Calculating the Number of Rotations for Offset Central Conductor Magnetization

[cataldo]

Hi all,
My name is aldo and I am mechanical enginner. My job is specialist in non-destructive testing in aeronautic sector. The typical method for to magnetize this components is circular magnetization by central or offset central conductor. Usually the central conductor is a copper bar. When I use the offset central conductor the specific ASTM says:"the distance along the part circumference (interior) that may be effectly examined shall be taken as approximately four times the diameter of the central conductor up to 360 degrees. The entire circumference shall be examined by rotating the part on the conductor, allowing for approximately a 10% magnetic filed overlap". Is there a mathematics formula that calculate the number exactly of rotation including in this formula also the 10% magnetic field overlap?I make the ratio between circumference of component and four times diameter of bar for to determinate the number of shots but i am not sure the 10% magnetic field overlap. I attached one picture
Tanks for your attention and best regards
Aldo

 

[marcusl]

EDIT: Sorry, I failed to carefully read the question.

Call D the interior diameter and d the wire diameter. The number of rotations is then

$$N=\frac{\pi D}{3.6 d}$$.

The factor 3.6 gives the 10% overlap. As skeleton says below, round N upwards to the nearest integer; the rotation angle is 360/N.

 

[skeleton]

If I understand the problem correctly (?), try this ...

My illustration does not coincide with my calculation numbers.

The illustration shows three coloured sectors; to complete the circle, four placements of the NDT would be required. Thus you would rotate the shaft 4 times, each with a 90 deg turn.

The calculation sheet matches different diameter data. Here only 2.9 placements are required; but this physically would necessitate 3 placements. Each placement would occur with a subsequent 120 deg rotation.

Adjust Ds and Dr to suit your problem.

 

[cataldo]

Hi skeleton and marcusl
Excuse me for late answer
Tanks for yours interest above my small problem...It has been cleared... I suggest this answers to my specialist of NDT...
Best regard
Cataldo