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# DIN 5480 Calculation of Distance Between Measuring Circles

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## DIN 5480 Calculation of Distance Between Measuring Circles

(OP)
I have a copy of DIN 5480-1 (March 2006) and DIN 5480-2 (May 2006). In DIN 5480-1 Figure 6 (page 19) is a data table to be shown on drawings. The example is for Hub DIN 5480-N120x3x38x9H Shaft DIN 5480-W120x3x38x8f.

I understand everything down to the last two rows of Figure 6:
HUB Max. dist. betw. measuring circles M2max 109.266
HUB Min. ref. dist. betw. measuring circles M2minRef (109.169)

SHAFT Max. ref. dimension over measuring circles M1maxRef (126.017)
SHAFT Min. dimension over measuring circles M1min 125.956

In Section 10.12 Quality Assurance (page 25) it states: "As an alternative, this (calculating inspection dimensions over and between measuring circles) can also be done using the deviation factors as described in DIN 5480-2.

The relevant tables in DIN 5480-2 for the example hub/shaft are Module m=3mm Table 22 Nominal Dimensions (page 28) and Table 23 Inspection Dimensions (page 29)

How are the data in Table 23 DM(2), M2, A*M2, DM(1), M1, A*M1, k, and Wk used to calculate the values shown in the last two rows of Figure 6?
Thank you.

### RE: DIN 5480 Calculation of Distance Between Measuring Circles

(OP)
Well, I think I figured it out.
For the hub, the numbers previously calculated from formulas in Table 6 page 23

Actual maximum space width emax = 6.361
Actual minimum reference space width emin = 6.305
Minumum effective space width evmin = 6.271

Using the formula from paragraph 10.12 on page 26

AM2=Ae x A*M2
From Tables 22 and 23 DIN 5480-2
A*M2=1.72
M2 nominal dimension 109.110
Space width=tooth thickness e2=s2=6.271

Starting with emax=6.361
AM2=(6.361-6.271)x1.72
AM2=0.1548
M2MAX=109.110+0.1548=109.265

emin=6.305
AM2=(6.305-6,271)x1.72
AM2=0.0585
M2minRef.=109.110+0.0585=109.169

Similarly for the shaft
Max effective tooth thickness svmax = 6.243
Actual maximum reference tooth thickness smax 6.220
Actual minimum tooth thickness smin = 6.180

Using the formula from paragraph 10.12 on page 26
AM1=AsxA*M1
From Tables 22 and 23 DIN 5480-2
A*M1=1.52
M1 nominal dimension 126.095
Space width=tooth thickness e2=s2=6.271

Starting with smax=6.220
AM1=(6.271-6.220)x1.52
AM1=0.0775
M1maxRef.=126.095-0.0775=126.0175

Smin=6.180
AM1=(6.271-6.180)x1.52
AM1=0.1383
M1min=126.095-0.1383=125.956

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