One rod 10 grams heavier on big end

[Tuner]

stroke is 8 inches now. that's a big engine
Whatever, I love you buddy. your kinda my hero
it's a shame we don't have better things to discuss at the moment.
You know one thing I never figured out is.....in a wet flow intake manifold it seems like with a sharp enough turn it bounces the fuel back into suspension, but a more gradual turn throws the fuel at the outside wall, and a very gradual turn it stays in suspension.
easy right?
But then there is accelerating and decelerating flow, surface to volume ratio, coarseness of atomization, temperature, speed difference between the wet flow and dry, and renolds number, and all that. easy to say, but exactly how sharp is sharp, in what circumstances, i still have never figured out.

It is not the stroke length .... it is the radius of the weight imbalance from the axis of rotation. How far the rod cap is from the center of the main bearing when the piston is at the bottom of the stroke. The circumference of the rod cap gyration is offset from the crank axis, inside the crank pin at TDC, and outside at BDC.

I confess to thinking of it as a balance knob on the rod cap where weight is usually removed for balancing, but the I beam rod is not so. In any case, the imbalance force depends on the actual radius of the excess weight from the axix of rotation.

[Chris_Hamilton]

I must confess Tuner, at first I thought you were serious.

[Rick!]

stroke is 8 inches now. that's a big engine
Whatever, I love you buddy. your kinda my hero
it's a shame we don't have better things to discuss at the moment.
You know one thing I never figured out is.....in a wet flow intake manifold it seems like with a sharp enough turn it bounces the fuel back into suspension, but a more gradual turn throws the fuel at the outside wall, and a very gradual turn it stays in suspension.
easy right?
But then there is accelerating and decelerating flow, surface to volume ratio, coarseness of atomization, temperature, speed difference between the wet flow and dry, and renolds number, and all that. easy to say, but exactly how sharp is sharp, in what circumstances, i still have never figured out.

It is not the stroke length .... it is the radius of the weight imbalance from the axis of rotation. How far the rod cap is from the center of the main bearing when the piston is at the bottom of the stroke. The circumference of the rod cap gyration is offset from the crank axis, inside the crank pin at TDC, and outside at BDC.

I confess to thinking of it as a balance knob on the rod cap where weight is usually removed for balancing, but the I beam rod is not so. In any case, the imbalance force depends on the actual radius of the excess weight from the axix of rotation.

Your intentions are good, your interpretation of the physics is not. And yes, I looked at that nomograph - it's what Civil Engineers use because they don't have to go through all the Calculus that Mechanical and Aero students did.
Using a fundamental imbalance formula, one can easily understand what it's calculating.
Force = mass x radius x (rotational velocity)
Using SI units;
mass = kilogram
radius = meter
rotational velocity = radians per second
The first thing you do is a dimensional analysis of the right hand side (RHS) to make sure the final units are in Newtons (force).
So, Force = mass x meter x radians²/seconds²
............= mass x meter/seconds² x radians, radians are unitless
and finally, Force = mass (m) x meter/seconds² → F = ma, the formula every one is familiar with.
Now, just substitute in values and calculate.
mass of imbalance = 10 grams = .01 kg
stroke of 400 sbc crankshaft = 3.75", radius = 1.875" (0.047625 meter)
rotational velocity = 6000RPM x 2π x 1 minute/60 seconds = 6000RPM x 2π/60
Finally, Force = .01kg x 0.047625 meter x (6000 x 2π/60)² = 188 Newtons or 42.3lbf (1 lbf = 4.448 Newtons)
If you calculate the acceleration is G's, you'll get around 1917, a wee bit less than the value you calculated.
The story about the circle the 10 grams makes around the crankshaft center is partly true, you still don't double the stroke. Rotational acceleration, or alpha (α) is based on the radius of the orbit, not the diameter, ever. This is the reason why the RPM term is squared → rotation velocity² x radius = alpha. This is a fundamental relationship that should not be misinterpreted.
To put things in perspective, the rotational bobweight on a SBC is right around 923 grams per this balance sheet, of which 10 grams is 1.1%. Just bolt the sumbitch in and do some Roadkill Garage burnouts. (the emoji signifies sarcasm)

[rfoll]

It's easy to overthink things, It can be a form of tunnel vision. We have all been there. A nice 383 isn't formula 1. When I bought my first scale, I started weighing everything. I was astounded at the difference in weight in sets of stock sbc rods, yet the engines run smooth. As a curiosity, I weighed a set of Vortec piston and rod, (powdered metal), assemblies. I found less than 2 grams difference in the whole set, most less than a gram. Like the man said, run it.

[modok]

For the 90 degree engine to be perfectly balanced it would need very long connecting rods in the same plane, not next to eachother.
So it's not perfect in any case, even if all the parts were perfect, so everything has a tolerance.
Overall consensus is IMO, If the balance factor is between 50 and 51% then it's about as good as it can get.

So depending on where it already was in the tolerance before, it could be balanced better than it was, although more likely it's slightly worse.

[Tuner]

People who take money for balancing engines are just frauds, that's the ticket, because you don't have to do anything but take the money, how much weight or imbalance doesn't matter because nobody can feel it anyway. Let it shake.

[smokie]

Isn't is half the stroke? 3.75/2=1.875. That's shows almost 1/2 pound out of balance.

[Dave Koehler]

4" radius
6500 rpm
10 grams
= 428.1 lbs of force