Official S65 Bearing Specification/Clearance Wiki

[Yellow Snow]

Originally Posted by swamp2

If I understand your objection correctly this is entirely incorrect. You are just talking about the error from the curvature of the bearing and not being at the true saddle of the bearing, right?

This problem is solved simply by Pythagoras formula. The height "error" from moving 1/10th (of a thousandths of an inch) is a whopping

0.0000000025" (about 6 x 10^-5 micron)

The reason being is obvious because the circle is basically flat here at these scales.

It takes moving about 0.02" to amount to an error of 1/10th (of a thousandths).

To me how the method is easily accurate is that one can manually "hunt" for the smallest possible number using the natural small spring pressure in the indicator, anywhere within a 40 thou band size will basically give the right answer.

Thankyou Swamp. You got it in one

[Bx Tpr]

So this

"So there you have it: ideal mains should be 0.0018 - 0.0029 and ideal rods should be 0.00153 - 0.00205. You're given 0.00125 clearance from the factory and some may be thinking of reducing it another 16-40% with coated bearings without sizing the journals because they believe it gives them extra protection. The extra protection is true if you kept all things the same including the bearing clearance; but reducing the bearing clearance by 16-40% to get the extra protection of the coating is not recommended."

And this




Will correct the bearing clearance issue?

What side of the rod was machined down? The fillet side or the side that faces the other rod?

[kawasaki00]

Originally Posted by Yellow Snow

Last time I looked, 90 Vickers was 53Hrb, not 85.

You are now saying the test is of the overall bearing, where as originally you were saying it was the surface coating. This is what will confuse people reading the thread.

What I'm trying to get across is that the Hrb measurement is accomplished with a 1/16th ball being penetrated into the material at 100 kg. The depth of penetration is then measured to give the Hrb value. (A simplified explanation)

If the soft layer is only half a thou (.0005") and mounted on a harder base substance. Please explain to me how you can get an accurate depth measurement which is many times the thickness of the layer that you are trying measure?

If you can't picture this, then imagine taking some 6082 aluminium at 2" thick and compare it's hardness reading to the exact same 6082 at .0005" thick which is mounted on a piece of hard steel.

Would you get the same result? Y/N

http://en.wikipedia.org/wiki/Hardness_comparison

I was going off this chart, I looked at the wrong side of it
When a bearing is worn till the point it starts to fall apart the top layer doesn't matter.
I never said the way I tested it was the most sophisticated way, but this simple test shows how they went from the soft lead copper bearing to the hard aluminum one.

[kawasaki00]

Originally Posted by Bx Tpr

So this

"So there you have it: ideal mains should be 0.0018 - 0.0029 and ideal rods should be 0.00153 - 0.00205. You're given 0.00125 clearance from the factory and some may be thinking of reducing it another 16-40% with coated bearings without sizing the journals because they believe it gives them extra protection. The extra protection is true if you kept all things the same including the bearing clearance; but reducing the bearing clearance by 16-40% to get the extra protection of the coating is not recommended."

And this

Will correct the bearing clearance issue?

What side of the rod was machined down? The fillet side or the side that faces the other rod?

That has nothing to do with bearing clearance. That is another aspect that is being looked at.

[Bx Tpr]

Originally Posted by kawasaki00

That has nothing to do with bearing clearance. That is another aspect that is being looked at.

So what are those clearance numbers for?

[Yellow Snow]

Originally Posted by kawasaki00

http://en.wikipedia.org/wiki/Hardness_comparison

I was going off this chart, I looked at the wrong side of it
When a bearing is worn till the point it starts to fall apart the top layer doesn't matter.
I never said the way I tested it was the most sophisticated way, but this simple test shows how they went from the soft lead copper bearing to the hard aluminum one.

But can you see how this would confuse readers into thinking the 07* bearing surfaces are massively harder than the 08* bearings?

Base material maybe, but surface material is still an open book until someone micro tests it..

Agree?

