I've got a 77 f-150 w/ 460 C6/ 3.00 9".
It has an Edelbrock performer manifold, 750 carb and double roller chain, CompCams 268H, lifters and valve springs, with Dynomax headers, 2 1/4" pipes with mustang flowmasters.
Anyway, My question is, what is the actual bore and stroke of the 460? The Chiltons and Haynes manuals are wrong because they only come up to about 422ci. They also say that the 460 has a shorter stroke that the 400ci.
Thanks, Jimmy

 

460 bore is 4.36" and stroke is 3.85"

 

That only comes up to 421.66 CI.
The formula is Bore x Stroke x # of cyl. x 3.14 = CI
4.36 x 3.85 x 8 x 3.14 = 421.66 or 422
4.00 x 3.00 x 8 x 3.14 = 301.44 or 302
4.00 x 3.50 x 8 x 3.14 = 351.68 or 351
4.00 x 4.00 x 8 x 3.14 = 401.92 or 400

They are all close except for the 460. Those are the same dimensions that I had.
Jimmy

 

Here is a formula I have and it works out.

0.7853982 x bore x bore x stroke x cylinders = CID

Which comes out to 459.84723 rounded off = 460.

 

Thanx, That works!!

 

I cut that out of a Hot Rod magazine eons ago and kept it.

 

I'm glad you saved it.
Thanks again,
Jimmy

 

here is a good site.
http://www.smokemup.com/utilities/calc/displacement.cfm


Dre'
1978 F-150, 460ci(750cfm,Edelbrock Performer,Dynomax Ceramic Coated headers,etc.)Auto-C6,373:1 - 9" Posi.

 

just to add something, the 400 does have a slightly longer stroke than the 460, i forget the #......

 

The 400 has a 4" bore and a 4" stroke.

 

The formula mathematically is pi X radius squared X height X number of cylinders.
This equates to 3.14 X (half bore size)squared X stroke X 8.
You get the same number as the formula. It is a volume formula of a cylinder.
Mark


Ford started it; Ford will finish it!

 

[updated:LAST EDITED ON 07-May-02 AT 02:23 AM (EST)]Huh ????? :-) Just kidding.