Calculate maximum number of balls for a Conrad ball bearing 2

[wzwqx2]

Does anyone know how to calculate the maximum number of balls that can be assembled into a deep groove ball bearing using a Conrad style assembly?

 

[electricpete]

Clarication in bold:
"I have provided in my attachment 9 Jan 11 0:02 a means to estimate Phi which I believe is correct subject to.."

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(2B)+(2B)' ?

 

[dinjin]

Electricpete,
I have T.A. Harris book Rolling Bearing Analysis, 2nd edition. Page 7 shows a similar illustration that is in
the url above by tbuelna except it has 7 balls instead of 5.

It also shows the equation similar to what tbuelna listed.
The illustration is titled "Diagram illustrating the method of assembly of a Conrad type deep-groove ball bearings."

Phi in this illustration is slightly greater than 180 degrees. It does not say the phi in the illustration is the same phi in the calculation.

I have conrad designed fill in my old porch glider but it did not have a cage to keep the balls spaced properly but was grease filled and worked fine and quiet for 50 years or more without a cage. I finally got rid of the glider as it was too expensive to replace the cushions.

I made up an excel sheet with the basic equation and found that pi seemed to work in the equation for approximately one half the number of balls for a full complement bearing.

The angular pitch of tight balls is 2 times the arcsine of
Ball Diameter divided by the mean Raceway Diameter. If this angle is Ta, then (Z-1) times Ta will be approximately 180 degrees.

The centerline shift of the eccentric rings is 1/2 the ball diamter. That might help.

 

[electricpete]

Working with Dor and Dir as outer and inner race radius, we know Dor = Dir + 2*Dball

If we have inner race centered in outer race, then we have uniform gap Dball all around (12:00 / 3:00 / 6:00 and 9:00 positions.)

If we temporarily ignore the presence of a land/shoulder on each side of the ring, we could push the inner race upward into full contact with the outer race. Now the gap between inner ring and outer ring moving around the bearing is:
0 at 12:00 position
1*Dball at 3:00 position
2*Dball at 6:00 position
1*Dball at the 9:00 position.
In this case, the angle Phi would be Pi because the ball can be inserted only to the 3:00 and 9:00 positions.

However, we ignored the presence of a land/shoulder to get there. To correct that we recognize we can no longer push the inner ring to contact with the outer ring. They are prevented from contact by the shoulders. So to adjust the picture we move the inner ring down slightly from previous positon (and we know the manufacturer still gave us enough room to install ball across the shoulder at the 6:00 position). This should have the effect of slightly increasing the clearance at 3:00 and 6:00 position compared to the previous case, so we have slightly more than 1*Dball gap at the 3:00 and 9:00 positions, and Phi would be slightly larger than Pi. But I’ll admit it might only be a small amount and probably not worth worrying about. I am happy to take your suggestion to use Pi as close enough.

The centerline shift of the eccentric rings is 1/2 the ball diamter

Where does this come from?


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(2B)+(2B)' ?

 

[dinjin]

The radial distance between the id of the outer ring and the od of inner ring is 1/2 the ball diameter. To shift the rings together at the extreme point requires that you shift the ring this distance.
I think it does make a differce whether the number of balls are odd or even for a formula to work.
For an odd number of balls you can enter 1/2 of the (Z-1) balls to each side and still have room for the last ball. However when an even number of balls are required, you may not be able to get the last ball in at that point.

 

[electricpete]

The radial distance between the id of the outer ring and the od of inner ring is 1/2 the ball diameter.

Why not 1x ball diameter?

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(2B)+(2B)' ?

 

[electricpete]

I would say the amount we shift is given by:
Offset = (Dor-Dir)/2 - OffsetOR - OffsetIR

Also Dor-Dir = 2*BallDiameter, so we could rewrite this as:
Offset = BallDiameter - OffsetOR - OffsetIR

This number will end up somewhere less than BallDiameter, but I see no reason to suspect exactly half BallDiameter unless it is just a general approximate thumbrule.

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(2B)+(2B)' ?

 

[dinjin]

I layed out the 20 and 10 inch diameter raceways in acad and conclude that the 32 and 16 one inch balls were ok.
I was surprised at the phi values. I assumed that the farmost position of any ball centerline would be at the equivalence of 1/4 the ball dia. to the right of the bearing centerline when the inner race was shifted 1/2 the ball diameter to the right of the bearing centerline.

phi brg Z
max. dia.
182.8651 20 32
185.7320 10 16
186.3695 9
187.1666 8
188.1921 7
189.5604 6
191.4783 5
194.3615 4
199.1881 3

From this I concluded that phi max. was equal to 180
plus 2 times the (arcsin (D/2 divided by d) times 180/pi)

 

[dinjin]

Electricpete,
Tbuelna's sketch should be turned 90 degrees to have a better understanding of how the balls are loaded in a Conrad bearing. You put the first ball in the top where the max lands are the greatest distance apart and shove the ball either to the left and then the next one to the right and they fall down into the bearing raceway. You keep inserting balls in the same manner until no more will go in the top. The first and second balls fall to the max possible position in the raceway before the rings are shifted to their final position. It is this max angle of phi that I was calculating in my analysis. I was surprised to find out that phi keeps increasing as the mean raceway gets smaller. When I was designing Conrad type bearings, I almost always used mean raceway diameter having a ratio
to the ball diameter above 16 to 1. I used the formula
of 1/2 the balls in a full complement bearing and used a plastic snap in cage design to keep the balls spaced almost evenly around the bearing raceway. We have not heard back from wzwqx2 (Automotive) so he may not understand what we were going thru as to analysis and only wanted a final answer or final equation. If max phi is known, you could divide phi by the tight ball pitch angle and it would give you the nuber of (Z-1) balls. Adding 1 to that number would be the max balls that could go into the raceways. In some cases, it is not possible to add 1 and still be able to get that last ball in at the gap. The formula was true except for one of the cases where the next to the last ball would not allow the last ball to be entered at the gap.

 

[tbuelna]

dinjin,

As you point out, the max number of balls that can be loaded between the offset races must take into account the fact that the races must eventually be shifted back to concentric positions with the complement of balls within the race grooves. So the max phi angle must allow for some minimum clearance between the balls and races.

I made another sketch to illustrate the condition.

Terry

 

[tbuelna]

In the sketch linked to my previous post, please pretend that the phi angle is centered on the outer race.

Terry

 

[dinjin]

tbuelna,
Your first illustration was actually more typical of a conrad type bearing. The second one is typical of what I was talking about in that you could not assemble the last ball in the gap between the lands and have it move to its extreme position. Although you can draw it there, you cannot put it in the assembly. I wish I knew how to post an acad drawing here or post it to another site so it could be opened or shared. To maximize the axial load of the bearing you want to minimize the id of the outer ring and maximize the od of the inner ring. This allows you to have a greater contact angle and not edge load the raceways. If it is strictly a radial bearing, it is not that critical.
If you subtract the od of the inner ring from the id of the outer ring it should be not less than the ball diameter. Your first illustration had these in the right proportions.
When the rings are shifted apart, the intersect of outer race nominal ball path and the inner race nominal ball path determines ball centerline at the max phi centerline. It is hard to define this without showing you a correct illustration. Thanks for posting the first illustration.