From: "Shannon Knoepflein" <kycshann@kyol.net>

I feel that is bunk.  I'm no ME thermo genius, but here is my simple
calculation.  We need to figure out how much heat is removed by the air
or the gas, and my simple mind says a specific heat capacity calculation
would give us this.  First, assumptions (estimates, lets not get too
caught up in exact numbers):

Peak is about 15:1 AFR
ROP, lets set it at 12:1 AFR
LOP, lets likewise set it at 18:1 AFR

On the ROP side, we have 12/1=15/x, or x=1.2, or 0.2 extra parts of
gasoline
On the LOP side, we have 18-15, or 3 extra parts of air

Energy = (temperature change)*(parts)*(specific heat capacity)

SHCoxygen=0.22
SHCgasoline=2.22

So, disregarding the temperature change as I really don't know what it
is, 0.22*3 parts air=0.66 and 2.22*0.2 parts gasoline=0.44

I'm really not sure if I'm anywhere close to making any sense here.  I'm
sure some ME can step in and set it straight.

However, here is the bottom line.  CHT's are lower for LOP, that is
proven, which also mean the valve seats and valves are cooler for LOP.
This means air carries away more heat than the gasoline, which to me
seems to disprove your theory below.  Now, running "leaner" or "not lean
enough" could certainly burn valves, as the highest CHT's (and valve
temp) occurs at about 50-70 ROP, so if you are running there at high
power, look out.  Running 70 LOP puts you way out of this danger zone.

Heat is heat, if the CHT's are cooler, the valve is cooler, and
something is carrying that heat away.  The only change is extra air
versus extra gasoline, so I have to believe the air works better, and
creates no danger to burning a valve.

Please someone that knows more shed some light on this.

Shannon