How do I calculate cubic inches from bore and stroke?
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
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.
3.1416 x (2.03 x 2.03) x 4 x 8 = 414 cu. in.
<small>[ October 14, 2002, 09:49 PM: Message edited by: P&DZs ]</small>
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)
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
<small>[ October 14, 2002, 10:47 PM: Message edited by: Brownie ]</small>
<strong>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</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">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.
<strong> </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Brownie:
<strong>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</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">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.</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Gotcha! Thanks. Had to think about that a minute, it makes perfect sense now.
Mike
<small>[ October 14, 2002, 11:42 PM: Message edited by: Brownie ]</small>
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<strong>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)</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">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" />
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<strong> </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Black Magic:
<strong>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)</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">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" /> </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">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.
<small>[ October 15, 2002, 11:31 AM: Message edited by: Old SStroker ]</small>
LAter <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />
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" />
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.
<small>[ October 15, 2002, 04:40 PM: Message edited by: BurnOut ]</small>
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






