I am wondering how many operators rotate parts of varying diameters at a constant surface speed to help reduce the number of saturation graphs they need to produce for each sized part, or if they have a part that requires peening from a small diameter out to a large diameter get on with the rate at which the part is covered, as the surface speed on a larger diameter part is greater than on a smaller diameter part. I would assume that the feed rate of the nozzle would have to reduce or part rotation speed slow down as the nozzle moves out.

We currently use 75mm per second as our starting point because of the flow rates and machinery we currently have.
We started at this because the guys used to do Almen strips by hand @ 1 second per pass. With automation and increased flow rates I believe you could easily double if not triple this rate.
Let me know how you go.

75mm per second converts to 4.5m per minute surface speed(Vc) (75x60=4500mm/sec)

To work out the RPM of a part.

RPM= (Vc X 1000)/(pi x DIA)

Where
Vc is velocity in Meters per minute

So a part 300mm in diameter has a required RPM of

RPM = (4.5x1000)/( pi x 300)
= 4500/942.5
=4.8 revolutions per minute

A part 50mm dia

RPM= (4.5x1000)/( pi x 50)
= 4500/157
=28.6 RPM

(If surface speed is in feet per minute substitute 12 for 1000)
By having all parts at the same surface speed I can use one saturation graph for any diameter part because all parameters are constant!
If the table speed is fixed at say 20 RPM you will need a different saturation graph for a change in diameter as the T1 point would change.
The intensity value would not be different just the time it takes to reach T1.

Roy

Last edited by Roy the kiwi; 11/10/10 08:35 PM. Reason: pi math symbol did not show