What does 122 pmm of lead mean? As a reality check I worked it out. My understanding is the the PPM is mass based (not volume).
Given that:
The engine holds 4.5 quarts or 4257ml. If the density is .866 g/ml I have 3686.6 grams of oil. If the lead is at 122 ppm, then there is at least 0.45 grams of lead missing from the rod bearings. Lead has a density of 11.43 grams per cc.
11.34*X = 0.45, X = .0397 cc's or 3.9 mm^3
There are four bearing sets that are about 18.8mm wide and they are about 48mm in diameter.
The surface area of one bearing is:
3.14*D*Width, 3.14*48*18.8= 2833mm^2
Four bearings: 11334mm^2
How thick a layer do I need to remove to account for 3.9mm^3 of lead in the oil? Assuming that all surfaces eroded evenly (I know that's not going to happen but I'm just trying to get an understanding of magnitude).
I get 0.000344mm. ACL says that the top layer is 0.013 thick. So its possible to get a 122 ppm reading and not have a large increase in clearance.
What happens if it was just one bearing and half of what was lost off of the bearing got trapped by the filter?
Now we have 7.8 mm^3 of lead missing from one bearing. That's a .0028 mm thick layer, so the shell would still have a 0.0102mm of lining left before the crank hits the nickel barrier layer that is behind the top surface.
The running clearance would have changed .0001".
My take is the I need to swap the bearings out, they are used up. I will inspect the rest of the engine, while I have it out and make a few minor upgrades.
King Bearings has a lot of good information of what fatigued bearings look like and what might be happening to mine.
Given that:
The engine holds 4.5 quarts or 4257ml. If the density is .866 g/ml I have 3686.6 grams of oil. If the lead is at 122 ppm, then there is at least 0.45 grams of lead missing from the rod bearings. Lead has a density of 11.43 grams per cc.
11.34*X = 0.45, X = .0397 cc's or 3.9 mm^3
There are four bearing sets that are about 18.8mm wide and they are about 48mm in diameter.
The surface area of one bearing is:
3.14*D*Width, 3.14*48*18.8= 2833mm^2
Four bearings: 11334mm^2
How thick a layer do I need to remove to account for 3.9mm^3 of lead in the oil? Assuming that all surfaces eroded evenly (I know that's not going to happen but I'm just trying to get an understanding of magnitude).
I get 0.000344mm. ACL says that the top layer is 0.013 thick. So its possible to get a 122 ppm reading and not have a large increase in clearance.
What happens if it was just one bearing and half of what was lost off of the bearing got trapped by the filter?
Now we have 7.8 mm^3 of lead missing from one bearing. That's a .0028 mm thick layer, so the shell would still have a 0.0102mm of lining left before the crank hits the nickel barrier layer that is behind the top surface.
The running clearance would have changed .0001".
My take is the I need to swap the bearings out, they are used up. I will inspect the rest of the engine, while I have it out and make a few minor upgrades.
King Bearings has a lot of good information of what fatigued bearings look like and what might be happening to mine.