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Effect of Longer Stroke?

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Old Jun 2, 2008 | 01:22 PM
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Top Ramen's Avatar
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Default Effect of Longer Stroke?

Hypothetical question - if you stroke a motor, all else being equal, does the timing need to change and, if so, how much? In practical terms, I know the increased displacement will require a bump in fuel, but would the timing map need to be changed?

Thanks in advance to you tuning gurus.
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Old Jun 3, 2008 | 09:08 AM
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Default Re: Effect of Longer Stroke? (Top Ramen)

Simple answer, yes.

Stroke and rod length determine dwell time at BDC and TDC, thus changing the timing "sweet spot."

- Derek
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Old Jun 3, 2008 | 11:27 AM
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Default Re: Effect of Longer Stroke? (D-Rob)

Thanks; I thought the change in dwell would affect timing. It would be minor though, right?
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Old Jun 3, 2008 | 01:33 PM
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Default Re: Effect of Longer Stroke? (Top Ramen)

Depends on what the old stroke is vs the new.

Also depends on RPM.

- Derek
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Old Jun 3, 2008 | 10:03 PM
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Default Re: Effect of Longer Stroke? (D-Rob)

kind of hard to determine what changes are necessary. typically, with increased stroke, you are also raising the compression and that right there limits the amount of advance you could use over the standard stroke.

"The shorter rod is going to have a shorter amount of time at TDC which to me means you could possibly throw more ignition at a high compression motor on pump gas and get away with it since the dwell time is shorter and the piston speeds away from TDC are going to be greater than that of the longer rod."

https://honda-tech.com/zerothread?id=2247744&page=1


Modified by 98vtec at 10:58 AM 6/4/2008
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Old Jun 3, 2008 | 10:43 PM
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Default

R/S ratio will determine (mechanically) the function relating cylinder pressure and the force vector pointing in the direction of crank rotation. (i.e. instantaneous torque)

I made up a really sweet excel spreadsheet that did a bunch of math with force vectors and the ideal gas law. I'll have to see if I can find it. I seem to recall that big stroke motors generally needed more advance to center the burn on the mechanically advantageous spot given comparable burn rates. (which isn't a perfect assumption...)
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