Hi guys,
I just want a clarification about this calculation. Somewhere i read that the no. of rolling elements in an antifriction bearing can be calculated by adding BPFO and BPFI. I checked some bearings and it was correct.But when i tried the theory, it was not matching.

BPFO = │Nr/2 (Si-So) (1- Rd Cos( Ø)/Pd)│

BPFI = │Nr/2 (Si-So) (1 + Rd Cos( Ø)/Pd)│

Where
Nr = No. of rolling elements.
Si = Inner race angular speed.
So = Outer race angular speed.
Rd = Rolling element diameter.
Pd = Bearing Pitch diameter.
Ø = Contact angle.

So BPFO + BPFI = │Nr/2 (Si-So) (1- Rd Cos( Ø)/Pd + 1 + Rd Cos( Ø)/Pd)│

= Nr/2 (Si-So) * 2

= │Nr (Si-So)│

For stationary Outer race,

BPFO + BPFI = │Nr * Si│

Here my question is how can we neglect the Si(angular speed of the inner race)? Can anybody tell whether my calculation is wrong or not?

I think maybe something is wrong with your equation.

Take a look on the left side here (the equations are in orders here)
http://www.update-intl.com/VibrationBook8g.htm

If you add them together you get the expected result: BFPO + BPFI = Nb (orders)

My question is, why is this so?

Take a ball bearing. Put a mark on the cage at the 12:00 position. Rotate the inner race through one complete revolution CW. You see the mark has moved to around 4:30... or 4.5/12 ~ 0.4 revolutions. Call this fraction number X ~ 0.4.

The units of X is cage revolutions/shaft revolution.

This number is of course FTF in orders. i.e.
FTF = X times machine speed
(cage revs/minute) = (cage revs/shaft rev) * (shaft revs/minute)

What is BPFO? Since there are Nb balls in one cage, the balls pass by a fixed point Nb times faster than FTF.
BFPO = FTF times Nb
(balls/minute) = (cage revs/minute) (balls/cage)

What is BPFI? the number of balls per minute past a given point on the inner race (which was rotating in our example). You have to know how far did the cage travel compared to the inner race. The inner race moved one full revolution (from 12:00 to 12:00) and the cage only moved X~0.4 revolutions (from 12:00 to 4:30), so the inner raced moved (1-X)~0.6 revolutions (it went further than cage by distance from 4:20 to 12:00).

So how to find BPFI? The logic is the same as BPFO except instead of X we use (1-X) as the relative movement of the cage compared to inner race per shaft revolution.

So the expression for BPFI thru the same logic ends up looking like the expression for BPFI except use (1-X) instead of X

We had
BFPO = Nb*FTF = Nb * X * Machine Speed
So
BFPI = Nb * (1-X) * Machine Speed

BFPO + BFPI = Nb * Speed * (X+1-X) = Nb*speed

Another way to say it.. start using the same X found above by experiement ~ 0.4.

One revolution, a given ball moves X revolutions with respect to outer ring and (1-X) revolutions with respect to the inner ring.

If we have speed = S revolutions per minute, that ball moves S*X revolutions per minute with respect to outer ring and X*(1-X) rpms with respect to inner ring.

If that were the only ball in the bearing, the impacting frequency would be S*X for the outer ring and S*(1-X) for the inner ring. But we have Nb balls so the impacting frequency is Nb times as fast. So the impacting frequency is Nb*S*X for the outer race and Nb*S*(1-X) for the inner ring.

Now if we add inner ring and outer ring impacting frequencies toegther again we see the total is Nb * S

=============
If you keep this experiment in mind (rotate the inner ring one rev and watch the distance of movement of the cage ~ 0.4 revs), then it is easy to visualize/remember all the following relatinships:

FTF~0.4*Speed
BPFO=Nb*FTF ~ Nb*0.4*Speed
BPFI=Nb*(1-FTF)~ Nb*0.6*Speed
BPFO+BPFI = Nb*Speed

This message has been edited. Last edited by: electricpete,

Pete,
It was really a nice explanation. Thanks.
So what we can derive from this is, the angular speed of the inner and outer race is not taken for the calculations. Let me ask another question. If we get a BPFO or BPFI from a bearing, is it always indicate that there is some problem in the outer or inner race?

I also liked that a lot Pete, thanks. Jenish, when I see just one fundamental x1 of BPFO or BPFI I know that something causes a nice sine shaped wave of that frequency. A spot fault, lik a spall in the race, cannot cause that, but an oval shape of the rings can, pending for instance a burr on the shaft under the inner ring or a falnge for the bearing housing that has a piece of packing material left so bolts skew the housing. A spot fault or even a series of spalls will cause rattle liek when you pass a railway crossing on the road. It causes a family of orders of that fault frequency. often, the inherent lowest ring resonance of the bearings ring/housing will tend to color the family to be highest around 500 to 1500 Hz. So an incipient damage from fatigue and aslo damage through raw dirt can be found in the approx. area x5 to x15 of either BPFO or BPFI or both, pending the exact nature of loads (direction and size) and damage from mounting and/or usage. As the damage grows around, the family tends to move downwards. if the bearing survives, you can say that you have formed a new bearing rolling race when just 1xRaceFault is left since it has deformed to a new oval shape. The inner box bearing in a press suction roll of a paper machine is a good example where that happens.