[SenorFunkyPants]

Originally Posted by JRV

1. Does anyone here or know anyone running a lighter oil (0W-40, 5W-40) consistently and has had this rod bearing issue? IDK if anyone would admit this but hope they would.
2. If switching to a lighter oil so it can squeeze in between the tight bearing clearances helps, didnt they recommend TWS because of the risk the oil film on the bearings would not hold during heavy load/high RPM? Sounds like a pick your poison scenario (BMW chose lesser of two evils, hence why TWS has decent cold flow for its SAE rating?) and no matter what maybe its best to replace the bearings once in its lifetime wether BMW admits an issue or not.
3. Is there anything on this subject matter that both sides agree on?

1. I think there are quite a few people running the lighter oils - trouble is there is no way of knowing if the oil is making things better or worse.
2. If you are under warranty, I would suggest keeping to the Castrol 10W-60. Out of warranty then you have to choose for yourself - if you are easy on your car and in a cool climate then the lighter oils should be fine.
3.

[regular guy]

Originally Posted by Yellow Snow

Thankyou Swamp. You got it in one

Don't thank him just yet...that wasn't what I was talking about. It's a question you still haven't addressed (along with all the others).

Still curious: what are your engine credentials, hardness scale credentials, math credentials? Aren't you that BMW diesel tuning guy? What are your credentials for S65/S85?

[swamp2]

Originally Posted by regular guy

Don't thank him just yet...that wasn't what I was talking about.

I'm very curious. I believe I followed you objection to this technique in this most recent post with the milling machine as well as others in the past. If we are both confused perhaps your explanation could use a different approach or a sketch.

Either way this issue is basically in the noise to me. The bearing thickness should contribute to the clearance but due to crush and other non-linear problems during installation one should not assume thinner bearing = more clearance and that there is a 1:1 relationship in them. Measuring in-situ is obviously the best idea and most accurate.

[regular guy]

Originally Posted by swamp2

I'm very curious. I believe I followed you objection to this technique in this most recent post with the milling machine as well as others in the past. If we are both confused perhaps your explanation could use a different approach or a sketch.

Either way this issue is basically in the noise to me. The bearing thickness should contribute to the clearance but due to crush and other non-linear problems during installation one should not assume thinner bearing = more clearance and that there is a 1:1 relationship in them. Measuring in-situ is obviously the best idea and most accurate.

I don't have time right now, but it's basically this:

Circle: 2.04655 diameter
X = r*cos(theta)
Y = r*sin(theta)

Move X by 0.0001 and you will see the results on Y. If this slip gauge spring is too strong to follow the contour of the circle and doesn't give the correct results, then I can see a small handful of possible errors introduced by using this technique. I've repeatedly asked him to explain how he guaranteed he was dead center and isn't even 1/20 of a degree off dead center, and he won't answer.

[Yellow Snow]

Originally Posted by regular guy

I don't have time right now, but it's basically this:

Circle: 2.04655 diameter
X = r*cos(theta)
Y = r*sin(theta)

Move X by 0.0001 and you will see the results on Y. If this slip gauge spring is too strong to follow the contour of the circle and doesn't give the correct results, then I can see a small handful of possible errors introduced by using this technique. I've repeatedly asked him to explain how he guaranteed he was dead center and isn't even 1/20 of a degree off dead center, and he won't answer.

I still don't know what you are babbling about.

Can you draw a sketch?

I explained the measurement method and I can't see what you are not grasping. Swamp seems to have got it straight away. Perhaps he can explain it better.

[catpat8000]

Originally Posted by regular guy

I don't have time right now, but it's basically this:

Circle: 2.04655 diameter
X = r*cos(theta)
Y = r*sin(theta)

Move X by 0.0001 and you will see the results on Y. If this slip gauge spring is too strong to follow the contour of the circle and doesn't give the correct results, then I can see a small handful of possible errors introduced by using this technique. I've repeatedly asked him to explain how he guaranteed he was dead center and isn't even 1/20 of a degree off dead center, and he won't answer.

Let me see if I can follow this.

