LS Valvetrain Limits Before it gets too unreliable?
Thread Starter
Teching In
Joined: Nov 2012
Posts: 17
Likes: 0
LS Valvetrain Limits Before it gets too unreliable? Hey guys I bought a forged bottom end package with a 3.0 stroke from a friend. Can't really change it but I was wondering if money was no concern what's the limit I can build a valvetrain to rev without heavy maint? Like Springs every 10-15k miles seems fine to me. This will mainly be a street car with track about twice a year. Lets just say my budget was unlimited what can I do? I've got one builder saying once it goes above 8000 it gets messy. I was hoping for 8500-9000. Using a dry sump of course. And before you want to question why I would want RPM's that high it's because my bottom end can handle it and I think a high RPM V8 sounds awesome. Just a personal project of mines not looking to break any records except with my wallet lol.
Thank you
Thank you
Thread Starter
Teching In
Joined: Nov 2012
Posts: 17
Likes: 0
Staging Lane
Joined: Feb 2005
Posts: 97
Likes: 2
From: Lynbrook, NY
The reciprocating valve train parts become more difficult to control (duh), but what is frequently misunderstood is that the required spring tension is not linear to the RPM - it's proportionate to the RPM^2: 9,000 isn't 12.5% more spring than 8,000, it's more than 26% higher.
Its the same with RPM vs. stroke length: the inertial load on the rod and journal vary (roughly) as the square of the RPM, so the obvious 3.000" vs. LS1 etc. 92mm isn't 3.622 / 3.000 = +27%, it's closer to +12% RPM at the same tensile load. 7,000 RPM > 7,848 RPM.
The math? Where
Z: piston acceleration, in f/s/s
N: RPM
S: stroke length, in inches
n: rod ratio
2189: a constant
Z = (N^2*S*(1+1/2n))/2189
Its the same with RPM vs. stroke length: the inertial load on the rod and journal vary (roughly) as the square of the RPM, so the obvious 3.000" vs. LS1 etc. 92mm isn't 3.622 / 3.000 = +27%, it's closer to +12% RPM at the same tensile load. 7,000 RPM > 7,848 RPM.
The math? Where
Z: piston acceleration, in f/s/s
N: RPM
S: stroke length, in inches
n: rod ratio
2189: a constant
Z = (N^2*S*(1+1/2n))/2189
Thread Starter
Teching In
Joined: Nov 2012
Posts: 17
Likes: 0
The reciprocating valve train parts become more difficult to control (duh), but what is frequently misunderstood is that the required spring tension is not linear to the RPM - it's proportionate to the RPM^2: 9,000 isn't 12.5% more spring than 8,000, it's more than 26% higher.
Its the same with RPM vs. stroke length: the inertial load on the rod and journal vary (roughly) as the square of the RPM, so the obvious 3.000" vs. LS1 etc. 92mm isn't 3.622 / 3.000 = +27%, it's closer to +12% RPM at the same tensile load. 7,000 RPM > 7,848 RPM.
The math? Where
Z: piston acceleration, in f/s/s
N: RPM
S: stroke length, in inches
n: rod ratio
2189: a constant
Z = (N^2*S*(1+1/2n))/2189
Its the same with RPM vs. stroke length: the inertial load on the rod and journal vary (roughly) as the square of the RPM, so the obvious 3.000" vs. LS1 etc. 92mm isn't 3.622 / 3.000 = +27%, it's closer to +12% RPM at the same tensile load. 7,000 RPM > 7,848 RPM.
The math? Where
Z: piston acceleration, in f/s/s
N: RPM
S: stroke length, in inches
n: rod ratio
2189: a constant
Z = (N^2*S*(1+1/2n))/2189
So Far if I'm revving to 8500 I'm thinking
3/8 pushrods
Titanium Valves coated
Shaft mounted rockers
Correct springs
What would be the first thing I'd be looking at breaking at that RPM?
Staging Lane
Joined: Feb 2005
Posts: 97
Likes: 2
From: Lynbrook, NY
I'd find a SpinTron (if you can afford it), or ask a cam pro (not a salesman, too bad Harold isn't with us) whether your lobes are stable with the spring load. The rocker ratio (if higher) is also relevant, since it adds spring load to the tappet and pushrod.
