I know the formula "bore * bore * stroke * .7854 * 8" is used to find the displacement on a V8 but... how do you factor in the difference between deck heights? Ex: GM lists the max bore and stroke for both short-deck and tall-deck MarkIV BowTie big block as 4.560(b) and 4.250(s) but the short-deck height is 9.800" and the tall-deck is 10.200". Shouldn't the tall-deck put out more c.i.? How would I find the max cubic inch for the tall-deck?

 

This is all you need.
(bore X 3.14) X stroke X # of cylinders
............^ that's pie, computer key bord dosen''t have the symble on it. It looks like ~ over top of 2 stright lines (up to down).
And to know the max bore and stroke the engine can take.
You can use this to find displacement of any bore with any stroke and any # of cylinders.

 

Thanks for the reply Oil Pan. I've never seen the method you gave but I tried it and came up with something different (I must've done something wrong). I'm only used to the usual B x B x S x .7845 x 8 method. Given the specs that GM gave, both the short-deck and tall-deck blocks equal out to the same cubic inch which doesn't seem right since the tall-deck has .400" more room per bore.

 

Oil pan is a little off... the area of a circle is Pi * r^2 (pi times the radius of the circle, squared).

The constant you've been using in your formula (0.7854) equates to the same thing, but using diameter (bore size) instead.

Using a bore of 4.030" (r = 0.5 x 4.030 = 2.015")...

4.030" x 4.030" x 0.7854 = 12.756 sq. in.

3.14159 x (2.015" x 2.105") = 12.756 sq. in.

Same thing. Then multiply by the number of pistons and the stroke of the crankshaft, and you have displacement in cubic inches.

As for your concern about the deck height, the formula is the same. A tall deck block will simply use longer rods and/or larger stroke because of the increased deck height. All you need to know is the size of the bore and the stroke of the crankshaft.

 

Thanks Jimlab. I know the bore (4.560) and stroke (4.250) of the Bow Tie Big Block which comes out to 555 c.i. but I thought perhaps somehow with the taller deck that 600 c.i. could be achieved. But now that I really think about it, you're right about the longer rod. Thanks again

 

The taller deck may provide more room for a longer stroke, but if you put a 3.75" throw crank in it, the displacement is determined by using 3.75". You don't take credit for the increased deck height unless you put in a longer throw crank - and you still use the crank throw dimension, not the deck height for the calculation.

Making the deck higher taller, the rods longer, moving the writst pin, etc. doesn't change the "displacement", which is defined at the volume displaced by the piston as it moves through the stroke. Bore and stroke defines the displacement, nothing else.

 

I learned that formula in 5th or 6th grade.
It is right just use it to bore and stroke a known engine, a 350 , 4'' bore and 3.480'' stroke and you get 349.67 cubic inches useing my math.

 

Try it with a bore size other than 4". 4" only happens to works because the radius (2) squared equals 4...

Your formula: (bore X 3.14) X stroke X # of cylinders
4.030" x 3.14 x 3.750" x 8 = 379.63 CID = 380?
4.030" x 3.14 x 3.875" x 8 = 392.28 CID = 392?

My formula: bore x bore x stroke x # cylinders x 0.7854
4.030" x 4.030" x 3.750" x 8 x 0.7854 = 382.67 CID = 383
4.030" x 4.030" x 3.875" x 8 x 0.7854 = 395.42 CID = 396

You have to use Pi r^2 with your formula for it to be accurate, otherwise it only works for a bore size that is a perfect square of the radius.

 

r^2 X pi thank you
I thougt I had to dosome thing with the r but I couldn't rember what. So I tryed it with out and it seamed to work.

 

when I want to find the displacement using different specs I just run it through DeskTop Dyno 2000 or this program http://www.oldwestracing.com/software/HighPerfMath.zip

 

>Here is another auto math site
http://www.smokemup.com/auto_math/index.php