LS1 block is 319-T5, heads are 356-T6.
Supercharged Northstar block and heads are 319-T7... references to the SC lower crank case have not mentioned the alloy.
I haven't found a page that specifically calls out the alloy used in the naturally aspirated Northstars, but I will @$$ume that it's 319-T7 just like the SC ones.

However, this site: http://www.key-to-metals.com/Article106.htm mentions that different alloys are preferred for sand vs die casting. The SC block is sand cast, but all lower crank cases and all N/A blocks are die cast... Hmm... I found another page that mentions 380 as being good for die casting and widely used in the automotive industry.

Vega 4 cylinder blocks were 390 aluminum.
Hypereutectic aluminum alloys have >12% Silicon.

Anyway...
This page: ( http://www.anidatech.com/hot.html#material ) lists coefficient for 380 aluminum at 1.21E-5/*F, but does not list 356-T6 for comparison.

Since that's the only number I've found, I'll use it.

P = piston position
V = piston velocity
A = piston acceleration

R = connecting rod length (5.943")
T = crank throw (1.654")
a = crank angle ATDC
S = RPS (141.7)

P(t) = Tcosa(t) + (R²-(Tsina(t))²)^1/2

V(t) = dP(t)/dt = (da(t)/dt)(dP(t)/da) = -(da(t)/dt)Tsina(t) - (da(t)/dt)T²sina(t)cosa(t) / (R²-(Tsina(t))²)^1/2

A(t) = dV(t)/dt = -(da(t)/dt)²Tcosa(t) - (d²a(t)/dt²)Tsina(t) -
((R²-(Tsina(t))²)^1/2 ((d²a(t)/dt²)T²sina(t)cosa(t) + (da(t)/dt)²T²cos²a(t) - (da(t)/dt)²T²sin²a(t)) - (da(t)/dt)²T⁴sin²a(t)cos²a(t)(R²-(Tsina(t))²)^-1/2) / (R²-(Tsina(t))²)

a(t) = 2 Pi S t
for 8500 RPM, S = 141.7
a(t) = 890.1t which is an angular velocity of 890 rad/s

I'm only concerned about acceleration at TDC, so I set t=0 (which means a(t) = 0) so that all the sin(a(t)) terms magically go away and all the cos(a(t)) terms go to 1. What remains is:

A(t)|_t=0 = (da(t)/dt)²(-T - T²/R)

da(t)/dt = 890
(da(t)/dt)² = 792,309.9

So piston acceleration at TDC in a Northstar at 8500 RPM is...

-1,675,201.25 in/sec²
-139,600.10ft/sec²
~ -4,362.50 g's

Edited 'cause I blew a sign the first time...

This is the "conventional" estimate of piston acceleration taking only stroke and angular velocity into account.
P(t) = Tcosa(t)

V(t) = -(da(t)/dt)Tsina(t)

A(t) = -((d²a(t)/dt²)Tsina(t)+(da(t)/dt)²Tcosa(t)

Again, the sine term goes to zero and the cosine term goes to 1 at TDC and we have

A(t)|_t=0 = -(da(t)/dt)²T

which from above is 792,309.9 * 1.654 = 1,310,480 in/sec²
or 3,412 g's, which differs significantly (22% low!) from the *correct* number above.

I picked up on the fact that I'd made a sign error by doing a sanity check and noticing that the number taking rod length into account was less than the one without... the opposite should be true. By not taking rod ratio length into account, the equation is implicitly for an infinitely long rod, and lengthening the rod reduces piston acceleration. With the sign error, lengthening the rod increased piston acceleration.

With a (estimated) 0.400 kg piston, this comes to 16,924 N of tension on the rod, which is in the neighborhood of 4200#. On a 0.240 in² cross section, this is a stress of about 17,600 psi.

This obviously neglects the mass of the rings, oil, pin and small end of the conrod, which I will add in later (when I have the balance sheet in front of me).

Piston weight is TBD but in the 375-400 g range
Stock pins weigh 112 g
Locks weigh 2 g
Rings weigh 30 g
Rod small end weighs 158 g

So total weight less the small end is 544 g. This leads to tension of ~5750#.
The tension from the mass of the small end can not be added in as easily, because that mass is distributed along the shank of the rod. Thus the rod is under greater tension close to the big end and less close to the small end. I am surprised that the rods are not tapered. The difference in stress leads to a difference in strain and fatigue life across the rod.
The small end's 158 g's translates to a strain of 1670# at the big end and 0# at the small end. Thus the average stress through the rod, and the number that should be used in the calculation of total rod stretch is 835#.

So the total rod tension for the purposes of calculating stress & strain is 6585#. Using a 0.240 cross section, this is a stress of 27,437 psi. With elastic modulus numbers between 30 & 33 Mpsi, this translates to a strain of 0.00091 - 0.00083 and elongation of 0.0049-0.0054". Wow. That's less than I expected...

But quench needs to be more than that. We need to add ~0.003-4 for bearing and piston pin clearances, which will drop to microinches with this much tension on things. Crank stretch needs to be taken into account, although that will surely be VERY minimal due to the MUCH larger cross sectional area of the sides of the crank throw. Piston stretch may account for a good chunk, but I'll need to talk to CP about that...

Yes, that's on my list, as well as oblonging of the big end and small end bores.