Are you saying:

- draw a circle with a radius of 2.04655 inches, centered at coordinates x=0, y=2.04655

- now look at an X displacement (along the x axis) of 0.0001 inches

- to find the corresponding Y position of this new X value in the original circle you would calculate:

theta = arccos(0.0001/2.04655) or
theta = 1.570747 rad

Then the Y displacement would be:
y = 2.04655 * sin(1.570747)
y = 2.0465499975

Then 2.04655-2.0465499975 = 2.49999976e-9
or 0.00000000249999975

This is basically the same value swamp obtained. So I guess my math agrees with swamp. It looks to me like a displacement of 0.0001 in the X direction is basically noise in the Y direction. What am I doing wrong?

Pat

[regular guy]

Originally Posted by catpat8000

Let me see if I can follow this.

Are you saying:

- draw a circle with a radius of 2.04655 inches, centered at coordinates x=0, y=2.04655

- now look at an X displacement (along the x axis) of 0.0001 inches

- to find the corresponding Y position of this new X value in the original circle you would calculate:

theta = arccos(0.0001/2.04655) or
theta = 1.570747 rad

Then the Y displacement would be:
y = 2.04655 * sin(1.570747)
y = 2.0465499975

Then 2.04655-2.0465499975 = 2.49999976e-9
or 0.00000000249999975

This is basically the same value swamp obtained. So I guess my math agrees with swamp. It looks to me like a displacement of 0.0001 in the X direction is basically noise in the Y direction. What am I doing wrong?

Pat

Diameter: 2.04655
Radius: 1.023275

X = R * Cos(theta)
Y = R * Sin(theta)
Where R = 1.023275, X = (R - 0.0001)

R = 1.023275
X = 1.023175

Solve for Cos(theta):
Cos(theta) = (1.023275 - 0.0001) / 1.023275
Cos(theta) = 0.99990277

Solve for theta:
theta = ArcCos(0.99990277)
theta = 0.01398049

Solve for Y:
Y = R * Sin(theta)
Y = 1.023275 * Sin(0.01398049)
Y = 1.023275 * 0.01398003
Y = 0.014305419

Pat, I'm pretty sure ARCCOS and ARCSIN are only defined [-1, 1], so your idea doesn't work.

[catpat8000]

Originally Posted by regular guy

Diameter: 2.04655
Radius: 1.023275

X = R * Cos(theta)
Y = R * Sin(theta)
Where R = 1.023275, X = (R - 0.0001)

R = 1.023275
X = 1.023175

Solve for Cos(theta):
Cos(theta) = (1.023275 - 0.0001) / 1.023275
Cos(theta) = 0.99990277

Solve for theta:
theta = ArcCos(0.99990277)
theta = 0.01398049

Solve for Y:
Y = R * Sin(theta)
Y = 1.023275 * Sin(0.01398049)
Y = 1.023275 * 0.01398003
Y = 0.014305419

Pat, I'm pretty sure ARCCOS and ARCSIN are only defined [-1, 1], so your idea doesn't work.

Well, I used a value for arccos which was between -1 and 1 so that should be fine. I did make an error and that error was that I used a radius of 2.04655 when that is really the diameter. But even so, when I re-calculate using a radius of 1.02328, it only means my result is roughly twice as big, which is still insignificant.

I don't want appear too firm on this one but, in this case, I think the error you are making is taking X = (R-0.0001).

If your bearing circle has its origin at 0, 0 then you are actually looking at an X displacement at the side of the bearing where a tangent to the bearing surface (circle) is essentially vertical. Hence you would see a large Y displacement with small X displacement.

As a side note, in my calculation, I assumed a bearing circle with origin at 0, R.

If only we could draw pictures in these forums...

[swamp2]

Ugh, I also hastily used r=2 (2 precisely since the extra digits don't really matter) of course the right thing is r=1. This changes the allowed band width to get accuracy of 1/10 (thousandth) to 0.03". It also doubles the other quoted figure to 0.000000005".

rg: Looks like we are all talking about the same thing but your trigonometry is a bit more screwed up than ours....

[regular guy]

Originally Posted by catpat8000

Well, I used a value for arccos which was between -1 and 1 so that should be fine. I did make an error and that error was that I used a radius of 2.04655 when that is really the diameter. But even so, when I re-calculate using a radius of 1.02328, it only means my result is roughly twice as big, which is still insignificant.

