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equation for displacement?

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Old 03-04-2007, 05:19 PM
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what is the equation for the displacement of an engine? is it bore times stroke times rod length or something like that?
Old 03-04-2007, 05:34 PM
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Area of the piston times the stroke length times the number of cylinders:

pi*(radius of piston)^2*stroke*8 (for an LS1)
Old 03-04-2007, 09:09 PM
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Too complicated for me, how about this

Bore X Bore X Stroke X 6.2832 for an 8 cylinder

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Old 03-04-2007, 09:33 PM
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thanks! I tried the bore x bore x stroke x 6.2832 for a 418 l92 and got 415?

Last edited by Bo Duke01; 03-04-2007 at 09:40 PM.
Old 03-05-2007, 07:10 AM
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Its....

Bore x Bore x Stroke x .7854 x # Cylinders
Old 03-05-2007, 11:15 AM
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this is the exact formula, alot of places claim a size close, I've seen 582's listed as 585 and 698's listed at 700 for whatever reason. Go by this formula, don't go by what they list. I did the last step for him and said it was for 8 cylinders only.

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Old 03-05-2007, 12:23 PM
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Your numbers will be slightly off if you don't use pi (you know, that whole 3.1415 number that is used when calculating the area of a circle). The numbers you are seeing are rounded off estimations of that (pi/4 = .7854-ish, (pi/4)*8 = 6.283-ish).
Old 03-05-2007, 12:38 PM
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The volume of a cylinder just equals the area of the base * height = ¶r²h

¶ = pi = 3.14159265

so: 3.14159 x (Bore/2)² x stroke = displacement of one cylinder in in³ if you enter bore and stroke in inches.

multiply this number by however many cylinders you have.

then take that number and multiply it by 0.0164 to get liters of displacement if you want to know that.
Old 03-05-2007, 12:42 PM
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I'm thinking the other guys just don't like geometry...
Old 03-05-2007, 01:43 PM
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Ehm... the stock parts are designed in metric. Meaning the Inches are also slightly off. The Aftermarket parts on the other hand.....
it depends i guess.
Old 03-05-2007, 03:07 PM
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Originally Posted by MeentSS02
I'm thinking the other guys just don't like geometry...
You're right, at least some of the guys don't like geometry. However, this forum is for advanced performance topics.

Calculating displacement (whether using imperial/English/fractional or metric units) is the same: surface area (Pi times the radius squared) times height times the number of cylinders. The units of the answer will change depending upon the units of measure used.

Steve
Old 03-05-2007, 03:12 PM
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Originally Posted by Steve Bryant
You're right, at least some of the guys don't like geometry. However, this forum is for advanced performance topics.
That was my point...only not stated so bluntly. I figured they would get it after a while...
Old 03-05-2007, 07:55 PM
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I'm way too old for all that, especially when you're in the 700's,

705 & 760 is close enough in my book.

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Old 03-05-2007, 10:12 PM
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Well, I'm a 1949 model myself which makes me 57 (58 in April). All I'm saying is that Bo Duke 01 who started this thread asked about the equation for displacement and since this is the advance performance tech forum, I would think that we would strive to help Bo or anyone understand the answer to his question. I've asked a lot of questions here and other places here on LS1Tech because I didn't know the answer and wanted to learn.

Steve
Old 03-06-2007, 01:49 PM
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Pi r^2 (aka ¶r².... aka pie arrr squared) -- is the area of a circle. (r being the radius, and Pi being the Pi constant -- 3.14159...)


if we take that area and multiply it by a height, then we have the area of a cylinder


if we take that area of one cylinder, and multiply it by however many cylinders we have, then we have the total area... also known as displacement.


this is how i remember, and think of it... i was in elementary school when they taught me pie arr squared... so its easy to remember from there.
Old 03-06-2007, 06:20 PM
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^^^ +1
Old 03-06-2007, 06:33 PM
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Maybe I should have integrated in cylindrical coordinates...
Old 03-06-2007, 07:17 PM
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Originally Posted by =Tac=
Its....

Bore x Bore x Stroke x .7854 x # Cylinders
Only LS1Tech Advanced Tech could complicate it beyond this.
Old 03-06-2007, 10:21 PM
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Originally Posted by andereck
Only LS1Tech Advanced Tech could complicate it beyond this.
I don't want to be rude or discourteous. However, It might be good to review the sticky for posting in this forum as opposed to some of the others https://ls1tech.com/forums/advanced-engineering-tech/396223-what-forum-please-read-before-posting.html.

I've participated in many of the other forums here at LS1Tech.com and in Performance-trucks.net over the last several years. Each has its own special emphasis and ground rules. Some of the discussions of cam dynamics, cylinder head aerodynamics, and other topics that are discussed here don't really belong anywhere else. However, many of these discussions have been pretty technical and advanced in nature and provide great opportunities for learning and exchanging ideas.

What thoughts do you folks who participate in this forum think about my line of reasoning?

All my best,

Steve
Old 03-06-2007, 10:27 PM
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Originally Posted by Steve Bryant
I don't want to be rude or discourteous. However, It might be good to review the sticky for posting in this forum as opposed to some of the others https://ls1tech.com/forums/showthread.php?t=396223.

I've participated in many of the other forums here at LS1Tech.com and in Performance-trucks.net over the last several years. Each has its own special emphasis and ground rules. Some of the discussions of cam dynamics, cylinder head aerodynamics, and other topics that are discussed here don't really belong anywhere else. However, many of these discussions have been pretty technical and advanced in nature and provide great opportunities for learning and exchanging ideas.

What thoughts do you folks who participate in this forum think about my line of reasoning?

All my best,

Steve
I'm thinking if you don't know how or don't want to use the number pi in a simple calculation of the area of a circle, you probably don't need to be posting in here. The sticky says advanced mathematics...although the idea of pi is about as basic as you get in the mathematical world.



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