LS1Tech.com

Brownie · 10-14-2002, 08:18 PM

I misplaced my engine calculations book.

How do I calculate cubic inches from bore and stroke?

i.e. 4.060" bore and 4.000" stroke == ??? ci

Also how do I calculate stroke based on rod length and crank info?

Thanks,

Mike

Old SStroker · 10-14-2002, 08:37 PM

Try this site:

http://murray.faithweb.com/formulas.htm#find_rear

Rod length has nothing to do with determining stroke.

Stroke is twice the amount that a rod journal centerline (on the crank) is offset from the main journal centerline.

P&DZs · 10-14-2002, 09:44 PM

Pi x (1/2 bore x 1/2 bore) x stroke x 8 = Displacement

3.1416 x (2.03 x 2.03) x 4 x 8 = 414 cu. in.

[ October 14, 2002, 09:49 PM: Message edited by: P&DZs ]

Black Magic · 10-14-2002, 09:59 PM

I use this formula =

Bore x Bore x Stroke x # of cylinders x .7854

I like using this formula because the .7854 is easy to remember and makes a little square on the calculator. (simple things for simple minds)

Brownie · 10-14-2002, 10:41 PM

Thanks Guys!

Old SStroker - thanks, I thought stroke was affected by rod length. I thought stroke had to do with the distance a piston traveled in the bore (due to crank offset & rod length). Thanks for setting me straight.

Mike

[ October 14, 2002, 10:47 PM: Message edited by: Brownie ]

Red2000SS · 10-14-2002, 11:29 PM

<blockquote>quote:<hr />Originally posted by Brownie:
Thanks Guys!

Old SStroker - thanks, I thought stroke was affected by rod length. I thought stroke had to do with the distance a piston traveled in the bore (due to crank offset & rod length). Thanks for setting me straight.

Mike<hr /></blockquote>Well you're right that the stroke is the distance travelled by the piston - it's just that rod length has no effect on this, as Old stroker pointed out. The stroke is entirely determined by the crankshaft.

Brownie · 10-14-2002, 11:34 PM

<blockquote>quote:<hr />Originally posted by Red2000SS:
 <blockquote>quote:<hr />Originally posted by Brownie:
Thanks Guys!

Old SStroker - thanks, I thought stroke was affected by rod length. I thought stroke had to do with the distance a piston traveled in the bore (due to crank offset & rod length). Thanks for setting me straight.

Mike<hr /></blockquote>Well you're right that the stroke is the distance travelled by the piston - it's just that rod length has no effect on this, as Old stroker pointed out. The stroke is entirely determined by the crankshaft.<hr /></blockquote>Gotcha! Thanks. Had to think about that a minute, it makes perfect sense now.

Mike

[ October 14, 2002, 11:42 PM: Message edited by: Brownie ]

v8 ute · 10-15-2002, 05:22 AM

<blockquote>quote:<hr />Originally posted by Black Magic:
I use this formula =

Bore x Bore x Stroke x # of cylinders x .7854

I like using this formula because the .7854 is easy to remember and makes a little square on the calculator. (simple things for simple minds)<hr /></blockquote>Your close, but no cigar.
I 'spose close enough is good enough for simple minds tho' <img border="0" title="" alt="[Wink]" src="gr_images/icons/wink.gif" />

Old SStroker · 10-15-2002, 07:10 AM

<blockquote>quote:<hr />Originally posted by v8 ute:
 <blockquote>quote:<hr />Originally posted by Black Magic:
I use this formula =

Bore x Bore x Stroke x # of cylinders x .7854

I like using this formula because the .7854 is easy to remember and makes a little square on the calculator. (simple things for simple minds)<hr /></blockquote>Your close, but no cigar.
I 'spose close enough is good enough for simple minds tho' <img border="0" title="" alt="[Wink]" src="gr_images/icons/wink.gif" /> <hr /></blockquote>Volume of a cylinder is diameter times diameter times length(stroke) times [Pi divided by 4].

Pi/4 is .785398163 so .7854 is 99.9997661% correct. That's about 8/10,000 (.0008)of a cubic inch on an LS1. That's close enough even for professional racing organizations, as well as being simple to remember.

The "little square" as a memory technique never occurred to me. That's elegant, Black Magic! Thanks.

[ October 15, 2002, 11:31 AM: Message edited by: Old SStroker ]

Dyno · 10-15-2002, 01:36 PM

Bore * Bore * Stroke * 6.2832 = CI for a 8 cylinder.

Bob2of3 · 10-15-2002, 01:42 PM

Well I sometimes forget where I left my calculator so THIS makes it alot easier on me <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" />

LAter <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />

ls1290 · 10-15-2002, 04:19 PM

An LS1 is made up of 8 cylinders. Volume of cylinder is

v = pi * ( radius/2 )^2 * height

So, the total volume of the motor be

CID = 8 * v

Or you can just use the following

Engine Calculator

<img border="0" alt="[Fluffy]" title="" src="graemlins/fluffy.gif" />

BurnOut · 10-15-2002, 04:32 PM

Brownie- the stroke DOES determine how far the piston moves up and down the bore, but the length of the rod has nothing to do with how far the piston travels, only "where" it travels.

In any event, my displacement calculation formula is as follows: (bore/2)^2 * pi * stroke * cylinder count. So your example of an 8 cylinder motor with a 4.06" bore with a 4" stroke would calculate like this: (4.06/2)^2 * 3.1415927 * 4 * 8 = 414.278 cubes. Really, all of the formulas people have recited are just different permutations of the same concept... gotta first find the area of the bore, multiply that by the stroke to get cylinder volume, then multiply that by the number of cylinders.

[ October 15, 2002, 04:40 PM: Message edited by: BurnOut ]

Brownie · 10-15-2002, 07:55 PM

Thanks guys! Great info.

I appreciate gettin me squared away.

Volume of a cylinder <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />

Stroke makes perfect sense now and I see how the rod length has nothing to do with the distance it travels.

Mike