I don't want appear too firm on this one but, in this case, I think the error you are making is taking X = (R-0.0001).

If your bearing circle has its origin at 0, 0 then you are actually looking at an X displacement at the side of the bearing where a tangent to the bearing surface (circle) is essentially vertical. Hence you would see a large Y displacement with small X displacement.

As a side note, in my calculation, I assumed a bearing circle with origin at 0, R.

If only we could draw pictures in these forums...

Originally Posted by swamp2

Ugh, I also hastily used r=2 (2 precisely since the extra digits don't really matter) of course the right thing is r=1. This changes the allowed band width to get accuracy of 1/10 (thousandth) to 0.03". It also doubles the other quoted figure to 0.000000005".

rg: Looks like we are all talking about the same thing but your trigonometry is a bit more screwed up than ours....

Why on earth are you guys using 0.0001 in a division problem as if 0.0001 were part of a ratio? This is a SUBTRACTION problem. Momentarily, I will post two more solution. 1) By my son who used pure algebra, 2) Brute force Excel spreadsheet. All three of mine come up with the same exact answer. Hold tight while I gather the Excel method together.

Swamp if my trig is screwed up, then correct it please.

[catpat8000]

Originally Posted by regular guy

Why on earth are you guys using 0.0001 in a division problem as if 0.0001 were part of a ratio? This is a SUBTRACTION problem. Momentarily, I will post two more solution. 1) By my son who used pure algebra, 2) Brute force Excel spreadsheet. All three of mine come up with the same exact answer. Hold tight while I gather the Excel method together.

Swamp if my trig is screwed up, then correct it please.

Let me try a different way of explaining my point of view. First, once again, my assumption is a unit circle with origin at x=0, y=1. This is a key point which I keep restating. This circle models the bearing.

Let's assume we are moving along the x axis by a certain amount, say 0.0001 inches. In order for the Y displacement along the circle to increase by the SAME AMOUNT as we move 0.0001 in the X dimension, we would need to be on a 45 degree slope. But we aren't. In the arena/scale in which we are measuring, we are durn near flat so the Y displacement MUST be less than the X displacement, not more.

The only way it could be more is if the slope is greater than 45 degrees. And the only place that will be true in my example is if we move out on the x axis a distance of 1. But in this case, we are no longer measuring what I thought we were discussing. Now we would be measuring the seam where the top and bottom bearing caps join.

So either you are working with a different picture than either swamp or myself, or we are all calculating different things. Or both. I'm sure you aren't mistaken and I am sure my model and my math is correct so we must have some sort of communication gap.

[regular guy]

Originally Posted by swamp2

Ugh, I also hastily used r=2 (2 precisely since the extra digits don't really matter) of course the right thing is r=1. This changes the allowed band width to get accuracy of 1/10 (thousandth) to 0.03". It also doubles the other quoted figure to 0.000000005".

rg: Looks like we are all talking about the same thing but your trigonometry is a bit more screwed up than ours....

Originally Posted by catpat8000

Let me try a different way of explaining my point of view. First, once again, my assumption is a unit circle with origin at x=0, y=1. This is a key point which I keep restating. This circle models the bearing.

Let's assume we are moving along the x axis by a certain amount, say 0.0001 inches. In order for the Y displacement along the circle to increase by the SAME AMOUNT as we move 0.0001 in the X dimension, we would need to be on a 45 degree slope. But we aren't. In the arena/scale in which we are measuring, we are durn near flat so the Y displacement MUST be less than the X displacement, not more.

The only way it could be more is if the slope is greater than 45 degrees. And the only place that will be true in my example is if we move out on the x axis a distance of 1. But in this case, we are no longer measuring what I thought we were discussing. Now we would be measuring the seam where the top and bottom bearing caps join.

So either you are working with a different picture than either swamp or myself, or we are all calculating different things. Or both. I'm sure you aren't mistaken and I am sure my model and my math is correct so we must have some sort of communication gap.

