In the attached video, there is a mark on the inner race (black), the outer race (black) and the cage (blue).

Initial condition: the three marks are lined up at 12:00.

Then I rotated the inner race CW through one full revolution.

Final condition: the outer race mark stays at 12:00, the inner race mark comes back to 12:00, and the cage mark is approx the 4:30 position.

The cage has moved CW by approx 0.375 revolutions (0.375 = 4.5/12).

The fraction 0.375 for this bearing is the cage frequency FTF in orders.

In other words
FTF = (0.375 cage rotations) / (one inner race rotation.) = 0.375

What is BPFO in orders? It is how many rolling elements pass a given point on the outer race during one revolution of the inner race. We have Nb (8) balls passing a given point on outer race per cage rotation, and FTF cage rotations per revolution of the inner race so BPFO = Nb*FTF. Saying it another way:

BPFO = [(Nb balls passing outer race) / (one cage rotations) ] * [(FTF cage rotations) / (one inner race rotation.)]
BPFO = [(FTF * Nb balls passing outer race / (one inner race rotation.)] = FTF * Nb
BPFO = [(0.375 * 8 balls passing outer race / (one inner race rotation.)] = 0.375 * 8

What is BPFI in orders? It is how many rolling elements pass a given point on the inner race in one revolution. The inner race moved 12 clock positions while the cage only moved 4.5 clock positions. The relative movement between inner race and cage was 12-4.5 = 7.5 clock positions. This is a fraction 0.675. So during one revolution of the inner race, it rolled past a fraction 0.675 = (1-FTF) of the balls. The number of balls passed by the inner race during one revolution is (1-FTF)*Nb.

BPFI = [(Nb balls past inner race) / (one inner race rotation relative to balls.)] * [(<1-FTF> inner race rotations relative to balls) / (one inner race rotation.)]
BPFI = Nb * (1-FTF) balls past inner race / inner race rotation
BFPI = Nb* (1-FTF) = 8*0.625

BFPO + BPFI = FTF*Nb + (1-FTF)*Nb = Nb
BFPO + BPFI = 0.375*8 + 0.625*8 = 8

Intuitively why does BPFO + BPFI = Nb? Because the fraction of a revolution from 12:00 CW to the blue (FTF) dot plus the fraction from the blue dot CW back 12:00 (1-FTF) is always one full revolution.

It may be tough to read my whole post. It may be worthless if you already understand this well. But for me, the most interesting and revealing part is just watching that blue spot move and realizing the fraction of a revolution we see (0.375) is FTF... everything else follows from there.

PICT3087.AVI (1,442 Kb, 198 downloads)

Hi pete,

I think i didn't understand what you tried to say on your post. You are right about the BFPO&BFPI and FTF relations (since you are accounting them in terms of orders).

I think that that the summations of the frequencies equals the number of rolling elements because they are derived in reference to the shaft frequency. I don't know if i explain myself but its something similar to the motion of planetary gears. The RPM's of the planetary are related via the diameteres and number of theeth of the gears.

Cool video, Pete

Thanks Pete. I dug through some old files and found a post we shared in 2003 that explains the same concept in your intuitive manner. Some people are satisfied with an algebraic explanation, but I need something more tactile and visual...which your explanation provides. A member with a user name "in2vibe" gave a similar explanation on the 2003 posting. It took your post to allow me to "re-remember" the concept. Thanks.

Pete,

following your line of thought...

if you can see FTF sidebands around BPFO, then could you see RPM-FTF sidebands around BPFI?

Mechanical pawn: where in venezuela are you?

Here is a semi-intuitive explanation.

David

Bearing_fault_freq.doc (30 Kb, 67 downloads)

Good job Pete. Whenever I do training or need to explain bearing faults, I mark up a bearing and rotate it. It really shows it clearly and an easy way to approximate the fault frequencies if you know the number of balls

My answer has nothing to do with the discussion, but to Elias: I'm from Maracaibo.

Does anyone know how to send private messages?. In my profile says something about it, but that's pretty much I know

Thanks.

This message has been edited. Last edited by: mechanical pawn,

I beleive the PM feature is not enabled in the boards.

If you think it would be useful, you might mention it on the "about the forums" forum.