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12-10-2013, 04:28 PM | #1563 | |
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12-10-2013, 04:34 PM | #1564 |
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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? Last edited by Bx Tpr; 12-10-2013 at 04:40 PM.. |
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12-10-2013, 04:47 PM | #1565 | |
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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.
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12-10-2013, 04:54 PM | #1566 | |
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12-10-2013, 04:58 PM | #1567 |
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12-10-2013, 04:58 PM | #1568 | |
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Base material maybe, but surface material is still an open book until someone micro tests it.. Agree? |
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12-10-2013, 05:01 PM | #1569 | |
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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. |
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12-10-2013, 05:39 PM | #1570 |
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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? |
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12-10-2013, 05:52 PM | #1571 | |
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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.
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12-10-2013, 05:59 PM | #1572 | |
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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. |
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12-10-2013, 06:11 PM | #1573 | ||
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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. |
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12-10-2013, 06:41 PM | #1574 | |
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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 |
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12-10-2013, 07:21 PM | #1575 | |
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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. |
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12-10-2013, 07:53 PM | #1576 | |
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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... Last edited by catpat8000; 12-10-2013 at 08:05 PM.. |
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12-10-2013, 09:42 PM | #1577 |
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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....
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12-10-2013, 09:50 PM | #1578 | ||
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Swamp if my trig is screwed up, then correct it please. |
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12-10-2013, 10:45 PM | #1579 | |
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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. |
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12-10-2013, 10:57 PM | #1580 | ||
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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. Last edited by regular guy; 12-10-2013 at 11:09 PM.. |
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12-10-2013, 11:03 PM | #1581 | |
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12-10-2013, 11:07 PM | #1582 |
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No, after I changed orientation on the circle, I see exactly why you did it. And I agree it is correct. Oh well.
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12-11-2013, 12:07 AM | #1583 | |||
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Earlier regular_guy, you mentioned doing a bearing test. Quote:
(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 Quote:
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12-11-2013, 12:29 AM | #1584 | |||
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