Well, we're all talking the same thing, and all our math is right (except I'm still not sure why Pat turned this into a ratio problem instead of a subtraction problem). Regardless, I was on (1, 0) of the unit circle instead of (0, +/-1). So the math was all right, but the answer was wrong because of my wrong starting location. So yes, I agree with you guys that starting at the correct location has much less effect on Y value than my original 0.0140 inch assertions.

Edit: Once I started doing the pure trig solution with the correct orientation, I can see exactly how you started with ARCCOS(0.0001 / Radius), and I agree it is correct.

[catpat8000]

Originally Posted by regular guy

Well, we're all talking the same thing, and all our math is right (except I'm still not sure why Pat turned this into a ratio problem instead of a subtraction problem). Regardless, I was on (1, 0) of the unit circle instead of (0, +/-1). So the math was all right, but the answer was wrong because of my wrong starting location. So yes, I agree with you guys that starting at the correct location has much less effect on Y value than my original 0.0140 inch assertions.

I assume since we have bottomed out and all agree on an answer now that you aren't really interested in hearing me explain my math. But if you want, I'll be happy to go into it!

[regular guy]

Originally Posted by catpat8000

I assume since we have bottomed out and all agree on an answer now that you aren't really interested in hearing me explain my math. But if you want, I'll be happy to go into it!

No, after I changed orientation on the circle, I see exactly why you did it. And I agree it is correct. Oh well.

[catpat8000]

Originally Posted by regular guy

No, after I changed orientation on the circle, I see exactly why you did it. And I agree it is correct. Oh well.

No harm, no foul. Now let's get back to fighting about the clearances on the different bearing sets.

Earlier regular_guy, you mentioned doing a bearing test.

Originally Posted by regular guy

Full set of virgin 088/089's (if you can still get them as discontinued): $375
Full set of 702/703's: $375
Two more sets of connecting rod bots: $230

Did you mean we would:
(0) measure the bearing thickness for each of the 16 bottom and top bearing caps
(1) install all 08* bearings in 8 rods and torque the rod bolts to full torque, then measure inside diameter for each rod
(2) install all 70* bearings in 8 rods and torque rod bolts to full torque, then measure inside diameter for each rod
(3) Compare

I'm trying to be explicit because I'm a big, fat n00b on this stuff and I want to understand what we're proposing. I assume the point of this is to determine whether bearing clearances have changed with the newer bearings?

Who would do the measurement? That seems important.

Then

Originally Posted by regular guy

Then to take this to the next level, after taking the measurements on virgin parts, a half set of each would be sent to Calico and WPC for coating/treatment. Then repeat the process when the parts return. Add another $300 in these expenses.

Why would we do this? What is this testing?

Pat

[regular guy]

Originally Posted by catpat8000

No harm, no foul. Now let's get back to fighting about the clearances on the different bearing sets.

Earlier regular_guy, you mentioned doing a bearing test.


Did you mean we would:
(0) measure the bearing thickness for each of the 16 bottom and top bearing caps
(1) install all 08* bearings in 8 rods and torque the rod bolts to full torque, then measure inside diameter for each rod
(2) install all 70* bearings in 8 rods and torque rod bolts to full torque, then measure inside diameter for each rod
(3) Compare

I'm trying to be explicit because I'm a big, fat n00b on this stuff and I want to understand what we're proposing. I assume the point of this is to determine whether bearing clearances have changed with the newer bearings?

I'm working on the proposal right now. Putting the finishing touches on it as we speak. Just spent 20 minutes looking for a 2.0000 inch calibration ring for a photo, but can't find it because it got "cleaned it" and now the prime suspect is asleep. I'm sure nobody else has this problem.

Who would do the measurement? That seems important.

I would do the measurements, just as I did before. Nobody has impeached a single one of my measurements yet, as I provide far too many photos and document every procedure and method for them to even try.

Why would we do this? What is this testing?

Pat

That's a great question that I'm asking myself. If the purpose is to pacify two people and think it will bring a concensus, then that's not going to happen. They will change the goal post anyways after the work is done and find a new reason to call BS on everything that has been accomplished. But if that's not the purpose, then there is something to be gained.