Cam question...

[Crustyfmb]

I've found a cheap set of SE 497 cams for sale and wondered if it's worth putting them in or not?
I've got an ultima ignition (single fire), HSR42 carb and S&S 2-1 exhaust already, but reading some articles/posts there seems to be contradicting info on whether there is any real benefit with stock 1200 heads.
I don't plan on spending all my time at top end, though I do like to play, and wouldn't want to loose out too much at the bottom/mid where most of my riding will be.
Any advice is welcome 👍

[kitabel]

The less efficient the heads are, the more benefit to hotter cams.
Is it fast enough now from 4,000 up?

[JB in NC]

Experiment is where the fun is. Go for it and report.

User warning: cams are a gateway drug.

Jim

[sportsterpaul]

Ran them in my 2000 with warmed up CV.40, Hi4 ignition and a Hooker 2:1 was one of my favorites ever.

[rejeanprimeau]

Check if the gear is made to fit in you bike model, if not you'll need a press to change it.

[aswracing]

9:1 is awfully soft for a 48 degree IVC. Those cams work best over the range of 10:1 to 10.5:1.

More info: https://www.hammerperf.com/ttvideos....mpressionratio

[kitabel]

I watched enough of the video to disagree. The speaker states that, since the intake valve is still open after BDC, only part of the stroke is captured.
At cranking, idle, and very low RPM this is generally true.
However, as the torque level rises, the entire stroke length is captured and VE approaches (and even passes) 100%, and static CR governs.
How is this possible? When the inertia of the intake charge is great enough to stop reversion. With a large port and valve, slow gas speed, and a large port volume as a buffer it occurs at higher speed than the reverse case, but it always happens.

I might also mention that DCR and CCP (closely related) do not predict engine power acccurately. Two different states of tune in the same engine (high static CR + late intake closing vs. low CR + early closing) may produce exactly the same DCR & CCP, but completely different manners and behavior.

[aswracing]

Originally Posted by kitabel

I watched enough of the video to disagree. The speaker states that, since the intake valve is still open after BDC, only part of the stroke is captured.
At cranking, idle, and very low RPM this is generally true.
However, as the torque level rises, the entire stroke length is captured and VE approaches (and even passes) 100%, and static CR governs.
How is this possible? When the inertia of the intake charge is great enough to stop reversion. With a large port and valve, slow gas speed, and a large port volume as a buffer it occurs at higher speed than the reverse case, but it always happens.

I might also mention that DCR and CCP (closely related) do not predict engine power acccurately. Two different states of tune in the same engine (high static CR + late intake closing vs. low CR + early closing) may produce exactly the same DCR & CCP, but completely different manners and behavior.

There''s so much wrong with this that it's a little bit difficult to know where to start. But in all fairness, part of it is my own fault, for not being explicit enough in the video. Something I've learned from doing these videos is that if you leave any room at all for misinterpretation, someone will do it. I don't do these from a script, I ad-lib them, and what seems obvious to me isn't always obvious to the observer. Most guys get it right, but a few go off in left field because of this. As an engineer, I should know better. When conveying information, you need to be exact, to avoid misunderstandings.

In any event, the opening comment in the above is one such misinterpretation ..

The speaker states that, since the intake valve is still open after BDC, only part of the stroke is captured.

I was referring to mechanical compression, not cylinder fill, when I said that only part of the compression stroke is used for compression. I should have used that term to avoid confusion. I didn't say only part is used for "capture" though. That's a misinterpretation. Anything not shoved out by reversion ends up getting captured, which I *thought* was obvious. You're basically conflating cylinder fill with compression, which I left the door open for that. My bad.

However, as the torque level rises, the entire stroke length is captured and VE approaches (and even passes) 100%, and static CR governs.

You've got the cause and effect backwards here. Cylinder fill increasing causes torque to increase, not the other way around. I actually mention this in the video.

When the inertia of the intake charge is great enough to stop reversion. With a large port and valve, slow gas speed, and a large port volume as a buffer it occurs at higher speed than the reverse case,[/B].

The video is explicit in talking about the ideal IVC being just before the point of reversion, perhaps you didn't watch that far. But you have the relationship between intake charge velocity and engine speed exactly backwards.

There's actually a well-understood number for the intake port velocity that delivers maximum cylinder fill. And there's a well understood process for figuring it out in any given motor at any given rpm.

but it always happens.

Not true. I once saw a motor being dyno'ed that would shoot a 3-foot stream of fuel out the mouth of the carb during a pull, over the entire rpm range. The IVC wasn't even that late, in the mid 50's. The lesson was that r/s ratio and IVC have to be matched. This particular motor was a short stroke, long rod arrangement, and the cam selection had not taken that into account. Pulled the IVC back and it ran like a bat out of hell.

I might also mention that DCR and CCP (closely related) do not predict engine power acccurately. Two different states of tune in the same engine (high static CR + late intake closing vs. low CR + early closing) may produce exactly the same DCR & CCP, but completely different manners and behavior.

This statement completely ignores the effect of rpm on power. You describe one motor configured for high rpm and another configured for low rpm and state that they'll have different power. Of course they will. Who would dispute that? RPM is half the power equation. Leave torque the same, and double the rpm, and you'll have double the power.