engine to increase displacement gives a negligible increase in displacement as hp is lost due to thinner cylinder walls.

[moper]

4.325 bore, 3.75 stroke. with a .040 overbore, actual displacement will increase to 448 cubic inches. .030 over is 446, .060 over is 451

[Bearcuda]

Hmmm, could it be .40 instead or is that way too much? What's the deal with those multiple numbers I got for mopar 340/440 for 1969?

[matt63]

Those multiple numbers are for the 3 engines listed in sequence ie the 4.04 is for the 340 and the 4.25 is for the 383 and the 4.32 is for the 440.

[Bearcuda]

Those multiple numbers are for the 3 engines listed in sequence ie the 4.04 is for the 340 and the 4.25 is for the 383 and the 4.32 is for the 440.

Oh yeah, duh.

[PlumCraZRT]

Oh yeah, duh.

 

[black71]

.785 x bore x bore x stroke x # cylinders
.785 x 4.36 x 4.36 x 3.75 x 8 = 447.676 cubic in.